Lectures for College Algebra

Instructor: Mr. Rivera, Adjunct Lecturer Department of Mathematics, The City College of New York

Welcome to College Algebra

This took way too long to make, and I don't get paid enough (at all) for this.

As a brief introduction, I am an adjunct lecturer and tutor at CCNY at the time this letter was written. When I was 19, I took my first job as an educator here, tutoring in the Marshak Building. In my first month, I fell flat on my face many times trying to help students understand and solve problems. However, with time and dedication, I was able to improve.

I expect people who read my lecture notes to struggle to learn concepts just a little bit as I have when I first took these classes and when I first tried to explain these concepts. Rest assured, that is perfectly normal. Through time and practice, you too can improve and use math and its logical tools in many situations.

Math is not just a bunch of equations or numbers. Math is a language, a thought process, and a form of understanding our world. The purpose of these lecture notes is to show you my thinking when I see problems or face a new concept I have not seen before. They are also intended to act as commentary as you read them when you wish. It is also a script for myself to walk you guys through the concepts, so expect the notes and my lessons to be very similar to one another.

They will be based on textbooks, which will be referenced on the syllabus. However, I know it is not fun for me to repeat a book. You can safely assume I will have my own twists on what we discuss.

If you wish to point out mistakes or seek clarity, or if you believe I can better approach a topic, do not hesitate to reach out to me at jrivera5@ccny.cuny.edu. I believe in improving and updating lectures to make the experience more entertaining for my students, and even for those who read my lectures outside of my class.

Good luck this semester and I do hope you enjoy your time at my lectures.

- Mr. Rivera, Adjunct Lecturer

About this Notebook

This is NOT a textbook. It is intended to be presented to students as lecture material with examples and clarity. As such, you should not expect homework problems to appear in these notes, as those are provided in separate workbooks and MYOPENMATH.

While this is a condensed version of a textbook used in this class, it should not be considered a substitute by any means. Your textbooks are still important resources, and their approaches may offer a style of teaching better suited than mine.

Likewise, this material is not a substitute for in-person lectures. Students, you will learn best through a lecturer, along with your classmates or peers, and missing these lectures will rob you of the best experience possible.

I encourage students to reach out to me or to their peers whenever you need help with a problem or a concept. I also apologize in advance for any errors that may remain in these notes. Please be wary for typos, as these notes were prepared under time constraints.

I would like to thank the students I have personally tutored at the Math and Physics Tutoring Center in the Marshak Building for helping me develop my skills as a communicator. This material has been shaped with you in mind as some of my most valuable references.

Lecture 1: June 1st, 2026

Syllabus and Introducing Ourselves

The Introduction to the course and the policies can be found on the syllabus. This lecture will be slightly shorter because of that. Good luck this semester and do not be afraid to ask questions. This is a team effort for everyone to lift each other up to the next level. It is entirely reasonable to ask a question and answer incorrectly. I do not bite and you will not lose points by doing this, but hopefully gain a better understanding.

The Language of Algebra (1.1)

We will begin with the Natural Numbers, denoted as $\mathbb{N}$ . Natural Numbers are defined similarly to counting numbers, which are always positive.

How many Natural Numbers do you think exist? The answer is infinitely many; in other words, the number is currently growing and will always continue to grow. The reason why is because if you can think of a number infinitely large, I am always able to think of a number larger than it by some random margin.

Within this collection of numbers, we can separate these into three separate categories: unit numbers, prime numbers, and composite numbers.

Unit Number: There is only a single unit number. That number is the number $1$ . The number $1$ only has $1$ divisor, unlike primes and composites.
Prime Numbers: We define prime numbers to be numbers that only have $2$ divisors: $1$ and the number itself. For example, $3$ is a prime number, since it can only be divided by $1$ and $3$ . There are infinitely many prime numbers, and there is only $1$ even prime: The number $2$ .
Composite Numbers: Composite numbers are numbers that can be divided by $3$ or more numbers. Take the number $4$ for example. It can be divided by $1$ and $4$ , but it can also be divided by $2$ , meaning $4$ is not a prime. There are infinitely many composite numbers.

Say we have three natural numbers $a$ , $b$ , and $c$ .

We define $c$ as being a sum of $a$ and $b$ if added together: $a + b = c$
The same can be said for a difference between two natural numbers: $c - a = b$
When it comes to multiplication, we define $c$ being a multiple of $a$ if there is a number $b$ that if multiplied with a, we receive $c$ as the product: $a \times b = c \quad \text{or} \quad a \cdot b = c \quad \text{or} \quad ab = c$

In turn, we can define $a$ and $b$ as divisors of $c$ , if $c$ is a multiple of $a$ and $b$ . Another name for our divisors can be factors. We will be seeing the word factor a lot throughout this semester so it is important we understand its definition.

Factor: A number or expression that is a divisor for a larger number or expression without a remainder.

Divisibility Tricks

There are a few neat tricks to help us see what is a factor of a larger number:
1. A number is divisible by $2$ if the last digit is even.
2. A number is divisible by $3$ if the sum of digits is also divisible by $3$ .
3. A number is divisible by $5$ if the last digit is $5$ or $0$ .
4. A number is divisible by $6$ if divisible by both $2$ and $3$ .
5. A number is divisible by $10$ if the last digit is $0$ .

With these tricks in mind, we can write composite numbers as a product of prime numbers. This process is called prime factorization. Take for example $6$ . $6$ has $4$ divisors: $1, 2, 3,$ and $6$ . $2$ and $3$ are prime numbers and $2 \times 3 = 6$ , meaning $2$ and $3$ are the prime factors of $6$ .

We can do this with larger numbers as well. Let's try it with a number like $18$ . We know $18 = 2 \times 9$ and we know $9 = 3 \times 3$ . Therefore, the prime factorization of $18$ is: $2 \times 3 \times 3 = 2 \cdot 3^2$

With prime factoring, we can now go into the Least Common Multiple (LCM).

Least Common Multiple (LCM): The smallest possible number that is a multiple of two separate numbers.

We can find the LCM through prime factorization. Let us utilize $6$ and $9$ and find the LCM between those two: 1. Primarily factor both: $6 = 2 \times 3$ and $9 = 3 \times 3$ . 2. We then take the highest power of each prime and multiply them all together. (We will talk more about powers very shortly this lecture to better understand what I mean). 3. Doing this, we result in $2 \cdot 3 \cdot 3$ . The product of these factors is $18$ . Therefore, the LCM of $6$ and $9$ is $18$ .

$\begin{aligned} 6 &= 2 \cdot 3 \\ 9 &= 3 \cdot 3 \end{aligned}$

Remember: Take each unique prime into consideration. If two numbers have a common prime factor, only use the highest power representation. $\text{LCM}(6,9) = 2 \cdot 3 \cdot 3 = 18$

Arithmetic Operations & Expressions

Now that we have covered some basic properties of Natural Numbers, we can look into their operations to get different numbers:
* Addition: We can add two numbers $a$ and $b$ to get their sum $c$ :
$a + b = c$ * Subtraction: We can subtract $a$ from $c$ to get their difference $b$ :
$c - a = b$ * Multiplication: We can multiply $x$ and $y$ to get their product $z$ :
$x \times y = xy = x(y) = z$ * Division: We can divide $z$ (the dividend) and $y$ (the divisor) to get the quotient $x$ :
$z \div y = x$

With these operations, we can manipulate numbers and expressions once we touch on equations and inequalities. Before we go further into what they are, we should go into what expressions we can have. An expression will always consist of at least one of these components: a constant and a variable.

Constant: A number that will always remain fixed. It can never change. It can sometimes be a letter to store a value like $C$ for example to denote a constant.
Variable: A number whose value can change. This can be defined as a letter $x$ or $y$ , which are the two standard letters in Algebra.

Equalities and Inequalities

The symbol "=" is known as the equal sign. This simply tells us that one expression (or number) is equal to another ( $x$ is $y$ ).

For inequalities, there are a total of $5$ symbols we need to go through:
* $x \neq y$ ( $x$ is not equal to $y$ )
* $x > y$ ( $x$ is strictly greater than $y$ )
* $x < y$ ( $x$ is strictly less than $y$ )
* $x \geq y$ ( $x$ is inclusively greater than or equal to $y$ )
* $x \leq y$ ( $x$ is inclusively less than or equal to $y$ )

Equation: Expressions or numbers connected by " $=$ ". With this, we can also state the inequalities from earlier can serve the same purpose as an " $=$ " with our desired expressions.

You can group numbers and expressions using parentheses $(\,)$ , brackets $[\,]$ , etc., to denote which specific operation to carry out first.

Exponents

We now touch on exponents. Let us go back to prime factoring briefly. We know the prime factor of $4$ is $2 \times 2$ . Let's say now I want to find out what $2 \times 2 \times 2 \times 2$ is, but I am too lazy to do actual multiplying. We know $2$ is the only number being used and it is being multiplied by itself $4$ times. We can make an exponent with $2$ as the base and $4$ as its power: $2 \times 2 \times 2 \times 2 = 2^4 = 16$

Base: A number or expression in which we multiply it by its respective power.
Power: The number (or sometimes expression, but that's for Precalc) of times we multiply the base by itself.

Looking back, we know $9 = 3 \times 3$ , which we can now write as $3^2$ (3 squared). This can be said for any second power. Looking at $2^2 = 4$ , we also know $4 \times 2 = 8$ . But we also know $2 \times 2 = 4$ , meaning $2 \times 2 \times 2 = 2^3 = 8$ (2 cubed). This can be said for any third power. We will visit other exponents much later in the semester. However, this is sufficient to understand prime factorization for now.

Order of Operations (P.E.M.D.A.S.)

With all of this information, we can now go into the order of operations. The concept is simple. We will be using what is called P.E.M.D.A.S.:
* P (Parentheses): We work around parentheses or brackets as mentioned earlier. We prioritize what operations we do in these groups.
* E (Exponents): We then turn our focus to exponents. We need to evaluate them to simplify them into a number.
* M (Multiplication): Next we multiply numbers together, this step is interchangeable with division.
* D (Division): Next we divide numbers from each other, this step is interchangeable with multiplication.
* A (Addition): Finally, we add it all up to get our final sum, simplifying our expression or number as much as possible. This step is interchangeable with subtraction.
* S (Subtraction): Finally, we subtract to get our final difference, simplifying our result completely. This step is interchangeable with addition.

Evaluating Expressions and Combining Like Terms

Now that we have our building blocks of math, we can move on to evaluating expressions when a variable is defined as a number. Say I have an expression $2x + 8$ and I have my variable $x$ defined as $3$ . This would then simplify to: $2(3) + 8 = 6 + 8 = 14$

These expressions can have several terms with multiple coefficients. We should know what we are talking about:

Term: A constant that is alone or paired up with some number of variables. For example, $x^2$ and $xy$ are both one term, but they are not the same, since $y \neq x$ in this case.
Coefficient: A leading constant that multiplies terms.

When we want to combine like terms, we want to take terms that are the same and add or subtract them.

Example: Simplify $5x^2 + 3x - 2 + 4x^2 - x + 7$ $\begin{aligned} &= (5x^2 + 4x^2) + (3x - x) + (-2 + 7) \\ &= 9x^2 + 2x + 5 \end{aligned}$

Notice how I separated the variable numbers from the constant numbers. Also notice that not all terms are put together equally. Some with higher exponents will be put together.

What we can also do is evaluate this expression for any number $x$ that I so choose. For example, let us evaluate the previous simplified expression $9x^2 + 2x + 5$ when $x = 2$ : $9(2^2) + 2(2) + 5 = 9(4) + 4 + 5 = 36 + 4 + 5 = 45$

On the Topic of Integers (1.2)

Now that we have gone pretty far into the natural numbers, allow us to expand our horizons a bit. We are now going to implement the number $0$ and all of the negative numbers. These numbers along with all of the natural numbers brings us the entire list of integers, normally denoted as $\mathbb{Z}$ .

Every natural number has a negative side or opposite on the number line (i.e., the negative side of $n$ is $-n$ for some natural number $n$ ) and they are both the same distance to $0$ . Another name for this is the additive inverse.

Zero cannot be positive or negative. It is the only neutral number in the entire number line.

Absolute Value

Now let's go over the concept of distance mentioned just now. Say I have $4$ and its negative, $-4$ . The distance between both numbers to $0$ is $4$ . The way we can write out this relationship is to use what is called the absolute value, denoted as $|n|$ for some integer $n$ . The absolute value of an integer will never be negative. This means we have our first critical inequality: $|n| \geq 0$

This works because there is no such thing as a negative distance travelled. As a thought experiment on this, ask yourself: Does it make sense to say I have travelled negative 2 miles? It sounds ridiculous, right? This is how people who work around numbers often think to verify if things truly makes sense. (This should remind you of Philosophy a little bit but I digress).

Therefore, $|-n| = |n| = n$ for any integer $n$ . Be careful however, some absolute values can be opposite. You can consider the absolute value similar to the P in PEMDAS.

Integer Operations

Remember those operations we did earlier with the natural numbers? Well, now they can be expanded to all of the integers, which can allow us to finally introduce the additive identity: Zero. Say I have some integer $a$ . Well, the additive identity allows us to preserve that integer: $a \pm 0 = a$ This was discovered by the Mayans and Indians separately, spreading across their respective hemispheres after further development.

I also wish to introduce a clear distinction when it comes to adding or subtracting integers. We already know the case of adding two positive numbers.

If we take $1 - 2$ , that will be equal to $-1$ . Another way of writing that would be: $1 + (-2) = -1$
We can even add and subtract negatives: $-1 + (-2) = -3 \quad \text{and} \quad -2 - (-1) = -2 + 1 = 1 - 2 = -1$

The question you likely think to yourself is: Why are we able to switch around signs of numbers like that when adding or subtracting? The reason why is because not only can we add and subtract negative numbers, but we can also multiply and divide them.

For example: $(-1) \times (-2) = 2$ . The reason why this works is I can factor out $(-1)$ from $2$ , since $(-1) \times (-1) = 1$ and $1 \times 2 = 2$ . And we know $1 \times (-1) = -1$ , so $4 \times (-2) = -8$ , and so on. This can be thought of similarly for division, say $8 \div (-4) = -2$ as a random example.

We can come up with a simple rule when multiplying and dividing: If the signs are the same, the result is positive, otherwise it is negative.

With this, we can also conclude: to find the opposite of an integer, simply multiply it by $-1$ . Like natural numbers mentioned earlier, we can evaluate expressions with integers now and perform order of operations (PEMDAS), giving us a lot more to consider.

To end our first lecture, I will leave you with a very inspirational quote:

"A NEGATIVE TIMES A NEGATIVE EQUALS A POSITIVE" > — Mr. Escalante, Stand and Deliver

Lecture 2: June 2nd, 2026

Now that we have covered integers, we can now expand our knowledge of the number line even further, by looking at numbers in between integers.

On the Topic of Fractions (1.3)

Say I have two integers, $p$ and $q$ and $q \neq 0$ . Now we are going to assume that $q$ is not able to divide $p$ , meaning there will be a remainder. If necessary, revisit long division of integers while considering a remainder as this will be much more important later on in the semester.

$\textbf{Example I: Divide } 17 \div 5$

$5 \times 3 = 15$

$17 - 15 = 2$

$17 \div 5 = 3 \text{ remainder } 2$

$\boxed{17 = 5(3) + 2}$

$\textbf{Example II: Divide } 19 \div 3$

$3 \times 6 = 18$

$19 - 18 = 1$

$19 \div 3 = 3 \text{ remainder } 1$

$\boxed{19 = 6(3) + 1}$

We can either write out the quotient as a remainder, which can be tedious work. Or we can leave $\frac{p}{q}$ in its simplest form. These numbers are called rational numbers, denoted as $\mathbb{Q}$ . This way of writing $p$ divided by $q$ is called a fraction. $p$ is known as the numerator or dividend, while $q$ is denoted as the denominator, or divisor. A fraction will be simplified if $p$ and $q$ have no more common factors amongst one another.

We can not allow the denominator to ever be zero, because we simply can't divide a number into zero groups. Think about that. I will be stressing this throughout the course. We call anything divided by 0 "undefined". If we have $\frac{0}{0}$ , the term will be "Indeterminate", a term you will learn more about in Precalc, so do not expect this often throughout my lectures.

$\frac{6}{8} = \frac{3}{4}, \quad \frac{-10}{15} = \frac{-2}{3}, \quad \frac{12}{4} = 3, \quad \frac{7}{1} = 7$

From the few equations we have from the previous page, each fraction is simplified so that numerator and denominator have no common factors. You will have a better idea on how we are able to simplify these fractions later.

Like integers, we can perform PEMDAS on rational numbers. Say I have 2 rational numbers $\frac{a}{b}$ and $\frac{c}{d}$ already simplified and all four of the numbers are unique integers. When I multiply, I simply take the product of the numerators and divide it by the product of the denominators. $\frac{a}{b} * \frac{c}{d} = \frac{ac}{bd}$ Note that $\frac{ac}{bd}$ can be simplified if ac and bd have common factors. More on this later in the lecture.

To briefly get this out of the way, $a * \frac{b}{c} = \frac{a}{1} * \frac{b}{c} = \frac{ab}{c}$ This is to simply deprive ourselves of any confusion we may run into as we go on in the lectures.

Now we can introduce ourselves to the multiplicative identity: 1. This can be obtained by multiplying a rational number $\frac{a}{b}$ by its multiplicative inverse $\frac{b}{a}$ . We can easily see the result be $\frac{a}{b} * \frac{b}{a} = \frac{ab}{ba} = 1$

We will explore $ab = ba$ later this lecture, so for now take my word for it. Now we will looking at adding and subtracting fractions, as they involve the same steps. Take my two rational numbers from earlier and say I want to add or subtract.

There are 3 types of fractions we want to observe. Those with the same denominator, those with different denominators, those with similar denominators.

Fractions with different denominators:

Say I have $a,b,c,d$ as integers, $b,d \neq 0$ such that we have the expression

$\frac{a}{b} \pm \frac{c}{d}$

The requirement is for the denominators to be exactly the same when we take the sum or difference. In other words, we need to find the least common denominator (LCD). The process is the same as the least common multiple as discussed in the previous lecture The only difference is us looking at the denominator instead of the numerator.

$\text{LCD}(\frac{1}{3},\frac{1}{2}) = \text{LCM}(2,3) = 6$

The strategy is to multiply one fraction by the other denominator in the form of 1. In worther words, we need an equivalent fraction, which now is a very good time to show you what it is. For example, $1=\frac{d}{d}$ and therefore $\frac{a}{b} = 1*\frac{a}{b} = \frac{d}{d}*\frac{a}{b} = \frac{ad}{bd}$ This is an equivalent fraction. Following this process, I can rewrite $\frac{c}{d}$ as $\frac{bc}{bd}$ . Now I can either add or subtract ad with bc, giving me the result $\frac{a}{b} \pm \frac{c}{d} = \frac{ad \pm bc}{bd}$

$\frac{2}{3} + \frac{1}{4} = \frac{2*4}{3*4} + \frac{1*3}{4*3} = \frac{8}{12} + \frac{3}{12} = \frac{11}{12}$

Fractions with the same denominator c

Now say I have $a,b,c$ as integers and $c \neq 0$ . With this, I have $\frac{a}{c} \pm \frac{b}{c}$ . Then I can just operate directly across the numerators since I already have the same denominator. $\frac{a \pm b}{c}$

$\frac{3}{7} + \frac{2}{7} = \frac{3+2}{7} = \frac{5}{7}$

Fractions with similar denominators:

Now say I have $a,b,c,n$ as integers, $c, n \neq 0$ as before but now I have $\frac{a}{c} \pm \frac{b}{nc}$ . Luckily we already know that nc is a multiple of c so we only need to edit one of the fractions. Using equivalent fractions, $\frac{a}{c} = 1*\frac{a}{c} = \frac{n}{n}*\frac{a}{c} = \frac{na}{nc}$ This prompts me to finish the job. $\frac{a}{c} \pm \frac{b}{nc} = \frac{na}{nc} \pm \frac{b}{nc} = \frac{na \pm b}{nc}$

$\frac{5}{6} - \frac{1}{12} = \frac{10}{12} - \frac{1}{12} = \frac{9}{12} = \frac{3}{4}$

All three of our results may be simplifiable, depending if $ab \pm bc$ has any common factors with $bd, c$ , or $nc$ to make an equivalent, simplified fraction.

Now we can finally move on to division. Take our two rational numbers from earlier and I make what is called a complex fraction. This is basically one fraction as a numerator, and another as a denominator.

$\frac{\frac{a}{b}}{\frac{c}{d}}$

It is a very good thing I have introduced the multiplicative inverse earlier, otherwise we would have been stuck. We know that dividing an integer by itself gives us 1. Another way of saying this taking our integer and multiplying it by its multiplicative inverse. In math terms: $1 = \frac{a}{a} = a * \frac{1}{a}$ . Which means, we can take $\frac{a}{b}$ and instead multiply it by the multiplicative inverse of $\frac{c}{d}$ That would end up being $\frac{d}{c}$ Therefore, $\frac{\frac{a}{b}}{\frac{c}{d}} = \frac{a}{b} * \frac{d}{c}$ . We turn this into a multiplication of fractions, which we have covered earlier. This method is called the Keep Change Flip method (KCF) Method and as you can see it is very useful to solve problems. Just a heads up that more complex problems can have us adding or subtracting fractions in the numerator and/or denominators which can be a problem with PEMDAS. We would have to address that on the spot before we actually address the complex fraction itself.

Example of Complex Fractions with KCF: $\frac{\frac{2}{3}}{\frac{4}{5}} = \frac{2}{3} * \frac{5}{4} = \frac{10}{12} = \frac{5}{6}$

Last thing I briefly want to touch up on is exponents. Say I have $(\frac{a}{b})^2$ Well, we know from earlier that we can multiple numbers by themselves some number of times. With that said, $(\frac{a}{b})^2 = \frac{a}{b} * \frac{a}{b} = \frac{a^2}{b^2}$ . We will see this again later in the semester to over exponents specifically.

$\left(\frac{3}{4}\right)^2 = \frac{3}{4} * \frac{3}{4} = \frac{9}{16}$

Like integers, we can simplify and even evaluate expressions by utilizing rational numbers.

$\frac{2x}{4} = \frac{x}{2} \quad \text{and} \quad \frac{6y^2}{3y} = 2y$

Before moving on, I want to reiterate the use of multiplying by $-1$ . If I have $\frac{a}{b}, b \neq 0$ , where I multiply by $-1$ , the following holds true:

$-1 * \frac{a}{b} = \frac{-a}{b} = \frac{a}{-b} = -\frac{a}{b}$

On the Topic of Real Numbers (1.5)

At last, we get to the real numbers. Real Numbers, denoted as $\mathbb{R}$ , hold all of the natural numbers, integers and rational numbers. However, it also holds nonrational, or irrational numbers. These irrational numbers, denonted as $\mathbb{I}$ , will not be able to be presented as a clean fraction with only integers. Since that is the case, they can't be expressed as a decimal with a pattern either. To give a pair of examples, the number $\pi$ or pi (sounds like pie) is an irrational number (3.14...). Another irrational number is Euler's number e (2.714...) Like rational numbers, irrational numbers can be operated under PEMDAS, meaning all of the real numbers can be operated under PEMDAS as well. Now that we have gone over all possible numbers on the number line, we can now go into their properties.

The commutative property for addition is a property of real numbers such that no matter what order of the numbers you put, you will always get the same result. Say I have three real numbers, $a,b$ and $c$ , such that $a + b = c$ , but we also know $c = b + a$ . Therefore, $a + b = b + a$ Next, the commutative property for multiplication is like that of addition. Say I take $d,e$ and $f$ as real numbers such that $de = f = ed$ , meaning $ed = de$ . This covers that $ab = ba$ example we went over a little while earlier.

A fair warning that this will not always hold true for subtraction or division.

$2 - 1 = 1 \neq 1 - 2 = (-1)$ $\frac{3}{4} = 3 \div 4 \neq \frac{4}{3} = 4 \div 3$

Now for the associative property. I take 3 real numbers, $a,b$ and $c$ again and no matter where I start, similar to the communative property, I will always get the same result. $a + b + c = a + (b + c) = (a + b) + c$ . The same can be done for multiplication. $abc = a(bc) = (ab)c$

This leads to the additive identity 0 and the multiplicative identity 1, which I have gone over earlier.

Two small notes on multiplication regarding 0: 1: If you take a real number and multiply by 0, you will get 0. $0*a=0$ 2: You can not divide by 0 because 0 does not have a multiplicative inverse. I will leave you to think on why that is the case.

We now cover the distributive property which at times can allow you to break PEMDAS if that process is too tedious. Take our 3 real numbers from earlier such that the expression $a(b + c)$ is possible. Then we can result in $a(b + c) = ab + ac$ . Similarly, $(b + c)a = ba + ca$ and we can do some manipulation if we wish thanks to the communative property.

And like rational numbers, we can evaluate and simplify real numbers that were variables as the situation calls for.

As a brief example, we have the expression $\frac{6x}{9x}$ , $x \neq 0$ , we can simplify by dividing numerator and denominator by $3x$ : $\frac{6x}{9x} = \frac{6x \div 3x}{9x \div 3x} = \frac{2}{3}$

How did I know to divide by 3 along with $x$ ? That will be discussed in a future lecture, but feel free to think and take a guess as to why.

Finally, for any real number $a$ , $0 * a = 0$ and for any real number $b \neq 0$ , $0 * \frac{1}{b} = \frac{0}{b} = 0$

Lecture 3: June 3rd, 2026

With everything we have covered with real numbers, we can now go over how to generally solve linear equations. We want to find solutions.

Strategizing Linear Equations (2.1)

Solution: A value that can be substituted for a variable to make a true statement in an equation.

To verify if a value we find is a solution to our equation, we simply substitute our variable for our value and evaluate both sides of the equation if necessary to see if the equation holds true.

As an example, we have: $2x + 3 = 11$

Suppose we want to verify if $x = 4$ is a solution. Substitute: $2(4) + 3 = 8 + 3 = 11$

Since both sides equal $11$ , the value $x = 4$ is a solution.

The primary focus for this section will be a linear equation. A linear equation is one that has a variable with some leading coefficient $a$ , added on with $b$ such that $y = ax + b$ . Where $a$ and $b$ are real numbers and $a \neq 0$ .

One of the best strategies for solving linear equations is to go backwards in the PEMDAS process if your equation is simple enough to do so.

Given the equation: $3x + 5 = 20$

Step 1: Undo addition by subtracting $5$ from both sides: $3x + 5 - 5 = 20 - 5 \implies 3x = 15$
Step 2: Undo multiplication by dividing both sides by $3$ : $\frac{3x}{3} = \frac{15}{3} \implies x = 5$

So, the solution is $x = 5$ .

Another strategy would be to first simplify both sides of your equation through PEMDAS. Then I would put all of my like terms on either side. For instance, I would put all of my variables on the left hand side of my equation and all of my constants on the right hand side, then utilize reverse PEMDAS. I can check my work by simply plugging in my result into my initial equation to verify if the solution is correct.

With arguably the best strategy in all of algebra, we now need to classify equations. There are three classifications for solutions of linear equations. They can be nonexistent, meaning there is no solution. Another name for this can be called a contradiction. Solutions can also be conditional. This means that there is a finite number of solutions for an equation and all other numbers are not solutions. Or they can be an identity, meaning every single number on the number line is a solution to our equation.

Examples of Classifications

Contradiction: $2x + 3 = 2x + 5$ Subtract $2x$ from both sides: $3 = 5 \quad \text{(False! No solution)}$
Conditional Solution: $3x - 7 = 2$ $3x = 2 + 7$ $3x = 9$ $x = 3 \quad \text{(One solution, } 1 < \infty \text{, conditional)}$
Identity: $4(x + 2) = 4x + 8$ $4x + 8 = 4x + 8$ $0 = 0 \quad \text{(True for all } x \text{, identity)}$

For the rest of the lecture, I will simply be going through a few problems of linear equations with fractions, in order for us to be a bit more comfortable with these types of problems. Another reason is to actually make sure we go back and address any issues any of you may have had over the first two lectures before the first quiz. If you were not present, you do have the text book and the work book to work off of. This is a chance to work a bit more together and be creative with what problems we want to create for ourselves and challenge ourselves, which is why I will only 1 example regarding fractions to give those of you absent an idea.

$0 = -4 + \frac{3x}{4} \implies 4 = \frac{3x}{4} \implies 16 = 3x \implies x = \frac{16}{3}$

Lecture 4: June 4th, 2026

For some reason, the website is not allowing me to put up certain images that I want you to see. All of what I paste in these lecture notes are personally coded onto a separate server. For now, the images will have to wait until the Brightspace is up and running. When that happens, I will upload the PDF version of my lecture notes and you will be able to see all of the images and the visual examples there. For now, I do apologize for the ugliness and lack of images. Hopefully this is the last lecture I have to do this with for the website.

Working with Formulas to Solve for a Variable (2.3)

When solving for variables, we can generally solve linear equations.
Say I have 3 real numbers, $a$ , $b$ , and $c$ and 2 variables $x$ and $y$ , such that

$ax + by = c.$

The general idea is to solve for $y$ in terms of $x$ :

$ax + by = c \implies by = c - ax \implies y = \frac{c - ax}{b}.$

This work can be applied to formulas you may have seen in high school geometry.

For example, the volume of a sphere is

$V = \frac{4}{3} \pi r^3,$

where $r$ is the radius.

The volume of a cylinder is

$V = \pi r^2 h,$

and we can solve for a height $h$ given a radius $r$ .

Its surface area $S$ can be defined as

$S = 2\pi(r^2 + rh).$

The area of a rectangle is

$A = w \cdot l,$

and we can solve for one variable given the other.

The area of a triangle is

$A = \frac{1}{2} b h,$

where $b$ is the base and $h$ is the height.

The area of a circle is

$A = \pi r^2,$

where $r$ is the radius.

We can also convert temperature between Celsius and Fahrenheit using

$C = \frac{5}{9}(F - 32),$

which can be treated as a linear equation.

Theorem

Theorem: A statement proven not by itself, but through reasoning and based on truths that already exist.
In other words, we deduce the statement using logic and known facts.

With that, I can now introduce one of the most popular theorems in all of mathematics.

The Pythagorean Theorem

The Pythagorean Theorem, discovered by Pythagoras, states that for any right triangle with legs $a$ and $b$ , the sum of their squares equals the square of the hypotenuse $c$ :

$a^2 + b^2 = c^2.$

The proof of this theorem is at the end of the book. When I prove things in that section, I show you why the statement is true.

Example

If a right triangle has hypotenuse $c = 5$ and height $b = 3$ , the base $a$ is

$a^2 + 3^2 = 5^2$

$a^2 = 16$

$a = \sqrt{16} = \pm 4.$

Negative solutions are mathematically valid, but in measurement, we use only positive values.

Perimeter Formulas

We primarily look at the perimeter for rectangles, triangles, and circles. Their derivations (where they come from) at this level are beyond the scope of this class, so I will simply present each of them.

Rectangle:

$P(\text{rectangle}) = 2l + 2w = 2(l + w)$

Triangle:

$P(\text{triangle}) = b + h + l$

Circle:

$P(\text{circle}) = 2\pi r$

The circumference is another name for the perimeter of a circle.

Solving Linear Inequalities (2.5)

Now that we have some introduction on what an equation is, we can now introduce ourselves to inequalities.

Like equations, we can write two expressions and compare them to each other. However, as the name suggests, the expressions will not always be equal.

Suppose we have 3 real numbers, $a$ , $b$ , and $c$ where $a \neq 0$ . We can write all of these inequalities as such:

$ax + b < c$

$ax + b \le c$

$ax + b > c$

$ax + b \ge c$

$ax + b \ne c$

As a friendly reminder, there are expressions that have the chance of being equal to another expression or number, which is why you see a little line under the alligator sign.

Inequalities have all of the properties of an equation, but an equation does not hold all properties of an inequality.

For example, we know that

$2 > 1.$

Multiply both sides by 2:

$4 > 2.$

Now divide both sides by $-2$ :

$-2 > -1.$

This statement is false. To make it true, we must reverse the inequality:

$-2 < -1.$

Rule: When we multiply or divide negative numbers on an inequality, we must always flip the alligator sign to preserve the statement.

Classifications of Inequalities

Like equations, inequalities can be conditional, an identity, or a contradiction.

Example of a Condition

$x + 3 > 5 \implies x > 2$

Example of an Identity

$x - x < 1 \implies 0 < 1$

Example of a Contradiction

$2x + 1 < 2x \implies 1 < 0$

Graphing Inequalities

Solve and graph:

$x + 2 < 1 \implies x < -1$

Solve and graph:

$x \ge 1$

I know this part is difficult without an image but try to visualize this as best you can. I promise when Brightspace comes up you will actually see what I am talking about if you did not come to my lecture today.

Open circle → endpoint not included
Closed circle → endpoint included

We will discuss this further when we study compound inequalities.

Lecture 5: June 8th, 2026

I am still waiting for Brightspace to get up and running. I apologize if I allude to any images and you are not able to see at this time. I promise the images are a real help to what I want you to visualize.

Solving Compound Inequalities (2.6)

Now that we are more familiar with inequalities, we can now start compounding them onto each other. I meant what I said about math being a language. When we compound inequalities, it would normally be two conditions. When we look at these compound inequalities, the algebraic language can often be a giveaway. We have AND inequalities and OR inequalities. Think of this line: You have less than two hours and you have at least an hour. Say I make some activity function where the output is time for $x$ activities I do. Say this is what comes of it:

$2x + 1 \geq 1$ AND $x - 4 < 2$ . These inequalities can be solved to be $x \geq 0$ AND $x < 6$ . This can be compounded together to say $0 \leq x < 6$ . This would be our solution to our compound inequality.

Similarly, we have OR statements that allow us to compound inequalities. $x > 4$ OR $x \leq -4$ .

Like regular inequalities, these can also be graphed on a number line.

$0 \le x < 3$

(Visualization: A line from 0 to 3, with a closed/solid circle at 0 and an open circle at 3)

$x \le -2 \quad \text{OR} \quad x > 2$

(Visualization: A line extending left from a closed circle at -2, and a line extending right from an open circle at 2)

Before we move on, say we want to compress these AND OR statements into what we call intervals.

Interval: A period of activity between two points on the number line. This can be used to measure averages in a certain time frame.

In short: It's a way of telling us what values we want to focus on. This leads us to interval notation. In short: We use parenthesis for stuff we do not include in our interval and brackets that we do include in our interval. For example, go back to our statement $0 \leq x < 6$ . This can be written in interval notation as: [0,6). As a special note: $\pm\infty$ would never be included in an interval because neither are numbers, they are ideas, meaning intervals can be infinitely long.

Each situation varies and will call for different intervals. However, you may notice in our earlier OR example that we have $x > 4$ OR $x \leq -4$ . We would have two intervals: ( $-\infty$ ,-4] OR (4, $\infty$ ). We need a way to combine these intervals. To do that, we would need a union of intervals, denoted as $\cup$ . Our final interval would then be: ( $-\infty$ ,-4] $\cup$ (4, $\infty$ ).

A fair note before we move on is there may be some AND OR statements that may carry no solutions, meaning there will be no interval for us to write in the first place. If every single number works, our interval would simply be $(-\infty,\infty)$ or the entire number line.

Solving Absolute Value Inequalities (2.7)

Before we continue on our rabbit hole of inequalities, we need to address equations of the absolute value. Recall that an absolute value of any number can NEVER be negative. What this means computationally is that $\left\lvert a \right\lvert \neq -a$ for any $a > 0$ Any absolute value of a variable equating to a negative number in anyway is an immediate contradiction.

Let's say now I have a few real numbers such that $a\left\lvert bx + c \right\rvert + d = e$ with $a \neq 0$ . I can simplify this to instead be $\left\lvert bx + c \right\rvert$ = $\frac{e - d}{a}$ The lead condition is $e > d$ or there is no solution.

The absolute value will have exactly two solutions, because the input can be either positive or negative, and still yield a positive value.

$bx + c = \frac{e - d}{a}$ $\text{OR}$ $bx + c = -\frac{e - d}{a} = \frac{d -e}{a}$

And then I can solve for both $x$ values as normal.

Now I want to explore a case where both sides of the equation carry the absolute value. Say I have $\left\lvert A \right\rvert = \left\lvert B \right\rvert$ In either case, the solution would be $A = \pm B$

Now with that covered, we can turn to inequalities. Say we have an algebraic expression $\mathbb{F}$ such that $\left\lvert \mathbb{F} \right\rvert < a$ , for some non-negative real number $a$ . Then I will have to possible solutions:

$\mathbb{F} < a$ AND $\mathbb{F} > -a$ . The reason why we flip this alligator is because $\mathbb{F}$ can be a negative value greater than $-a$ since we know $\mathbb{F}$ can be a positive value less than $a$ . We can then express our solution as $-a < \mathbb{F} < a$ .

This can be done with the other inequality as well. When doing it with greater than signs, we can not make that same connection. Recall the AND OR statements from earlier this lecture. The less than sign can indicate the AND statement while the greater than sign can indicate the OR statement, and you can solve accordingly. here is an example of each case. As said before, $\leq \text{and} \geq$ also work.

Example 1 (Less Than → AND Statement)

$|x - 3| < 5$

Rewrite as a compound inequality:

$-5 < x - 3 < 5$

Add 3 to all three parts:

$-2 < x < 8$

Since this is a less than inequality, the solution is an AND statement:

$-2 < x \quad \text{AND} \quad x < 8$

Interval Notation:

$(-2,\,8)$

(Visualization: A line from -2 to 8, with open circles at both endpoints)

Example 2 (Greater Than → OR Statement)

$|x - 3| > 5$

Rewrite as two separate inequalities:

$x - 3 < -5 \quad \text{OR} \quad x - 3 > 5$

Solve each:

$x < -2 \quad \text{OR} \quad x > 8$

Since this is a greater than inequality, the solution is an OR statement.

Interval Notation:

$(-\infty,-2) \cup (8,\infty)$

(Visualization: A line extending left from an open circle at -2, and a line extending right from an open circle at 8)

Lecture 6: June 9th, 2026

Now that we are going deeper into algebra, we need to start talking about graphs. Graphs are visual representations of our expressions and they can come in many various shapes and forms. So many, that we will not be able to cover all types this semester, as a special class of functions are reserved for the next semester of mathematics. Graphs in math are typically done in a rectangular plane, which you will be getting to know very well throughout your math classes after this.

Graphing Linear Equations in 2 Variables (3.1)

Throughout the previous lecture, you have only been graphing along the number line. This is the lecture where you will begin to make your own in 2D. You have already seen what it is like to see graphs on a single line with their intervals, so they can be omitted from this lecture. We will only be going over 2D graphing.

We are already familiar with a point $(x,y)$ , but we have never tried plotting it on a graph. What we do first is we take a grid with some unit of measurement to provide better accuracy for our graphs.

For this, I drew a graph. If you want to follow along, I suggest turning to page 38 in your book to see a clear grid.

You see the $O$ at the very center of the graph? That $O$ stands for the point of origin. It is the point $(0,0)$ . The four quadrants help us map what is positive and what is negative. All four quadrants are separated by both axes of $x$ and $y$ .

We will now briefly go over the four quadrants (Q):

Q1: $x > 0$ AND $y > 0$
Q2: $x < 0$ but $y > 0$
Q3: $x < 0$ AND $y < 0$
Q4: $x > 0$ but $y < 0$
$x = 0$ : y-axis
$y = 0$ : x-axis

As a few brief examples: * The point $(1,1)$ is in Q1 * The point $(-1,-7)$ is in Q3 * The point $(0,1)$ is on the $y$ -axis

You will be able to start seeing this better as you put this into practice.

We now return to our linear equation $Ax + By = C$ , where $A$ and $B$ are non-zero numbers. This is a linear equation, as this can be manipulated to rather be:

$y = \frac{C - Ax}{B} = -\frac{A}{B}x + \frac{C}{B} = mx + b$

This final arrangement is known as the slope-intercept form. We will go over finding the slope $m$ and the $y$ -intercept $b$ in the next lecture.

Our point $(x_0,y_0)$ is a solution to our linear equation, as long as when we substitute $x_0$ and $y_0$ into our linear equation, the result is a true statement. As you have seen in the previous lecture, we see our linear equation is a straight line and every point on the line is guaranteed to be a solution.

Let's explore briefly what if there is no rise or no run: * If there is no rise, then there is simply no slope ( $m = 0$ ), meaning our once linear equation now becomes a constant equation $y = b$ . * Similarly, if there is no run, we will have an undefined slope, as we cannot divide by $0$ . This means we would have a vertical line at some constant $x$ value. Note, this is NOT a function, as $x$ has more than one $y$ value as a result.

Take for example $y = 3$ . This has no slope and our function will always output $3$ (a horizontal line). If we take $x = -2$ however, then we would have a vertical line at that specific $x$ value.

Now we will look at intercepts. To find the $x$ -int, we need to set $y$ = 0 and to find the $y$ -int, we need to set $x$ =0. We know the $y$ -intercept crosses the $y$ -axis and the $x$ -intercepts cross the $x$ -axis.

To solve for them: 1. Set $x = 0$ then evaluate for $y$ to get the $y$ -intercept. 2. Set $y = 0$ then solve for $x$ to get the $x$ -intercept.

Now as to actually plot such lines, what we do is we can make ourselves a table of $x$ and $y$ values. Then for each $x$ , we plug it into some expression and then evaluate in order to identify what $y$ is. After, we then plot our point on our rectangular coordinate system. We do this for several points, then we draw a line across. That is it!

As a very simple example, say I have the function:

$y = 2x + 1$

What I will do is evaluate my equation for $y$ given $x$ values ranging from $-2$ to $2$ .

x	y = 2x + 1	Ordered Pair (x, y)
-2	-3	(-2, -3)
-1	-1	(-1, -1)
0	1	(0, 1)
1	3	(1, 3)
2	5	(2, 5)

Next, I plot these points on a coordinate grid, and finally, I simply draw a straight line passing directly across these points to visualize our trend.

Lecture 7: June 10th, 2026

I am posting this a night early, because I was invited to a Mets game for the night before this lecture. Better to get ahead of this now than to worry about it when I get home.

Now that we are beginning to familiarize ourselves with these linear graphs, it is time to know how we are able to mathematically interpret them.

On the Topic of Slopes (3.2)

The slope can determine the trend our linear equation takes as it is a constant rate of change. The equation is $y = mx + b$ where $y$ is our output, $m$ is our slope, $x$ is our input, and $b$ is our initial value. We will touch more on what an initial value is shortly.

Before finding slopes, we must discuss ordered pairs, or points. A point in short is for some $x$ value we insert, we get back a $y$ value. This can be presented as $(x,y)$ , where $y$ is the output for some input $x$ .

To find the slope, we need to take the rise of our line, or the difference of two random $y$ -values on a graph, and divide it by the run, or the change of the $x$ -values we used to find our two $y$ values mentioned earlier. This then gives us a very practical formula:

$m = \frac{y_2 - y_1}{x_2 - x_1}$

Note: This holds true for any two points $(x_1,y_1)$ and $(x_2,y_2)$ where $x_2 \neq x_1$ .

If $m$ is positive, then our line will be pointing upward at a slanted angle.
If negative, it will be pointing downward at a slanted angle.
If there is no rise ( $y_2 - y_1 = 0$ ) then the line will be horizontal.
If there is no run ( $x_2 - x_1 = 0$ ) then the slope will be undefined, resulting in a vertical line.

With the slope defined, let's talk about our initial value. Say I have $y = mx + b$ . If $x = 0$ , then $y = b$ since $m(0)=0$ . Our value $b$ is going to be on the $y$ -axis directly, crossing over between positive and negative $x$ values, which is what we call the $y$ -intercept ( $y$ -int).

We have already talked about the lines and slope-intercept form. We will briefly talk about intercepts.

Say I have 3 real numbers such that $Ax + By = C$ . What I can do is find the $y$ -int by setting $x = 0$ . Then $By = C$ , leading us to $y = \frac{C}{B}$ . Now let us talk about the $x$ -intercept ( $x$ -int).

$x$ -intercept: $x$ values that force our output $y$ to be $0$ , such that $(x,0)$ exists. These points sit only on the $x$ -axis.

So if $y = 0$ , then $Ax = C$ implying our $x$ -int will be $x = \frac{C}{A}$ . Therefore, our intercepts are:

$\left(0, \frac{C}{B}\right) \quad \text{and} \quad \left(\frac{C}{A}, 0\right)$

Many real world applications utilize linear modeling the way we will be doing so in examples. This can be done with conversions and modeling linear trends. As shown throughout, we are able to even graph these lines based off the intercepts alone..

We will now briefly touch on parallel lines (denoted as $\parallel$ ) and perpendicular lines (denoted as $\perp$ ).

Say I have two equations:

$y_1 = m_1x + b_1$
$y_2 = m_2x + b_2$
Parallel: $y_1$ and $y_2$ are parallel ( $y_1 \parallel y_2$ ) if and only if they have the same slope: $m_1 = m_2$
Perpendicular: $y_1$ and $y_2$ are perpendicular to each other ( $y_1 \perp y_2$ ) if and only if one of their slopes is the negative reciprocal of the other: $m_2 = -\frac{1}{m_1}$ (You can easily solve for $m_1$ in terms of $m_2$ to verify going the other way).

Parallel Example

Consider the two lines: $y_1 = 2x + 1$ $y_2 = 2x - 3$

Both lines have slope $m = 2$ . Since $m_1 = m_2$ , the lines are parallel ( $y_1 \parallel y_2$ ).

Perpendicular Example

Consider the line: $y_1 = 2x + 1$

The slope is $m_1 = 2$ . The negative reciprocal of $2$ is: $m_2 = -\frac{1}{2}$

Now consider a third line: $y_3 = -\frac{1}{2}x + 4$

Since $m_3 = -\frac{1}{m_1}$ , the lines are perpendicular ( $y_1 \perp y_3$ ).

Finding Line Equations (3.3)

Recall from the previous section where we first defined $m$ to be $\frac{y_2 - y_1}{x_2 - x_1}$ for any two points $(x_1,y_1), (x_2,y_2)$ . Well, we can simplify this equation just a small bit by multiplying the denominator on both sides of the equation, yielding:

$y_2 - y_1 = m(x_2 - x_1)$

Now we change the second point to an open variable point $(x,y)$ , meaning we only need $1$ concrete fixed point instead of two, yielding:

$y - y_1 = m(x - x_1)$

This equation is known as the point-slope form for some point $(x_1,y_1)$ and slope $m$ . If we were to isolate and solve for $y$ , we would end up with:

$y = mx + (y_1 - mx_1)$

We can set $b = (y_1 - mx_1)$ since this combination is purely composed of real numbers (constants). This yields what we have been working with from the beginning:

$y = mx + b$

This is what we call slope-intercept form.

Example

Let's take a slope $m = 3$ going through a point $(x_1,y_1) = (4,5)$ . I can utilize point-slope form to write it as:

$y - 5 = 3(x - 4)$ $y - 5 = 3x - 12 \implies y = 3x - 7$

When solving for a slope from scratch, we can always refer to our first definition of the slope $m$ from the previous section, but usually, you will have the slope already provided or implied.

With everything that we know now of parallel and perpendicular lines, we can even find these line equations, given other line equations. What we do is we select two random points, preferably the intercepts, and solve for our $m$ . Then we can find our line given the point we need our new line to pass through. This can be done in either structural setup!

Lecture 8: June 11th, 2025

We are now ready to talk about functions this lecture. We will then discuss graphing said functions in the next lecture. We have been building ourselves up to this lecture since the beginning and this will only be the absolute tip of the iceberg for all of Mathematics.

On the Topic of Functions and Relations (3.5)

Now that we are more comfortable with inputs and outputs, we can finally talk about functions. They are a fundamental part of how the real world works and you will be seeing real world examples every single day, no matter where you look.

For example, pretend you are on an escalator and you rise 1 step for each step you take, and you are initially 2 steps above. I can predict how tall I will be on this escalator by utilizing linear functions. I know my output to be y and my slope m = 1. I also know I am initially 2 steps above already. So I can define my new height as:

y = x + 2

where x is what step I am currently on and y is how many steps high I am.

With many functions, we can come up with relations.

Relation: A set (or collection) of ordered pairs (x, y). This prompts us to discover domains and ranges of our function.
Domain: All possible x values that make the function possible. Another name for x can be the independent variable.
Range: All possible y values that come from our domain. Another name for y can be the dependent variable.

Example 1: Linear Function

f(x) = 2x - 5

There are no restrictions on x.

Domain: (-∞, ∞)

Since a linear function can produce any real number output,

Range: (-∞, ∞)

Example 2: Rational Function

f(x) = 1 / (x - 3)

The denominator cannot be zero:

x - 3 \neq 0 \Rightarrow x \neq 3

Domain: (-∞, 3) ∪ (3, ∞)

The function can never equal zero (because the numerator is 1).

Range: (-∞, 0) ∪ (0, ∞)

Example 3: Square Root Function

f(x) = \sqrt(x - 2)

The expression inside the square root must be non-negative:

x - 2 \geq 0 \Rightarrow x \geq 2

Domain: [2, ∞)

Since square roots are always non-negative,

Range: [0, ∞)

Mapping

Mapping: A visual or tabular way to show how each input x is paired with an output y in a relation or function.

Example 1: Mapping for a Function

Consider the function:

f(x) = x 2

We can create a mapping table:

x	f(x) = y
-2	4
-1	1
0	0
1	1
2	4

Mapping Diagram:

x = {-2, -1, 0, 1, 2} y = {0, 1, 4}

-2 → 4, -1 → 1, 0 → 0, 1 → 1, 2 → 4

Here, each x value is mapped to its corresponding y value.

Defining a Function

With all of these pieces of information, we can finally give a solid definition of what a function is.

Function: A relation assigning x values in the domain to a single y value in the range.

With this definition, we can also make this statement: If an x value has more than one y value, we will not have a function.

In terms of notation, x (our input) is our domain value. f is the name of our function, and f(x) acts the same way as y, denoting our range from our inputs along our domain.

Notation:

x = input (domain)
f = function
f(x) = y = output (range)

Example 1: A Function

Consider the relation: f(x) = x²

x	f(x)
-2	4
-1	1
0	0
1	1
2	4

✓ Each x has exactly one y value. Therefore, this relation is a function.

Example 2: Not a Function

Consider the relation given by the circle: x² + y² = 4

x	y
-2	0
-1	±√3
0	±2
1	±√3
2	0

Notice that for x = 1, there are two y values: y = √3 and y = -√3.
Therefore, this relation is not a function.

We can verify logically by confirming if one input has either a single, or multiple outputs. If an input has more than one output, the relation is not a function.

With this, we can plug in any input we want and attempt to evaluate it through our function to obtain our output. Also notice how when x = 0, we have two y values. We can verify this graphically through what we call the Vertical Line Test.

The Vertical Line Test: Take a vertical line and scan it from left to right of your graph. If one x value has more than one y value, your graph is not a function!

We will go over why not all inputs will yield an output later in the semester, but for now, they will.

Evaluating Algebraic Inputs

Before the lecture ends, I want to introduce one last concept regarding inputs and outputs. As an example, take:

f(x) = x 2 + 4x + 2

Remember: x is my input.

Say I want my new input to be x = x + 5 and I want to apply this to my new function. All I have to do is make that substitution in this function:

f(x + 5) = (x + 5) 2 + 4(x + 5) + 2 = x 2 + 10x + 25 + 4x + 20 + 2 = x 2 + 14x + 47

What I can also do is add to my output to get a different output. Say I want f(2) + 5. I can just take the output and add 5:

f(2) + 5 = (2) 2 + 4(2) + 2 + 5 = 19

There are cases where I can do both at the same time, but I digress, since it is simply adding one step for a change in input. You will be getting more familiar with these types of problems in Precalc, but it is worth introducing them to you lightly now, while we are still covering functions from an algebraic standpoint, before we look into functions' graphs.

Before moving on, I wish to reiterate how important functions are in the real world. It allows us to solve practical problems and at times accurately predict a few events. You can find examples in a lot more places than you realize.

Lecture 9: June 15th, 2026

With an introduction to functions, we now turn to visualizing them in greater detail. We will also be looking at systems of linear equations.

Graphing Functions (3.6)

We now talk about a function's graph. We know that we can define y as a function of x. In other words, y = f(x), where f is our function we input x into to obtain our output. This yields points of (x, y) = (x, f(x)).

We already discussed that the domain and range of a linear function is every single real number on the number line.

Domain: (-∞, ∞)
Range: (-∞, ∞)

We also know the domain of a constant function and its range. We take y = b.

Domain: (-∞, ∞)
Range: {b} Note that the curly brackets show this is the ONLY number where the range is presented.

Starting off simple, we will start off with the Identity function: y = f(x) = x. We already know its domain and range and y-intercept since it is linear and b = 0 and since there is no coefficient in front of the x, m = 1.

x	x
-3	-3
-2	-2
-1	-1
0	0
1	1
2	2
3	3

Graph of y = x

[Graph: Linear function passing through the origin with a slope of 1, spanning from x = -3 to 3 and y = -3 to 3]

Domain: (-∞, ∞)
Range: (-∞, ∞)

Now we are going to make things a bit more interesting. Let us take the function f(x) = x². We know its domain will obviously be the entire number line but what about its range? Well, we know that any number we put in will always yield a non-negative number, unless x = 0. We can even check with a table.

x	x²
-3	9
-2	4
-1	1
0	0
1	1
2	4
3	9

Graph of y = x²

[Graph: U-shaped parabola opening upwards with vertex at (0,0), reaching up to y = 9 at x = -3 and x = 3]

Domain: (-∞, ∞)
Range: [0, ∞)

We can now take it up a notch and say we look at f(x) = x³.

Graph of y = x³

[Graph: S-shaped cubic curve passing through the origin, increasing sharply into quadrant 1 and decreasing into quadrant 3]

We make our table in order to justify the curve of our graph.

x	x³
-3	-27
-2	-8
-1	-1
0	0
1	1
2	8
3	27

Domain: (-∞, ∞)

Range: (-∞, ∞)

Looking at both the table and the graph again, we can easily see both the domain and range to be the same as our linear equation. I will allow you to ponder why before we move on.

Finally, we will be looking at f(x) = |x| and √x. You should be able to make a table for each. As for square root of x, I suggest you start memorizing the squares of the first 10 numbers in order to make your lives easier and go a bit quicker. I will leave them in a table which you will see very shortly.

Graph of y = |x|

[Graph: V-shaped absolute value graph with the vertex at (0,0), perfectly symmetric across the y-axis]

x	\|x\|
-5	5
-4	4
-3	3
-2	2
-1	1
0	0
1	1
2	2
3	3
4	4
5	5

Domain: (-∞, ∞)
Range: [0, ∞)

Graph of y = √x

[Graph: Smooth curve starting at (0,0) and rising gradually to the right into the first quadrant]

x	√x
0	0
1	1
4	2
9	3
16	4
25	5
36	6
49	7
64	8
81	9
100	10

Domain: [0, ∞)

Range: [0, ∞)

You will look at different forms of these functions in greater detail in Precalc. I will however leave you with their name:

Transformations

We have been working a while with linear equations by themselves and we have been able to solve for their intercepts and even plot their graphs. We are now going to make things more interesting by working on a system of these linear equations.

Solving Systems of Linear Equations with two Variables (4.1)

This course will only pair two linear equations of two variables together to make simple systems.

System of Linear Equations: Multiple equations paired together, being solutions under the same condition of being true, if they both have the solution (x, y).

To put it in perspective, it will not only be solving for one variable in terms of the other, but to make sure that a solution satisfies both equations. Like a single linear equation, we plug in a potential solution (x, y) into our system to verify if the statement does hold true.

To get us started, I will have two linear equations and put them together as a system. Say I have six random real numbers that are not 0, and variables x, y to represent our point. With this, we can make our random system to be

Ax + By = C
Gx + Hy = K

The process will seem scary at first, but after some practice it will be simple. First we can solve for one variable in terms of the other. Normally we would start with the easier of the two equations, but I will work on the top equation first. I will solve for the variable y in terms of x.

Ax + By = C

By = C - Ax

y = ^{(C - Ax)}⁄_B

Now that I solved for y in terms of x in my first equation, I can now turn to my second equation. The trick here is to substitute our y in the second equation with what we solved for in the first equation.

Gx + Hy = Gx + H( ^{(C - Ax)}⁄_B )

K = Gx + H( ^{(C - Ax)}⁄_B )

K = Gx + ^HC⁄_B - ^HA⁄_Bx

K - ^HC⁄_B = Gx - ^HA⁄_Bx

K - ^HC⁄_B = x(G - ^HA⁄_B)

x = ^{(K - ^HC⁄_B)} ⁄ _{(G -
^HA⁄_B)} = ^{(BK - HC)} ⁄ _{(GB
- HA)}

Now that we have solved for our x, we can solve for our y by plugging x into the equation where we solved for y earlier.

y = ^{( C - A( ^{(BK - HC)}⁄_{(GB - HA)} )
)} ⁄ _B

And we are done. We have successfully solved for our first linear equation. Now I can imagine you thinking: What in the world did I do and why use random letters? This looks scary, I know. The reason why I introduce the process this way is to introduce to you how people are prone to proving things using any value with certain conditions.

The other reason why I introduce it to you this way, is to show that there are two conditions that must hold, otherwise there will be no solution:

1) B ≠ 0
2) GB ≠ HA

The reason why these conditions are so strict is because we can not divide by 0 under any circumstances. The value will just simply not be definable. I wanted to do what I did in order to introduce these conditions to you in order for you to watch out for any surprises. Rest assured: This is only one way to solve a system of equations and there is no one size fits all. There are multiple algorithms to solve systems of equations and I will even introduce one other way now. Just note: Different algorithms have different conditions, so take care of what you do when you see this in a real world background.

I can imagine you thinking a lot when I introduced all of this, and it can easily be a bit overwhelming. To bring us back down, I am going to give a trio of examples when a system of linear equations do and do not have solutions so we can get a feel for the computations. I will also be implementing graphs so you can see how you can sometimes tell how to identify them by simply looking at a graph.

Remember earlier this semester with identities, conditionals and contradictions? Well they can be applied here as well. I will be presenting one of each and we can move on to what I believe is the second most interesting thing you will learn in this book of lectures.

All three examples will be simple integers, but do note that rational numbers do have a chance of showing up, so it is crucial that you get a little practice in with those.

A Conditional System

½x + ⅓y = 2
&frac33;⁄₄x - ½y = 1

Multiply the first equation by 6:

3x + 2y = 12

Multiply the second equation by 4:

3x - 2y = 4

Add the equations:

6x = 16

x = ⁸⁄₃

Substitute back:

3(⁸⁄₃) + 2y = 12

8 + 2y = 12

2y = 4

y = 2

(x, y) = (⁸⁄₃, 2)

[Graph: Two intersecting lines. One blue line sloping down representing ½x + ⅓y = 2, one red line sloping up representing ¾x - ½y = 1. They intersect at a single distinct point (*)]

Notice how the lines perfectly intersect at only one point and they go their separate ways.

A contradiction of a System

x + y = 2
x + y = 5

I can take y = 2 - x and plug that into our second equation.

x + 2 - x = 2 = 5

This statement is obviously false, so there is no solution to our system of linear equations for any point (x, y). This is called an inconsistent system.

[Graph: Two parallel lines that never touch. Blue line below passing through y = 2, Red line above passing through y = 5]

Notice how these two lines are parallel, meaning they will never touch.

An Identity in a System

x + y = 4
2x + 2y = 8

Same song and dance.

y = 4 - x

Plug it into our second equation.

2x + 2(4 - x) = 2x + 8 - 2x = 8 = 8

This will always be true for any combination of (x, y), resulting this to be an identity.

[Graph: A single line line path on which both equations reside. A solid blue line overlapping perfectly with a dashed green line]

Here I want you to realize that the second line is joined in with the first line. If you notice, the second equation is simply double the first. There is a special area of this called linear dependence, but I digress as that is a different class altogether. However, this class of lines is called Coincident Lines. One last thing I want to note: For infinitely many solutions, you can write it as an ordered pair: (x,y) = (x, mx + b). Where you can solve for y as a line and substitute said line.

As promised, the other way of doing this is what is known as the elimination method. In short, we can multiply the system with certain factors in order to have coefficients of one variable opposite to each other. We then add these equations and then solve for the other. I will only do a single example in these notes to give you an idea.

3x + y = 5
2x - 3y = 7

In most cases of math, there is more than one correct method to solve this system through the elimination method. For this example, I will be eliminating the y-variable.

3x + y = 5 ⇒ 3(3x + y) = 3(5) ⇒ 9x + 3y = 15

I now have a new system of equations to solve.

9x + 3y = 15
2x - 3y = 7

What I will do now is add these two equations together, eliminating the y variable, allowing me to solve for x. Doing this will leave me with the following:

9x + 3y = 15
+
2x - 3y = 7

11x + 0y = 22

⇒ 11x = 22 ⇒ x = 2 ⇒ y = -1

(x, y) = (2, -1)

And we can plug our proposed solution into our original system to verify whether it is a valid solution.

Of course, if all else fails, we can simply solve both equations for y and plot them on the same graph and find the point where they intersect.

Lecture 10: June 16th, 2026

This will be one of the most satisfying sections I will be teaching in this class, because of how nicely things clean up. You will very shortly understand what I mean by that when we start working with exponents.

Properties of Exponents and a Review of Scientific Notation (5.2)

Remember at the very start of the semester we had our statement 2 · 2 · 2 = 2³ = 8. We can just say this is two cubed or two to the third power. The two at the bottom tells us what we are multiplying, the number at the top tells us how many times we multiply our number by. They are called the base and the exponent or power respectively.

Let's make this a little bit more interesting now. We have a very basic equation 8 · 4 = 32. Here is the interesting part however: 32 = 2⁵. We also know 2 · 2 = 2² = 4. So we can write 8 · 4 = 2 · 2 · 2 · 2 · 2 = 2⁵. But we can also write 32 = 8 · 4 = 2³ · 2² = 2⁵. With all of this, we have:

8 · 4 = 32 = 2⁵ = 2³ · 2²

I want you to specifically pay attention to our most recent equation. Notice anything strange about it? Notice how the exponents add to 5, which ends up being our resulting exponent. Let's generalize it a bit more.

Let's say we have our old fashioned variable x and we want to square it. To square something, we know we just take a number and multiply it by itself.

x · x = x²

Simple right? Now what if I tell you we can write x = x¹? We can rewrite our equation to now be x¹ · x¹ = x². The thing is: 1 + 1 = 2. And we also know 2 to be our resulting exponent. In case you have not noticed yet, the exponents of the same base add with each other! We can write a rule out of this.

For some base a and two exponents m and n:

a^m · aⁿ = a^m+n

Another rule we can come up with is for any number or expression with a power of 1, the result will just be the original expression.

a¹ = a for some base a

This is where things start to get more interesting now. We can do this for multiplying, but what about dividing? Before we get into dividing, we must revisit the multiplicative inverse and additive identity.

Recall the additive identity 0 such that 0 + m = m. We also know the additive identity can be obtained by adding a number with its additive inverse, or opposite. Given that fact, we can make the following equation: m + (-m) = 0. Now say we apply the power of 0 to a base: a⁰, where a ≠ 0. This would result in the multiplicative identity 1. Now we can make this rule:

a⁰ = 1 for any base a ≠ 0

Now recall we can obtain the multiplicative identity 1 by multiplying a by its reciprocal ¹⁄_a to get 1. That would mean we can come up with this equation:

a · (1/a) = 1 = a⁰ = a¹ · a^x = a^1+x

We can then rewrite this as 0 = 1 + x. We can immediately see that x = -1. With this, we can make a rule:

1/a = a^-1 for some base a ≠ 0

With this, now we can get into some division.

I have x³ and I want to divide it by x². We know this to be x if we factor out both of our expressions fully, but let's see what happens if we apply what we have just learnt.

x³ / x² = x³ · x^-2 = x^3-2 = x¹ = x

We can now come up with a new rule for subtracting! For any base a,

a^m / aⁿ = a^m-n

Expanding further into negative exponents, think back to our multiplicative inverse of ^a⁄_b from back in Lecture 2. We can now define this algebraically like so:

(a/b)^-1 = b/a

If you notice, we can expand this relationship through the dividend and the divisor at the same time.

(a/b)^-1 = a^-1 / b^-1 = (1/a) / (1/b) = b · (1/a) = b/a

With this revelation, we can make a new rule that will make a fine addition to our collection.

(a/b)^m = a^m / b^m, b ≠ 0

As an example, recall earlier, way back in Lecture 2, where we were squaring three quarters:

(3/4)² = 9/16

Now what happens if I were to square ⁹⁄₁₆? The idea is, we can write this in a more favorable condition.

(9/16)² = ((3/4)²)² = (3/4)² · (3/4)² = (3/4)⁴

You can see the exponents multiplying to get the product of 4! With this, we have a new rule:

(aⁿ)^m = a^mn

We can also write ³⁄₄ = 3 · ¹⁄₄, allowing us to rewrite (3/4)⁴ = 3⁴ · (1/4)⁴.

This means we can have a and b take the same power separately and we would retain the same result! In other words, we have a new rule:

(ab)^m = a^m b^m

We also can get (1/4)⁴ = 1 / 4⁴. With this, we can write a new rule:

(a/b)^m = a^m / b^m

This is all very overwhelming which I completely understand. It will become a bit more satisfying once you start doing practice problems. In the meantime however, I am going to summarize it in one nice table. This may be very useful to you later so hopefully you do not take this for granted.

Property	Rule	Example
Product of Powers	a^m · aⁿ = a^m+n	2³ · 2⁴ = 2⁷
Quotient of Powers	a^m / aⁿ = a^m-n	5⁶ / 5² = 5⁴
Power of a Power	(a^m)ⁿ = a^mn	(3²)⁴ = 3⁸
Power of a Product	(ab)ⁿ = aⁿ bⁿ	(2 · 5)³ = 2³ · 5³
Power of a Quotient	(a/b)ⁿ = aⁿ / bⁿ	(3/4)² = 9/16
Zero Exponent	a⁰ = 1	7⁰ = 1
Negative Exponent	a^-n = 1 / aⁿ	2^-3 = 1/8
Exponent of 1	a¹ = a	9¹ = 9

There is one last rule in exponents, but that will come much later in the course. In the meantime we are going to visit one of the biggest concepts in all of the mathematical sciences.

We are going to take a brief look at scientific notation, which deals a lot with decimals. The biggest thing to note is that every single decimal deals with some power of 10ⁿ.

For example, I have 150, which I can write as 1.5 · 10² or 1.5E2. Another example would be 0.03 = 3 · 10^-2 = 3E-2. You can see if our number is greater than 1, we will have a positive exponent of 10. If it is less than 1, we see a negative exponent of 10. Let's explore this a bit more.

Scientific Notation: A number expressed as a multiple to a power of 10.

This can be written as a · 10ⁿ or aEn, where n is our exponent typically written as an integer, and 1 ≤ |a| < 10.

The general idea is: for each time you move the decimal to the right, you take your number and divide by 10, and for each time you move it to the left, you multiply by 10. Be sure to add the necessary zeros.

Move decimal to the left:

45000.0 = 4.5 × 10⁴

Move decimal to the right:

0.0032 = 3.2 × 10^-3

To get the number without the power of 10, simply move the decimal in the opposite direction.

Move decimal to the right (positive exponent):

4.5 × 10⁴ = 45000

Move decimal to the left (negative exponent):

3.2 × 10^-3 = 0.0032

Adding and Subtracting Polynomials (5.1)

This section should be short and sweet. We can finally talk about polynomial expressions. For a while now we have been working with linear expressions, generally in the form mx + b. This is called a binomial, where a "nomial" is an expression with some number of terms.

For example: A monomial is simply a one-term expression. It can have multiple variables multiplied to make one single term. Our constant b, and a random product a(x²)y(z³) are both monomials.

A binomial is an expression with two terms, hence "bi". For example, mx + b is a binomial.

Now we can talk about a trinomial, which is an expression with 3 terms. For example, we have x² + 2x + 1.

We can keep going with 4 or more terms, but that can be exhausting. Instead, we generalize 4 or more terms as a polynomial, which is an expression with four or more terms. Now we need to talk about their degrees. And I do not mean by 360 degrees or temperature, or even diplomas.

The Degree of a Polynomial: The highest exponent on the polynomial (polynomial degree).

For example, I have x² + 2x + 1. The degree of this polynomial would be 2.

Even numbers can be polynomials. 7 is a monomial and its degree is 0, since it has no variables to go along with it.

The Degree of a Term: The sum of exponents from each variable. Example: The degree of 2x³y⁴ is 7.

The degree of a term will always determine the degree of any polynomial. The highest degree always wins.

Like linear expressions, we are able to combine like terms, just like we did all the way back in the very first lecture.

4x + 3 + 6x - 2 = (4x + 6x) + (3 - 2) = 10x + 1

Needless to say, this can happen with higher degree polynomials. Now we can finally talk about polynomial functions.

Polynomial Function: A function defined by a polynomial. For example: f(x) = 4x³ + 3x² + 3x + 43 is a polynomial function. We can simply evaluate various x values to yield various outputs.

Just like numbers and expressions, we can add and subtract functions. The rule is generally simple and very easy to understand:

If I have f(x) and g(x) as polynomial functions,

(f ± g)(x) = f(x) ± g(x)

Say I have f(x) = x² and g(x) = 30

(f + g)(x) = x² + 40
(f - g)(x) = x² - 40

Like other functions, we can plug in various inputs to evaluate for various outputs. This is just simple evaluation and doing math, so we will just end the lecture here.

Like other functions, we can plug in various inputs to evaluate for outputs. This is simple evaluation.

Lecture 11: June 17th, 2026

This took multiple hours and for some reason, the website has not been cooperative. I sorry if it looks shorter than normal, as I mainly carried the essentials of this lecture over to Alium.

Now that we are going further along with polynomials, it is time we start showing more of the properties that they hold.

Multiplying Polynomials (5.3)

This is where the distributive property finally starts to be put forth at the center stage. Recall the distributive property:

a(b + c) = ab + ac = ba + ca = (b + c)a

Staying on the topic of functions, in addition to adding and subtracting polynomial functions, we can also multiply them.

(f · g)(x) = f(x) · g(x)

This rule becomes especially important as we increase the degree of polynomials.

Products with Monomials:

We distribute a single term across another expression. For example:

3x(3x + 4) = 9x² + 12x

Products with Binomials:

We distribute each term in one binomial across the other:

(3x + 2)(x + 1) = 3x(x + 1) + 2(x + 1)

= 3x² + 3x + 2x + 2 = 3x² + 5x + 2

This is often remembered as the FOIL method.

(3x + 2)(x + 1) = (3x)(x) + (3x)(1) + (2)(x) + (2)(1)

= 3x² + 5x + 2

Products with Polynomials:

(2x² + 3x - 1)(x + 4)

= 2x³ + 8x² + 3x² + 12x - x - 4

= 2x³ + 11x² + 11x - 4

Special Products:

Binomial Squares Pattern:

(a ± b)² = a² ± 2ab + b²

Conjugate Pairs / Difference of Squares:

(a + b)(a - b) = a² - b²

Function Multiplication Example

If f(x) = x + 3 and g(x) = x³ + 3:

(f · g)(x) = (x + 3)(x³ + 3)

= x⁴ + 3x³ + 3x + 3

Dividing Polynomials (5.4)

Dividing Monomials:

(54a²b²) ÷ (6ab⁴) = 9a / b² = 9ab^-2

Dividing a Polynomial by a Monomial:

(24x⁴y² - 42x) ÷ (2xy)

= 12x³y - 21y^-1

Polynomial Long Division:

We divide polynomials the same way we divide numbers.

Example:

(x² + 9x + 20) ÷ (x + 5)

Result:

x + 4

With Remainder:

(x² + 9x + 22) ÷ (x + 5)

= x + 4 + 2/(x + 5)

The remainder is divided by the original divisor.

Like multiplication, I can use functions in division! (f/g)(x) = f(x)/g(x)

Lecture 12: June 18th, 2026

Factoring will be the cornerstone for what is to come for the rest of the course. This can be seen as another form of dividing.

Grouping and the Greatest Common Factor (6.1)

Recall the word factor as a noun rather than a verb. A factor is a term that can be multiplied with another that will yield a larger product. Sometimes, these terms can share common factors, allowing us to better organize expressions and equations. With that, we can discuss what a Greatest Common Factor is.

Greatest Common Factor: The greatest term shared between expressions. Examples: GCF(4, 6) = 2, GCF(x², x³) = x²

We have seen a bit of this process all the way in our very first lecture when we were working on prime factorization.

As an example, take 25x⁴, 35x³, and 20x². What I am able to do is expand it out in terms of prime factorization.

25x⁴ = 5 · 5 · x · x · x · x

35x³ = 5 · 7 · x · x · x

20x² = 5 · 2 · 2 · x · x

With that, we can identify that 5, and a pair of x factors make up the most common factors. Therefore, the Greatest Common Factor of the three would be 5x². Now that you have been introduced to this, we shall move on.

Recall the distributive property a(b + c) = ab + ac. I know that the GCF(ab, ac) = a, so I can factor out a from both b and c: ab + ac = a(b + c)

As a friendly example, take x² + x. I know the GCF(x², x) = x, so I can factor out x from my expression.

x² + x = x(x + 1)

We have been looking at monomials to factor, now let's look at binomials. As an example, say I want to factor 8n(n - 4) + 5(n - 4).

The GCF(8n(n - 4), 5(n - 4)) = (n - 4) ⇒ 8n(n - 4) + 5(n - 4) = (n - 4)(8n + 5)

Sometimes, factoring polynomials will not be nearly as easy as the example we just did. Luckily, we were able to come up with a strategy to make factoring easier. That strategy is called grouping.

Say I have xy + 8y + 3x + 24. What I can do is I can group certain common factors amongst each other. The neat part about this is the fact no matter what path I choose to take in grouping, it will always lead to the same, correct answer.

For this example, I will be grouping the y values together away from those without y values. This can be done with parentheses.

xy + 8y + 3x + 24 = (xy + 8y) + (3x + 24)

What I can do next is take the GCF between my two groups.

GCF(xy, 8y) = y and GCF(3x, 24) = 3

With that, I can factor out my GCFs from their respective groups.

(xy + 8y) + (3x + 24) = y(x + 8) + 3(x + 8)

And now you notice it is the same situation as our earlier example, allowing us to finish factoring completely.

y(x + 8) + 3(x + 8) = (x + 8)(y + 3) = xy + 3x + 8y + 24

Lecture 13: June 23rd, 2026

Now that we have an idea of factoring through GCF and grouping, we can now turn to trinomials. Binomials are so easy, so we will simply move on.

Factoring Trinomials (6.2)

We will first start by going over trinomials with the form x² + bx + c for some real numbers b and c. (This form is known as the quadratic trinomial) The idea is I want to find two numbers m and n such that (m + n) = b and mn = c. With that in mind, I can rewrite my expression like so:

x² + bx + c = x² + (m + n)x + mn = x² + mx + nx + mn

Now we can simply factor by grouping and GCF at the same time:

x² + mx + nx + mn = (x² + mx) + (nx + mn) = x(x + m) + n(x + m)

Our GCF is now x + m:

x(x + m) + n(x + m) = (x + m)(x + n)

Like earlier, there is more than one correct path to factoring such trinomials.

Now let's make this interesting with an example. Recall from the previous lecture I had the trinomial x² + 9x + 20. We already know the factors to be (x + 5)(x + 4) but now we are going to show that this is the case without the need for division.

The first step is to separate the middle term so I can factor with the first and last terms. A fair warning that this will vary by case. 5 and 4 sum to 9 and multiply to 20. With that said,

x² + 9x + 20 = x² + (4 + 5)x + 4 · 5 = x² + 4x + 5x + 20

What I can do next is group out the first two terms from the last two terms.

(x² + 4x) + (5x + 20)

The GCF of the first group is x and the GCF of the other is 5:

x(x + 4) + 5(x + 4)

Finally, I can factor out x + 4 from both x and 5, resulting in:

(x + 5)(x + 4)

Going back to the first step, I could have also grouped the first and third term differently. The idea is that there is more than one correct route to get the correct answer. You can have different ideas of grouping, but the point is to get back to the correct answer. This can be confirmed simply by multiplying our factors together through the distributive property.

Let's take a look at two more examples before we start writing another rule.

I want to factor out x² - 11x + 28. So now I need two numbers that add to -11 and multiply to 28. Luckily, -7 and -4 will do the trick:

x² - 11x + 28 = (x - 7)(x - 4)

You can check for yourself simply by multiplying them together.

By now, you should be seeing a pattern. That pattern being that for every trinomial with a degree of 2 that is factorable, we always have (x ± a)(x ± b) as a result.

Another example would be x² + x - 2. Well I know 2 + (-1) = 1 and 2 · (-1) = -2 so I would end up with the factor:

(x - 1)(x + 2)

You can see that factors can be positive or negative. To help us determine signs, we have two rules to help.

Say I have some trinomial x² + bx ± c where b = m + n and c = mn. Here are the two rules:

1) If c is positive, then the factorization will be (x ± m)(x ± n)
2) If c is negative, then the factorization will be (x ± m)(x ∓ n). It will vary by case for when m is positive or negative.

Now this is all fine as long as the coefficient of x² is 1. Now what if we have a coefficient a to have ax²?

We can now expand the trinomial to be ax² + bx + c. As much as I want to say there is a formula for this, it sadly comes down to trial and error. Depending on how large the numbers are, you can have many different answers. Even worse: Only one combination is correct.

Say I have 6x² - 13x + 5. I know 1 and 5 are the only factors of 5, since 5 is a prime number. Now we need to consider 6. The number 6 has 6 · 1 and 3 · 2. We have a total of four possible answers:

1) (x - 5)(6x - 1) = 6x² - 31x + 5 → Really??
2) (6x - 5)(x - 1) = 6x² - 11x + 5 → Nope
3) (3x - 5)(2x - 1) = 6x² - 13x + 5 → Correct!
4) (2x - 5)(3x - 1) = 6x² - 17x + 5 → Not quite

This can look intimidating, because the only way for us to find the correct answer is to simply take the product of each.

We do however, have a trick that can help. Consider the trinomial 12x² - 11xy - 5y². The general rule is: If you cannot find the GCF in your trinomial, then your factor should not have any GCF either. In other words, you cannot factor any further than your binomial products that we have been looking at this lecture.

12x² - 17xy - 5y²
Possible factors	Product
(x - y)(12x + 5y)	12x² - 7xy - 5y²
(x - 5y)(12x + y)	12x² - 59xy - 5y²
(2x - y)(6x + 5y)	12x² + 8xy - 5y²
(2x - 5y)(6x + y)	12x² - 28xy - 5y²
(3x - y)(4x + 5y)	12x² + 11xy - 5y² Not quite
(3x + y)(4x - 5y)	12x² - 13xy - 5y²
(3x - 5y)(4x + y)	12x² - 17xy - 5y² Winner!
(4x - y)(3x + 5y)	12x² + 17xy - 5y²

You can see several possible factors not working, because they have GCF's, when our initial trinomial had none.

This has been occurring for thousands of years, until the 1800s. Mathematicians have found a very lucky alternative method, which is not a formula. This is known as the "AC Method".

Let's try again with our trinomial 12x² - 11xy - 5y². We already know it carries no GCF. What I do first is take the product of a and c, based on the standard trinomial ax² + bx + c.

a = 12 and c = -5, so 12 · -5 = -60. Now I need to find two numbers that add to -11 and multiply to -60. We can go for -15 and 4. Now I can rewrite my trinomial.

12x² - 11xy - 5y² = 12x² - 15xy + 4xy - 5y²

We just dragged the unfamiliar back into familiar territory, because we can group these terms! As we discussed earlier, we have multiple paths, all leading to the same answer! The path I will choose is putting multiples of 4 in one group.

12x² - 15xy + 4xy - 5y² = (12x² + 4xy) - (15xy + 5y^2)

We can now factor out the GCF from each expression and finish the job:

(12x² + 4xy) - (15xy + 5y²) = 4x(3x + y) - 5y(3x + y) = (3x + y)(4x - 5y)

Now you may wonder to yourselves: What if the AC Method does not work when solving equations? If that is the case, then we consider the polynomial to be prime, meaning we would need to utilize other methods of solving for x directly, which will be discussed later in this course.

As an example, x² + 9x + 7 is prime because there are no pairs of numbers that add to 9 and multiply to 7, so the polynomial is prime.

Now we're going to kick it up a notch and introduce a trinomial with a higher degree.

Let's say I have x⁴ + 7x² + 12. This looks scary, but we do have a solution. We can just substitute u = x²:

x⁴ + 7x² + 12 = u² + 7u + 12

This just becomes a very simple trinomial to factor:

u² + 7u + 12 = (u + 4)(u + 3) = (x² + 4)(x² + 3)

Sometimes the expression we choose to substitute will be a binomial. I can just as easily have the expression:

(x - 2)² + 7(x - 2) + 12

I can just let u = x - 2 and we can obtain the result:

u² + 7u + 12 = (u + 4)(u + 3) = (x - 2 + 4)(x - 2 + 3) = (x + 2)(x + 1)

Of course, you can just expand the trinomial and factor it using the various methods we just learned, but where is the fun in that? Now that you have gone into factoring, things will not be as simple anymore. I cannot stress enough for you to practice these methods in your workbook and your homeworks.

Lecture 14: June 23rd, 2026

Factoring Special Products (6.3)

This section is going to serve a bit as a cool off, considering the dense content we covered in the previous lecture.

Recall from two lectures ago when we saw (a ± b)² = a² ± 2ab + b² for real numbers a and b. This is what we call perfect square trinomials. Well, we are also able to factor our perfect squares for any two real numbers.

a 2 \pm 2ab + b 2 = (a \pm b) 2

Easy enough right? As a brief example, consider x² ± 2x + 1. I have a = x and b = 1. So with that in mind,

x 2 \pm 2x + 1 = (x \pm 1) 2

There are other examples where it will get a bit more complicated, so we are going to explore a little bit more.

I can expand these binomial expressions to have a coefficient. This can be factored to:

m 2 x 2 \pm 2mbx + b 2 = (mx \pm b) 2

As an example, say I have 4x² - 20x + 25. I have a = 2x and b = 5. As -2ab = -2(2x)(5) = -20x, we can factor this to:

4x 2 - 20x + 25 = (2x - 5) 2

There may be cases where all three terms of a trinomial will have a GCF, so we can factor that out first in order to make the quadratic more pleasant to look at, and see if it resembles a perfect square.

Difference of Squares

Now let's move on to a difference of squares. Recall that if we have two real numbers a and b and we make conjugate pairs out of them, then we can produce a difference of squares from them:

(a + b)(a - b) = a 2 - b 2

Just like before, this is shown to be factorable going the other way:

a 2 - b 2 = (a - b)(a + b)

Note: This does not work for a sum of squares (a² + b²), because it is more often than not a prime binomial.

We are going to take it up a notch this time and work on a factor in a factor.

Say I have 16x⁴ - 81. I have a = 4x⁴ and b = 9. So:

16x 4 - 81 = (4x 2 - 9)(4x 2 + 9)

Now the question is: Are we done? The answer is no. We still have the first binomial to factor. Luckily, we can treat a = 2x and b = 3. With that in mind,

16x 4 - 81 = (4x 2 - 9)(4x 2 + 9) = (2x - 3)(2x + 3)(4x 2 + 9)

Now we are done, since we cannot factor this expression any further. Remember to look for the GCF regularly to make the special products more apparent.

Cubes

Now we move onto cubes. Say I have a and b as real numbers so we have a³ ± b³. In either case, the expression can be factored into:

a 3 \pm b 3 = (a \pm b)(a 2 \mp ab + b 2)

As a brief example, say I have 6x³y + 48y⁴. Before I move further, I can take the GCF of this expression, which is 6y. For the remaining sum of cubes, we use a = x and b = 2y:

6x 3 y + 48y 4 = 6y(x 3 + 8y 3) = 6y(x + 2y)(x 2 - 2xy + 4y 2)

Just as a heads up: We are capable of using multiple methods of factoring that we have learned in the previous lecture to compound the amount of factoring we may have to do. It is important to practice these techniques and work together to have a better grasp on what we have covered so far.

Lecture 15: June 29th, 2026

On the Topic of Polynomial Equations (6.5)

With everything we have covered, we are finally going into polynomial equations. As a brief heads up, the degree of our polynomials will be the degree of our polynomial equation. You will understand what that means as we move into this section.

Say we have a trinomial ax² + bx + c where the coefficients are real numbers with a ≠ 0. Our polynomial equation would end up being:

ax² + bx + c = 0

This particular equation is known as a quadratic equation.

With this, we are going to introduce a property to make solving quadratic equations easier:

Zero Product Property: For a · b = 0, a and/or b = 0

Here is a more satisfying example. Say I have a quadratic equation x² + 5x = -6. Using our factoring skills, we now know:

x² + 5x = -6 ⇒ x² + 5x + 6 = 0 ⇒ (x + 3)(x + 2) = 0

This tells me the following: EITHER

x + 3 = 0 ⇒ x = -3

x + 2 = 0 ⇒ x = -2

This can be done with more complex quadratic expressions and there are times where the degree will be higher than 2.

Take 16y² = 2y(16y² + 1) = 32y³ + 2y ⇒ 32y³ - 16y² + 2y = 0. I can factor out 2y from the expression, yielding:

32y³ - 16y² + 2y = 2y(16y² - 8y + 1) = 2y(4y - 1)² = 0

y = 0 or y = 1/4

We can check by simply plugging in our values back into our initial equation.

Notice that we will not always have the case where one side of the equation will be zero. It will be our job to set up the equation like that so we are able to return to the standard practices.

Now we turn back to functions yet again. Say I have a polynomial function f(x). For that function, if there is some x value such that f(x) = 0, then x is a zero of the function. We are more familiar with the term x-intercept. So in our words:

x is an x-int for our function f(x)

Say I have 2x² - 7x + 5. I can factor this and equate it to zero to find my x-ints:

2x² - 7x + 5 = (2x - 5)(x - 1) = 0

So I have x = 1 and x = 5/2. These are my zeroes for my function f(x) and they are also my x-int. To find the y-int, x = 0 leading to f(0) = 5.

We can now start applying what we learned with a couple of word problems.

The product of two consecutive odd numbers is 255 and I need to find the integers. Let n denote the first odd number and n + 2 the second. The product makes 255, so:

n(n + 2) = 255

n² + 2n = 255 ⇒ n² + 2n - 255 = 0 ⇒ (n + 17)(n - 15) = 0

This means that n must be -17 or 15. Moving to (n + 2 = -15 or 17). We can take the product of each: (-15)(-17) = 15 · 17 = 255.

Another example can be in finding a length for a matching area. Say I have a rectangle with a length of l and a width of l + 1 and the area is given as 30 ft².

So:

l(l + 1) = 30 ⇒ l² + l - 30 = 0 = (l + 6)(l - 5)

⇒ l = -6, 5

We have two possible solutions. However, length can never be negative so -6 cannot be a solution.

l = 5 ⇒ l + 1 = 6

Finally, say I have a right triangle with a length x, a height x + 1, and a hypotenuse of length 5.

Using the Pythagorean Theorem, I know a² + b² = c². With that in mind:

x² + (x + 1)² = x² + x² + 2x + 1 = 2x² + 2x + 1 = 25 ⇒ 2x² + 2x - 24 = 0

We can factor out 2 to get the equation:

x² + x - 12 = 0 = (x + 4)(x - 3)

Leaving us with x = -4, 3. If x = 3, then x + 1 = 4 and if x = -4, then x + 1 = -3, and we can simply check by applying them to Pythagoras' equation:

(±3)² + (±4)² = 9 + 16 = 25 = 5²

We are now entering the point of the course where we are going to start seeing things that we have discussed way back when come back into light and mix with other topics we have learned more recently.

Multiplying and Dividing Rational Expressions (7.1)

To begin, recall the rational numbers, denoted as Q, and the operations we have done with those. As it turns out, polynomial functions are capable of sharing the same properties as these numbers. With that, we are able to begin forming rational expressions.

For some polynomial functions P(x) and Q(x), assuming Q(x) ≠ 0, we can take the quotient of the two to make our new function:

R(x) = P(x) / Q(x)

For a few brief examples, here are a few rational expressions:

4x / (5y²), (x + 1) / (x² + 1), (x² + 2x + 1) / (x - 1)

Now we are going to look at where our function R(x) is undefined. We take R(x) = P(x) / Q(x). We know we can't divide by 0. That can only mean Q(x) ≠ 0, which means there may be certain x values that force this inequality to be false.

Say I have a rational expression (a + 10) / (a² + 4a + 3)

Since the denominator can never equal 0, a² + 4a + 3 = (a + 3)(a + 1) ≠ 0

This means that a ≠ -3 and a ≠ -1

Another example would be (y - 1) / (3y + 2). Again, denominators can never be zero, so

3y + 2 ≠ 0 ⇒ y ≠ -2/3

Now recall what it means to simplify. For an expression to be completely simplified in this case, there must be no common factors between the numerator and denominator.

(x + 2) / (x + 3), (y + 3) / (y² - 4)

Recall now for some real numbers a, b, c, where b and c ≠ 0, this statement holds true:

(ac) / (bc) = (a/b) · (c/c) = (a/b) · 1 = a/b

This can also be extended into our polynomial expressions.

Say I want to simplify this rational expression:

(x² - x - 2) / (x² - 3x + 2)

To do so, I must factor both numerator and denominator in order to determine what can cancel with each other:

(x² - x - 2) / (x² - 3x + 2) = [(x + 1)(x - 2)] / [(x - 1)(x - 2)] = [(x + 1)/(x - 1)] · [(x - 2)/(x - 2)] = [(x + 1)/(x - 1)] · 1 = (x + 1)/(x - 1)

For a bit more clarity, here is another example:

(x² - 3x - 10) / (x² + x - 2)

(x² - 3x - 10) / (x² + x - 2) = [(x + 2)(x - 5)] / [(x + 2)(x - 1)] = (x - 5)/(x - 1)

Notice how I skipped a few steps. I know that terms will cancel out, so I may as well save a bit of time and just mark them instead of going step by step.

As a brief break, say I have expressions a and b, a ≠ b, such that I can have the expression a - b

We now consider its additive inverse: -(a - b) = -a + b = b - a, (Note that the last line in the equation is true because of the commutative property) then we can take their ratio to get -1.

(a - b) / (b - a) = -(b - a) / (b - a) = -1 · [(b - a)/(b - a)] = -1

Take the following rational expression:

(x² + x - 2) / (1 - x²)

We simply factor and cancel out like binomials, since we already know these steps in the process.

(x² + x - 2) / (1 - x²) = [(x + 2)(x - 1)] / [(1 - x)(1 + x)]

We know 1 + x = x + 1 thanks to the commutative property. As for 1 - x, we can simply factor -1 from both terms.

1 - x = -(x - 1) ⇒ [(x + 2)(x - 1)] / [(1 - x)(1 + x)] = [(x + 2)(x - 1)] / [-(x - 1)(x + 1)] = (x + 2) / [-(x + 1)] = -(x + 2) / (x + 1)

Now let's get into multiplying rational expressions together. Take p, q, r, s, where q and s ≠ 0, then I can make this statement:

(p/q) · (r/s) = (p/s) · (r/q) (Commutative Property) = (pr) / (qs)

As a simple example, take the rational expressions (2x² + 5x - 12) / (x² - 16) and (x² - 8x + 16) / (2x² - 13x + 15) and I want to take the product. The general idea is to first factor each expression as much as possible before we do anything else.

(2x² + 5x - 12) / (x² - 16) = [(x + 4)(2x - 3)] / [(x + 4)(x - 4)]

(x² - 8x + 16) / (2x² - 13x + 15) = (x - 4)² / [(2x - 3)(x + 5)]

Now we can take the product between the two.

[(2x² + 5x - 12) / (x² - 16)] · [(x² - 8x + 16) / (2x² - 13x + 15)] = (x - 4) / (x - 5)

Taking the same four polynomial expressions mentioned earlier, we can also divide them. The additional condition we have now is r ≠ 0

(p/q) ÷ (r/s) = (p/q) / (r/s) = (p/q) · (s/r) = (ps) / (qr)

This is possible due to our good rule KCF (Keep-Change-Flip), along with the fact we can have a multiplicative inverse.

Take (x³ - 8) / (3x² - 16x + 16) and (x² - 4) / 6

(x³ - 8) / (3x² - 16x + 16) = [(x - 2)(x² + 2x + 4)] / [(x - 4)(3x - 4)]

(x² - 4) / 6 = [(x - 2)(x + 2)] / 6

We can now utilize the KCF method.

[(x³ - 8) / (3x² - 16x + 16)] ÷ [(x² - 4) / 6] = [6(x² + 2x + 4)] / [(x - 2)(x - 4)(3x - 4)]

Looking back at the start of this section we now turn to our function R(x) = P(x) / Q(x)

We know Q(x) ≠ 0, which means we can solve for certain x values that make this statement false.

For example, let R(x) = (2x² - 10x) / (4x² - 16x - 20) = [2x(x - 5)] / [4(x - 5)(x + 1)]

We know the denominator can't equal 0, so x ≠ 5, -1. -1 is simple, but as you can see from our fraction, x - 5 cancels out. So we ask ourselves why does x = 5 not work? The reason why is if we take R(5), we end up with 0/0. This fraction is indeterminate because we can not take 0 and divide it into 0 groups, R(5) is not definable without other forms of verification, which are beyond the scope of this class.

The domain would end up being (-∞, -1) ∪ (-1, 5) ∪ (5, ∞)

And like the expressions we have seen above, we can do some manipulation to simplify our functions and make them easier to read.

We can have two scenarios regarding rational functions:

R(x) = f(x) · g(x)

R(x) = f(x) / g(x)

As an example, let f(x) = x + 2 and g(x) = (x² - 4) / 5

For the first scenario:

f · g = (x + 2) · [(x² - 4) / 5] = [(x² - 4)(x + 2)] / 5

For the other scenario:

f / g = (x + 2) / [(x² - 4) / 5] = [5(x + 2)] / (x² - 4) = [5(x + 2)] / [(x + 2)(x - 2)] = 5 / (x - 2)

Lecture 16: June 30th, 2026

Adding and Subtracting Rational Expressions (7.2)

Now we go to adding and subtracting rational expressions. Once again, I am going to go ALL the way back to the second lecture and bring back a few rules to serve as a reference.

Say I have expressions a,b,c,d and , with c,d ≠ 0. With all of this, I can make the following equations possible.

Rational Expressions with the Same Denominator

(a/c) \pm (b/c) = ((a \pm b)/c)

Example:

((x+1)/y) + ((2x-3)/y) = (((x+1) + (2x-3))/y) = ((3x - 2)/y)

Rational Expressions with Different Denominators

(a/b) \pm (c/d) = ((ad \pm bc)/bd)

Example:

(x/y) + (2/z) = (((x * z) + (2 * y))/yz) = ((xz + 2y)/yz)

I have derived (Or presented to you in better detail at least) all of these equations for rational numbers all the way back in the second lecture. If you are interested in seeing how they were made, consider that a very good reference.

Opposing Denominators

Now we are going to look just a bit closer at opposing denominators. This is considered to be between expressions with the same and similar denominators, since they are ALMOST the exact same. However, we have a very quick work around for this.

Example:

Say I take

((x^2 - 5x)/(x^2 - 4)) - ((6x - 6)/(4 - x^2))

Well I know that 4 - x^2 = -(x^2 - 4). With that, I can start making some things happen as follows:

((x^2 - 5x)/(x^2 - 4)) - ((6x - 6)/(4 - x^2)) = ((x^2 - 5x)/(x^2 - 4)) - (-(6x - 6)/(x^2 - 4))

= ((x^2 - 5x)/(x^2 - 4)) + ((6x - 6)/(x^2 - 4)) = ((x^2 - 5x + 6x - 6)/(x^2 - 4)) = ((x^2 + x - 6)/(x^2 - 4))

Now I can simplify by factoring:

((x^2 + x - 6)/(x^2 - 4)) = (((x + 3)(x - 2))/((x - 2)(x + 2))) = ((x + 3)/(x + 2)) * ((x - 2)/(x - 2)) = 1 * ((x + 3)/(x + 2)) = ((x + 3)/(x + 2))

I do admit this problem was a bit more loaded compared to others we have seen so far, but this was done in order to build confidence in our ability to solve these problems.

Finding the Least Common Denominator (LCD)

For rational expressions with different denominators, we want to find the LCM of their denominators, or the Least Common Denominator (LCD) between the two fractions. The equation I have presented already addresses that. However, it is important to keep it as simple as possible. With that said, I strongly recommend you factor first before you apply this special formula.

Example: say we want to simplify

((x^2 - 4)/(x+2)) + (3x/(x^2 - 4))

First, factor expressions where possible: $x^2 - 4 = (x-2)(x+2)$ So the expression becomes: $(((x-2)(x+2))/(x+2)) + (3x/((x-2)(x+2)))$
Simplify fractions where possible: $((x-2)/1) + (3x/((x-2)(x+2))) = (x-2) + (3x/((x-2)(x+2)))$
Find the Least Common Denominator (LCD), which is: $(x-2)(x+2)$
Rewrite each fraction with the LCD: $(((x-2)(x+2))/((x-2)(x+2))) + (3x/((x-2)(x+2))) = (((x-2)(x+2) + 3x)/((x-2)(x+2)))$
Simplify the numerator: $(x-2)(x+2) + 3x = x^2 - 4 + 3x = x^2 + 3x - 4 = (x + 4)(x - 1)$

Final Simplified Expression:

(((x + 4)(x - 1))/((x-2)(x+2)))

Rational Functions

Finally, we get to rational functions. We can let a rational function R(x) be a simple sum or difference of other rational functions.

R(x) = F(x) \pm G(x)

Let

F(x) = ((x+1)/(x-2)), G(x) = (3/(x-2))

Then their sum is a rational function:

R(x) = F(x) + G(x) = ((x+1)/(x-2)) + (3/(x-2))

Since the denominators are the same, we can combine the numerators:

R(x) = (((x+1) + 3)/(x-2)) = ((x+4)/(x-2))

Similarly, for a difference:

R(x) = F(x) - G(x) = ((x+1)/(x-2)) - (3/(x-2)) = (((x+1) - 3)/(x-2)) = ((x-2)/(x-2)) = 1 (x \neq 2)

Lecture 17

We will now talk about simplifying complex rational expressions. All it basically is simplifying on steroids. It only takes what we have done so far and puts it into one section with multiple steps, all of which we have already seen before hand.

Simplifying Complex Rational Expressions (7.3)

The reason why we call it complex is because we are once again going to be working with complex fractions, which we have covered all the way back in our second lecture.

Recall: For a complex fraction,
$((a/b) / (c/d)) = (a/b) * (d/c)$

The overall goal is to utilize the LCD, which we can do through factoring if a certain component of the expression allows us to. We can then get a single rational expression over another and then utilize our KCF method. I will show you one example which is a lot stronger compared to the workbook problems as a demonstration, then we went through workbook examples in class by students' choice.

Simplify:

\frac{\frac{3}{x - 4} + \frac{5}{x + 4}}{\frac{7}{x + 4} - \frac{2}{x^{2} - 16}}

Factor the denominator:

x^{2} - 16 = (x - 4) (x + 4)

Simplify the complex numerator:

\frac{3}{x - 4} + \frac{5}{x + 4} = \frac{3 (x + 4) + 5 (x - 4)}{(x - 4) (x + 4)}

= \frac{3 x + 12 + 5 x - 20}{(x - 4) (x + 4)} = \frac{8 x - 8}{(x - 4) (x + 4)} = \frac{8 (x - 1)}{(x - 4) (x + 4)}

Simplify the complex denominator:

\frac{7}{x + 4} - \frac{2}{x^{2} - 16} = \frac{7 (x - 4) - 2}{(x - 4) (x + 4)}

= \frac{7 x - 28 - 2}{(x - 4) (x + 4)} = \frac{7 x - 30}{(x - 4) (x + 4)}

Divide the simplified parts (Keep-Change-Flip):

\frac{\frac{8 (x - 1)}{(x - 4) (x + 4)}}{\frac{7 x - 30}{(x - 4) (x + 4)}} = \frac{8 (x - 1)}{(x - 4) (x + 4)} \cdot \frac{(x - 4) (x + 4)}{7 x - 30}

= \frac{8 (x - 1)}{7 x - 30}

There are three values of x that will make our expression undefined:

x \neq \pm 4, x \neq \frac{30}{7}

We now move on to Equations and Inequalities. Like polynomial equations, we are able to apply rational expressions to the same forms of algebra.

Solving Rational Equations (7.4)

We can easily define a rational equation as an equation with rational expressions. Nothing too special, which is why I am not going to really emphasize that as much as I have before, since we are simply extending what we know into the new stuff.

As an example of a rational equation, we have y + 1 = (y + 6) / (y^2 - 36) and I would have to solve for some y.

We know the first condition is that y can NOT be ±6. The reason why is because if so, our denominator on the right hand side of the equation would be 0, forcing our expression to be undefined.

We can however, factor and simplify this equation a bit to see a better picture.

(y + 6) / (y^2 - 36) = (y + 6) / ((y + 6)(y - 6)) = 1 / (y - 6) = y + 1 \Rightarrow (y + 1)(y - 6) = 1

I can then expand this equation to make this into a trinomial equation and equate it to 0.

1 = y^2 - 5y - 6 \Rightarrow y^2 - 5y - 7 = 0

This equation can not be solved using what we have learned so far, as our trinomial is not factorable. We can verify that it is prime by seeing the AC method will not yield two numbers that add to -5 and multiply to -7. There is a method we can use, but we will revisit this problem specifically near the end of the semester.

There are multiple ways of solving rational equations, especially utilizing the tools we have learned throughout the semester so far. This is where the beauty of "One size does not fit all coming in". You can afford variation in your work, and you will have the same correct answer as the person sitting beside you. Just be sure to check.

Let's look at an example that we can solve now.

1 - (2 / x) = 15 / x^2

I have several routes to choose from when it comes from this problem. I personally prefer to get rid of the denominators first chance I get.

1 - (2 / x) = 15 / x^2 \Rightarrow x^2(1 - (2 / x)) = x^2 * (15 / x^2) \Rightarrow x^2 - (2x^2 / 2x) = x^2 - 2x = 15

Subtract the 15 over on both sides to get a quadratic equation.

x^2 - 2x - 15 = 0

Factor our trinomial through any means we have learned so far.

(x - 5)(x + 3) = 0 \Rightarrow x - 5 = 0 OR x + 3 = 0 \Rightarrow x = 5 OR x = -3

We have two possible solutions for our equation. We can simply check by plugging our solutions into our original equation to verify the equation stays true.

These equations can have no solutions, a finite number of solutions, as you have seen above, or infinitely many. You can think back to contradictions, conditions and identities from earlier if that helps you with our reasoning.

Examples of Classifications

Contradiction: $(1 / x) + 2 = (1 / x) + 5$ Subtract (1/x) from both sides: $2 = 5$ This is a contradiction, so the equation has no solution.
Identity: $(x + 3) / (x + 1) = (x + 3) / (x + 1) \Rightarrow 1 = 1$ This is true for all x ≠ -1 (to avoid division by zero). This is an identity.

Before we move on, I want to introduce to you to extraneous solutions, which are solution candidates that turn out to not be solutions of our original equation. As an example, we will solve for y.

(y / (y + 6)) = 4 + (72 / (y^2 - 36)) \Rightarrow y = 4(y + 6) + (72(y + 6)) / ((y + 6)(y - 6)) = 4(y + 6) + 72 / (y - 6)

We now have a simpler equation to work with:

y = 4(y + 6) + 72 / (y - 6) \Rightarrow y(y - 6) = 4(y^2 - 36) + 72 \Rightarrow y^2 - 6y = 4y^2 - 72

We now move things to one side of the equation, while the other side will simply be 0.

3y^2 + 6y - 72 = 3(y^2 + 2y - 24) = 3(y + 6)(y - 4) = 0

This leads me to believe y = -6 or y = 4. However, y = -6 is extraneous because if we plug it into our original function, both sides of the equation end up undefined.

Now that we are more comfortable with equations, we can finally move onto functions. When we utilize rational functions, we can find the domain of a function and utilize outputs from f(x) to solve for an input x.

As an example, consider the rational function:

f(x) = (2x + 3) / (x - 1)

The domain is all real numbers except where the denominator is zero:

x - 1 \neq 0 \Rightarrow x \neq 1

So the domain is:

(-\infty, 1) \cup (1, \infty)

Suppose we want to find x such that:

f(x) = 5

Set up the equation:

(2x + 3) / (x - 1) = 5

Multiply both sides by (x - 1) (valid because x ≠ 1):

2x + 3 = 5(x - 1)

Simplify:

2x + 3 = 5x - 5

3 + 5 = 5x - 2x

8 = 3x

x = 8/3

Check that x = 8/3 ≠ 1, so it is in the domain. Thus, the solution is:

x = 8/3

By now when looking at equations, we can see ourselves solving for numbers. Recall the slope of a function m, can be solved utilizing point-slope form for any given point.

m = (y - 5) / (x - 4)

Where the point is (4,5) and I want to solve for y, in terms of my variables x and m. Utilizing what we know, this should be straightforward, given that this is a two step problem.

m = (y - 5) / (x - 4) \Rightarrow m(x - 4) = mx - 4m = y - 5 \Rightarrow y = mx + (5 - 4m)

We can go a bit more aggressive with rationals here to solve.

(2 / x) + (1 / 3) = 1 / y

We can utilize what we know, but for my approach, I am going to first combine the two terms on the left.

1 / y = (2 / x) + (1 / 3) = (3(2) + 1(x)) / 3x = (x + 6) / 3x = 1 / y

A nice technique I have not yet went through is this: We do have the option of simply multiplying both denominators on both sides and then dividing (x + 6) on both sides to solve for y. I can do all of this instead in one single step: Flip the fractions over!

y = 3x / (x + 6)

This is very useful and it saves a step. The only condition is, it must be one fraction that is equal to the other. With that said: It is crucial that you simplify it first to utilize this step to its full effect.

Lecture 18: July 2nd, 2026

7.6 Solving Rational Inequalities

Now that we have a better look at equalities, we are going to look back at inequalities. You have already seen what inequalities can do to linear expressions back in lecture 6. If you want a refresher on the 5 signs of inequality, refer to the very first lecture in section 1.1. Below is an example of all 5. In each, we are going to solve the inequality for $x$ , plot the solutions on a number line, and present the solutions in interval notation.

1. Less than → AND Statement (<)

(x - 2) / (x + 1) < 1

Subtract 1:

(x - 2) / (x + 1) - 1 < 0 \Rightarrow [x - 2 - (x + 1)] / (x + 1) < 0 \Rightarrow -3 / (x + 1) < 0

Multiply both sides by -1 (reversing the inequality sign):

3 / (x + 1) > 0 \Rightarrow x + 1 > 0 \Rightarrow x > -1

Since this is a less than inequality, the solution behaves like an AND condition.

Interval Notation: $(-1, \infty)$

2. Greater than → OR Statement (>)

(2x + 1) / (x - 3) > 0

Critical points (found through set-to-zero evaluations):

2x + 1 = 0 \Rightarrow x = -½, x - 3 = 0 \Rightarrow x = 3

Test intervals:

(-\infty, -½), (-½, 3), (3, \infty)

The sign chart analysis gives:

x < -½ OR x > 3

Interval Notation: $(-\infty, -½) \cup (3, \infty)$

3. Less than or equal to → Includes Zero (≤)

(x - 4) / (x + 2) \leq 0

Critical points:

x - 4 = 0 \Rightarrow x = 4, x + 2 = 0 \Rightarrow x = -2

Solution set verification via interval comparison:

-2 < x \leq 4

Interval Notation: $(-2, 4]$

4. Greater than or equal to → Includes Zero (≥)

(3x - 6) / (x - 1) \geq 0

Critical points:

3x - 6 = 0 \Rightarrow x = 2, x - 1 = 0 \Rightarrow x = 1

Sign chart evaluation:

x < 1    \text{or}    x ≥ 2

Interval Notation: $(-\infty, 1) \cup [2, \infty)$

5. Not equal to → Excluded Values (≠)

(x + 5) / (x - 2) \neq 0

Numerator restriction: $x + 5 \neq 0 \Rightarrow x \neq -5$

Denominator restriction: $x - 2 \neq 0 \Rightarrow x \neq 2$

Interval Notation: $(-\infty, -5) \cup (-5, 2) \cup (2, \infty)$

Application: Economic Cost Analysis

Consider the total cost function:

C(x) = 4x + 15

Where $C$ represents total cost for $x$ items produced, factoring in a $15 maintenance fee. The average cost function $c(x)$ is:

c(x) = C(x) / x = (4x + 15) / x = 4 + 15/x

To find how many items must be produced so that the average cost is at most 7 dollars:

c(x) \leq 7 \Rightarrow 4 + 15/x \leq 7 \Rightarrow 15/x \leq 3 \Rightarrow x \geq 5

This means at least 5 items must be produced to keep the average cost at or below 7 dollars.

8.1 Simplifying Expressions with Roots

Recall that $22 = 4$ ("two squared"). Thus, 4 is the square of 2. We phrase these algebraic relationships explicitly as follows:

If $n 2 = m$ :
1)

n

is a Square Root of

m

.
2)

m

is the Square of

n

The radical notation $\sqrtm$ targets a non-negative real value, where $m$ is called the radicand. Formally:

Given

n 2 = m

⇒

\sqrtm = \pmn

for

n \geq 0

Numerical Illustrations:

$n 2 = 9 \Rightarrow n = \pm3$ , hence $\sqrt9 = \pm3$ . (Check: $(3) 2 = (-3) 2 = 9$ )
$n 2 = 16 \Rightarrow n = \pm4$ , hence $\sqrt16 = \pm4$ . (Check: $(4) 2 = (-4) 2 = 16$ )

Warning: If $m < 0$ , then $\sqrtm$ is not a real number. For instance, there is no real number solution to $n 2 = -1$ .

Generalization: The n-th Root

If $a n = b$ , we state that $a$ is the n-th root of $b$ , written as:

n√b = a    (\text{where } n ≥ 2, n \in \mathbb{N})

Properties of n-th Roots based on Index parity:

n is an EVEN root: Shares identical restrictions with basic square roots.
- If $b \geq 0 \Rightarrow n \sqrtb$ is a real number. (e.g., $4 \sqrt16 = \pm2$ )
- If $b < 0 \Rightarrow n \sqrtb$ is NOT a real number. (e.g., $4 \sqrt-16$ has no real solution)
n is an ODD root: ⁿ√b will ALWAYS yield a real number.
- Positive value: $3 \sqrt27 = 3$
- Negative value: $3 \sqrt-8 = -2$ (Since $(-2) 3 = -8$ )

Mental Estimations (Without a Calculator)

By mapping values to a foundational baseline of perfect squares, bounds can be quickly established.

Number	Square
1	1
2	4
3	9
4	16
5	25
6	36
7	49
8	64
9	81
10	100
11	121
12	144

Estimation Examples:

Estimate √10: Since $9 < 10 < 16$ ⇒ $\sqrt9 < \sqrt10 < \sqrt16$ ⇒ $3 < \sqrt10 < 4$ .
Estimate √50: Since $49 < 50 < 64$ ⇒ $\sqrt49 < \sqrt50 < \sqrt64$ ⇒ $7 < \sqrt50 < 8$ .

Variable Expressions and Core Identities

Simplifying expressions containing variable components demands parsing through distinct mathematical conditions:

1) When $n$ is ODD:

n \sqrta n = a

2) When $n$ is EVEN:

n \sqrta n = |a|

Odd Case Example: $3 \sqrt(-2) 3 = 3 \sqrt-8 = -2$
Even Case Example: $\sqrt(-3) 2 = \sqrt9 = 3 = |-3|$

Generalizing for powers with multi-factors within an even index: $n \sqrta nm = n \sqrt(a m) n = |a m |$

Example: $\sqrt4x 6 = |2x 3 | = 2|x 3 |$ (Absolute values are vital here because if $x < 0$ , $x 3$ would evaluate out negatively).

8.2 Simplifying Radical Expressions

Product Property of n-th Roots:
For real numbers

n \sqrta

and

n \sqrtb

n \sqrta \cdot n \sqrtb = n \sqrtab

A radical expression $n \sqrta$ is in its simplest radical form if the radicand $a$ contains no factors that can be written as perfect powers of $b n$ .

Square Root (n = 2): $\sqrt18 = \sqrt(9 \cdot 2) = 3\sqrt2$ | $\sqrt(x 2 y) = x\sqrty$
Cube Root (n = 3): $3 \sqrt54 = 3 \sqrt(27 \cdot 2) = 3 3 \sqrt2$ | $3 \sqrt(x 3 y 2) = x 3 \sqrty 2$
Fourth Root (n = 4): $4 \sqrt80 = 4 \sqrt(16 \cdot 5) = 2 4 \sqrt5$ | $4 \sqrt(x 4 y 3) = x 4 \sqrty 3$

Quotient Property of n-th Roots

If $b \neq 0$ , and $n$ is a natural number such that $n \geq 2$ :

n \sqrt(a / b) = n \sqrta / n \sqrtb

Numerical Example: $3 \sqrt(8 / 27) = 3 \sqrt8 / 3 \sqrt27 = 2 / 3$
Variable Example: $4 \sqrt(x 8 / y 4) = 4 \sqrtx 8 / 4 \sqrty 4 = x 2 / y$

Lecture 19: July 6th, 2026

Remember last lecture where things were starting to look oddly familiar? Well, we are going to discuss exactly why that is the case. After that, we are going to go through their practical mathematical operations.

8.3 Simplifying Rational Exponents

Let's look at an algebraic equation to get the ball rolling. Take $(2 x) 3 = 2$ . We know thanks to the established rules of exponents that $(2 x) 3 = 2 3x$ . With that in mind:

(2 x) 3 = 2 3x = 2 1

This can be simplified directly because both sides share the same base of two. Since the bases are equal, their exponents must match:

3x = 1 \Rightarrow x = ⅓

Substituting this value back in gives us $2 3 \cdot ⅓ = (2 3) ⅓ = 8 ⅓ = 2$ . We also know from arithmetic that $3 \sqrt8 = 2$ . This can only mean one thing:

2 ⅓ = 3 \sqrt2

The Fundamental Identity:
For some real number

a

and an index

n \geq 0

a 1/n = n \sqrta

Example 1 (Numerical example):

16 ¼ = 4 \sqrt16 = 2

Example 2 (Variable example):

x ⅓ = 3 \sqrtx

This is a colossal revelation because we can now expand our working knowledge of roots by pairing them directly with exponential properties gathered earlier in the course.

Let's say we want to take this fractional root and raise it to an integer power $m$ . We can look at this relationship in two distinct ways:

$a m/n = (a 1/n) m = (n \sqrta) m$
$a m/n = (a m) 1/n = n \sqrta m$

Putting this all together for integers $m$ and $n$ (where $n \neq 0$ ):

a m/n = n \sqrta m = (n \sqrta) m

Example 1 (Numerical example):

8 ⅔ = 3 \sqrt8 2 = (3 \sqrt8) 2 = (2) 2 = 4

Example 2 (Variable example):

x 5/2 = \sqrtx 5 = (\sqrtx) 5

The Complete Rules of Exponents

This marks the 9th and final rule regarding exponents. Our comprehensive reference table is now complete, opening up a much wider path of algebraic functions to solve.

Property	Rule	Example
Product of Powers	$a m \cdot a n = a m+n$	$23 \cdot 2 4 = 2 7$
Quotient of Powers	$a m / a n = a m-n$	$56 / 5 2 = 5 4$
Power of a Power	$(a m) n = a mn$	$(3 2) 4 = 3 8$
Power of a Product	$(ab) n = a n b n$	$(2 \cdot 5) 3 = 2 3 \cdot 5 3$
Power of a Quotient	$(a / b) n = a n / b n$	$(3 / 4) 2 = 9 / 16$
Zero Exponent	$a 0 = 1$	$70 = 1$
Negative Exponent	$a -n = 1 / a n$	$2 -3 = 1 / 8$
Exponent of 1	$a 1 = a$	$91 = 9$
Fractional Exponent	$a m/n = n \sqrta m = (n \sqrta) m$	$8 ⅔ = 3 \sqrt8 2 = (3 \sqrt8) 2 = 4$

Lecture 20: July 9th, 2026

8.4 Adding, Subtracting, and Multiplying Radical Expressions

When working with radical expressions, addition and subtraction follow the exact same structural rules as combining like terms. Instead of variable matching, we match and combine Like Radicals (radicals sharing the exact same index and radicand).

Introductory Examples

Example 1: $3\sqrt2 + 5\sqrt2 = 8\sqrt2$

Example 2: $7\sqrt5 - 2\sqrt5 = 5\sqrt5$

Example 3 (Simplify first): $4\sqrt3 + 6\sqrt12 = 4\sqrt3 + 6(2\sqrt3) = 4\sqrt3 + 12\sqrt3 = 16\sqrt3$

Example 4 (Variables): $2x\sqrty + 5x\sqrty = 7x\sqrty$

Often, the expressions given are not initially in simplest radical form. Bringing them to their simplest form before performing arithmetic operations allows hidden like radicals to reveal themselves cleanly.

Example 1: Simplify

\sqrt50 + 3\sqrt18

Simplify individual components:

\sqrt50 = \sqrt(25 \cdot 2) = 5\sqrt2, \sqrt18 = \sqrt(9 \cdot 2) = 3\sqrt2

Combine like terms:

5\sqrt2 + 3(3\sqrt2) = 5\sqrt2 + 9\sqrt2 = 14\sqrt2

Example 2: Simplify

2\sqrt75 - \sqrt27

Simplify individual components:

\sqrt75 = \sqrt(25 \cdot 3) = 5\sqrt3, \sqrt27 = \sqrt(9 \cdot 3) = 3\sqrt3

Combine like terms:

2(5\sqrt3) - 3\sqrt3 = 10\sqrt3 - 3\sqrt3 = 7\sqrt3

Example 3: Simplify

3x\sqrt32 + 4x\sqrt8

Simplify individual components:

\sqrt32 = \sqrt(16 \cdot 2) = 4\sqrt2, \sqrt8 = \sqrt(4 \cdot 2) = 2\sqrt2

Combine like terms:

3x(4\sqrt2) + 4x(2\sqrt2) = 12x\sqrt2 + 8x\sqrt2 = 20x\sqrt2

Multiplying Radicals

Using exponential properties, we can verify that taking a power of a product works identically inside radical frameworks:

(ab) 1/n = a 1/n \cdot b 1/n \Rightarrow n \sqrta \cdot n \sqrtb = n \sqrtab

Example 1: $\sqrt3 \cdot \sqrt12 = \sqrt(3 \cdot 12) = \sqrt36 = 6$

Example 2: $3 \sqrt2 \cdot 3 \sqrt16 = 3 \sqrt(2 \cdot 16) = 3 \sqrt32 = 2 3 \sqrt4$

Example 3 (Coefficients): $2\sqrt5 \cdot 3\sqrt2 = (2 \cdot 3)\sqrt(5 \cdot 2) = 6\sqrt10$

Example 4 (Variables): $\sqrtx \cdot \sqrt4x 3 = \sqrt(4x 4) = 2x 2$

Polynomial Distributions and Special Products

We can multiply complex radical expressions by adopting polynomial expansion steps (such as FOIL distributing or leveraging binomial identities):

Special Polynomial Core Products:
1) Perfect Square Binomials:

(a \pm b) 2 = a 2 \pm 2ab + b 2

2) Difference of Squares:

(a + b)(a - b) = a 2 - b 2

Example 1 (Standard FOIL Distribution):

(\sqrt2 + 3)(\sqrt2 - 5) = \sqrt2\cdot\sqrt2 - 5\sqrt2 + 3\sqrt2 - 15

= 2 - 2\sqrt2 - 15 = -13 - 2\sqrt2

Example 2 (Square of a Sum):

(\sqrt3 + 2) 2 = (\sqrt3) 2 + 2(2)(\sqrt3) + 2 2

= 3 + 4\sqrt3 + 4 = 7 + 4\sqrt3

Example 3 (Difference of Squares Identity):

(\sqrt5 + 3)(\sqrt5 - 3) = (\sqrt5) 2 - 3 2 = 5 - 9 = -4

8.5 Dividing Radical Expressions

Division mirrors our product properties directly. Radicals can be consolidated into or expanded out of fractions using the following core conversion rule:

n \sqrt(a / b) = n \sqrta / n \sqrtb (where b \neq 0)

Example 1: $\sqrt50 / \sqrt2 = \sqrt(50 / 2) = \sqrt25 = 5$

Example 2: $\sqrtx 5 / \sqrtx 2 = \sqrt(x 5 / x 2) = \sqrtx 3 = x\sqrtx$

Example 3 (Term Split Splitting):

(\sqrt18 + \sqrt8) / \sqrt2 = (\sqrt18 / \sqrt2) + (\sqrt8 / \sqrt2) = \sqrt9 + \sqrt4 = 3 + 2 = 5

Rationalizing the Denominator

In standard algebra, fractional evaluations are not considered fully simplified if the denominator contains irrational roots. We remove radicals from denominators through a clean procedural mechanism called rationalizing the denominator.

Three Rules For a Simplified Radical Expression:

The radicand contains no perfect exponential factor groups matching the index.
No fractional components remain tucked directly inside a radical.
No radicals exist inside the denominator.

Case A: Monomial Denominators (Single-Term Radicals)

Multiply both the numerator and denominator by a targeted factor that transforms the inner radical into a clean, perfect power matching the index:

Example 1 (Square Root):

(5 / \sqrt2) \cdot (\sqrt2 / \sqrt2) = 5\sqrt2 / 2

Example 2 (Cube Root):

(2 / 3 \sqrt5) \cdot (3 \sqrt25 / 3 \sqrt25) = 2 3 \sqrt25 / 3 \sqrt125 = 2 3 \sqrt25 / 5

Example 3 (Variable Variable Identity):

(1 / 3 \sqrtx 2) \cdot (3 \sqrtx / 3 \sqrtx) = 3 \sqrtx / 3 \sqrtx 3 = 3 \sqrtx / x

Case B: Binomial Denominators (Two-Term Radicals)

When the denominator contains a two-term algebraic setup, multiply the top and bottom by its conjugate pair. Recall that the conjugate pair of a binomial expression $(ax + b)$ is simply $(ax - b)$ . This triggers a difference of squares expansion, completely clearing the troublesome roots.

Example 1:

3 / (\sqrt2 + 1) \cdot (\sqrt2 - 1) / (\sqrt2 - 1) = 3(\sqrt2 - 1) / [(\sqrt2) 2 - 1 2]

= 3(\sqrt2 - 1) / (2 - 1) = 3(\sqrt2 - 1)

Example 2:

5 / (3 + \sqrt7) \cdot (3 - \sqrt7) / (3 - \sqrt7) = 5(3 - \sqrt7) / (9 - 7) = (15 - 5\sqrt7) / 2

Example 3 (Dual Radicals):

2 / (\sqrt3 + \sqrt5) \cdot (\sqrt3 - \sqrt5) / (\sqrt3 - \sqrt5) = 2(\sqrt3 - \sqrt5) / (3 - 5)

= 2(\sqrt3 - \sqrt5) / -2 = -(\sqrt3 - \sqrt5) = \sqrt5 - \sqrt3