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College Algebra

Explore the path DeGatchi took to get to the position he is at with Math and AI. A truly rare display of documentation in notoriously difficult topics. If you want to learn the fundamentals of Math this is precisely where you want to be. Similar to DeGatchi's crypto career log, it runs through the entire journey from start to finish! Read, learn and laugh, fren. I hope you can take the leap I did to conquer your wildest dreams... - DeGatchi 28/04/2024 00:08

27/04 - College Algebra Prerequisities, Part 1

”The laws of nature are written in the language of mathematics.” - Galileo


Before starting anything:

  1. A term can be a number, variable or product of numbers and variables, separates by plus or minus signs, e.g. 3x2+5x+23x^2 + 5x + 2 has 3 terms.
  2. An expression is a combination of terms, which includes numbers, variables (letters that represent numbers), and arithmetic operations like addition, subtraction, multiplication, and division, e.g. 3x2+5x+23x^2 + 5x + 2 is an expression.
  3. A product is the result of multiplication.
  4. The quotient is the result of division.
  5. A factor is a number or algebraic expression that divides another number or expression evenly.

For the retards, like myself, lets start with a quotient: the number returned from division, e.g. 15 is the quotient in 453=15\dfrac{45}{3} = 15.

What if we wanted to find the multiplicative inverse, aka the reciprocal? By swapping the numerator and denominator — since we know the answer is 15 aka 151\dfrac{15}{1}, 115\dfrac{1}{15}

A product, on the otherhand, is the number or expression resulting from the multiplication of two or more numbers or expressions, e.g. 10x2+15x=5x(2x+3)10x^2 + 15x = 5x(2x + 3) where 5x(2x+3)5x(2x + 3) is the equivalent expression that is a product! This example contained factoring a polynomial (more on these later), which means finding an equivalent expression that is a product.


A set is a collection of objects that can be determined by {}\{\}, e.g. the roster method looks like {1,2,3,...}\{1, 2, 3, ...\}.

Then we can move onto the intersection of sets. These are elements found in both sets, e.g. in A={1,2,3}A=\{1,2,3\} and B={3,4,5}B=\{3,4,5\} we see 3 is the common real number. We can say this as AB={3}A \cap B = \{3\}. Think of this as a bridge where the number crosses from one side to the other!

Next, there is the use of the union ABA \cup B. Can you guess what this is? It’s the set of elements that are members of set AA or BB e.g. {1,2,3}{3,4,5}={1,2,3,4,5}\{1,2,3\} \cup \{3,4,5\} = \{1,2,3,4,5\}! Think of this as a cup where we want one of each number flavour. We don’t want to overdose on number flavours so we add only one of each.

Real Numbers

Real numbers (R\Reals) are, what I consider as, the letters of English alphabet. If we don’t know what letters there are how can we form words (the analogy for expressions and formulas)?


The properties of real numbers are what we can do with said letters. What are the rules of the game so we can perform? These are super important to know. Without these you cannot go on to modify expressions, move terms around and solve equations — mastering these helps with calculus hardcore (from my experience).

  1. Commutative: Changing the order doesn’t affect the sum; think as going to the store back and forth — the destination should be the same.
    • 13+7=7+1313 + 7 = 7 + 13
    • 25=52\sqrt{2} \cdot \sqrt{5} = \sqrt{5} \cdot \sqrt{2}
  2. Associative: Changing grouping doesn’t affect the product; think as if i give you a pistol and you gave me an ak we can still shoot each other (lmao).
    • 2(3x)=(23)x=6x-2(3x) = (-2 \cdot 3)x = -6x
  3. Distributive: Multiplying the outside var by the inner parentheses member(s); we’d use this to factor the like terms to make it simpler.
    • a(b+c)=ab+aca(b+c) = a \cdot b + a \cdot c
    • 3(4x2y+6)=34x(3)2y36-3(4x - 2y +6) = -3 \cdot 4x - (-3) \cdot 2y - 3 \cdot 6
  4. Inverse: The inverse gives the same identity; we use this to simplify negative expressions.
    • Additive Inverse: a+(a)=0a + (-a) = 0 and (a)+a=0(-a) + a = 0
    • Multiplicitive Inverse: a1a=1,0a \cdot \dfrac{1}{a} = 1, \not = 0 and 1aa=1,0\dfrac{1}{a} \cdot a = 1, \not = 0
  5. Identity: Removing either 0 or 1 bc it doesn’t change anything; think of getting rid of your ex.
    • Additive Identity: a+0=aa + 0 = a
    • Multiplicitive Identity: a1=aa * 1 = a


There are subsets, think of as architypes in a game, that have different properties — try noticing the pattern between them as you go down.

Make a name for the new anime homie that assassinates his trbie, n’wiri (to remember the order: N, W, I, R, I)!

N\N Natural numbers

positive whole numbers, not including zero


N0\N_0 Whole numbers

positive whole numbers, including zero. The 0 can be used with any other set to tell us 0 is included.


Z\Z Integers

positive and negative whole numbers and zero


Q\mathbb{Q} Rational (Quotient) numbers

rational numbers that can be expressed as the quotant (divide result) of 2 integers

{17=171,3,0,2,3,23=0.6666}\{-17 = \dfrac{-17}{1}, -3, 0, 2, 3, \dfrac{-2}{3} = -0.6666\}

Irational numbers

all numbers where the decimals are neither terminating or repeating, so not quotient of integers

{21.414214,π3.142,π21.571}\{ \sqrt{2} \approx 1.414214, \pi \approx 3.142, -\dfrac{\pi}{2} \approx -1.571\}

R\R Real Numbers

the set that contains all possible rational and irrational numbers

π,2,3,...\pi, \sqrt{2}, \sqrt{3}, ...

C\Complex Complex Numbers

are any number of the form a+iba + ib, where a,bRa, b \in \R.

This means a and b are elements of set R\R

E.g. ...,i,2i,3+5i,......, -i, 2 - i, \sqrt{3} + 5i, ...

Everything is a rational number, aside from irrational numbers.


  • Q0\mathbb{Q}_0^- means any number in the set rational number that’s negative, inclduing 0.
  • Q+\mathbb{Q}^+ means any number in the set rational number that’s positive, not including 0.
  • Any set can repllace Q here.

Set Builder Notation

The condition xAx \in A and xBx \notin B means that x must belong to A but x must not belong to B.

A B=AB={xxA and xB}A \ B = A - B = \{x | x \in A \text{ and } x \notin B \}

Absolute Value

The absolute value a|a| really means how far away are we from 0? If -3 is the answer then a=3|a| = 3, for 3 it’s the exact same!

28/04 - College Algebra Prerequisities, Part 2


Exponential notation looks like bnb^n, where bb is the base and nn is the exponent. For example, b3b2=(bbb)(bb)=b7b^3 \cdot b^2 = (b \cdot b \cdot b)(b \cdot b) = b^7

There are a few rules we must understand, so we can modify expressions at will:

Product Rule

The product is the result of multiplying!

bmbn=bm+nb^m \cdot b^n = b^{m+n}

For example, lets use this expression

(5x2)(3x46x3+5x8)(5x^2)(3x^4 - 6x^3 + 5x - 8)

When using the distributive property w/ (5x2)(5x^2) to (3x4)(3x^4) you should see think of it as

53=155 \cdot 3 = 15

And, with our exponents,

x2x4=x2+4=x6x^2 \cdot x^4 = x^{2+4} = x^6


x2x4=xxxxxx=x6x^2 \cdot x^4 = x \cdot x \cdot x \cdot x \cdot x \cdot x = x^6

Then the final version is

=15x6= 15x^6

Then you can distribute and simplify the rest :)

Quotient Rule

The quotient is the output after dividing!

bmbn=bmn,b0\dfrac{b^m}{b^n} = b^{m-n}, b \not = 0

If we have


Then we can think of it as

bbbbbbbb\dfrac{b \cdot b \cdot b \cdot b \cdot b }{b \cdot b \cdot b}

And we can cancel out like terms

bbbbbbbb\dfrac{\cancel{b \cdot b \cdot b} \cdot b \cdot b }{\cancel{b \cdot b \cdot b}}

To be left with

bb=b2b \cdot b = b^2

Negative Exponent Rule

Typically in algebra we want the exponent to be positive and so we use this rule to get it to that state.

bn=bn1=1bnb^{-n} = \dfrac{b^{-n}}{1} = \dfrac{1}{b^n}

For example,


When we 4 - 7 we get -3 so

x4x7=x31\dfrac{x^4}{x^7} = \dfrac{x^{-3}}{1}

But when when visually seeing the exponents before this negative exponent calculation

x4x7=xxxxxxxxxxx\dfrac{x^4}{x^7} = \dfrac{x \cdot x \cdot x \cdot x}{x \cdot x \cdot x \cdot x \cdot x \cdot x \cdot x}

Then when we cancel the like terms

=xxxxxxxxxxx= \dfrac{\cancel{x \cdot x \cdot x \cdot x}}{\cancel{x \cdot x \cdot x \cdot x} \cdot x \cdot x \cdot x}

We’re left with

=1xxx=1x3= \dfrac{1}{x \cdot x \cdot x} = \dfrac{1}{x^3}

So that must mean that for any negative exponent on top we can inverse it and remove the subtraction operator applied to it!

x4x7=x31=1x3\dfrac{x^4}{x^7} = \dfrac{x^{-3}}{1} = \dfrac{1}{x^3}

Zero-Exponent Rule

When the exponent is 0 then the result is 1!

To create the mental model, think that we have

b2b2=bbbb=b22=b0=1\dfrac{b^2}{b^2} = \dfrac{\cancel{b \cdot b}}{\cancel{b \cdot b}} = b^{2-2} = b^0 = 1

What if we had


Then everything in the parentheses will equate to 1, therefore

31=3-3 \cdot 1 = -3

Power Rule

(bm)n=bmn(b^m)^n = b^{mn}


(22)3=(22)(22)(22)=22+2+2(2^2)^3 = (2^2) \cdot (2^2) \cdot (2^2) = 2^{2+2+2}

Products To Powers

Think of this as distributing the power to everything contained in the parentheses as a standalone.

(ab)n=anbn(ab)^n = a^nb^n


(3x)2=32x2(3x)^2 = 3^2x^2

Quotients To Powers

Same as the products to power, except with rationals.

(ab)2=a2b2\left(\dfrac{a}{b}\right)^2 = \dfrac{a^2}{b^2}

I want to reiterate the concepts we just saw with this example, it will build your intuition ALOT more surprisingly,

5x2y38x4y5\dfrac{5x^{-2}}{y^{-3}} \cdot \dfrac{8x^4}{y^{-5}}

I know what you’re thinking, “how the fuck am I going to solve this?!”

Remember, from the negative exponent rule, that we can inverse a variable if we also do the same for the exponent’s operator. Therefore, lets get rid of the negative exponents

=5y3x28x4y51= \dfrac{5y^3}{x^2} \cdot \dfrac{8x^4y^{5}}{1}

Lets get the product

=5y3+5x4x2= \dfrac{5y^{3+5}x^4}{x^2}

And remember the quotient rule for the xs

=5y3+5xxxxxx= \dfrac{5y^{3+5} \cdot x \cdot x \cdot \cancel{x \cdot x}}{\cancel{x \cdot x}}


=5y8x2= 5y^{8} x^2

Lets take it a step further and try this one,

24xy27x2÷36x2y345xy4\dfrac{24xy}{27x^{-2}} \div \dfrac{36x^2y^{-3}}{45xy^4}

There is the concept called [keep, change, flip] when dealing with rationals and multiplication/division. Essentially we keep one side of the expression, change the operator to the opposite (div to mul, or reverse) and flip the other side’s numerator and denominator. So lets do it here,

24xy27x245xy436x2y3\dfrac{24xy}{27x^{-2}} \cdot \dfrac{45xy^4}{36x^2y^{-3}}

Now we can

64xyx29395xy4y366x2\dfrac{6 \cdot 4 xy \cdot x^2}{9 \cdot 3} \cdot \dfrac{9 \cdot 5 x y^4 \cdot y^3}{6 \cdot 6 x^2}

Now we can look at it like this

6699x2x24xy35xy76\cancel{\dfrac{6}{6} \cdot \dfrac{9}{9} \cdot \dfrac{x^2}{x^2}} \cdot \dfrac{4xy}{3} \cdot \dfrac{5xy^7}{6}

Then simplifying further

=225x2y8332= \dfrac{\cancel{2} \cdot 2 \cdot 5 x^2 y^8}{3 \cdot 3 \cdot \cancel{2}} =10x2y89= \dfrac{10 x^2 y^8}{9}

What about for

x+25=78\dfrac{x + 2}{5} = \dfrac{7}{8}

We can cross multiple the 5 * 7 and 8 distribute the x + 2

35=8x+1635 = 8x + 16

Remove 16 from both sides

19=8x19 = 8x 198=238=x\dfrac{19}{8} = 2\dfrac{3}{8} = x


Lets disect what a square root (sqrt) w/ the radical expression an=b\sqrt[n]{a} = b

  • nn is the radical index. If n is: odd then b, even then |b|.
  • \sqrt{} is the radical
  • aa is the radicand, e.g. 4=2\sqrt{4} = 2 as 22=42^2 = 4, where 4 is the principal sqrt.

Think of it as radi, [cal, and, index].

The principal nth root of a real number a an=b\sqrt[n]{a} = b means that bn=ab^{n} = a


The perfect square is when a root’ed number returns an integer. If we have 36=6\sqrt{36} = 6, would be 6, bc multiplying itself (6^2), “perfectly” fits into 36.

If there is no perfect square, there will be factors of it. A factor is a number or expression that can be multiplied by another factor to get a product.

There is the greatest sqrt factor, referring to the largest perfect square that divides the number evenly, e.g. 50x=252x\sqrt{50x} = \sqrt{{25 \cdot 2x}} since 2525 can be factored further we get 25=45252x=452x=202x25 = 4 \cdot 5 \therefore \sqrt{{25 \cdot 2x}} = 4\cdot5\sqrt{2x} = 20\sqrt{2x}

Another example is 500=2520=2520=520\sqrt{500} = \sqrt{25 \cdot 20} = \sqrt{25} \sqrt{20} = 5 \sqrt{20}. Since 20 contains a perfect square factor, 4, we need to simplify futher: 520=545=545=(52)5=1055 \sqrt{20} = 5 \sqrt{4 \cdot 5} = 5 \sqrt{4} \sqrt{5} = (5 \cdot 2) \sqrt{5} = 10\sqrt{5}.

The aim is to get the lowest numbers to form the original number to make the expression less complex, e.g. for 36 we’d get 18218 \cdot 2 but we can go futher and do 929 \cdot 2, then to get 36 we have the lowest factorisation w/ 3222=363^2 \cdot 2^2 = 36

Combining Radicals

Now with out new-found knowledge we can fight the mini boss battle: 450x632x4 \sqrt{50x} - 6\sqrt{32x}

450x632x=(4x)524 \sqrt{50x} - 6\sqrt{32x} = (4 \cdot x) \sqrt{5} \sqrt{2}

Expression Rules

  1. Product: The sqrt of a product is the product of sqrts, ab=ab\sqrt{ab} = \sqrt{a} \cdot \sqrt{b}
  2. Quotient: If a (nominator) and b (the denominator) are nonnegative real numbers and b != 0, then: ab=ab\sqrt{\dfrac{a}{b}} = \dfrac{\sqrt{a}}{\sqrt{b}}. The sqrt of a quotient is the quotient of the sqrts.

Add + Sub Radicals

  • Add: 411+211=(4+2)11=6114\sqrt{11} + 2\sqrt{11} = (4+2)\sqrt{11} = 6\sqrt{11}
  • Sub: 5x75x=15x75x=(17)5x=65x\sqrt{5x} - 7 \sqrt{5x} = 1\sqrt{5x} - 7 \sqrt{5x} = (1-7) \sqrt{5x} = -6\sqrt{5x}

Rationalising Denominators

This concept is where things started to click for me. Essentially, any number divided by itself is 1, e.g. 33=1\dfrac{\sqrt{3}}{\sqrt{3}} = 1.

So, 2 approximations that look entirely different, e.g. 13=33\dfrac{1}{\sqrt{3}} = \dfrac{\sqrt{3}}{3}, can mean the same thing.

  1. let’s break this down 13\dfrac{1}{\sqrt{3}}
=131\\ = \dfrac{1}{\sqrt{3}} \cdot 1 =1333\\ = \dfrac{1}{\sqrt{3}} \cdot \dfrac{\sqrt{3}}{\sqrt{3}}
  1. here’s the fascinating part: =39\\ = \dfrac{\sqrt{3}}{\sqrt{9}}
  2. simplyfiying further =33\\ = \dfrac{\sqrt{3}}{3}
  3. therefore 13=33\\ \dfrac{1}{\sqrt{3}} = \dfrac{\sqrt{3}}{3}


Radical expressions that involve the sum and difference of the same two terms are called conjugates (joint together).

The general rule for multiplying conjugates is:

(a+b)(ab)=(a)2(b)2=ab(\sqrt{a} + \sqrt{b})(\sqrt{a} - \sqrt{b}) = (\sqrt{a})^2 - (\sqrt{b})^2 = a - b

Why is this? We apply the Distributive property into:

(ab)(ab)=(aa)+(a(b))+(ba)+((b)b)(\sqrt{a} \cdot \sqrt{b}) (\sqrt{a} - \sqrt{b}) = (\sqrt{a} \cdot \sqrt{a}) + (\sqrt{a} \cdot (-\sqrt{b})) + (\sqrt{b} \cdot \sqrt{a}) + ((-\sqrt{b}) \cdot \sqrt{b})

Make this more readable:

=a(ab)+(ab)b= a - (\sqrt{ab})+(\sqrt{ab})-b

The like terms cancel out:

=(ab)+(ab)= - (\sqrt{ab})+(\sqrt{ab})

Which leaves us with:

=ab= a -b

Lets look at another example, hx+hx\dfrac{h}{{\sqrt{x+h}} - \sqrt{x}} then the conjugate of the denominator is x+h+x\sqrt{x + h} + \sqrt{x} (the opposite).

As an example, 75+3\dfrac{7}{5 + \sqrt{3}}

=75+35353\\ = \dfrac{7}{5 + \sqrt{3}} \cdot \dfrac{5 - \sqrt{3}}{5 - \sqrt{3}}

Rember how this is =1. Now we convert the two denominators to (a)2(b)2(\sqrt{a})^2 - (\sqrt{b})^2

=7(53)52(3)2\\ = \dfrac{7(5 -\sqrt{3})}{5^2 - (\sqrt{3})^2}

Simplify the denominator

=7(53)253\\ = \dfrac{7(5 -\sqrt{3})}{25 - 3}

Rational Exponents

Lets take a look at two expressions: (712)2=7122=71=7(7^{\frac{1}{2}})^2 = 7^{\frac{1}{2} \cdot 2} = 7^1 = 7 and (7)2=77=49=7(\sqrt{7})^2 = \sqrt{7} \cdot \sqrt{7} = \sqrt{49} = 7

From this we see that: 7127^{\frac{1}{2}} means 7\sqrt{7}

And so we can make the definition, as long as n >= 2:

a1n=ana^\frac{1}{n} = \sqrt[n]{a}

And, as long as a != 0,

a1n=1a1n=1a3a^{-\frac{1}{n}} = \dfrac{1}{a^{\frac{1}{n}}} = \dfrac{1}{\sqrt[3]{a}}

But what about rationals where the numeration > 1?

a23=(a13)2=(a2)13a^\frac{2}{3} = (a^\frac{1}{3})^2 = (a^2)^\frac{1}{3}


a23=(a3)2=a23a^\frac{2}{3} = (\sqrt[3]{a})^2 = \sqrt[3]{a^2}

And we take this further with a new definition,

amn=(an)m=amna^{\frac{m}{n}} = (\sqrt[n]{a})^m = \sqrt[n]{a^m}

And if amna^{-\frac{m}{n}} is a nonzero real number, then

amn=1amna^{-\frac{m}{n}} = \dfrac{1}{a^\frac{m}{n}}

Notice how from the previous definition, a1n=ana^\frac{1}{n} = \sqrt[n]{a}, n is the denominator and the only thing that changed was the numerator m. The reason why () come into play in the latter is because adding the exponent of 1 to something doesn’t change anything! So really, the latter is the general definition of rational exponents!

Lets do an example:

325316x34=(3216)(x53x34)\dfrac{32^\frac{5}{3}}{16x^\frac{3}{4}} = \left(\dfrac{32}{16}\right)\left(\dfrac{x^\frac{5}{3}}{x^\frac{3}{4}}\right) =2x5334= 2x^{\frac{5}{3} - \frac{3}{4}}

Now, if you’re mathematically illiterate like myself, we need to simplify this mofo right here. It’s a bit confusing but I’ve done the hard work for the tribe:

We begin by finding the common denominator w/ 3 and 4: 12, essentially the lowest number w/ both numbers — like factoring.

So for 5334{\dfrac{5}{3} - \dfrac{3}{4}} we’re really just multiplying the numerator 5 and denominator 3 by the opposite expression’s denominator 4:

53=5433=2012\dfrac{5}{3} = \dfrac{5\cdot4}{3\cdot3} = \dfrac{20}{12}

And do for the other expression with the former,

34=3343=912\dfrac{3}{4} = \dfrac{3\cdot3}{4\cdot3} = \dfrac{9}{12}

Then we get

=2012912=20912=2x1211= \dfrac{20}{12} - \dfrac{9}{12} = \dfrac{20 - 9}{12}= 2x^{\frac{12}{11}}

We finally got through the core sht! Only [polynomials, factoring, trig] to go (fck me it takes ages to write about this stuff, but damn is it sticking).

02/05 - Polynomials!

Gm, welcome to learning about pole-pee-no-my-balls.

As always, lets learn the Vocabulary first, wtf are nomials:

  • Monomial is a single term, 7x37x^3
  • Binomial: is 2 terms, 7x3+2x27x^3 + 2x^2
  • Trinomial: is 3 terms, 7x3+2x2+57x^3 + 2x^2 + 5
  • Polynomial: is anything beyond 3 terms, 7x3+2x2+5x+697x^3 + 2x^2 + 5x + 69

We describe polynomials w/ negatives w/ parentheses to be explicit (like a rap song)

7x39x2+13x67x^3 - 9x^2 + 13x - 6

As, 7x3+(9x2)+13x+(6)7x^3 + (-9x^2) + 13x + (- 6)

I trip up a lot with the - bullshit so this is quite important to note: axn=(axn)-ax^n = (-ax^n).

The n represents the degree. We order polynomials in cronological order from left to right in terms of degrees. E.g. 7x37x^3 will be ordered before 7x27x^2. The degree of a polynomial is the greatest of the degress of all its terms.

Taking a look at an algebraic expression w/ a polynomial in x (expression of vars + coefficients combined using addition, subtraction and multiplication): anxn+an1xn1+an2xn2+...+a1x+a0a_{n}x^n + a_{n-1}x^{n-1} + a_{n-2}x^{n-2} + ... + a_1x + a_0

  • The degree of the polynomial is nn. This is important because it is the biggest rate of change.
  • The leading coefficient (numerical or constant factor that multiplies a var) (the term w/ the highest exponent/power) is ana_n
  • The constant term is a0a_0

Polynomial Sub, Add, Mul

Subtracting polynomials by combining like terms (variables and their exponent powers are the same), e.g.

9x313x3=(9+13)x3=4x3-9x^3 - 13x^3 = (-9 + 13)x^3 = 4x^3

We are using the the distributive property to factor out the common term, x3x^3

Adding polynomials: 5x4+6x4=11x45x^4 + 6x^4 = 11x^4

However we cannot Simplify 5x4+3x25x^4 + 3x^2 because there are no like terms!

Multiplying polynomials:


Combining like terms,

=(56)(x4x4)= (5\cdot6)(x^4 \cdot x^4) =30x4+4=30x8= 30x^{4+4} = 30x^8

But what would we do if there is no monomials? E.g,

(2x+3)(x2+4x+5)(2x + 3) (x^2 + 4x + 5)

Well we would use the distributive property (our lord and saviour!):

Distributing 2x

2x(x2+4x+5)=2x3+8x2+10x2x(x^2 + 4x + 5) = 2x^3 + 8x^2 + 10x

Then distributing 3

3(x2+4x+5)=3x2+12x+153(x^2 + 4x + 5) = 3x^2 + 12x + 15

Then combining like terms

=2x3+(3x2+8x2)+(10x+12x)+15= 2x^3+ (3x^2 + 8x^2) + (10x + 12x) + 15


=2x3+112+22x+15= 2x^3+ 11^2 + 22x + 15

Product Of Two Binomials

Once again, we use the distributive property to find the product of two binomials. For example,

(3x5)(4x+5)(3x - 5) (4x + 5) =3x(4x)+3x(5)+(5(4x))+(5(5))= 3x(4x) + 3x(5) + (-5(4x)) + (-5(5)) =12x2+15x+(20x)+(25)= 12x^2 + 15x + (-20x) + (-25) =12x2+5x25= 12x^2 + -5x -25

Similar to multiplying conjugates, we find the Product of the Sum and Difference of Two Terms as follows: (A+B)(AB)=A2B2(A + B)(A - B) = A^2 - B^2 because the two terms cancel eachother out (BA)(A(B))=ABAB=0(B \cdot A) (A \cdot (- B)) = AB - AB = 0

Squaring & Cubing Binomials

(A+B)3=A3+3A2B+3AB2+B3(A + B)^3 = A^3 + 3A^2B + 3AB^2 + B^3

How do we break this down? Lets expand the exponents:

(A+B)3=(A+B)(A+B)(A+B)(A + B)^3 = (A + B) (A + B) (A + B)

Lets distribute the first 2:

(A+B)(A+B)=A2+AB+AB+B2=A2+2AB+B2(A + B) (A + B) = A^2 + AB + AB + B^2 = A^2 + 2AB + B^2

Now we distribute the third:

=(A2+2AB+B2)(A+B)= (A^2 + 2AB + B^2) (A + B) =(A3+2A2B+2AB2+B3)= (A^3 + 2A^2B + 2AB^2 + B^3)

03/05 - Factoring Polynomials


Factoring a polynomial containing the sum of monomials means finding an expression that is a product:

10x2+15x=5x(2x+3)10x^2 + 15x = 5x(2x+3)

The goal of factoring a polynomial is: to use one or more factoring techniques until each of the polynomial’s factors, except possibily for a monomial factor, is prime (a natural number, greater than 1, that is not the product of two smaller natural numbers) or irreducible, aka being factored completely.

The greatest common factor (GCF) is the largest positive integer that divides two or more integers without leaving a remainder.

18x3+27218x^3 + 27^2

9 factors into 18 (9*2) and 27 (9*3),

=9x2(2x)+9x2(3)= 9x^2(2x) + 9x^2(3)

What about for,

x2(x+3)+5(x+3)x^2 (x+3) + 5(x + 3)

We see that (x+3)(x + 3) is the common factor in both terms, so we can reverse the distributive property into:

(x+3)(x2+5)(x + 3) (x^2 + 5)

Factoring By Grouping

Some polynomials only have 1 as the greatest common factor. But by grouping the terms together we can actually create the common factor.

For example, these terms have the common factor of 1,

x3+4x2+3x+12x^3 + 4x^2 + 3x + 12

If we group them together then we may be able factor out something that is common

First, the common factor is x2x^2 with:

(x3+4x2)=x2(x+4)(x^3 + 4x^2) = x^2(x+4)

The second, with the common factor of 33, becomes

(3x+12)=3(x+4)(3x + 12) = 3(x + 4)

Combining them together we get

=x2(x+4)+3(x+4)= x^2(x + 4) + 3(x + 4)

Notice the common factor again? It’s (x+4)(x + 4). We use this as one term then the other would be the remaining vars, (x2+3)(x^2 + 3). Therefore,

=(x+4)(x2+3)= (x + 4)(x^2 + 3)

Lets do one that isn’t so intuitive

2x2+32=02x^2 + 3 - 2 = 0

Since there isn’t any factors that equate to -2 when multiplying them together and add up to 3 we multiply -2 by our coefficient 2 from 2x22x^2 to create -4 for factoring.

Now we can factor 4 and -1 to create

2x2+4x1x2=02x^2 + 4x -1x - 2 = 0

Notice how the result doesn’t change!!

Now we can factor by grouping since there are 4 terms

2x(x+2)1(x+2)=02x(x + 2) - 1 (x + 2) = 0

Then we factor out x + 2

(x+2)(2x1)=0(x + 2)(2x - 1) = 0

And now we can solve the expression

x+2=0x + 2 = 0 x=2x = -2


2x1=02x - 1 = 0 2x=12x = 1 x=12x = \dfrac{1}{2}

Factoring Trinomials

Lets try factor a trinomial in two variables:

2x27xy+3y22x^2 -7xy + 3y^2

First we need to find the first 2 terms that create 2x22x^2 when distributed

(2x+..)(x+..)(2x + ..) (x + ..)

Then we need to figure out the remaining two terms that whose product is 3y23y^2

(y)(3y)=(y)(3y)(y)(3y) = (-y)(-3y)

Finally, we need to make sure the sum of the far left and right products is equal to 7xy-7xy

(2x+y)(x+3y)(2x + -y) (x + -3y)

Then we can verify but distributing it

=(2xx)+(2x3y)+(yx)+(y3y)= (2x \cdot x) + (2x \cdot -3y) + (-y \cdot x) + (-y \cdot -3y)


=2x26xyxy4y= 2x^2 - 6xy - xy - 4y

And, we’re back where we started!

=2x27xy4y= 2x^2 - 7xy - 4y

Factoring Difference Of Two Squares

A2B2=(A+B)(AB)A^2 - B^2 = (A + B)(A - B)

For example,

81x24981x^2 - 49 =81x249= \sqrt{81x^2} - \sqrt{49} =(9x)272= (9x)^2 - 7^2 =(9x+7)(9x7)= (9x + 7)(9x - 7)

Repeated Factorisation

This is when we can factorise a term further after factorising the expression.

For example,

x481x^4 - 81 =x481= \sqrt{x^4} - \sqrt{81} =(x2)292= (x^2)^2 - 9^2

Difference of squares

=(x2+9)(x29)= (x^2 + 9)(x^2 - 9)

Factoring further step

=(x2+9)(x29)= (x^2 + 9)(x^2 - \sqrt{9})

Notice that we now have a square subtracting another square, (x232)(x^2 - 3^2). We can factor further w/ the difference of two squares again (A2B2=(A+B)(AB)A^2 - B^2 = (A + B)(A - B)) only because it’s subtracting. We cannot do the same for (x2+9)(x^2 + 9) because its adding, which is not the difference of two squares requirement, A2B2A^2 - B^2. If we were expand, we’d be doing the opposite of factoring.

=(x2+9)(x232)= (x^2 + 9)(x^2 - 3^2) =(x2+9)(x+3)(x3)= (x^2 + 9)(x + 3)(x - 3)

This took me a while to understand this, so don’t beat yourself up — I was seriously lost lmao. The main point is the subtraction enables the difference of two squares factorisation, whereas addition does not.

Factoring Perfect Square Trinomials

Before we start, a perfect square trinomial is a trinomial that can be factored into two identical binomial factors, e.g. x2+6x+9=(x+3)(x+3)=(x+3)2x^2 + 6x + 9 = (x+3)(x+3) = (x+3)^2.

There are ways to factor a percect square trinomial. Notice how the first sign is reflected as the sign in the parentheses.

  1. A2+2AB+B2=(A+B)2A^2 + 2AB + B^2 = (A + B)^2
  2. A22AB+B2=(AB)2A^2 - 2AB + B^2 = (A - B)^2

We identify a perfect square trinomial via:

  1. The first and last terms being squares of monomials or integers
  2. The middle term is twice the product of the expressions being squared in the first and last terms


25x260x+3625x^2 - 60x + 36

Break down the terms w/ sqrts

=25x260x+36= \sqrt{25x^2} - 60x + \sqrt{36}

Now we have the sqrt product

=(5x)260x+62= (5x)^2 - 60x + 6^2

Verify the middle term is twice the product of the outer two

60x=2(5x)(6)60x = 2(5x)(6)

It is, therefore

=(5x6)2= (5x - 6)^2

Factoring the Sum and Difference of Two Cubes

Factoring the Sum of Two Cubes

A3+B3=(A+B)(A22AB+B2)A^3 + B^3 = (A + B)(A^2 - 2AB + B^2)

Factoring the Difference of Two Cubes

A3B3=(AB)(A2+2AB+B2)A^3 - B^3 = (A - B)(A^2 + 2AB + B^2)

Notice how the first parentheses uses the same sign as the 2nd term and then the 2nd parentheses use the opposite of the 2nd term’s sign.

For example,

64x312564x^3 - 125 =6431253=(4x)353= \sqrt[3]{64} - \sqrt[3]{125} = (4x)^3 - 5^3 =6431253=4353= \sqrt[3]{64} - \sqrt[3]{125} = 4^3 - 5^3 =(45)(42+2(4)(5)+52)= (4 - 5)(4^2 + 2(4)(5) + 5^2)

Factoring Polynomial Strategy

Factoring is super important to grasp as it’s how we can simplify expressions drastically. And since polynomials are extremely common in mathematics we want to be well equiped with all tools at our disposal, which is why algebra is so god damn important — good luck doing anything beyond this if you don’t truly undersatnd algebra.

  1. If there is a common factor, factor out the greatest common factor
  2. Check how many terms there are then:
    1. 2 terms:
      • Difference of two squares: A2B2=(A+B)(AB)A^2 - B^2 = (A + B)(A - B)
      • Sum of two cubes: A3+B3=(A+B)(A22AB+B2)A^3 + B^3 = (A + B)(A^2 - 2AB + B^2)
      • Difference of two cubes: A3B3=(AB)(A2A2AB+B2)A^3 - B^3 = (A - B)(A^2 A 2AB + B^2)
    2. 3 terms: If perfect square trinomial use the following, otherwise trial and error:
      • A2+2AB+B2=(A+B)2A^2 + 2AB + B^2 = (A + B)^2
      • A22AB+B2=(AB)2A^2 - 2AB + B^2 = (A - B)^2
    3. 4+ terms, try factoring by grouping
  3. If more than 1 term in the factored polynomial can be further factored, go further until factored completely

Factoring Fractional Exponents

x(x+1)34+(x+1)14x(x + 1)^{-\frac{3}{4}} + (x+1)^{\frac{1}{4}}

We have the greatest common factor (x+1)34(x + 1)^{-\frac{3}{4}}. Express each term w/ the GCF

=(x+1)34x+(x+1)34(x+1)= (x + 1)^{-\frac{3}{4}}x + (x + 1)^{-\frac{3}{4}}(x+1)

Factor out GCF

=(x+1)34[x+(x+1)]=(x+1)34[2x+1]= (x + 1)^{-\frac{3}{4}} [x + (x + 1)] = (x + 1)^{-\frac{3}{4}} [2x + 1]

Then we use the previously discussed negative quotient rule, bn=1bnb^{-n} = \dfrac{1}{b^n}. Remember we get rid of the negative sign on the rational exponent and switch from multiplying to dividing!

=2x+1(x+1)34= \dfrac{2x + 1}{(x + 1)^{\frac{3}{4}}}

05/05 - Rational expressions

Speaking of rationals, we’re almost at the end of our retardo-prep for algebra! Only a fraction of the way left (lmao, kill me)!

Wtf is a rational expression? The quotient (product of division) of two polynomials, e.g. xx21\dfrac{x}{x^2 - 1}

Since rational expressions are division and division by zero is undefined we must establish the domain, e.g. for xx2\dfrac{x}{x - 2} our domain would be x2x \not = 2 otherwise the rational could be 0, and everyhing will crash and burn. And so our full expression is xx2,x2\dfrac{x}{x - 2}, x \not = 2

Simplifying Rational Expressions

A rational express is only simplified if its numerator and denominator have no common factors other than 1 or -1

  1. Factor the numerator and the denominator completely
  2. Divide both the numerator and the denominator by any commmon factors

For example,

x2+6x+5x225\dfrac{x^2 + 6x + 5}{x^2 - 25}

Look for how we can factor further — sqrt into perfect square

=x2+6x+5x225= \dfrac{x^2 + 6x + 5}{x^2 - \sqrt{25}}

Denominator difference of two squares

=x2+6x+5x252= \dfrac{x^2 + 6x + 5}{x^2 - 5^2}

Factor; reverse distribute

=x2+6x+5(x+5)(x5)= \dfrac{x^2 + 6x + 5}{(x + 5)(x - 5)}

Turn quadratic numerator into binomial expression (we find 2 numbers that add up to the middle term 6x). We can actually cancel out like terms here, (x+5)=1(x + 5) = 1, and make them both 1

=(x+5)(x+1)(x+5)(x5)= \dfrac{(x + 5)(x + 1)}{(x + 5)(x - 5)}

Since we canceled 5 out from the numerator and and the -5 from our simplified denominator (from 25), our domain is added to our expression

=(x+1)(x5),x{5,5}= \dfrac{(x + 1)}{(x - 5)}, x \not = \{-5, 5\}

Multiplying Rational Expressions

x7x1x213x21\dfrac{x-7}{x-1} \cdot \dfrac{x^2 - 1}{3x-21}

Difference of two squares and apply factor the common factor

=x7x1(x+1)(x1)3(x7)= \dfrac{x-7}{x-1} \cdot \dfrac{(x + 1)(x - 1)}{3(x - 7)}

Notice how we have the same terms on both sides, (x7)(x - 7) and (x1)(x - 1), so we can cancel them out (bc it doesn’t change the final result)

=x+13= \dfrac{x + 1}{3}

Since The denominator has factors of x - 1 and x - 7 then domain is x != [1, 7]

=x+13,x1,x7= \dfrac{x + 1}{3}, x \not = 1, x \not = 7

Dividing Rational Expressions

x22x8x29/x4x+3\dfrac{x^2 - 2x - 8}{x^2 - 9} / \dfrac{x - 4}{x + 3}

Swap the sign to multiply and swap numerator and denominator

=x22x8x232x+3x4= \dfrac{x^2 - 2x - 8}{x^2 - 3^2} \cdot \dfrac{x + 3}{x - 4}

Difference of two squares + factoring a quadratic

=(x4)(x+2)(x+3)(x3)x+3x4= \dfrac{(x - 4)(x + 2)}{(x + 3)(x - 3)} \cdot \dfrac{x + 3}{x - 4}

Common terms in both rationals; we can cancel them and they get removed from our domain (x != [-3, 3, 4]). Remember we can’t have the rational = 0.

=x+2x3,x{3,3,4}= \dfrac{x + 2}{x - 3}, x \not = \{-3, 3, 4\}

Add & Subtract Rational Expressions

Rational numbers that have no common factors in their denominators can be added or subtracted w/ one of the following properties:


ab+cd=ab+bcbd,b{0,d}\dfrac{a}{b} + \dfrac{c}{d} = \dfrac{ab + bc}{bd}, b \not = \{0, d\}


abcd=abbcbd,b{0,d}\dfrac{a}{b} - \dfrac{c}{d} = \dfrac{ab - bc}{bd}, b \not = \{0, d\}

Notice how they cross divide for the numerators and the denominators they multiply themselves.


Example w/ the same denominator

5x+1x294x2x29\dfrac{5x + 1}{x^2 - 9} - \dfrac{4x - 2}{x^2 - 9}

Combine the terms into a single expression bc of the same denominator

=5x+1(4x2)x29= \dfrac{5x + 1 - (4x -2)}{x^2 - 9}

Remove the parentheses and change the sign of each term within

=5x+14x+2x29= \dfrac{5x + 1 - 4x + 2}{x^2 - 9}

We can factor 9 into a perfect square w/ sqrt

=5x+14x+2x232= \dfrac{5x + 1 - 4x + 2}{x^2 - 3^2}

difference of two squares!

=x+3(x+3)(x3)= \dfrac{x + 3}{(x + 3)(x - 3)}

Remove like terms + express the domain

=1x3,x{3,3}= \dfrac{1}{x - 3}, x \not = \{-3, 3\}

Simple enough, right?

Lets up the ante and try something that will probably be spotted in the wird more often. One w/ no common factors in the denominators?

x+22x34x+3\dfrac{x + 2}{2x-3} - \dfrac{4}{x+3}

Cross multiply numerators and multiply both denominators

=(x+2)(x+3)(2x3)4(2x3)(x+3)= \dfrac{(x+2)(x+3) - (2x - 3)4}{(2x-3)(x+3)}

We don’t open the parentheses bc that was the variable we were given with

=x2+5x+6(8x12)(2x3)(x+3)= \dfrac{x^2 + 5x + 6 - (8x - 12)}{(2x-3)(x+3)}

We can remove the parentheses and change sign of each term within

=x2+5x+68x+12(2x3)(x+3)= \dfrac{x^2 + 5x + 6 - 8x + 12}{(2x-3)(x+3)}

Simplify and specify domain,

  1. 2x3=02x−3=0 leads to x=32x = \dfrac{3}{2}​

    because 2x3=02x − 3 = 0 equals 2x=32x = 3 then further x=32x = \dfrac{3}{2}

  2. x+3=0x+3=0 leads to x=3x=3x= −3x = −3

=x23x+18(2x3)(x+3),x{32,3}= \dfrac{x^2 - 3x + 18}{(2x-3)(x+3)}, x \not = \{\dfrac{3}{2}, -3\}

Least Common Denominator

It is the smallest positive integer that is divisible by all the denominators in a set of fractions.

Lets do an example,

x+3x2+x2+2x21\dfrac{x + 3}{x^2 + x - 2} + \dfrac{2}{x^2 - 1}

First, we need to find the least common denominator by factoring the denominators

  1. x2+x2=(x+2)(x1)x^2 + x - 2 = (x + 2)(x - 1) because x2+x+2=(x2)(x+1)x^2 + x + 2 = (x - 2)(x + 1)
  2. x21=(x+1)(x1)x^2 - 1 = (x + 1)(x - 1)


=x+3(x+2)(x1)+2(x+1)(x1)= \dfrac{x + 3}{(x + 2)(x - 1)} + \dfrac{2}{(x + 1)(x - 1)}

We can see the only factor not in the first denominator is (x+1)(x + 1) therefore the least common denominator is (x+2)(x1)(x+1)(x + 2)(x - 1)(x + 1)

So we add it onto both,

=x+3(x+2)(x1)x+1x+1+2(x+1)(x1)x+2x+2= \dfrac{x + 3}{(x + 2)(x - 1)} \cdot \dfrac{x + 1}{x + 1} + \dfrac{2}{(x + 1)(x - 1)} \cdot \dfrac{x + 2}{x + 2}

Rearrange the right denominator to match the left

=(x+3)(x+1)(x+2)(x1)(x+1)+2(x+2)(x+2)(x1)(x+1)= \dfrac{(x + 3)(x + 1)}{(x + 2)(x - 1)(x + 1)} + \dfrac{2(x + 2)}{(x + 2)(x - 1)(x + 1)}

Keep in mind this isn’t actually changing the outcome, only the appearance. But we can start to sum these up.

=(x+3)(x+1)+2(x+2)(x+2)(x1)(x+1)= \dfrac{(x + 3)(x + 1) + 2(x + 2) }{(x + 2)(x - 1)(x + 1)}

Then simplify it

=x2+4x+3+2x+4(x+2)(x1)(x+1)= \dfrac{x^2 + 4x + 3 + 2x + 4}{(x + 2)(x - 1)(x + 1)}

Combine like terms + establish the domain

=x2+6x+7(x+2)(x1)(x+1),x{2,1,1}= \dfrac{x^2 + 6x + 7}{(x + 2)(x - 1)(x + 1)}, x \not = \{-2, 1, -1\}

Complex Rational Expressions

Aka complex fractions have numerators or denominators containing one or more rational expressions, e.g. 1+1x11x\dfrac{1 + \dfrac{1}{x}}{1 - \dfrac{1}{x}} or 1x+h1xh\dfrac{\dfrac{1}{x + h} - \dfrac{1}{x}}{h}

For example, let’s simplify

1+1x11x\dfrac{1 + \dfrac{1}{x}}{1 - \dfrac{1}{x}}

We start by getting the lowest common denominator 1(x)1(x)

=11+1x111x= \dfrac{\dfrac{1}{1} + \dfrac{1}{x}}{\dfrac{1}{1} - \dfrac{1}{x}}

The goal here is to have the denominators match to make everything easier. So whatever we don’t have for one denominator we want to add the opposite one. Remember anything you do to the denominator you do to the numerator.

In this case the LCD x1x \cdot 1 is applied by adding xx to the left fraction, 11 to the right fraction (remains the same).

=1x1x+11x11x1x11x1= \dfrac{\dfrac{1 \cdot x}{1 \cdot x} + \dfrac{1 \cdot 1}{x \cdot 1}}{\dfrac{1 \cdot x}{1 \cdot x} - \dfrac{1 \cdot 1}{x \cdot 1}}

Do the calculation

=xx+1xxx1x= \dfrac{\dfrac{x}{x} + \dfrac{1}{x}}{\dfrac{x}{x} - \dfrac{1}{x}}

The denominators are now the exact same we can merge the fractions together

=x+1xx1x= \dfrac{\dfrac{x + 1}{x}}{\dfrac{x - 1}{x}}

Invert and multiply; Division by a fraction is equivalent to multiplication by its reciprocal.

The reciprocal of a fraction is created by swapping its numerator and denominator.

For example, the reciprocal of ab\dfrac{a}{b} is ba\dfrac{b}{a}, provided that a and b are non-zero.

=x+1xxx1= \dfrac{x + 1}{x} \cdot \dfrac{x}{x - 1}

Remove like terms that cross, x.


Beause multiplication is commutative, the ordering doesn’t matter!

So, x+1xxx1\dfrac{x + 1}{x} \cdot \dfrac{x}{x - 1} can be represented as x+1x1xx\dfrac{x + 1}{x - 1} \cdot \dfrac{x}{x} which is the same as x+1x11\dfrac{x + 1}{x - 1} \cdot 1

And we finally get

=x+1x1= \dfrac{x + 1}{x - 1}

I want to show another example that I personally struggled for 2 weeks on, causing me go down the rabbit hole of why I’m so stupid.

1x+h1xh\dfrac{\dfrac{1}{x + h} - \dfrac{1}{x}}{h}

The first thing we should do is forget about the h\dfrac{}{h} so we can focus on the numerator, 1x+h1x\dfrac{1}{x + h} - \dfrac{1}{x}

Next we need make the denominators the same.

We do this by asking ourselves “what is the least common denominator?” (the smallest number that can be used as the denominator for all fractions in a set). What would make the denominators the same?

If we multiply the left by x and the right by x + h we get x(x+h)x(x+h)

(1)x(x+h)x=xx(x+h)\dfrac{(1)x}{(x + h)x} = \dfrac{x}{x(x+h)}

and the right side,

(1)x+h(x)x+h=x+hx(x+h)\dfrac{(1) x+h}{(x) x+h} = \dfrac{x+h}{x(x+h)}

now we can merge the two, remembering the subtraction operator

=xx(x+h)x+hx(x+h)= \dfrac{x}{x(x+h)} - \dfrac{x+h}{x(x+h)}

After merging

=x(x+h)x(x+h)= \dfrac{x - (x+h)}{x(x+h)}

Here’s a small detail but -(x+h) actually means -1(x+h), and so

=x1(x+h)x(x+h)= \dfrac{x - 1(x+h)}{x(x+h)}

We can apply the distributive property

=xxhx(x+h)= \dfrac{x - x - h}{x(x+h)}

Notice the xs cancel out in the numerator

=0hx(x+h)=hx(x+h)= \dfrac{0 - h}{x(x+h)} = \dfrac{- h}{x(x+h)}

Now we can add the h\dfrac{}{h}

=hx(x+h)h= \dfrac{\dfrac{- h}{x(x+h)}}{h}

We can reverse the division of h into multiplication into it’s numerator x(x+h)x(x+h)