27/04  College Algebra Prerequisities, Part 1
”The laws of nature are written in the language of mathematics.”  Galileo
Intro
Before starting anything:
 A term can be a number, variable or product of numbers and variables, separates by plus or minus signs, e.g. $3x^2 + 5x + 2$ has 3 terms.
 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. $3x^2 + 5x + 2$ is an expression.
 A product is the result of multiplication.
 The quotient is the result of division.
 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 $\dfrac{45}{3} = 15$.
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 $\dfrac{15}{1}$, $\dfrac{1}{15}$
A product, on the otherhand, is the number or expression resulting from the multiplication of two or more numbers or expressions, e.g. $10x^2 + 15x = 5x(2x + 3)$ where $5x(2x + 3)$ 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.
Sets
A set is a collection of objects that can be determined by $\{\}$, e.g. the roster method looks like $\{1, 2, 3, ...\}$.
Then we can move onto the intersection of sets. These are elements found in both sets, e.g. in $A=\{1,2,3\}$ and $B=\{3,4,5\}$ we see 3
is the common real number. We can say this as $A \cap B = \{3\}$. Think of this as a bridge where the number crosses from one side to the other!
Next, there is the use of the union $A \cup B$. Can you guess what this is? It’s the set of elements that are members of set $A$ or $B$ e.g. $\{1,2,3\} \cup \{3,4,5\} = \{1,2,3,4,5\}$! 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 Number ($\Reals$)
Real numbers ($\Reals$) 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)?
Subsets
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
)!
 Natural numbers: nums we use for counting $\{1,2,3,...\}$
 Whole numbers: natural numbers w/ 0, bc it felt lonely $\{0,1,2,3,...\}$
 Integers: whole numbers that also went backwards $\{3,2,1,0,1,2,3,...\}$
 Rational numbers: rational numbers that can be expressed as the quotant (divide result) of 2 integers $\{17 = \dfrac{17}{1}, 3, 0, 2, 3, \dfrac{2}{3} = 0.6666\}$
 Irational numbers: all numbers where the decimals are neither terminating or repeating, so not quotient of integers $\{ \sqrt{2} \approx 1.414214, \pi \approx 3.142, \dfrac{\pi}{2} \approx 1.571\}$
Everything is a rational number, aside from irrational numbers (duh).
Real Number Properties
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).
 Commutative: Changing the order doesn’t affect the sum; think as going to the store back and forth — the destination should be the same.
 $13 + 7 = 7 + 13$
 $\sqrt{2} \cdot \sqrt{5} = \sqrt{5} \cdot \sqrt{2}$
 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).
 $2(3x) = (2 \cdot 3)x = 6x$
 Distributive: Multiplying the outside var by the inner parentheses member(s); we’d use this to factor the like terms to make it simpler.
 $a(b+c) = a \cdot b + a \cdot c$
 $3(4x  2y +6) = 3 \cdot 4x  (3) \cdot 2y  3 \cdot 6$
 Inverse: The inverse gives the same identity; we use this to simplify negative expressions.
 Additive Inverse: $a + (a) = 0$ and $(a) + a = 0$
 Multiplicitive Inverse: $a \cdot \dfrac{1}{a} = 1, \not = 0$ and $\dfrac{1}{a} \cdot a = 1, \not = 0$
 Identity: Removing either
0
or1
bc it doesn’t change anything; think of getting rid of your ex. Additive Identity: $a + 0 = a$
 Multiplicitive Identity: $a * 1 = a$
Absolute Value
The absolute value $a$ really means how far away are we from 0
? If 3
is the answer then $a = 3$, for 3
it’s the exact same!
28/04  College Algebra Prerequisities, Part 2
Exponents
Exponential notation looks like $b^n$, where $b$ is the base and $n$ is the exponent. For example, $b^3 \cdot b^2 = (b \cdot b \cdot b)(b \cdot b) = b^7$
There are a few rules we must understand, so we can modify expressions at will:
Product Rule
The product is the result of multiplying!
$b^m \cdot b^n = b^{m+n}$For example, lets use this expression
$(5x^2)(3x^4  6x^3 + 5x  8)$When using the distributive property w/ $(5x^2)$ to $(3x^4)$ you should see think of it as
$5 \cdot 3 = 15$And, with our exponents,
$x^2 \cdot x^4 = x^{2+4} = x^6$BECAUSE
$x^2 \cdot x^4 = x \cdot x \cdot x \cdot x \cdot x \cdot x = x^6$Then the final version is
$= 15x^6$Then you can distribute and simplify the rest :)
Quotient Rule
The quotient is the output after dividing!
$\dfrac{b^m}{b^n} = b^{mn}, b \not = 0$If we have
$\dfrac{b^5}{b^3}$Then we can think of it as
$\dfrac{b \cdot b \cdot b \cdot b \cdot b }{b \cdot b \cdot b}$And we can cancel out like terms
$\dfrac{\cancel{b \cdot b \cdot b} \cdot b \cdot b }{\cancel{b \cdot b \cdot b}}$To be left with
$b \cdot b = b^2$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.
$b^{n} = \dfrac{b^{n}}{1} = \dfrac{1}{b^n}$For example,
$\dfrac{x^4}{x^7}$When we 4  7
we get 3
so
But when when visually seeing the exponents before this negative exponent calculation
$\dfrac{x^4}{x^7} = \dfrac{x \cdot x \cdot x \cdot x}{x \cdot x \cdot x \cdot x \cdot x \cdot x \cdot x}$Then when we cancel the like terms
$= \dfrac{\cancel{x \cdot x \cdot x \cdot x}}{\cancel{x \cdot x \cdot x \cdot x} \cdot x \cdot x \cdot x}$We’re left with
$= \dfrac{1}{x \cdot x \cdot x} = \dfrac{1}{x^3}$So that must mean that for any negative exponent on top we can inverse it and remove the subtraction operator applied to it!
$\dfrac{x^4}{x^7} = \dfrac{x^{3}}{1} = \dfrac{1}{x^3}$ZeroExponent Rule
When the exponent is 0
then the result is 1
!
To create the mental model, think that we have
$\dfrac{b^2}{b^2} = \dfrac{\cancel{b \cdot b}}{\cancel{b \cdot b}} = b^{22} = b^0 = 1$What if we had
$3(7xy^6)^0)$Then everything in the parentheses will equate to 1
, therefore
Power Rule
$(b^m)^n = b^{mn}$E.g.
$(2^2)^3 = (2^2) \cdot (2^2) \cdot (2^2) = 2^{2+2+2}$Products To Powers
Think of this as distributing the power to everything contained in the parentheses as a standalone.
$(ab)^n = a^nb^n$E.g.
$(3x)^2 = 3^2x^2$Quotients To Powers
Same as the products to power, except with rationals.
$\left(\dfrac{a}{b}\right)^2 = \dfrac{a^2}{b^2}$I want to reiterate the concepts we just saw with this example, it will build your intuition ALOT more surprisingly,
$\dfrac{5x^{2}}{y^{3}} \cdot \dfrac{8x^4}{y^{5}}$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
$= \dfrac{5y^3}{x^2} \cdot \dfrac{8x^4y^{5}}{1}$Lets get the product
$= \dfrac{5y^{3+5}x^4}{x^2}$And remember the quotient rule for the x
s
Aka
$= 5y^{8} x^2$Lets take it a step further and try this one,
$\dfrac{24xy}{27x^{2}} \div \dfrac{36x^2y^{3}}{45xy^4}$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,
Now we can
$\dfrac{6 \cdot 4 xy \cdot x^2}{9 \cdot 3} \cdot \dfrac{9 \cdot 5 x y^4 \cdot y^3}{6 \cdot 6 x^2}$Now we can look at it like this
$\cancel{\dfrac{6}{6} \cdot \dfrac{9}{9} \cdot \dfrac{x^2}{x^2}} \cdot \dfrac{4xy}{3} \cdot \dfrac{5xy^7}{6}$Then simplifying further
$= \dfrac{\cancel{2} \cdot 2 \cdot 5 x^2 y^8}{3 \cdot 3 \cdot \cancel{2}}$ $= \dfrac{10 x^2 y^8}{9}$What about for
$\dfrac{x + 2}{5} = \dfrac{7}{8}$We can cross multiple the 5 * 7
and 8
distribute the x + 2
Remove 16
from both sides
Radicals
Lets disect what a square root (sqrt) w/ the radical expression $\sqrt[n]{a} = b$
 $n$ is the radical index. If
n
is: odd thenb
, even thenb
.  $\sqrt{}$ is the radical
 $a$ is the radicand, e.g. $\sqrt{4} = 2$ as $2^2 = 4$, where
4
is the principal sqrt.
Think of it as radi
, [cal, and, index]
.
The principal nth root of a real number a $\sqrt[n]{a} = b$ means that $b^{n} = a$
Radical Factorisation
The perfect square is when a root’ed number returns an integer. If we have $\sqrt{36} = 6$, 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. $\sqrt{50x} = \sqrt{{25 \cdot 2x}}$ since $25$ can be factored further we get $25 = 4 \cdot 5 \therefore \sqrt{{25 \cdot 2x}} = 4\cdot5\sqrt{2x} = 20\sqrt{2x}$
Another example is $\sqrt{500} = \sqrt{25 \cdot 20} = \sqrt{25} \sqrt{20} = 5 \sqrt{20}$. Since 20
contains a perfect square factor, 4
, we need to simplify futher: $5 \sqrt{20} = 5 \sqrt{4 \cdot 5} = 5 \sqrt{4} \sqrt{5} = (5 \cdot 2) \sqrt{5} = 10\sqrt{5}$.
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 $18 \cdot 2$ but we can go futher and do $9 \cdot 2$, then to get 36
we have the lowest factorisation w/ $3^2 \cdot 2^2 = 36$
Combining Radicals
Now with out newfound knowledge we can fight the mini boss battle: $4 \sqrt{50x}  6\sqrt{32x}$
$4 \sqrt{50x}  6\sqrt{32x} = (4 \cdot x) \sqrt{5} \sqrt{2}$Radical Expression Rules
 Product: The sqrt of a product is the product of sqrts, $\sqrt{ab} = \sqrt{a} \cdot \sqrt{b}$
 Quotient: If
a
(nominator) andb
(the denominator) are nonnegative real numbers andb != 0
, then: $\sqrt{\dfrac{a}{b}} = \dfrac{\sqrt{a}}{\sqrt{b}}$. The sqrt of a quotient is the quotient of the sqrts.
Add + Sub Radicals
 Add: $4\sqrt{11} + 2\sqrt{11} = (4+2)\sqrt{11} = 6\sqrt{11}$
 Sub: $\sqrt{5x}  7 \sqrt{5x} = 1\sqrt{5x}  7 \sqrt{5x} = (17) \sqrt{5x} = 6\sqrt{5x}$
Rationalising Denominators
This concept is where things started to click for me. Essentially, any number divided by itself is 1
, e.g. $\dfrac{\sqrt{3}}{\sqrt{3}} = 1$.
So, 2 approximations that look entirely different, e.g. $\dfrac{1}{\sqrt{3}} = \dfrac{\sqrt{3}}{3}$, can mean the same thing.
 let’s break this down $\dfrac{1}{\sqrt{3}}$
 here’s the fascinating part: $\\ = \dfrac{\sqrt{3}}{\sqrt{9}}$
 simplyfiying further $\\ = \dfrac{\sqrt{3}}{3}$
 therefore $\\ \dfrac{1}{\sqrt{3}} = \dfrac{\sqrt{3}}{3}$
Conjugates
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:
$(\sqrt{a} + \sqrt{b})(\sqrt{a}  \sqrt{b}) = (\sqrt{a})^2  (\sqrt{b})^2 = a  b$Why is this? We apply the Distributive property into:
$(\sqrt{a} \cdot \sqrt{b}) (\sqrt{a}  \sqrt{b}) = (\sqrt{a} \cdot \sqrt{a}) + (\sqrt{a} \cdot (\sqrt{b})) + (\sqrt{b} \cdot \sqrt{a}) + ((\sqrt{b}) \cdot \sqrt{b})$Make this more readable:
$= a  (\sqrt{ab})+(\sqrt{ab})b$The like terms cancel out:
$=  (\sqrt{ab})+(\sqrt{ab})$Which leaves us with:
$= a b$Lets look at another example, $\dfrac{h}{{\sqrt{x+h}}  \sqrt{x}}$ then the conjugate of the denominator is $\sqrt{x + h} + \sqrt{x}$ (the opposite).
As an example, $\dfrac{7}{5 + \sqrt{3}}$
$\\ = \dfrac{7}{5 + \sqrt{3}} \cdot \dfrac{5  \sqrt{3}}{5  \sqrt{3}}$Rember how this is =1
. Now we convert the two denominators to $(\sqrt{a})^2  (\sqrt{b})^2$
Simplify the denominator
$\\ = \dfrac{7(5 \sqrt{3})}{25  3}$Rational Exponents
Lets take a look at two expressions: $(7^{\frac{1}{2}})^2 = 7^{\frac{1}{2} \cdot 2} = 7^1 = 7$ and $(\sqrt{7})^2 = \sqrt{7} \cdot \sqrt{7} = \sqrt{49} = 7$
From this we see that: $7^{\frac{1}{2}}$ means $\sqrt{7}$
And so we can make the definition, as long as n >= 2
:
And, as long as a != 0
,
But what about rationals where the numeration > 1
?
Thus,
$a^\frac{2}{3} = (\sqrt[3]{a})^2 = \sqrt[3]{a^2}$And we take this further with a new definition,
$a^{\frac{m}{n}} = (\sqrt[n]{a})^m = \sqrt[n]{a^m}$And if $a^{\frac{m}{n}}$ is a nonzero real number, then
$a^{\frac{m}{n}} = \dfrac{1}{a^\frac{m}{n}}$Notice how from the previous definition, $a^\frac{1}{n} = \sqrt[n]{a}$, 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:
$\dfrac{32^\frac{5}{3}}{16x^\frac{3}{4}} = \left(\dfrac{32}{16}\right)\left(\dfrac{x^\frac{5}{3}}{x^\frac{3}{4}}\right)$ $= 2x^{\frac{5}{3}  \frac{3}{4}}$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 ${\dfrac{5}{3}  \dfrac{3}{4}}$ we’re really just multiplying the numerator 5
and denominator 3
by the opposite expression’s denominator 4
:
And do for the other expression with the former,
$\dfrac{3}{4} = \dfrac{3\cdot3}{4\cdot3} = \dfrac{9}{12}$Then we get
$= \dfrac{20}{12}  \dfrac{9}{12} = \dfrac{20  9}{12}= 2x^{\frac{12}{11}}$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 polepeenomyballs.
As always, lets learn the Vocabulary first, wtf are nomials:
 Monomial is a single term, $7x^3$
 Binomial: is 2 terms, $7x^3 + 2x^2$
 Trinomial: is 3 terms, $7x^3 + 2x^2 + 5$
 Polynomial: is anything beyond 3 terms, $7x^3 + 2x^2 + 5x + 69$
We describe polynomials w/ negatives w/ parentheses to be explicit (like a rap song)
$7x^3  9x^2 + 13x  6$As, $7x^3 + (9x^2) + 13x + ( 6)$
I trip up a lot with the 
bullshit so this is quite important to note: $ax^n = (ax^n)$.
The n
represents the degree. We order polynomials in cronological order from left to right in terms of degrees. E.g. $7x^3$ will be ordered before $7x^2$. 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): $a_{n}x^n + a_{n1}x^{n1} + a_{n2}x^{n2} + ... + a_1x + a_0$
 The degree of the polynomial is $n$. 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 $a_n$
 The constant term is $a_0$
Polynomial Sub, Add, Mul
Subtracting polynomials by combining like terms (variables and their exponent powers are the same), e.g.
$9x^3  13x^3 = (9 + 13)x^3 = 4x^3$We are using the the distributive property to factor out the common term, $x^3$
Adding polynomials: $5x^4 + 6x^4 = 11x^4$
However we cannot Simplify $5x^4 + 3x^2$ because there are no like terms!
Multiplying polynomials:
$(5x^4)(6x^4)$Combining like terms,
$= (5\cdot6)(x^4 \cdot x^4)$ $= 30x^{4+4} = 30x^8$But what would we do if there is no monomials? E.g,
$(2x + 3) (x^2 + 4x + 5)$Well we would use the distributive property (our lord and saviour!):
Distributing 2x
Then distributing 3
Then combining like terms
$= 2x^3+ (3x^2 + 8x^2) + (10x + 12x) + 15$Simplifying
$= 2x^3+ 11^2 + 22x + 15$Product Of Two Binomials
Once again, we use the distributive property to find the product of two binomials. For example,
$(3x  5) (4x + 5)$ $= 3x(4x) + 3x(5) + (5(4x)) + (5(5))$ $= 12x^2 + 15x + (20x) + (25)$ $= 12x^2 + 5x 25$Similar to multiplying conjugates, we find the Product of the Sum and Difference of Two Terms as follows: $(A + B)(A  B) = A^2  B^2$ because the two terms cancel eachother out $(B \cdot A) (A \cdot ( B)) = AB  AB = 0$
Squaring & Cubing Binomials
$(A + B)^3 = A^3 + 3A^2B + 3AB^2 + B^3$How do we break this down? Lets expand the exponents:
$(A + B)^3 = (A + B) (A + B) (A + B)$Lets distribute the first 2:
$(A + B) (A + B) = A^2 + AB + AB + B^2 = A^2 + 2AB + B^2$Now we distribute the third:
$= (A^2 + 2AB + B^2) (A + B)$ $= (A^3 + 2A^2B + 2AB^2 + B^3)$03/05  Factoring Polynomials
Factoring
Factoring a polynomial containing the sum of monomials means finding an expression that is a product:
$10x^2 + 15x = 5x(2x+3)$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.
$18x^3 + 27^2$9
factors into 18 (9*2)
and 27 (9*3)
,
What about for,
$x^2 (x+3) + 5(x + 3)$We see that $(x + 3)$ is the common factor in both terms, so we can reverse the distributive property into:
$(x + 3) (x^2 + 5)$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
,
If we group them together then we may be able factor out something that is common
First, the common factor is $x^2$ with:
$(x^3 + 4x^2) = x^2(x+4)$The second, with the common factor of $3$, becomes
$(3x + 12) = 3(x + 4)$Combining them together we get
$= x^2(x + 4) + 3(x + 4)$Notice the common factor again? It’s $(x + 4)$. We use this as one term then the other would be the remaining vars, $(x^2 + 3)$. Therefore,
$= (x + 4)(x^2 + 3)$Lets do one that isn’t so intuitive
$2x^2 + 3  2 = 0$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 $2x^2$ to create 4
for factoring.
Now we can factor 4
and 1
to create
Notice how the result doesn’t change!!
Now we can factor by grouping since there are 4
terms
Then we factor out x + 2
And now we can solve the expression
$x + 2 = 0$ $x = 2$and
$2x  1 = 0$ $2x = 1$ $x = \dfrac{1}{2}$Factoring Trinomials
Lets try factor a trinomial in two variables:
$2x^2 7xy + 3y^2$First we need to find the first 2 terms that create $2x^2$ when distributed
$(2x + ..) (x + ..)$Then we need to figure out the remaining two terms that whose product is $3y^2$
$(y)(3y) = (y)(3y)$Finally, we need to make sure the sum of the far left and right products is equal to $7xy$
$(2x + y) (x + 3y)$Then we can verify but distributing it
$= (2x \cdot x) + (2x \cdot 3y) + (y \cdot x) + (y \cdot 3y)$Simplify
$= 2x^2  6xy  xy  4y$And, we’re back where we started!
$= 2x^2  7xy  4y$Factoring Difference Of Two Squares
$A^2  B^2 = (A + B)(A  B)$For example,
$81x^2  49$ $= \sqrt{81x^2}  \sqrt{49}$ $= (9x)^2  7^2$ $= (9x + 7)(9x  7)$Repeated Factorisation
This is when we can factorise a term further after factorising the expression.
For example,
$x^4  81$ $= \sqrt{x^4}  \sqrt{81}$ $= (x^2)^2  9^2$Difference of squares
$= (x^2 + 9)(x^2  9)$Factoring further step
$= (x^2 + 9)(x^2  \sqrt{9})$Notice that we now have a square subtracting another square, $(x^2  3^2)$. We can factor further w/ the difference of two squares again ($A^2  B^2 = (A + B)(A  B)$) only because it’s subtracting. We cannot do the same for $(x^2 + 9)$ because its adding, which is not the difference of two squares requirement, $A^2  B^2$. If we were expand, we’d be doing the opposite of factoring.
$= (x^2 + 9)(x^2  3^2)$ $= (x^2 + 9)(x + 3)(x  3)$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. $x^2 + 6x + 9 = (x+3)(x+3) = (x+3)^2$.
There are ways to factor a percect square trinomial. Notice how the first sign is reflected as the sign in the parentheses.
 $A^2 + 2AB + B^2 = (A + B)^2$
 $A^2  2AB + B^2 = (A  B)^2$
We identify a perfect square trinomial via:
 The first and last terms being squares of monomials or integers
 The middle term is twice the product of the expressions being squared in the first and last terms
E.g,
$25x^2  60x + 36$Break down the terms w/ sqrts
$= \sqrt{25x^2}  60x + \sqrt{36}$Now we have the sqrt product
$= (5x)^2  60x + 6^2$Verify the middle term is twice the product of the outer two
$60x = 2(5x)(6)$It is, therefore
$= (5x  6)^2$Factoring the Sum and Difference of Two Cubes
Factoring the Sum of Two Cubes
$A^3 + B^3 = (A + B)(A^2  2AB + B^2)$Factoring the Difference of Two Cubes
$A^3  B^3 = (A  B)(A^2 + 2AB + B^2)$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,
$64x^3  125$ $= \sqrt[3]{64}  \sqrt[3]{125} = (4x)^3  5^3$ $= \sqrt[3]{64}  \sqrt[3]{125} = 4^3  5^3$ $= (4  5)(4^2 + 2(4)(5) + 5^2)$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.
 If there is a common factor, factor out the greatest common factor
 Check how many terms there are then:
 2 terms:
 Difference of two squares: $A^2  B^2 = (A + B)(A  B)$
 Sum of two cubes: $A^3 + B^3 = (A + B)(A^2  2AB + B^2)$
 Difference of two cubes: $A^3  B^3 = (A  B)(A^2 A 2AB + B^2)$
 3 terms: If perfect square trinomial use the following, otherwise trial and error:
 $A^2 + 2AB + B^2 = (A + B)^2$
 $A^2  2AB + B^2 = (A  B)^2$
 4+ terms, try factoring by grouping
 2 terms:
 If more than 1 term in the factored polynomial can be further factored, go further until factored completely
Factoring Fractional Exponents
$x(x + 1)^{\frac{3}{4}} + (x+1)^{\frac{1}{4}}$We have the greatest common factor $(x + 1)^{\frac{3}{4}}$. Express each term w/ the GCF
$= (x + 1)^{\frac{3}{4}}x + (x + 1)^{\frac{3}{4}}(x+1)$Factor out GCF
$= (x + 1)^{\frac{3}{4}} [x + (x + 1)] = (x + 1)^{\frac{3}{4}} [2x + 1]$Then we use the previously discussed negative quotient rule, $b^{n} = \dfrac{1}{b^n}$. Remember we get rid of the negative sign on the rational exponent and switch from multiplying to dividing!
$= \dfrac{2x + 1}{(x + 1)^{\frac{3}{4}}}$05/05  Rational expressions
Speaking of rationals, we’re almost at the end of our retardoprep 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. $\dfrac{x}{x^2  1}$
Since rational expressions are division and division by zero is undefined we must establish the domain, e.g. for $\dfrac{x}{x  2}$ our domain would be $x \not = 2$ otherwise the rational could be 0, and everyhing will crash and burn. And so our full expression is $\dfrac{x}{x  2}, x \not = 2$
Simplifying Rational Expressions
A rational express is only simplified if its numerator and denominator have no common factors other than 1
or 1
 Factor the numerator and the denominator completely
 Divide both the numerator and the denominator by any commmon factors
For example,
$\dfrac{x^2 + 6x + 5}{x^2  25}$Look for how we can factor further — sqrt into perfect square
$= \dfrac{x^2 + 6x + 5}{x^2  \sqrt{25}}$Denominator difference of two squares
$= \dfrac{x^2 + 6x + 5}{x^2  5^2}$Factor; reverse distribute
$= \dfrac{x^2 + 6x + 5}{(x + 5)(x  5)}$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, $(x + 5) = 1$, and make them both 1
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
Multiplying Rational Expressions
$\dfrac{x7}{x1} \cdot \dfrac{x^2  1}{3x21}$Difference of two squares and apply factor the common factor
$= \dfrac{x7}{x1} \cdot \dfrac{(x + 1)(x  1)}{3(x  7)}$Notice how we have the same terms on both sides, $(x  7)$ and $(x  1)$, so we can cancel them out (bc it doesn’t change the final result)
$= \dfrac{x + 1}{3}$Since The denominator has factors of x  1
and x  7
then domain is x != [1, 7]
Dividing Rational Expressions
$\dfrac{x^2  2x  8}{x^2  9} / \dfrac{x  4}{x + 3}$Swap the sign to multiply and swap numerator and denominator
$= \dfrac{x^2  2x  8}{x^2  3^2} \cdot \dfrac{x + 3}{x  4}$Difference of two squares + factoring a quadratic
$= \dfrac{(x  4)(x + 2)}{(x + 3)(x  3)} \cdot \dfrac{x + 3}{x  4}$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
.
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:
Addition
$\dfrac{a}{b} + \dfrac{c}{d} = \dfrac{ab + bc}{bd}, b \not = \{0, d\}$Subtraction
$\dfrac{a}{b}  \dfrac{c}{d} = \dfrac{ab  bc}{bd}, b \not = \{0, d\}$Notice how they cross divide for the numerators and the denominators they multiply themselves.
Examples
Example w/ the same denominator
$\dfrac{5x + 1}{x^2  9}  \dfrac{4x  2}{x^2  9}$Combine the terms into a single expression bc of the same denominator
$= \dfrac{5x + 1  (4x 2)}{x^2  9}$Remove the parentheses and change the sign of each term within
$= \dfrac{5x + 1  4x + 2}{x^2  9}$We can factor 9
into a perfect square w/ sqrt
difference of two squares!
$= \dfrac{x + 3}{(x + 3)(x  3)}$Remove like terms + express the domain
$= \dfrac{1}{x  3}, x \not = \{3, 3\}$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?
$\dfrac{x + 2}{2x3}  \dfrac{4}{x+3}$Cross multiply numerators and multiply both denominators
$= \dfrac{(x+2)(x+3)  (2x  3)4}{(2x3)(x+3)}$We don’t open the parentheses bc that was the variable we were given with
$= \dfrac{x^2 + 5x + 6  (8x  12)}{(2x3)(x+3)}$We can remove the parentheses and change sign of each term within
$= \dfrac{x^2 + 5x + 6  8x + 12}{(2x3)(x+3)}$Simplify and specify domain,

$2x−3=0$ leads to $x = \dfrac{3}{2}$
because $2x − 3 = 0$ equals $2x = 3$ then further $x = \dfrac{3}{2}$

$x+3=0$ leads to $x= −3x = −3$
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,
$\dfrac{x + 3}{x^2 + x  2} + \dfrac{2}{x^2  1}$First, we need to find the least common denominator by factoring the denominators
 $x^2 + x  2 = (x + 2)(x  1)$ because $x^2 + x + 2 = (x  2)(x + 1)$
 $x^2  1 = (x + 1)(x  1)$
Therefore,
$= \dfrac{x + 3}{(x + 2)(x  1)} + \dfrac{2}{(x + 1)(x  1)}$We can see the only factor not in the first denominator is $(x + 1)$ therefore the least common denominator is $(x + 2)(x  1)(x + 1)$
So we add it onto both,
$= \dfrac{x + 3}{(x + 2)(x  1)} \cdot \dfrac{x + 1}{x + 1} + \dfrac{2}{(x + 1)(x  1)} \cdot \dfrac{x + 2}{x + 2}$Rearrange the right denominator to match the left
$= \dfrac{(x + 3)(x + 1)}{(x + 2)(x  1)(x + 1)} + \dfrac{2(x + 2)}{(x + 2)(x  1)(x + 1)}$Keep in mind this isn’t actually changing the outcome, only the appearance. But we can start to sum these up.
$= \dfrac{(x + 3)(x + 1) + 2(x + 2) }{(x + 2)(x  1)(x + 1)}$Then simplify it
$= \dfrac{x^2 + 4x + 3 + 2x + 4}{(x + 2)(x  1)(x + 1)}$Combine like terms + establish the domain
$= \dfrac{x^2 + 6x + 7}{(x + 2)(x  1)(x + 1)}, x \not = \{2, 1, 1\}$Complex Rational Expressions
Aka complex fractions have numerators or denominators containing one or more rational expressions, e.g. $\dfrac{1 + \dfrac{1}{x}}{1  \dfrac{1}{x}}$ or $\dfrac{\dfrac{1}{x + h}  \dfrac{1}{x}}{h}$
For example, let’s simplify
$\dfrac{1 + \dfrac{1}{x}}{1  \dfrac{1}{x}}$We start by getting the lowest common denominator $1(x)$
$= \dfrac{\dfrac{1}{1} + \dfrac{1}{x}}{\dfrac{1}{1}  \dfrac{1}{x}}$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 $x \cdot 1$ is applied by adding $x$ to the left fraction, $1$ to the right fraction (remains the same).
$= \dfrac{\dfrac{1 \cdot x}{1 \cdot x} + \dfrac{1 \cdot 1}{x \cdot 1}}{\dfrac{1 \cdot x}{1 \cdot x}  \dfrac{1 \cdot 1}{x \cdot 1}}$Do the calculation
$= \dfrac{\dfrac{x}{x} + \dfrac{1}{x}}{\dfrac{x}{x}  \dfrac{1}{x}}$The denominators are now the exact same we can merge the fractions together
$= \dfrac{\dfrac{x + 1}{x}}{\dfrac{x  1}{x}}$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 $\dfrac{a}{b}$ is $\dfrac{b}{a}$, provided that a
and b
are nonzero.
Remove like terms that cross, x
.
Why?
Beause multiplication is commutative, the ordering doesn’t matter!
So, $\dfrac{x + 1}{x} \cdot \dfrac{x}{x  1}$ can be represented as $\dfrac{x + 1}{x  1} \cdot \dfrac{x}{x}$ which is the same as $\dfrac{x + 1}{x  1} \cdot 1$
And we finally get
$= \dfrac{x + 1}{x  1}$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.
$\dfrac{\dfrac{1}{x + h}  \dfrac{1}{x}}{h}$The first thing we should do is forget about the $\dfrac{}{h}$ so we can focus on the numerator, $\dfrac{1}{x + h}  \dfrac{1}{x}$
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 $x(x+h)$
and the right side,
$\dfrac{(1) x+h}{(x) x+h} = \dfrac{x+h}{x(x+h)}$now we can merge the two, remembering the subtraction operator
$= \dfrac{x}{x(x+h)}  \dfrac{x+h}{x(x+h)}$After merging
$= \dfrac{x  (x+h)}{x(x+h)}$Here’s a small detail but (x+h)
actually means 1(x+h)
, and so
We can apply the distributive property
$= \dfrac{x  x  h}{x(x+h)}$Notice the x
s cancel out in the numerator
Now we can add the $\dfrac{}{h}$
$= \dfrac{\dfrac{ h}{x(x+h)}}{h}$We can reverse the division of h
into multiplication into it’s numerator $x(x+h)$
We can cancel the h
s out bc they’re both products w/ common factors
And just to make it cleaner
$= \dfrac{1}{x(x+h)}$Jeeeeez, what a ride! I hope you finally understand how this works :) This shit took me 2 god damn weeks to ravel my head around — went into how factorisation, fraction cancelation and simplificaiton.
14/05 — Simplifying
Factoring
Factoring is the process of changing a sum to a product.
For example, lets start w/ the 4 term sum $ax + ay + bx + by$
We factor it into a 2 term sum, bc we still have the addition operator
$= a(x + y) + b(x + y)$Then we complete the factorisation by turning it into a product, distinguished by the multiplication operator.
$=(x + y)(a + b)$Reducing Fractions
$\dfrac{x(x  2)  15}{(x2)(x + 3)}$We can only reduce if both numerator and denominator are products. You cannot reduce until they’re products.
The subtraction operation makes the numerator a sum, the denominator is a product. We cannot cancel out the $x2$ bc one is a sum and the other is a product, we need the common elements to be factors. So how do we convert the numerator into a product?
First we distribute the x
Then factor the trinomial in the numerator, creating a product in the numerator
$= \dfrac{(x  5)(x + 3)}{(x2)(x + 3)}$