Math - Chapter 5

Factoring

Monomial common factors

Monomial = 1 term

factor out a monomial out of a set of terms

6x² - 14x + 8

the biggest number that can
divide all of these terms is 2

divide everything by 2

6x² - 14x + 8
___________
2

2(3x² - 7x + 4)

you cannot change the equation

that's why the 2 is put in the
front of the brackets

if you ere to expand it you would
end up with the same equation as before

nothing was changed

only simplified

no biggest common variable

2 is the only thing common
among all 3 terms

Binomial common factors

Binomial = 2 terms

A binomial factor that is shared by all the terms of a polynomial

common binomial among a polynomial

7(x-3) - 2x (x-3)

(x-3) is the greatest common factor

(x-3) [ 7(x-3) 2x (x-3)]
______ - ______
(x-3) (x-3)

divide everything by (x-3)

put it outside the bracket

∴ equation not changed

(x-3)(7 - 2x)

When (x-3) is removed from the terms

(7-2x)

Group common factors

4 terms separated into 2 groups

find common factor for both groups

produces a binomial common factor

2wx + 10w +7x + 35
l_________l l______l
group one Group two

Group one ->
2 and w = greatest common factor

2w(x+5)

2w(x+5) + 7(x+5)

binomial common factor

(x+5)(2w+7)

refer to binomial common factor

Group 2 ->
7 is the greatest common factor

7(x+5)

Factoring Quadratic expressions (a≠1)

ax² + bx + c

Find 2 integers that when multiplied together is equal to a X c

Those same two integers have
to add up to b

a(x+r)(x+s)

3x² + 8x +4

__ x __ = 12 ( a X c)
__ + __ = 8 (b)

Those two integers are
2 and 6

2 x 6 = 12 (a X c)
2 + 6 = 8 (b)

Break down b into 2x and 6x
and add that to the equation

3x² + 2x + 6x +4

Group factor

refer to group factoring

x(3x+2) + 2(3x+2)

(3x+2)(x+2)

Where a, b, c are integers

Factoring Quadratic expressions (a=1)

ax² + bx + c

where a, b, and c are integers

in this case a is equal to one

find 2 integers that when
multiplied is equal to c

those 2 integers have to also
equal to b when added together

a(x+r)(x+s)

ax² + 5x +6

always see if you can common
factor something out first

in this case no

__ x __ = 6 (c)
__ + __ = 5 (b)

Those two integers are 2 and 3

2 x 3 = 6 (c)
2 + 3 = 5 (b)

r = 2
s = 3

can be switched around

r ans s can equal to
which ever integer

(x+2)(x+3)

Expanding

difference of squares

(a+b)(a-b)

a² - b²

all term a = to the same thing
all term b = to the same thing

(4x+5y)(4x-5y)

simplifies to

(4x)² - (5y)²

16x² - 25y²

part of factoring as well

Multiply binomials

2 factors of 2 terms being
multiplied to each other

(x-4)(x-5)

methods to do this

its like applying the
distributive property twice

multiply each term in the first
factor by each in the second

FOIL

F - First
O - Outer
I - Inner
L - Last

distributive property

multiply a single term
by 2 or more terms

used to expand
equations/expressions

squaring binomials

(a+b)²

Short form

can be solved in 2 ways

using Formula

a²+2ab+b²

(a+b)²

a² - 2ab+b²

(a-b)²

Expanding

(a+b)²

(a+b)(a+b)

FOIL it

a²+ab+ab+b²

a²+2ab+b²

Subtopic

The -2 is multiplied to
everything in the bracket

Expanded

done so to get rid of the -2

The biggest number that is common
among these terms is 4

both can be divided by 4

divide everything by 4 and add brackets around the divided terms

write the 4 in front of the brackets

simplified

used to simplify
equations/expressions

done so by finding the greatest common
factor in an expression/equation

2x -> a
3 -> b

(2x)² + 2(2x)(3) + (3)²

(2x+3)(2x+3)

4x² + 6x + 6x + 9

4x² + 12x + 9

(3x+4)(8x+5)

24x² + 15x + 32x + 20

24x² + 47x + 20

Perfect square trinomials

Square root of a and the square root of c
all multiplied by 2 is equal to b

4x² + 12x + 9

Square root of 4 -> 2
square root of 9 is -> 3

2(2)(3)

= 12 = b

∴ This is a perfect square trinomial

(2x+3)²