# Quadratic Relations Lesson

### By: Cindy Cheng

## Common Factoring

- factoring is the opposite of expanding
- if every term of a polynomial is divisible by the same constant, the constant is called a
__common factor__

Example:

ab+bc=a(b+c)

_ _ _ _ _ _ _ _ _ _> common factor

4x+20=4(x+5)

- a polynomial is not considered to be completely factored until the
__greatest common factor__(G.C.F) has been factored out

Example:

4x+20=2(2x+10) → not completely factored

4x+20=4(x+5) → completely factored

Try it out:

a) 3x+9

b)12x-16

c)24+16x-8x²

## Factoring By Grouping

bracket off first 2 and last 2 terms

=**(**x³-2x²**)**+**(**8x+16**)**

factor the brackets

=x²(x+2)+8(x+2)

since (x+2) is the same, you can bring it together

=(x+2)(x²+8)

## Factoring Simple Trinomials

**7**x+

**6**

find two numbers that multiplies into 6 and adds to 7

P:6

S:7

the 2 numbers are 6 and 1.

=(x+6)(x+1)

## Factoring Complex Trinomials

nothing that multiplies to -5 and adds to 3. So you have to multiply a(2) with c(-5) which would be your third term

P: -10

S: 3

decompose the middle term

=2x²**-2x+5x**-5

factor by grouping

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

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

## Factoring Difference of squares

if both terms can be squared and the second term is a negative, it can easily be factored.

=(x+9)(x-9)

## solving quadratic equations by factoring

factor off the equation

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

there will be 2 answers that equal x in the end.

x+1=0 and x+2=0

x=-1 and x=-2

## completing the square

block off the first 2 terms

=(3x²+24x)-11

factor out the A

=3(x²+8x)-11

add "zero" inside the brackets. To get zero-- (middle term/2)²

=3(x²+8x+16-16)-11

bring out the negative

=3(x²+8x+16)-11-48

=3(x²+8x+16)-59

factor the rest

=3(x+4)²-59