## Factoring quadratics

Factoring/factorizing/factorization is the process of finding the factors. For quadratics (a degree-two polynomials: $ax^2+bx+c$) it looks like the following:

$$ax^2+bx+c=(ex+f)(gx+h)$$

Numbers **f** and **h** are called polynomial factors.

### Simple version (a=1)

For $ax^2+bx+c$ where $a=1 (e.g. x^2+7x+6)$ you have to find 2 numbers (polynomial factors **f, h**) that are factors of **c** and if added together you will get **b**.

In $x^2+7x+6$, c is 6. Number 6 has the following factors: 2×3 ; 6×1. Just the sum of the factors 6 and 1 is equal to b (7). So our f and h are 6 and 1.

$$x^2+7x+6=(x+6)(x+1)$$

Let's prove it:

$$(x+6)(x+1)=x^2+x+6x+6=x^2+7x+6$$

The situation is different with some of the terms being negative:

$$x^2−6x+8$$

8 is positive, so the factors will always be either both positive (+) or both negative (−): 2×4 ; (−2)×(−4) ; 1×8 ; (−1)×(−8)

If the middle coefficient is negative (−6) we have to use negative factors. (−6)=(−2)+(−4). So our f and h are: −2 and −4:

$$x^2−6x+8=(x−2)(x−4)$$

Let's prove it:

$$(x−2)(x−4)=x^2−4x−2x+8=$$

$$x^2−6x+8$$

### Difficult version (a≠1)

For $ax^2+bx+c$ where a≠1 (e.g. $2x^2+13x+6$) you have to find 2 numbers that are factors of **a×c** and add to give you **b**.

$$2x^2+13x+6$$

$$a×c=2×6=12$$

Let's find factors of 12: 2×6 ; 3×4 ; 12×1

Next, find the pair of factors that adds to

**b** (13), it's only the last one: b=13=12+1 .

Now, split the middle and finish by grouping (sometimes we need to take two tries to find which of the two middle addends should go first):

$$2x^2+12x+x+6=2x(x+6)+(x+6)=$$

$$=(x+6)(2x+1)$$

Let's prove it:

$$(x+6)(2x+1)=2x^2+x+12x+6$$

Another example with negative terms:

$$3x^2+10x−8$$

$$a×c=3×(−8)=(−24)$$

Let's find factors of (−24): (−24)×1 ; (−1)×24 ; (−3)×8 ; 8×(−3) ; 12×(−2) ; ...

We have the pair that adds up to b (10): 10=12+(−2)

Split the middle term:

$$3x^2+10x−8=3x^2+12x−2x−8=$$

$$=3x(x+4)−2(x+4)=(3x−2)(x+4)$$

Let's prove it:

$$(3x−2)(x+4)=3x^2+12x−2x−8=$$

$$=3x^2+10x−8$$