## 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:

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.

Let's prove it:

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

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:

Let's prove it:

### 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**.

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):

Let's prove it:

Another example with negative terms:

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:

Let's prove it: