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:
$$(3x−2)(x+4)=3x^2+12x−2x−8=$$
$$=3x^2+10x−8$$
