The purpose of factoring a quadratic is to turn the quadratic into a product of 2 binomials.

## Method 1

Method 1 starts with factoring the product of the roots. Let the quadratic we are factoring be $ax^2+bx+c=0$. When factored, it will be in the form of $$a(x-r)(x-s)=0,$$ where $r$ and $s$ are the roots of the quadratic, and where $r+s=-b$ and $r\cdot s=c$.

### Example

$x^2-8x+12$

Since the coefficient on the $x^2$ term is $1$, we know are quadratic factors in the form of $(x-r)(x-s)$. We know that the factor pairs of 12 are $(1,12), (2,6), (3,4), (4,3), (6,2), (1,12), (-1,-12), (-2,-6), (-3,-4), (-4,-3), (-6,-2),$ and $(-12,-1).$ We can find that only $(2,6)$ and $(6,2)$ satisfy our equations $r+s=-b$ and $r\cdot s=c$, so the factored form of $x^2-8x+12$ is $(x-2)(x-6)$.

### Limitations

This method cannot be used to factor quadratics with complex or irrational roots.

## Method 2

Method 2 starts by using the sum. Let the quadratic we are factoring be $ax^2+bx+c=0$. When factored, it will be in the form of $$a(x-r)(x-s)=0,$$ where $r$ and $s$ are the roots of the quadratic, and where $r+s=-b$ and $r\cdot s=c$.

### Example

$x^2-8x+13$

We know that $r+s=-(-8)$, so we can set $r=4-u$ and $s=4+u$. Then, we get that $r\cdot s= 16-u^2 = 13$, giving us that $u^2=3$, or $u=\pm \sqrt{3}$. Because we have both $4-u$ and $4+u$ as our roots, it doesn't matter which one is plugged in, giving us that the factored form of $x^2-8x+13$ is $\left(x-\left(4+\sqrt{3}\right)\right)\left(x-\left(4-\sqrt{3}\right)\right)$.

None known.