# How do you factor  x^4 - 8x^3 + 12x^2?

Jul 16, 2016

${x}^{2} \left(x - 6\right) \left(x - 2\right)$

#### Explanation:

Take out the common factor of ${x}^{2}$ first.

${x}^{2} \left({x}^{2} - 8 x + 12\right)$

Find factors of 12 which add to 8.
6 x 2 = 12 and 6 + 2 = 8. The factors we need are 6 and 2.
The signs are both negative.

${x}^{2} \left(x - 6\right) \left(x - 2\right)$

Jul 16, 2016

${x}^{2} \left(x - 6\right) \left(x - 2\right)$

#### Explanation:

The first step in factorising is to take out the common factor ${x}^{2}$

$\Rightarrow {x}^{2} \left({x}^{2} - 8 x + 12\right)$

Now we require to factorise the quadratic inside the bracket.

For the standard quadratic function $\textcolor{red}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{a {x}^{2} + b x + c} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

Consider the factors which multiply to give ac and sum to give b.

For ${x}^{2} - 8 x + 12$

a = 1 , b = -8 and c = 12

Require factors of product $a c = 1 \times 12 = 12$ which sum to -8

In this case these are -6 and -2 as product = 12 and sum = -8

$\Rightarrow {x}^{2} - 8 x + 12 = \left(x - 6\right) \left(x - 2\right)$

$\Rightarrow {x}^{4} - 8 {x}^{3} + 12 {x}^{2} = {x}^{2} \left(x - 6\right) \left(x - 2\right)$