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

Mar 12, 2018

The factored form is $\left({x}^{2} + 4\right) \left(x - 2\right)$.

#### Explanation:

You have to factor by grouping:

$\textcolor{w h i t e}{=} {x}^{3} - 2 {x}^{2} + 4 x - 8$

$= \textcolor{b l u e}{{x}^{2}} \cdot x - \textcolor{b l u e}{{x}^{2}} \cdot 2 + 4 x - 8$

$= \textcolor{b l u e}{{x}^{2}} \left(x - 2\right) + 4 x - 8$

$= \textcolor{b l u e}{{x}^{2}} \left(x - 2\right) + \textcolor{red}{4} \cdot x + \textcolor{red}{4} \cdot - 2$

$= \textcolor{b l u e}{{x}^{2}} \left(x - 2\right) + \textcolor{red}{4} \left(x - 2\right)$

$= \left(\textcolor{b l u e}{{x}^{2}} + \textcolor{red}{4}\right) \left(x - 2\right)$

Mar 12, 2018

This factors to $\left({x}^{2} + 4\right) \left(x - 2\right)$

#### Explanation:

Starting with the left-hand side, we see ${x}^{3}$; lets factor that first.

Based on the middle terms, I would believe that the factors are uneven (NOT ${x}^{\frac{3}{2}}$) This is because that if it WAS even, there wouldn't be two terms in the middle of the equation. I begin to write the factors as such:

$\left({x}^{2} + a\right) \left(x + b\right)$

Now, expand the above and back-substitute the values of a & b from the starting equation:

$\left({x}^{2} + a\right) \left(x + b\right) = {x}^{3} + b {x}^{2} + a x + a b$

Comparing the middle coefficients, it would seem that b=-2 and a=4. This is confirmed with the final term $a b = 4 \cdot \left(- 2\right) = - 8$

Therefore, the factored solution is:
$\left({x}^{2} + 4\right) \left(x - 2\right)$