# How do you factor: y= x^3 - 4x^2 + 4x - 16 ?

Mar 12, 2016

Factor by grouping to find:

$y = {x}^{3} - 4 {x}^{2} + 4 x - 16$

$= \left({x}^{2} + 4\right) \left(x - 4\right)$

$= \left(x - 2 i\right) \left(x + 2 i\right) \left(x - 4\right)$

#### Explanation:

Factor by grouping:

$y = {x}^{3} - 4 {x}^{2} + 4 x - 16$

$= \left({x}^{3} - 4 {x}^{2}\right) + \left(4 x - 16\right)$

$= {x}^{2} \left(x - 4\right) + 4 \left(x - 4\right)$

$= \left({x}^{2} + 4\right) \left(x - 4\right)$

That's as far as you can go with Real coefficients: $\left({x}^{2} + 4\right)$ has no linear factors with Real coefficients since ${x}^{2} + 4 \ge 4 > 0$ for all $x \in \mathbb{R}$.

If you allow Complex coefficients then we can go a little further:

$= \left({x}^{2} - {\left(2 i\right)}^{2}\right) \left(x - 4\right)$

$= \left(x - 2 i\right) \left(x + 2 i\right) \left(x - 4\right)$