# How do you factor completely x^3 - 4x^2 + 4x - 16?

Dec 31, 2015

Factor by grouping to find:

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

#### Explanation:

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

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

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

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

If we allow Complex coefficients...

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