# How do you factor y^3-8y+2y^2-16?

Jun 21, 2015

=color(red)((y+2) (y^2-8)

#### Explanation:

${y}^{3} - 8 y + 2 {y}^{2} - 16$

Factorising by grouping:
Forming two groups of two terms:

color(red)(y^3-8y)+color(blue)(2y^2-16
$= y \left({y}^{2} - 8\right) + 2 \left({y}^{2} - 8\right)$
=color(red)((y+2) (y^2-8)

Jun 21, 2015

${y}^{3} - 8 y + 2 {y}^{2} - 16$

$= \left(y + 2\right) \left({y}^{2} - 8\right)$

$= \left(y + 2\right) \left(y - 2 \sqrt{2}\right) \left(y + 2 \sqrt{2}\right)$

#### Explanation:

First factor by grouping

$f \left(y\right) = {y}^{3} - 8 y + 2 {y}^{2} - 16$

$= \left({y}^{3} - 8 y\right) + \left(2 {y}^{2} - 16\right)$

$= y \left({y}^{2} - 8\right) + 2 \left({y}^{2} - 8\right)$

$= \left(y + 2\right) \left({y}^{2} - 8\right)$

The remaining quadratic term can be factorized by treating it as a difference of squares with irrational coefficients:

$f \left(y\right) = \left(y + 2\right) \left({y}^{2} - {\left(\sqrt{8}\right)}^{2}\right)$

$= \left(y + 2\right) \left(y - \sqrt{8}\right) \left(y + \sqrt{8}\right)$

...using the difference of squares identity:

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

Finally note that $\sqrt{8} = \sqrt{{2}^{2} \cdot 2} = 2 \sqrt{2}$

$f \left(y\right) = \left(y + 2\right) \left(y - 2 \sqrt{2}\right) \left(y + 2 \sqrt{2}\right)$