# How do you factor 10y^3 + 20y^2 - 6y?

Apr 11, 2018

$2 y \left(5 {y}^{2} + 10 y - 3\right) = 2 y \left(y + 1 - \frac{2 \sqrt{10}}{5}\right) \left(y + 1 + \frac{2 \sqrt{10}}{5}\right)$

#### Explanation:

Given: $10 {y}^{3} + 20 {y}^{2} - 6 y$

Find the greatest common factor (GCF) of each term:

$10 {y}^{3} = \textcolor{red}{2} \cdot 5 \cdot \textcolor{red}{y} \cdot y \cdot y$

$20 {y}^{2} = \textcolor{red}{2} \cdot 2 \cdot 5 \cdot \textcolor{red}{y} \cdot y$

$- 6 y = - \left(\textcolor{red}{2} \cdot 3 \cdot \textcolor{red}{y}\right)$

GCF is $\textcolor{red}{2 y}$

Factor the GCF: $2 y \left(5 {y}^{2} + 10 y - 3\right)$

$x = \frac{- B \pm \sqrt{{B}^{2} - 4 A C}}{2 A}$ where $A {x}^{2} + B x + C = 0$
$y = \frac{- 10 \pm \sqrt{{10}^{2} - 4 \cdot 5 \cdot \left(- 3\right)}}{2 \cdot 5} = - \frac{10}{10} \pm \frac{\sqrt{160}}{10}$
$y = - 1 \pm \frac{\sqrt{16} \sqrt{10}}{10} = - 1 \pm \frac{4 \sqrt{10}}{10} = - 1 \pm \frac{2 \sqrt{10}}{5}$
$10 {y}^{3} + 20 {y}^{2} - 6 y = 2 y \left(y + 1 - \frac{2 \sqrt{10}}{5}\right) \left(y + 1 + \frac{2 \sqrt{10}}{5}\right)$