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

Oct 3, 2016

$6 {y}^{3} + 3 {y}^{2} - 3 y = \textcolor{g r e e n}{\left(3 y\right) \left(y + 1\right) \left(2 y - 1\right)}$

#### Explanation:

Extracting the obvious common factor color(green)(""(3y)) should be fairly obvious.
The more difficult problem is whether the remaining factor color(brown)(""(2y^2+y-1)) has any factors.

Treating color(brown)(""(2y^2+y-1)) as a quadratic of the equation $a {y}^{2} + b y + c = 0$
and applying the quadratic formula: $y = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$
we would get
$\textcolor{w h i t e}{\text{XXX}} y = \frac{- 1 \pm \sqrt{{1}^{2} - 4 \left(2\right) \left(- 1\right)}}{2 \left(2\right)}$
$\textcolor{w h i t e}{\text{XXXX}} = \frac{- 1 \pm 3}{4}$
$\textcolor{w h i t e}{\text{XXXX}} = - 1 \mathmr{and} + \frac{1}{2}$
This implies color(brown)(""(2y^2+y-1)) has (incomplete) factors $\textcolor{g r e e n}{\left(y + 1\right) \left(y - \frac{1}{2}\right)}$

Multiplying $\left(y + 1\right) \left(y - \frac{1}{2}\right)$ gives $\textcolor{b l u e}{{y}^{2} + \frac{1}{2} y - \frac{1}{2}}$
which implies that an additional factor of $\textcolor{g r e e n}{2}$ is required.
So color(brown)(2y^2+y-1) = color(green)(2(y+1)(y-1/2)
or$\textcolor{w h i t e}{\text{XXXXXXX}} = \textcolor{g r e e n}{\left(y + 1\right) \left(2 y - 1\right)}$