# How do you factor 16 + 20y - 6y^2?

May 16, 2015

$16 + 20 y - 6 {y}^{2} = - 2 \left(3 {y}^{2} - 10 y - 8\right)$

$3 {y}^{2} - 10 y - 8$ is of the form $a {y}^{2} + b y + c$, with $a = 3$, $b = - 10$ and $c = - 8$.

The general quadratic solution tells us that $a {y}^{2} + b y + c = 0$ when

$y = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

So substituting our values of $a$, $b$ and $c$, that gives

$y = \frac{10 \pm \sqrt{{10}^{2} - 4 \cdot 3 \cdot \left(- 8\right)}}{2 \cdot 3}$

$= \frac{10 \pm \sqrt{100 + 96}}{6}$

$= \frac{10 \pm \sqrt{196}}{6}$

$= \frac{10 \pm 14}{6}$

That is for $y = 4$ and $y = - \frac{2}{3}$

So $\left(y - 4\right)$ and $\left(3 y + 2\right)$ are both factors of $3 {y}^{2} - 10 y - 8$

In summary $16 + 20 y - 6 {y}^{2} = - 2 \left(y - 4\right) \left(3 y + 2\right) = 2 \left(4 - y\right) \left(3 y + 2\right)$