How do you simplify (8y^3-16y^2)/(2y^3-2y^2-4y)?

Jun 16, 2015

You can start by eliminating extra $y$ factors.

$\frac{8 {y}^{3} - 16 {y}^{2}}{2 {y}^{3} - 2 {y}^{2} - 4 y}$

$= \frac{\cancel{y} \left(8 {y}^{2} - 16 y\right)}{\cancel{y} \left(2 {y}^{2} - 2 y - 4\right)}$

Now how about those $2$ factors?
$= \frac{\cancel{2} \left(4 {y}^{2} - 8 y\right)}{\cancel{2} \left({y}^{2} - y - 2\right)}$

And any canceling? Yup.

$= \frac{4 y \cancel{\left(y - 2\right)}}{\cancel{\left(y - 2\right)} \left(y + 1\right)}$

$= \frac{4 y}{y + 1}$