# How do you simplify (-49y^4 + 42y^3)/ (7y)?

May 9, 2015

First write it as:
$\frac{- 49 {y}^{4}}{7 y} + \frac{42 {y}^{3}}{7 y} = - 7 {y}^{4} / y + 6 {y}^{3} / y =$
Then use the fact that: ${a}^{m} / {a}^{m} = {a}^{m - n}$
So:$- 7 {y}^{4 - 1} + 6 {y}^{3 - 1} =$
$= - 7 {y}^{3} + 6 {y}^{2}$

May 9, 2015

The answer is $- {y}^{2} \left(7 y - 6\right)$ or multiply times $- 1$ to get ${y}^{2} \left(6 - 7 y\right)$ .

Simplify $\frac{- 49 {y}^{4} + 42 {y}^{3}}{7 y}$ .

Factor out $7 {y}^{3}$ .

$\frac{- 7 {y}^{3} \left(7 y - 6\right)}{7 y}$

Simplify.

$\frac{- \cancel{7} {y}^{3}}{\cancel{7} y} \left(7 y - 6\right)$ =

$- {y}^{3 - 1} \left(7 y - 6\right)$ =

$- {y}^{2} \left(7 y - 6\right)$

You can multiply times $- 1$ to get ${y}^{2} \left(6 - 7 y\right)$. (optional)