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

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Answer 1 · 2015-05-09T03:55:28

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

Answer 2 · 2015-05-09T04:41:52

The answer is $-y^2(7y-6)$ or multiply times $-1$ to get $y^2(6-7y)$ .

Simplify $(-49y^4+42y^3)/(7y)$ .

Factor out $7y^3$ .

$(-7y^3(7y-6))/(7y)$

Simplify.

$(-cancel7y^3)/(cancel7y)(7y-6)$ =

$-y^(3-1)(7y-6)$ =

$-y^2(7y-6)$

You can multiply times $-1$ to get $y^2(6-7y)$ . (optional)

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