Answer
$-16y^3 \left( y-4 \right)$
Work Step by Step
$\bf{\text{Solution Outline:}}$
First, get the $GCF$ of the given expression, $
-16y^4+64y^3
.$ Then, divide the given expression and the $GCF.$ Express the answer as the product of the $GCF$ and the resulting quotient. Finally, factor out the $-1$.
$\bf{\text{Solution Details:}}$
The $GCF$ of the constants of the terms $\{
-16,64
\}$ is $
16
.$ The $GCF$ of the common variable/s is the variable/s with the lowest exponent. Hence, the $GCF$ of the common variable/s $\{
y^4,y^3
\}$ is $
y^3
.$ Hence, the entire expression has $GCF=
16y^3
.$
Factoring the $GCF=
16y^3
,$ the given expression is equivalent to
\begin{array}{l}\require{cancel}
16y^3 \left( \dfrac{-16y^4}{16y^3}+\dfrac{64y^3}{16y^3} \right)
.\end{array}
Using the Quotient Rule of the laws of exponents which states that $\dfrac{x^m}{x^n}=x^{m-n},$ the expression above simplifies to
\begin{array}{l}\require{cancel}
16y^3 \left( -y^{4-3}+4y^{3-3} \right)
\\\\=
16y^3 \left( -y^{1}+4y^{0} \right)
\\\\=
16y^3 \left( -y+4(1) \right)
\\\\=
16y^3 \left( -y+4 \right)
.\end{array}
Factoring out $-1$ from the second factor results to
\begin{array}{l}\require{cancel}
-16y^3 \left( y-4 \right)
.\end{array}