Answer
$4a^{-7/5} \left( -a^{}+4 \right)$
Work Step by Step
Factoring out the given factor, $
4a^{-7/5}
,$ then the given expression, $
-4a^{-2/5}+16a^{-7/5}
,$ is equivalent to
\begin{array}{l}\require{cancel}
4a^{-7/5} \left( \dfrac{-4a^{-2/5}}{4a^{-7/5}}+\dfrac{16a^{-7/5}}{4a^{-7/5}} \right)
\\\\=
4a^{-7/5} \left( -a^{-\frac{2}{5}-\left(-\frac{7}{5}\right)}+4a^{-\frac{7}{5}-\left(-\frac{7}{5}\right)} \right)
\\\\=
4a^{-7/5} \left( -a^{-\frac{2}{5}+\frac{7}{5}}+4a^{-\frac{7}{5}+\frac{7}{5}} \right)
\\\\=
4a^{-7/5} \left( -a^{\frac{5}{5}}+4a^{0} \right)
\\\\=
4a^{-7/5} \left( -a^{1}+4(1) \right)
\\\\=
4a^{-7/5} \left( -a^{}+4 \right)
.\end{array}
