Answer
$\dfrac{2(x-2)}{x-5}$
Work Step by Step
Factoring the expressions and then cancelling the common factor/s between the numerator and the denominator, then the given expression, $
\dfrac{2x^2+12x-32}{x^2+16x+64}\cdot\dfrac{x^2+10x+16}{x^2-3x-10}
$, simplifies to
\begin{array}{l}\require{cancel}
\dfrac{2(x^2+6x-16)}{(x+8)(x+8)}\cdot\dfrac{(x+8)(x+2)}{(x-5)(x+2)}
\\\\=
\dfrac{2(\cancel{x+8})(x-2)}{(\cancel{x+8})(\cancel{x+8})}\cdot\dfrac{(\cancel{x+8})(\cancel{x+2})}{(x-5)(\cancel{x+2})}
\\\\=
\dfrac{2(x-2)}{x-5}
.\end{array}
