Step 1 :
To factor $ 2x^{3}-x^{2}-32x+16 $ we can use factoring by grouping:
Group $ \color{blue}{ 2x^{3} }$ with $ \color{blue}{ -x^{2} }$ and $ \color{red}{ -32x }$ with $ \color{red}{ 16 }$ then factor each group.
$$ \begin{aligned} 2x^{3}-x^{2}-32x+16 = ( \color{blue}{ 2x^{3}-x^{2} } ) + ( \color{red}{ -32x+16 }) &= \\ &= \color{blue}{ x^{2}( 2x-1 )} + \color{red}{ -16( 2x-1 ) } = \\ &= (x^{2}-16)(2x-1) \end{aligned} $$Step 2 :
Rewrite $ x^{2}-16 $ as:
$$ x^{2}-16 = (x)^2 - (4)^2 $$Now we can apply the difference of squares formula.
$$ I^2 - II^2 = (I - II)(I + II) $$After putting $ I = x $ and $ II = 4 $ , we have:
$$ x^{2}-16 = (x)^2 - (4)^2 = ( x-4 ) ( x+4 ) $$