Answer
$\dfrac{-8}{\sqrt{y}+4}=-\dfrac{8(\sqrt{y}-4)}{y-16}$
Work Step by Step
$\dfrac{-8}{\sqrt{y}+4}$
Multiply the numerator and the denominator by the conjugate of the denominator and simplify:
$\dfrac{-8}{\sqrt{y}+4}=\dfrac{-8}{\sqrt{y}+4}\cdot\dfrac{\sqrt{y}-4}{\sqrt{y}-4}=\dfrac{-8(\sqrt{y}-4)}{(\sqrt{y})^{2}-4^{2}}=...$
$...=-\dfrac{8(\sqrt{y}-4)}{y-16}$
