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 of this fraction by the conjugate of the denominator and simplify if possible:
$\dfrac{-8}{\sqrt{y}+4}=\dfrac{-8}{4+\sqrt{y}}\cdot\dfrac{4-\sqrt{y}}{4-\sqrt{y}}=\dfrac{-8(4-\sqrt{y})}{4^{2}-(\sqrt{y})^{2}}=...$
$...=\dfrac{-8(4-\sqrt{y})}{16-y}=\dfrac{8(\sqrt{y}-4)}{16-y}=-\dfrac{8(\sqrt{y}-4)}{y-16}$
