Answer
$\displaystyle \frac{(y-4)(y^{2}+4)}{y-1}$
Work Step by Step
Dividing with $\displaystyle \frac{P}{Q}$ equals multiplying with the reciprocal, $\displaystyle \frac{Q}{P}.$
$ \displaystyle \frac{y^{2}-16}{1}\div\frac{y^{2}+3y-4}{y^{2}+4}=\frac{y^{2}-16}{1}\cdot\frac{y^{2}+4}{y^{2}+3y-4}\qquad$
... factor what you can
$\left[\begin{array}{lll}
y^{2}-16= & ... & y^{2}+3y-4=\\
=(y-4)(y+4) & & =(y+4)(y-1)
\end{array}\right]$
$=\displaystyle \frac{(y-4)(y+4)}{1}\cdot\frac{y^{2}+4}{(y+4)(y-1)}\qquad$... divide out the common factors
$=\displaystyle \frac{(y-4)(1)}{1}\cdot\frac{y^{2}+4}{(1)(y-1)}\qquad$
= $\displaystyle \frac{(y-4)(y^{2}+4)}{y-1}$
