Answer
$x=-1$ (multiplicity 2) and $x=4$ (multiplicity 2)
$g(x)=(x-4)^2(x+1)^2$
Work Step by Step
Step 1. List the possible rational zeros:
$\frac{p}{q}=\pm1,\pm2,\pm4,\pm16$
Step 2. Start with easy numbers; use synthetic division to test for zeros as shown in the figure. We have: $x=-1,4$
Step 3. With the quotient $x^2-3x-4=(x-4)(x+1)$, we have $g(x)=(x-4)^2(x+1)^2$
Step 4. We conclude that the zeros are $x=-1$ (multiplicity 2) and $x=4$ (multiplicity 2).
