Answer
$-3$ and $2$ (multiplicity 2)
Work Step by Step
Step 1. Given $f(x)=x^3-x^2-8x+12$, we can list possible rational zeros as $\pm1,\pm2,\pm3,\pm4,\pm6,\pm12$
Step 2. There are 2 sign changes in $f(x)$ indicating that there could be 2 or 0 positive real zeros.
Step 3. $f(-x)=-x^3-x^2+8x+12$, there is 1 sign change in $f(-x)$ indicating that there must be 1 negative real zero.
Step 4. Use synthetic division to find one real zeros as shown in the figure.
Step 5. The resulting quotient gives $x^2-4x+4=0$ or $(x-2)^2=0$ which gives $x=2$ (multiplicity 2)
Step 6. The zeros are $-3$ and $2$ (multiplicity 2)
