Answer
$\{-4,1,1,1\}$
Work Step by Step
Step 1. Given $f(x)=x^4+x^3-9x^2+11x-4$, we can list possible rational zeros as $\pm1,\pm2,\pm4$
Step 2. There are 3 sign changes in $f(x)$ indicating that there could be 3 or 1 positive real zeros.
Step 3. $f(-x)=x^4-x^3-9x^2-11x-4$, 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 zero as shown in the figure.
Step 5. The resulting quotient gives $x^3-3x^2+3x-1=0$ or $(x-1)^3=0$ which gives $x=1$ (multiplicity 3)
Step 6. The zeros are $\{-4,1,1,1\}$
