Answer
$f(x)=2x^4-2$
Work Step by Step
Step 1. Given three zeros as $x=\pm1, i$, we can find all the 4 zeros as $x=\pm1,\pm i$
Step 2. Write the polynomial as
$f(x)=c(x+1)(x-1)(x+i)(x-i)=c(x^2-1)(x^2+1)=c(x^4-1)$
where $c$ is a constant
Step 3. Using the condition $f(3)=160$, we have $(3^4-1)c=160$, $80c=160$ and $c=2$
Step 4. We find the polynomial as $f(x)=2x^4-2$
