Answer
$x \in \{-1, 4\}$
$f(x)=(x+1)^2(x-4)^2$
Work Step by Step
See The Rational Zero Theorem:
If $\frac{p}{q}$ is a zero of the polynomial $f(x) $with integer coefficients, then $p$ is a factor of the constant term, $a_{0}$, and $q$ is a factor of the leading coefficient, $a_{n}$.
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$f(x)=x^4-6x^{3}+x^2+24x+16$
a. Candidates for zeros, $p/q$
$p:\qquad \pm 1, \pm 2, \pm4, \pm8, \pm16$
$q:\qquad \pm 1$
$\displaystyle \frac{p}{q}:\qquad \pm 1, \pm 2, \pm4, \pm8, \pm16$
b. Try for $x=4:$
$\begin{array}{lllll}
\underline{4}| & 1 & -6 & 1 & 24& 16\\
& & 4 & -8 & -28& -16\\
& -- & -- & -- & --\\
& 1 & -2 & -7 & -4& |\underline{0}
\end{array}$
$4$ is a zero,
$f(x)=(x-4)(x^3-2x^{2} -7x-4)$
Try for $x=-1$
$\begin{array}{lllll}
\underline{-1}| & 1 & -2 & -7 & -4\\
& & -1 & 3 & 4\\
& -- & -- & -- & --\\
& 1 & -3 & -4 & |\underline{0}
\end{array}$
$-1$ is a zero
$f(x)=(x-4)(x+1)(x^2-3x-4)$
c. Factorize the trinomial $x^2-3x-4$...
factorize to find a number that the product will be $-4\times1=-4$...
whose sum is $-3$.
$x^2-3x-4=x^2+x-4x-4$
$(x+1)(x-4)$
The zeros of f satisfy $f(x)=0$
$f(x)=(x+1)^2(x-4)^2=0$
$x \in \{-1, 4\}$
