Answer
$x\displaystyle \in\left\{-1, \frac{3- \sqrt {13}}{2}, \frac{3+ \sqrt {13}}{2}, 4\right\}$
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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$f(x)=x^4-6x^{3}+4x^{2}+15x+4$
Candidates for zeros, $\frac{p}{q}:$
$p:\qquad \pm 1, \pm 2, \pm4$
$q:\qquad \pm 1$
$\displaystyle \frac{p}{q}:\qquad \pm 1, \pm 2, \pm4$
Try for $x=-1:$
$\begin{array}{lllll}
\underline{-1}|& 1 & -6 & 4 & 15 & 4\\
& & -1 & 7& -11& -4\\
& -- & -- & -- & --\\
& 1 & -7 & 11& 4 & |\underline{0}
\end{array}$
$-1$ is a zero,
$f(x)=(x+1)(x^3-7x^2+11x+4)$
Try for $x=4$:
$\begin{array}{lllll}
\underline{4}|& 1 & -7 & 11 & 4\\
& & 4& -12& -4\\
& -- & -- & -- & --\\
& 1 & -3& -1 & |\underline{0}
\end{array}$
$4$ is a zero,
$f(x)=(x+1)(x-4)(x^2-3x-1)$
Solving for the trinomial, using the quadratic formula for the quadratic equation of $ax^2+bx+c$, $x=\frac{-b\pm \sqrt {b^2-4ac}}{2a}$.
in this case, $x^2-3x-1$, $x=\frac{3\pm \sqrt {-3^2-4 \times 1\times (-1)}}{2\times 1}=\frac{3\pm \sqrt {13}}{2}$.
Therefore, the real zeros of the equations are:
$x\displaystyle \in\left\{-1, \frac{3- \sqrt {13}}{2}, \frac{3+ \sqrt {13}}{2}, 4\right\}$
