Answer
$x\displaystyle \in\left\{ \frac{1- \sqrt {5}}{2}, \frac{1+ \sqrt {5}}{2},3\right\}$
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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$f(x)=x^4-7x^{3}+14x^{2}-3x-9$
Candidates for zeros, $\frac{p}{q}:$
$p:\qquad \pm 1, \pm 3, \pm9$
$q:\qquad \pm 1$
$\displaystyle \frac{p}{q}:\qquad \pm 1, \pm 3, \pm 9$
Try for $x=3:$
$\begin{array}{lllll}
\underline{3}|& 1 & -7 & 14 & -3 & -9\\
& & 3 & -12& 6& 9\\
& -- & -- & -- & --\\
& 1 & -4 & 2& 3 & |\underline{0}
\end{array}$
$3$ is a zero,
$f(x)=(x-3)(x^3-4x^2+2x+3)$
Try for $x=3$:
$\begin{array}{lllll}
\underline{3}|& 1 & -4 & 2 & 3\\
& & 3& -3& -3\\
& -- & -- & -- & --\\
& 1 & -1& -1 & |\underline{0}
\end{array}$
$3$ is a zero,
$f(x)=(x-3)^2(x^2-x-1)$
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, $x^2+x-1$, $x=\frac{1\pm \sqrt {-1^2-4 \times 1\times (-1)}}{2\times 1}=\frac{1\pm \sqrt {5}}{2}$.
Therefore, the real zeros of the equations are:
$x\displaystyle \in\left\{ \frac{1- \sqrt {5}}{2}, \frac{1+ \sqrt {5}}{2},3\right\}$
