Answer
zeros:$\displaystyle \qquad -1,\ 4,\ \frac{3\pm\sqrt{13}}{2}$
Work Step by Step
$P(x)=x^{4}-6x^{3}+4x^{2}+15x+4,$
($2$ sign variations, we expect $2$ or $0$ real zero)
$P(-x)=x^{4}+6x^{3}+4x^{2}-15x+4,$
($2$ sign variations, we expect $2$ or $0$ negative real zeros)
$a_{0}=4 \quad $p: $\pm 1,\pm 2,\pm 4$
$a_{n}=1,\qquad $q: $\pm 1,$
Possible rational zeros $\displaystyle \frac{p}{q}:\quad$
with synthetic division, we try$ \pm$1, $\pm$2,...
$\begin{array}{r|rrrrrrrrrrr}\hline
-1 & 1 & -6 & 4 & 15 & 4 & & \\
& & -1 & 7 & -11 & -4 & & \\
\hline & 1 & -7 & 11 & 4 & 0 & & \\
& & & & & & & \\
4 & 1 & -7 & 11 & 4 & & & \\
& & 4 & -12 & -4 & & & \\
\hline & 1 & -3 & -1 & 0 & & & \\
\end{array}$
$P(x)=(x+1)(x-4)(x^{2}-3x-1)$
Find the zeros of the second factor with the quadratic formula
$x=\displaystyle \frac{3\pm\sqrt{3^{2}-4(1)(-1)}}{2(1)}=\frac{3\pm\sqrt{13}}{2}$
zeros:$\displaystyle \qquad -1,\ 4,\ \frac{3\pm\sqrt{13}}{2}$
