Answer
$P(x)=(x-2)^2(x+3)(2x-1)(x^{2}+1)..$
zeros: $\quad -3, \ 1/2,\ \ 2, $
Work Step by Step
$P(x)=2x^{6}-3x^{5}-13x^{4}+29x^{3}-27x^{2}+32x-12,$
($5$ sign variations, we expect $5,3,$ or $1$ real zeros)
$P(-x)=2x^{6}+3x^{5}-13x^{4}-29x^{3}-27x^{2}-32x-12,$
($1$ sign variations, we expect $1$ negative real zero)
$a_{0}=-12 \quad $p: $\pm 1,\pm 2,\pm 3,\pm 4,\pm 6,\pm 12$
$a_{n}=2,\qquad $q: $\pm 1,\pm 2$
Possible rational zeros $\displaystyle \frac{p}{q}:\quad$
with synthetic division, we try$ \pm$1, $\pm$2,...
$\begin{array}{r|rrrrrrrrrrr}\hline
2 & 2 & -3 & -13 & 29 & -27 & 32 & -12 \\
& & 4 & 2 & -22 & 14 & -26 & 12 \\
\hline & 2 & 1 & -11 & 7 & -13 & 6 & 0 \\
& & & \swarrow & & & & \\
2 & 2 & 1 & -11 & 7 & -13 & 6 & \\
& & 4 & 10 & -2 & 10 & -6 & \\
\hline & 2 & 5 & -1 & 5 & -3 & 0 & \\
& & & \swarrow & & & & \\
-3 & 2 & 5 & -1 & 5 & -3 & & \\
& & -6 & 3 & -6 & 3 & & \\
\hline & 2 & -1 & 2 & -1 & 0 & & \\
& & & \swarrow & & & & \\
1/2 & 2 & -1 & 2 & -1 & & & \\
& & 1 & 0 & 1 & & & \\
\hline & 2 & 0 & 2 & 0 & & & \\
\end{array}$
$P(x)=(x-2)(x-2)(x+3)(x-\displaystyle \frac{1}{2})(2x^{2}+2).$
the last term has no real zeros.
$ P(x)=(x-2)(x-2)(x+3)(x-\displaystyle \frac{1}{2})\cdot$2$(x^{2}+$1$)$
$P(x)=(x-2)(x-2)(x+3)(2x-1)(x^{2}+1)..$
zeros: $\quad -3, 1/2, 2, $
