Answer
zeros:$\displaystyle \qquad \frac{1}{2},\ \ \ \frac{1\pm\sqrt{3}}{2}$
Work Step by Step
$P(x)=4x^{3}-6x^{2}+1,$
($2$ sign variations, we expect $3$ or $1$ positive real zero)
$P(-x)=-4x^{3}-6x^{2}+1$
($1$ sign variations, we expect $1$ negative real zeros)
$a_{0}=1 \quad $p: $\pm 1,$
$a_{n}=4,\qquad $q: $\pm 1,\pm 2,\pm 4$
Possible rational zeros $\displaystyle \frac{p}{q}:\quad$
with synthetic division, we try$ \pm$1, $\pm 1/2$...
$\begin{array}{r|rrrrrrrrrrr}\hline
1/2 & 4 & -6 & 0 & 1 & & & \\
& & 2 & -2 & -1 & & & \\
\hline & 4 & -4 & -2 & 0 & & & \\
\end{array}$
$P(x)=(x+\displaystyle \frac{1}{2})(4x^{2}-4x-2)$
Find the zeros of the second factor with the quadratic formula
$x=\displaystyle \frac{4\pm\sqrt{(-4)^{2}-4(4)(-2)}}{2(4)}=\frac{4\pm\sqrt{48}}{8}=\frac{4\pm 4\sqrt{3}}{8}=\frac{1\pm\sqrt{3}}{2}$
zeros:$\displaystyle \qquad \frac{1}{2},\ \ \ \frac{1\pm\sqrt{3}}{2}$