Answer
zero: $2,$
$P(x)=(x-2)^{3}$
Work Step by Step
$P(x)=x^{3}-6x^{2}+12x-8$
($3$ sign variations, we expect $1$ or $3$ positive real zeros)
$P(-x)=-x^{3}-6x^{2}-12x-8$
( no sign variations, we expect no negative real zeros)
$a_{0}=-8 \qquad $p: $\pm 1,\pm 2,\pm 4,\pm 8,$
$a_{n}=1,\qquad $q: $\pm 1, $
Possible rational zeros $\displaystyle \frac{p}{q}:\quad \pm 1,\pm 2,\pm 4,\pm 8,$
Test $x=1 $ by synthetic division:
$\left[\begin{array}{lllllll}
\underline{1|} & 1 & -6 & 12 & -8 & & \\
& & 1 & -5 & -7 & & \\
& -- & -- & -- & -- & & \\
& 1 & -5 & -7 & \underline{|-15} & \Rightarrow & P(1)\neq 0,
\end{array}\right.$
$1$ is not a zero, test $x=2$
$\left[\begin{array}{lllllll}
\underline{2|} & 1 & -6 & 12 & -8 & & \\
& & 2 & -8 & 8 & & \\
& -- & -- & -- & -- & & \\
& 1 & -4 & 4 & \underline{|0} & \Rightarrow & P(2)=0,
\end{array}\right.$
$x=2$ is a zero, $(x-2)$ is a factor of $P(x).$
The bottom row gives coefficients of the quotient.
$P(x)=x^{3}-3x-4=(x-2)(x^{2} -4x+4)$
We can factor the second parentheses:
recognize a square of a difference of x and 2
$P(x)=(x-2)(x-2)^{2}=(x-2)^{3}$
So, the zero is 2