Answer
$P(x)=(x-2)(x-2)^{2}=(x-2)^{3}$
Zeros: $ \ 2\ \ \ $
Work Step by Step
$P(x)=x^{3}-6x^{2}+12x-8$
$P(-x)=-x^{3}-6x^{2}-12x-8$
Decscart's rule of signs:
P(x) has 3 sign changes $\Rightarrow$ 3 or 1 positive zeros.
P(-x) has 0 sign change $\Rightarrow$ 0 negative zeros.
Rational Zeros Theorem:
candidates for p:$\quad \pm 1,\pm 2,\pm 4,\pm 8$
candidates for q:$\quad \pm 1,$
Possible rational zeros $\displaystyle \frac{p}{q}$:$\quad \pm 1,\pm 2,\pm 4,\pm 8$
Testing with synthetic division,
$\left.\begin{array}{l}
2 \ \ |\\ \\ \\ \end{array}\right.\begin{array}{rrr}
1 &-6& 12 &-8\\\hline
&2 & -8 &8\\\hline
1& -4 & 4&|\ \ 0\end{array}$
$P(x)=(x-2)(x^{2}-4x+4)$
For the trinomial, recognize a perfect square.
$P(x)=(x-2)(x-2)^{2}=(x-2)^{3}$
Zeros: $ \ 2\ \ \ $
