CNF or DNF, $p(x)=0 \implies f=1$ Write either a $CNF$ or a$\;DNF$ of the expression $f$ such that
$f=0\iff (x_2x_1x_0)_{(2)}$ is a zero point of the polynomial:
$$p(x)=x^3-8x^2+12x=x(x-2)(x-6)$$
$$p(x)=0\iff x\in \{0,2,6\}$$
$$\begin{array}{|c|c|c|c|}
\hline
x_2& x_1 & x_0 & f=1 \\ \hline
0 &0 &0 & 1\\ \hline
0 &0  &1 &0\\ \hline
0 &1  &0 &1\\ \hline
0 & 1 &1 &0\\  \hline
1 &0 &0 &0\\ \hline
1 &0  &1 &0\\ \hline
1 &1  &0 &1\\ \hline
1 & 1 &1 &0\\  \hline
\end{array}$$
$DNF:\overline{x_2}\cdot\overline{x_1}\cdot\overline{x_0}+\overline{x_2}\cdot x_{1}\cdot\overline{x_0}+x_2\cdot x_1\cdot \overline{x_0}$
Is this correct?

It's correct, and if you want to find the Minimal form, you can use a k-map:
$$\boxed{\begin{array}{ccccc}
&x_2'x_1'&x_2'x_1&x_2x_1&x_2x_1'\\
x_0'&\color{orange}1&\color{orange}1&1&0\\
x_0&0&0&0&0\end{array}}
\boxed{\begin{array}{ccccc}
&x_2'x_1'&x_2'x_1&x_2x_1&x_2x_1'\\
x_0'&1&\color{red}1&\color{red}1&0\\
x_0&0&0&0&0\end{array}}$$
$$\color{orange}{x_2'x_0'}+\color{red}{x_1x_0'}\tag*{Minimal DNF form}$$
$$\boxed{\begin{array}{ccccc}
&x_2'x_1'&x_2'x_1&x_2x_1&x_2x_1'\\
x_0'&1&1&1&0\\
x_0&\color{orange}0&\color{orange}0&\color{orange}0&\color{orange}0\end{array}}
\boxed{\begin{array}{ccccc}
&x_2'x_1'&x_2'x_1&x_2x_1&x_2x_1'\\
x_0'&1&1&1&\color{red}0\\
x_0&0&0&0&\color{red}0\end{array}}
$$
$$(\color{orange}{x_0})'(\color{red}{x_2x_1'})'=x_0'(x_2'+x_1)\tag*{Minimal CNF form}$$
