Answer
By checking if $x=1$ is a zero of the polynomial.
Work Step by Step
We know that the factor theorem states that any real number, say a, is a zero of a polynomial $p\left( x \right)$ if and only if, $\left( x-a \right)$ is a factor of the polynomial $p\left( x \right)$.
Now, we find $p\left( 1 \right)$
For $x=1$,
$\begin{align}
& {{\left( 1 \right)}^{3}}-2{{\left( 1 \right)}^{2}}-11\left( 1 \right)+12=1-2-11+12 \\
& =\left( 1+12 \right)+\left( -2-11 \right) \\
& =13-13 \\
& =0
\end{align}$
Thus, this shows that 1 is a zero of the polynomial and by the factor theorem, $\left( x-1 \right)$ is a factor of the polynomial ${{x}^{3}}-2{{x}^{2}}-11x+12$.
