<s>[INST] Show that $1$ and $2$ are zeros of the following polynomial 
Show that $1$ and $2$ are zeros of the polynomial $P(x)=x^4-2x^3+5x^2-16x+12$ and
   hence that $(x-1)(x-2)$ is a factor of $P(x)$

 [/INST] The is an application of the Factor Theorem. 
If $\operatorname{f}(x)$ is a polynomial and $\operatorname{f}(p)=0$, then $x-p$ is a factor of $\operatorname{f}(x)$. In particular: if $\operatorname{f}(1)=0$ then $x-1$ is a factor of $\operatorname{f}(x)$ and if $\operatorname{f}(2)=0$ then $x-2$ is a factor of $\operatorname{f}(x)$. Hence to show that $\operatorname{f}(x)$ has $(x-1)(x-2)$ as a factor, we must show that both $\operatorname{f}(1)=0$ and $\operatorname{f}(2)=0$. Notice that:
\begin{array}{ccccc}
\operatorname{f}(1) &=& 1^4 - 2 \times 1^3 + 5 \times 1^2 - 16 \times 1 +12 &=& 1 - 2 + 5 - 16 + 12 &=& 0 \\
\operatorname{f}(2) &=& 2^4 - 2 \times 2^3 + 5 \times 2^2 - 16 \times 2 +12 &=& 16 - 16 + 20 - 32 + 12 &=& 0
\end{array}
It follows that both $x-1$ and $x-2$ divide $\operatorname{f}(x)$ and so their product $(x-1)(x-2)$ divides $\operatorname{f}(x)$.
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