Question:
If $p(x)=x^{3}-3 x^{2}+2 x$, find $p(0), p(1), p(2)$. What do you conclude?
Solution:
$p(x)=x^{3}-3 x^{2}+2 x$
Putting $x=0$ in $(1)$, we get
$p(0)=0^{3}-3 \times 0^{2}+2 \times 0=0$
Thus, $x=0$ is a zero of $p(x)$.
Putting $x=1$ in (1), we get
$p(1)=1^{3}-3 \times 1^{2}+2 \times 1=1-3+2=0$
Thus, $x=1$ is a zero of $p(x)$.
Putting $x=2$ in (1), we get
$p(2)=2^{3}-3 \times 2^{2}+2 \times 2=8-3 \times 4+4=8-12+4=0$
Thus, $x=2$ is a zero of $p(x)$.
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