# If p(x)=x3−3x2+2x,

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 $p(x)$.