Question
If $p(x) = x^3 - 3x^2 + 2x$, find $p(0), p(1), p(2)$. What do you conclude?
✓
Answer
$p(x)=x^3-3 x^2+2 x \ldots \text { (1) Putting } x=0 \text { in (1), }$
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 $p(x)$.
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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