Question:
If $p(x)=x^{3}+x^{2}-9 x-9$, find $p(0), p(3), p(-3)$ and $p(-1)$. What do you conclude about the zero of $p(x) ?$ Is 0 a zero of $p(x)$ ?
Solution:
$p(x)=x^{3}+x^{2}-9 x-9$
Putting $x=0$ in $(1)$, we get
$p(0)=0^{3}+0^{2}-9 \times 0-9=0+0-0-9=-9 \neq 0$
Thus, x = 0 is not a zero of p(x).
Putting x = 3 in (1), we get
$p(3)=3^{3}+3^{2}-9 \times 3-9=27+9-27-9=0$
Thus, x = 3 is a zero of p(x).
Putting x = –3 in (1), we get
$p(-3)=(-3)^{3}+(-3)^{2}-9 \times(-3)-9=-27+9+27-9=0$
Thus, x = –3 is a zero of p(x).
Putting x = –1 in (1), we get
$p(-1)=(-1)^{3}+(-1)^{2}-9 \times(-1)-9=-1+1+9-9=0$
Thus, $x=-1$ is a zero of $p(x)$.