Use the Factor Theorem to determine whether $ g(x) $ is a factor of $ p(x) $ in each of the following cases:
(i) $ p(x)=2 x^{3}+x^{2}-2 x-1, g(x)=x+1 $
(ii) $ p(x)=x^{3}+3 x^{2}+3 x+1, g(x)=x+2 $
(iii) $ p(x)=x^{3}-4 x^{2}+x+6, g(x)=x-3 $
To do:
We have to find whether polynomial $g(x)$ is a factor of polynomial $p(x)$ in each of the given cases.
Solution:
We know that, if $g(x)$ is a factor of $p(x)$, then the remainder will be zero.
(i) \( p(x)=2 x^{3}+x^{2}-2 x-1, g(x)=x+1=x-(-1) \)
So, the remainder will be $p(-1)$.
$p(-1) = 2 (-1)^{3}+(-1)^{2}-2 (-1)-1$
$= 2(-1)+1 +2-1$
$=-2+3-1$
$=0$
Therefore, $g(x)$ is a factor of polynomial $p(x)$.
(ii) \( p(x)=x^{3}+3 x^{2}+3 x+1, g(x)=x+2=x-(-2) \)
So, the remainder will be $p(-2)$.
$p(-2) = (-2)^{3}+3 (-2)^{2}+3 (-2)+1$
$= -8+3(4)-6+1$
$=-14+12+1$
$=-14+13$
$=-1$
$≠ 0$
Therefore, $g(x)$ is not a factor of polynomial $p(x)$.
(iii) \( p(x)=x^{3}-4 x^{2}+x+6, g(x)=x-3 \)
So, the remainder will be $p(3)$.
$p(3) =(3)^{3}-4 (3)^{2}+(3)+6$
$= 27-4(9) +3+6$
$=36-36$
$=0$
Therefore, $g(x)$ is a factor of polynomial $p(x)$.
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