# 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) $p(x)=x^{3}-4 x^{2}+x+6, g(x)=x-3$

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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 $p(x)$.

Updated on 10-Oct-2022 13:39:07