(x+1) is not one of the factors of ......... | vedclass

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(D) According to the Factor Theorem,$(x+1)$ is a factor of a polynomial $p(x)$ if and only if $p(-1) = 0$.For option $A$: $p(-1) = (-1)^3 + 2(-1)^2 +

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$p(x)=x^{3}-1$

Vedclass · Answer

(D) According to the Factor Theorem,$(x+1)$ is a factor of a polynomial $p(x)$ if and only if $p(-1) = 0$.For option $A$: $p(-1) = (-1)^3 + 2(-1)^2 + 2(-1) + 1 = -1 + 2 - 2 + 1 = 0$. Thus,$(x+1)$ is a factor.For option $B$: $p(-1) = (-1)^3 - 2(-1) - 1 = -1 + 2 - 1 = 0$. Thus,$(x+1)$ is a factor.For option $C$: $p(-1) = (-1)^3 + 2(-1)^2 - 1 = -1 + 2 - 1 = 0$. Thus,$(x+1)$ is a factor.For option $D$: $p(-1) = (-1)^3 - 1 = -1 - 1 = -2$. Since $p(-1) 
eq 0$,$(x+1)$ is not a factor of $p(x) = x^3 - 1$.

$(x+1)$ is not one of the factors of .........

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