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(A) According to the Factor Theorem,$g(x) = x - a$ is a factor of $p(x)$ if $p(a) = 0$.Here,$g(x) = x + 1$,which can be written as $x - (-1)$. So,$a =

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Yes,$g(x)$ is a factor of $p(x)$.

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(A) According to the Factor Theorem,$g(x) = x - a$ is a factor of $p(x)$ if $p(a) = 0$.Here,$g(x) = x + 1$,which can be written as $x - (-1)$. So,$a = -1$.We evaluate $p(-1)$:$p(-1) = 2(-1)^3 + (-1)^2 - 2(-1) - 1$$p(-1) = 2(-1) + 1 + 2 - 1$$p(-1) = -2 + 1 + 2 - 1$$p(-1) = 0$Since $p(-1) = 0$,by the Factor Theorem,$g(x)$ is a factor of $p(x)$.

Use the Factor Theorem to determine whether $g(x)$ is a factor of $p(x)$ in each of the following cases: $p(x) = 2x^3 + x^2 - 2x - 1$,$g(x) = x + 1$.

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