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(B) polynomial $p(x)$ is a multiple of $g(x)$ if and only if $g(x)$ divides $p(x)$ completely,which means the remainder must be $0$.According to the R

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No

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(B) polynomial $p(x)$ is a multiple of $g(x)$ if and only if $g(x)$ divides $p(x)$ completely,which means the remainder must be $0$.According to the Remainder Theorem,if we divide $p(x)$ by $g(x) = 2x + 1$,we set $g(x) = 0$ to find the value of $x$:$2x + 1 = 0 \implies x = -\frac{1}{2}$.Now,calculate $p(-\frac{1}{2})$:$p(-\frac{1}{2}) = 2(-\frac{1}{2})^3 - 11(-\frac{1}{2})^2 - 4(-\frac{1}{2}) + 5$$= 2(-\frac{1}{8}) - 11(\frac{1}{4}) + 2 + 5$$= -\frac{1}{4} - \frac{11}{4} + 7$$= \frac{-1 - 11 + 28}{4} = \frac{16}{4} = 4$.Since the remainder is $4$,which is not equal to $0$,$p(x)$ is not a multiple of $g(x)$.

Check whether $p(x)$ is a multiple of $g(x)$ or not:
$p(x) = 2x^3 - 11x^2 - 4x + 5, \quad g(x) = 2x + 1$

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Check whether $p(x)$ is a multiple of $g(x)$ or not:$p(x) = 2x^3 - 11x^2 - 4x + 5, \quad g(x) = 2x + 1$