If p(x)=2x3−11x2−4x+5 and g

$p(x)=2 x^{3}-11 x^{2}-4 x+5$

$g(x)=2 x+1=2\left(x+\frac{1}{2}\right)=2\left[x-\left(-\frac{1}{2}\right)\right]$

Putting $x=-\frac{1}{2}$ in $p(x)$, we get

$p\left(-\frac{1}{2}\right)=2 \times\left(-\frac{1}{2}\right)^{3}-11 \times\left(-\frac{1}{2}\right)^{2}-4 \times\left(-\frac{1}{2}\right)+5=-\frac{1}{4}-\frac{11}{4}+2+5=-\frac{12}{4}+7=-3+7=4 \neq 0$

Therefore, by factor theorem, (2x + 1) is not a factor of p(x).

Hence, g(x) is not a factor of p(x).