Check whether $g(x)$ is a factor of $p(x)$ or not,where $p(x) = 8x^3 - 6x^2 - 4x + 3$ and $g(x) = \frac{x}{3} - \frac{1}{4}$.

(YES) According to the Factor Theorem,$g(x)$ is a factor of $p(x)$ if $p(a) = 0$,where $a$ is the zero of $g(x)$.
First,find the zero of $g(x)$:
$g(x) = \frac{x}{3} - \frac{1}{4} = 0$
$\frac{x}{3} = \frac{1}{4}$
$x = \frac{3}{4}$
Now,evaluate $p\left(\frac{3}{4}\right)$:
$p\left(\frac{3}{4}\right) = 8\left(\frac{3}{4}\right)^3 - 6\left(\frac{3}{4}\right)^2 - 4\left(\frac{3}{4}\right) + 3$
$= 8 \times \frac{27}{64} - 6 \times \frac{9}{16} - 3 + 3$
$= \frac{27}{8} - \frac{54}{16} - 3 + 3$
$= \frac{27}{8} - \frac{27}{8} - 3 + 3 = 0$
Since $p\left(\frac{3}{4}\right) = 0$,by the Factor Theorem,$g(x)$ is a factor of $p(x)$.