जाँच कीजिए कि $g(x), p(x)$ का एक गुणनखंड है या नहीं, जहाँ $p(x)=8 x^{3}-6 x^{2}-4 x+3$ और $g(x)=\frac{x}{3}-\frac{1}{4}$ है।

$g(x)=\frac{x}{3}-\frac{1}{4}=0$ gives $x=\frac{3}{4}$

$g(x)$ will be a factor of $p(x)$ if $p\left(\frac{3}{4}\right)=0$ (Factor theorem)

Now,$p\left(\frac{3}{4}\right)=8\left(\frac{3}{4}\right)^{3}-6\left(\frac{3}{4}\right)^{2}-4\left(\frac{3}{4}\right)+3$

$=8 \times \frac{27}{64}-6 \times \frac{9}{16}-3+3=0$

Since, $p\left(\frac{3}{4}\right)=0,$ So, $g(x)$ is a factor of $p(x)$