Using factor theorem, show that g(x) is a factor of p(x), when

Let:

$p(x)=x^{3}-8$

Now,

$g(x)=0 \Rightarrow x-2=0 \Rightarrow x=2$

By the factor theorem, (x – 2) is a factor of the given polynomial if p(2) = 0.
Thus, we have:

$p(2)=\left(2^{3}-8\right)=0$

Hence, $(x-2)$ is a factor of the given polynomial.