Question

Check whether p(x) is a multiple of g(x) or not where p(x)=x³-6x+1 , g(x)=2-3x

Answer 1

Answer:

No, $p(x)$ is not a multiple of $g(x)$.

Step-by-step explanation:

If $p(x)$ is to be proved a multiple of $g(x)$, then it means to prove $p(x)$ a factor of $p(x)$.

So, let's use Factor Theorem:

The Zero of $x$:

$2-3x=0$

$-3x= -2$

$3x = 2$

$x=\frac{2}{3}$

By Factor Theorem:

$p(x)={x}^3-6x+1$

$p(\frac{2}{3})={(\frac{2}{3})}^3-6(\frac{2}{3})+1$

$p(\frac{2}{3})=\frac{8}{27}-4+1$

$p(\frac{2}{3})=\frac{8-108+27}{27}$

$p(\frac{2}{3})=\frac{-73}{27}$