Check whether p(x) is a multiple of g(x) or n | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) polynomial $p(x)$ is a multiple of $g(x)$ if and only if $g(x)$ divides $p(x)$ completely,which means the remainder must be $0$.To find the remain

Vedclass · Accepted Answer

No,it is not a multiple.

Vedclass · Answer

(B) polynomial $p(x)$ is a multiple of $g(x)$ if and only if $g(x)$ divides $p(x)$ completely,which means the remainder must be $0$.To find the remainder when $p(x)$ is divided by $g(x) = 2 - 3x$,we set $g(x) = 0$:$2 - 3x = 0 \implies 3x = 2 \implies x = \frac{2}{3}$.Now,we evaluate $p\left(\frac{2}{3}ight)$:$p\left(\frac{2}{3}ight) = \left(\frac{2}{3}ight)^{3} - \left(\frac{2}{3}ight) + 1$$= \frac{8}{27} - \frac{2}{3} + 1$$= \frac{8 - 18 + 27}{27} = \frac{17}{27}$.Since the remainder is $\frac{17}{27} 
eq 0$,$p(x)$ is not a multiple of $g(x)$.

Check whether $p(x)$ is a multiple of $g(x)$ or not,where $p(x) = x^{3} - x + 1$ and $g(x) = 2 - 3x$.

Similar Questions

Factorise $144 x^{2}-289 y^{2}$.

Factorise $6x^{3}-23x^{2}+29x-12$.

For what value of $m$ is $x^{3}-2 m x^{2}+16$ divisible by $x+2$?

Factorise : $2x^{3}-3x^{2}-17x+30$

If $x+1$ is a factor of the polynomial $2x^2+kx$,then the value of $k$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module