Question

If $(x-1)$ is a factor of polynomial $f\left(x\right)$ but not of $g\left(x\right)$, it must be a factor of which of the following polynomials?

• A. $f\left(x\right)g\left(x\right)$
• B. $-f\left(x\right)+g\left(x\right)$
• C. $f\left(x\right)-g\left(x\right)$
• D. $\left\{f\right(x)+g(x\left)\right\}g\left(x\right)$

Answer 1

The correct option is A $f\left(x\right)g\left(x\right)$

$f\left(x\right)=(x-1)q\left(x\right)$ but $q\left(x\right)\ne (x-1)r\left(x\right)$

As $f\left(x\right)q\left(x\right)=(x-1)q\left(x\right)q\left(x\right)\therefore (x-1)$ is a factor of it.

But in $f\left(x\right)\pm q\left(x\right)=(x-1)q\left(x\right)\pm q\left(x\right)$ as $q\left(x\right)$ is not divisible

by $(x-1)\therefore$ we can't take out common factor from it & hence

$\left[f\right(11)\pm q(x\left)\right]$ not divisible by $(x-1)$