Question

Let $f\left(x\right)=5x^3+px+q$, where $p$ and $q$ are real numbers. When $f\left(x\right)$ is divided by $x^2+x+1$, the remainder is $0$. Then the value of $p-q$ is

• A. $0$
• B. $5$
• C. $-5$
• D. $1$

Answer 1

The correct option is B $5$

$f\left(x\right)=5x^3+px+q$
We know that
$x^2+x+1=0$ has roots $\omega ,{\omega }^2$ where $\omega$ is the cube root of unity.

So, $f\left(\omega \right)=5{\omega }^3+p\omega +q=0$
$\Rightarrow 5+p\omega +q=0\cdots \left(1\right)f\left({\omega }^2\right)=5{\omega }^6+p{\omega }^2+q=0\Rightarrow 5+\frac{p}{\omega }+q=0\cdots \left(2\right)$
Using equation $\left(1\right)$ and $\left(2\right)$, we get
$p=0$
$\Rightarrow q=-5\therefore p-q=5$