Question

The value of q such that $x–3$ divides the polynomial $p\left(x\right)=x^3–9x^2+qx–15$ exactly, is

• A. 14
• B. 23
• C. -20
• D. 18

Answer 1

The correct option is B 23

Given that $(x–3)$ divides the polynomial $p\left(x\right)=x^3–9x^2+qx–15$ exactly. This implies that the remainder will be zero i.e., $(x–3)$ is a factor of $p\left(x\right)$.
$\therefore p\left(3\right)=0$ (By factor theorem)
$\Rightarrow (3{)}^3-9(3{)}^2+q\left(3\right)-15=0$
$\Rightarrow 27-81+3q-15=0$
$\Rightarrow 3q-69=0$
$\Rightarrow q=\frac{69}{3}=23$
Hence, the correct answer is option (2).