Question

If the remainder when $x^3+2x^2+kx+3$ is divided by x-3 is 21, find the zeroes of $x^3+2x^2+kx-18$

• A. -3, -3, 2
• B. -3, -2, 3
• C. -3, 2, 3
• D. -2, 3, 3

Answer 1

The correct option is B -3, -2, 3

Solution: P(3) = 48 + 3k =21

$\Rightarrow$ K =-9

Hence, $x^3+2x^2-9x+3=(x-3)$ x Quotient + 21

$\Rightarrow x^3+2x^2-9x-18=9x-3)$ x Quotient

$\Rightarrow$ Quotient = $\frac{x^3+2x^2-9x-18}{x-3}$


$$\begin{array}{c}{x}^2+5x+6\\ x-3x^3+2x^2-9x-18\\ \underset{–}{x^3-2x^2}2\\ 5x^2-9x\\ \underset{–}{5x^2-15x}\\ 6x-18\\ \underset{–}{6x-18}\\ -5\end{array}$$

Factorizing the quotient, $x^2+5x+6=x^2+3x+2x+6=x(x+3)+2(x+3)=(x+2)(x+3)$

Hence, the factors of $x^3+2x^2-9x-18$ are $x-3$, $x+2$ and $x+3$
$\Rightarrow$ the zeroes are -3,-2,3.