Question

If $a{x}^3+9x^2+4x-1$ is divided by $(x+2)$, the remainder is $-6$; then the value of ${}^{\prime }{a}^{\prime }$ is

• A. $-3$
• B. $-2$
• C. $0$
• D. $\frac{33}{8}$

Answer 1

The correct option is D $\frac{33}{8}$

According to remainder theorem, remainder $=f(-2)$.
Thus, $-6=a\times (-2{)}^3+9\times (-2{)}^2+4\times (-2{)}^1-1$
$\therefore$ $-6=-8\times a+36-8-1$
$\therefore$ $a=\frac{33}{8}$