Question

Find the value of $a$ if the division of $a{x}^3+9x^2+4x-10$ by $(x+3)$ leaves a remainder $5$.

Answer 1

Determine the value of $a$

Let $p\left(x\right)=a{x}^3+9x^2+4x-10$

If $f\left(x\right)$ is divided by $x+c$, then the remainder is $f\left(-c\right)$.$[\because x+c=x-(-c\left)\right]$

The divisor here is $x+3$.

$\begin{array}{rcl}x+3& =& x-\left(-3\right)\left[\mathrm{Form}x-c\mathrm{with}c=-3\right]\end{array}$

So, by the remainder theorem, when $p\left(x\right)$ is divided by $x+3$, then the remainder is $p\left(-3\right)$.

$\begin{array}{rcl}p\left(-3\right)& =& 5\\ a{\left(-3\right)}^3+9{\left(-3\right)}^2+4\left(-3\right)-10& =& 5\\ -27a+81-12-10& =& 5\\ -27a& =& -54\\ a& =& 2\end{array}$

Hence, the value of $a$ is $2$.