Question

If $a{x}^3+3x^2+bx-3$ has a factor $(2x+3)$ and leaves remainder $-3$ when divided by $(x+2)$, find the values of $a$ and $b$. With these values of $a$ and $b$, factorise the given expression.

Answer 1

Let

$p\left(x\right)=a{x}^3+3x^2+bx-3$

$g\left(x\right)=2x+3=0\Rightarrow x=-\frac{3}{2}$

$f\left(x\right)=x+2=0\Rightarrow x=-2$

Given

$g\left(x\right)$ is a factor of $f\left(x\right)$

$\therefore$ By factor theorem,

$p(-\frac{3}{2})=0\Rightarrow a{(-\frac{3}{2})}^3+3{(-\frac{3}{2})}^2+b(-\frac{3}{2})-3=0\Rightarrow a(-\frac{27}{8})+3\left(\frac{9}{4}\right)+b(-\frac{3}{2})-3=0\Rightarrow -\frac{27a}{8}+\frac{27}{4}-\frac{3b}{2}-3=0\Rightarrow \frac{(-27a+54-12b)}{8}=3\Rightarrow -27a+54-12b=24\Rightarrow -3(9a+4b)=24-54=-30\Rightarrow 9a+4b=10...\left(i\right)$

Also, $p\left(x\right)$ when divided by $f\left(x\right)$ leaves a remainder $-3$

$\therefore$ By remainder theorem,

$p(-2)=-3\Rightarrow a(-2{)}^3+3(-2{)}^2+b(-2)-3=-3\Rightarrow -8a+12-2b=0\Rightarrow 8a+2b=12\Rightarrow 4a+b=6...\left(ii\right)$

Solving $\left(i\right)$ and $\left(ii\right)$, we get

$a=2$ and $b=-2$

Hence $p\left(x\right)=2x^3+3x^2-2x-3$

$=x^2(2x+3)-(2x+3)==(2x+3)(x^2-1)=(2x+3)(x+1)(x-1)$