Question

Find the value of a and b when the polynomial $\text{f(x)}={\text{a(x)}}^3+3{\text{x}}^2+\text{bx}–3$ is exactly divisible by $(2\text{x}+3$) and leaves a remainder $-3$ when divided by $\text{(x+2)}$.​

Answer 1

Step 1: Form the equations
Given: $\text{f(x)}={\text{a(x)}}^3+3{\text{x}}^2+\text{bx}–3$

From remainder theorem, we know that when a polynomial $\text{f(x)}$ is divided by $\text{(x-a)}$, then the remainder is $\text{f(a)}$.
$\text{f(x)}$ is exactly divisible by $\text{(x-1)}$. So, remainder = $0$.
$\therefore \text{f}\left(\begin{array}{c}-\frac{3}{2}\end{array}\right)=0$
$a{\left(\begin{array}{c}-\frac{3}{2}\end{array}\right)}^3+3{\left(\begin{array}{c}-\frac{3}{2}\end{array}\right)}^2+b\left(\begin{array}{c}-\frac{3}{2}\end{array}\right)-3=0$
$a\left(\begin{array}{c}-\frac{27}{8}\end{array}\right)+3\left(\begin{array}{c}\frac{9}{4}\end{array}\right)-\frac{3b}{2}-3$
$-27\text{a}+30-12\text{b}=0$
$27\text{a}-12\text{b}+30$-(i)

Also, $\text{f(x)}$ leaves remainder $-3$ when divided by $\text{(x+2)}$.
$\therefore \text{f(-2)}=-3$
$\text{a}(-2{)}^3+3(-2{)}^2+\text{b}(-2)-3=-3$
$-8\text{a}+12-2\text{b}-3=-3$
$-8\text{a}-2\text{b}=-12$
$\text{b}=6-4\text{a}$-(ii)

Step 2: Find the value of a and b
Put the value of b from equation

$27\text{a}=-12(6-4\text{a})+30$
$27\text{a}-72+483+30$
$-21\text{a}=-42$
$\text{a}=2$

Put the value of a in equation (ii)
$\text{b}=6-4\left(2\right)$
$\text{b}=-2$

Hence, $\text{a}=2$ and $\text{b}=-2$.