Question

If $x^3+a{x}^2+bx+6$ is divisible by $(x-2)$ and leaves remainder $3$ when divided by $(x-3)$ then find the value of $a$ and $b$.

Answer 1

Let the given polynomial be $p\left(x\right)=x^3+a{x}^2+bx+6$

Given $p\left(x\right)$ is divisible by $(x–2)$, hence $p\left(2\right)=0$

Put $x=2$ in $p\left(x\right)$

$p\left(2\right)=2^3+a{\left(2\right)}^2+b\left(2\right)+6$

$\Rightarrow 0=8+4a+2b+6$

$\Rightarrow 4a+2b=–14$

$\Rightarrow 2a+b=–7$ .........$\left(1\right)$

It is also given that when $p\left(x\right)$ is divided by $(x–3)$ the remainder is $3$

That $p\left(3\right)=3$

Put $x=3$ in $p\left(x\right)$

$p\left(3\right)=3^3+a{\left(3\right)}^2+b\left(3\right)+6$

$\Rightarrow 3=27+9a+3b+6$

$\Rightarrow 9a+3b=–30$

$\Rightarrow 3a+b=–10$ .....$\left(2\right)$

Subtract $\left(2\right)$ from $\left(1\right)$

$\Rightarrow 3a+b-2a-b=-10+7$

$\therefore a=–3$

Substitute $a=–3$ in $\left(1\right)$, we get

$2(–3)+b=–7$

$\therefore b=–1$