Question

The polynomial $a{x}^3+b{x}^2+x–6$ has $(x+2)$ as a factor and leaves a remainder 4 when divided by $(x–2)$. Find $a$ and $b$.

• A. a=0, b=2
• B. a=0, b=4
• C. a=2, b=4
• D. a=2, b=2

Answer 1

The correct option is A a=0, b=2

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

By using factor theorem, $(x+2)$ can be a factor of $p\left(x\right)$ only when $p(-2)=0$
$p(-2)=a{(-2)}^3+b{(-2)}^2+(-2)-6=0$
$\Rightarrow -8a+4b-8=0$
$\therefore -2a+b=2...\left(i\right)$

Also when $p\left(x\right)$ is divided by $(x-2)$ the remainder is 4.

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

Adding equations (i) and (ii), we get
$(-2a+b)+(2a+b)=2+2$
$\Rightarrow 2b=4\Rightarrow b=2$

Putting $b=2$ in (i) we get

$-2a+2=2$
$\Rightarrow -2a=0\Rightarrow a=0$
Hence,
$a=0andb=2$