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. $b=2$
• C. $a=2$
• D. $b=0$

Answer 1

Answer: (A), (B)

The correct options are
A $a=0$
B $b=2$

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

Since (x+2) is a factor of p(x), then by Factor theorem $p(-2)=0$
$\Rightarrow a(-2{)}^3+b(-2{)}^2+(-2)-6=0$
$\Rightarrow -8a+4b-8=0$

$\Rightarrow -2a+b=2$ ...(i)

Also when p(x) is divided by (x-2) the remainder is 4, therefore by Remainder theorem $p\left(2\right)=4$
$\Rightarrow a(2{)}^3+b(2{)}^2+2-6=4$
$\Rightarrow 8a+4b+2-6=4$

$\Rightarrow 8a+4b=8$
$\Rightarrow 2a+b=2$ ...(ii)

Adding equation (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, $a=0$ and $b=2$