Question

If $2x^3+a{x}^2+bx-6$ has (x-1) as a factor and leaves reminder 2 when divided by (x-2), find the value of a and b respectively.

• A. -8,12
• B. 8,-12
• C. -4,10
• D. 4,-10

Answer 1

The correct option is A -8,12

$f\left(x\right)=2x^3+a{x}^2+bx-6$

$(x-1)$ is a factor

$\therefore f\left(1\right)=0\Rightarrow 2+a+b=6$

$\Rightarrow a+b=4\to 1$

and $f\left(x\right)$ leaves remainder $2$ on being divided by $(x-2)$

$\therefore f\left(2\right)=2$

$\Rightarrow 2\times 2^3+a\times 2^2+b\times 2-6=2$

$\Rightarrow 4a+2b+16-6=2$

$\Rightarrow 4a+2b=2-10=-8$

$\Rightarrow 2a+b=-4$

Solving $1$ and $2$

$2a+b=-4$

$a+b=4$

$---$

$a=-8$

$a+b=4$

$-8+b=4$

$b=4+8=12$

$\therefore a=-8;b=12$