Question

If $(x-3)(x-2)$ is the HCF of the polynomials $P\left(x\right)=(x^2-2x-3)(2x^2+ax-2)andQ\left(x\right)=(x^2+x-2)(3x^2+bx-3)$, then $a$ and $b$ are equal to

• A. 3 and -8
• B. 8 and -3
• C. 2 and -8
• D. -8 and 2

Answer 1

The correct option is A 3 and -8

$P\left(x\right)=(x^2-2x-3)(2x^2+ax-2)=(x+1)(x-3)(2x^2+ax-2)$
As, $HCF=(x-3)(x+2)$
$\therefore$ P(-2) = 0
or $x=-2$ is the zero.
$\Rightarrow$ [(-2 + 1)(-2 - 3) {2 $\times$ 4 + a (-2) - 2}] = 0
$\Rightarrow (-1)(-5)(8-2a-2)=0$
$\Rightarrow 5(6-2a)=0$
$\Rightarrow a=3$
Also, $Q\left(x\right)=(x^2+x-2)(3x^2+bx-3)=(x-1)(x+2)(3x^2+bx-3)$
As, $HCF=(x-3)(x+2)$
$\therefore$ Q(3) = 0
or $x=3$ is the zero.
$\Rightarrow$ (3 - 1) (3 + 2) (3 $\times$ 9 + 3b - 3) = 0
$\Rightarrow \left(2\right)\left(5\right)(27+3b-3)=0$
$\Rightarrow 10(24+3b)=0$
$\Rightarrow b=-8$