Question

If $(x-3)(x-2)$ is the HCF of the polynomials $x^2-2x-3\left)\right(2x^2+ax-2)$ and $(x^2+x-2)(3x^2+bx-3)$, then $a\&b$ is

• A. $8\&-3$
• B. $-8\&2$
• C. $2\&-8$
• D. $3\&-8$

Answer 1

The correct option is D $3\&-8$

Let,
$P\left(x\right)=(x^2-2x-3)(2x^2+ax-2)=(x+1)(x-3)(2x^2+ax-2)$

As, HCF is $(x-3)(x+2)$
$x+2=0\Rightarrow x=-2$

$x=-2$ is the zero.
$\therefore P(-2)=0$
$P(-2)=(-2+1)(-2-3)2\times 4+a(-2)-2$
$(-1)(-5)(8-2a-2)=0$
$5(6-2a)=0$
$a=3$

Let,
$Q\left(x\right)=(x^2+x-2)(3x^2+bx-3)=(x-1)(x+2)(3x^2+bx-3)$

As, HCF is $(x-3)(x+2)$
Similarly $Q\left(3\right)=0$

$Q\left(3\right)=(3-1)(3+2)(3\times 9+3b-3)\left(2\right)\left(5\right)(27+3b-3)=010(24+3b)=0b=-8$