Question

For what values of $a$ and $b$ the polynomials $p\left(x\right)=(x^2+5x+6)(x^2+2x-a)$ and $q\left(x\right)=(x^2-x-2)(x^2+7x+b)$ have $(x+2)(x-2)$ as their HCF.

Answer 1

Given : $p\left(x\right)=(x^2+5x+6)(x^2+2x-a)$ and $q\left(x\right)=(x^2-x-2)(x^2+7x+b)$ have HCF $h\left(x\right)=(x+2)(x-2)$

$\therefore x=-2,2$ are the roots of both $p\left(x\right)$ and $q\left(x\right)$

Now, $x^2+5x+6=x^2+3x+2x+6=(x+3)(x+2)$ ....... $\left(i\right)$

Now, $p\left(x\right)=(x^2+5x+6)(x^2+2x-a)=h\left(x\right)f\left(x\right)$, where $f\left(x\right)$ is the factor of $p\left(x\right)$

$⟹(x^2+5x+6)(x^2+2x-a)=(x+2)(x-2)f\left(x\right)$

$⟹(x+3)(x+2)(x^2+2x-a)=(x+2)(x-2)f\left(x\right)$

$⟹(x+3)(x^2+2x-a)=(x-2)f\left(x\right)$

Take $x=2$

$⟹(2^2+2\times 2-a)=0$

$⟹a=8$ ..... $\left(ii\right)$

Also, $q\left(x\right)=(x^2-x-2)(x^2+7x+b)=h\left(x\right)g\left(x\right)$, where $g\left(x\right)$ is another factor of $q\left(x\right)$

$⟹(x-2)(x+1)(x^2+7x+b)=(x+2)(x-2)f\left(x\right)$

$⟹(x+1)(x^2+7x+b)=(x+2)f\left(x\right)$

Take $x=-2$

$⟹\left(\right(-2{)}^2+7(-2)+b)=0$

$⟹4-14+b=0$

$⟹b=10$

Hence, $a=8$ and $b=10$