Question

If the HCF of the polynomials $p\left(x\right)=(x-3)(2x^2-ax+2)$ and $q\left(x\right)=(x-2)(x^2-bx-6)$ is $h\left(x\right)=x^2-5x+6$, find the values of $a$ and $b$.

Answer 1

$p\left(x\right)=(x-3)(2x^2-ax+2)$ ........ $\left(i\right)$

and $q\left(x\right)=(x-2)(x^2-bx-6)$ ........ $\left(ii\right)$

$h\left(x\right)=x^2-5x+6$ is HCF of $p\left(x\right)$ and $q\left(x\right)$

$h\left(x\right)=x^2+5x+6$

$=x^2-3x-2x+6$

$=x(x-3)-2(x-3)$

$=(x-3)(x-2)$

$\therefore x=3,2$ are the roots of $p\left(x\right),q\left(x\right)$ and $h\left(x\right)$

$p\left(x\right)=(x-3)(2x^2-ax+2)=(x-3)(x-2)f\left(x\right)$, for some $f\left(x\right)$

$⟹(2x^2-ax+2)=(x-2)f\left(x\right)$

Take $x=2$

$⟹2(2{)}^2-a(2)+2=0$

$⟹8-2a+2=0$

$⟹a=5$

Similarly, $q\left(x\right)=(x-3)(x-2)g\left(x\right)$, for some factor $g\left(x\right)$ of $q\left(x\right)$

$⟹(x^2-bx-6)=(x-3)g\left(x\right)$

Take $x=3$

$⟹3^2-b\left(3\right)-6=0$

$⟹3-3b=0$

$⟹b=1$

Hence, $a=5$ and $b=1$