Question

If the HCF of the polynomials $p\left(x\right)=(x+1)(x^2+ax+4)$ and $q\left(x\right)=(x+4)(x^2+bx+2)$ is $h\left(x\right)=x^2+5x+4$, find the values of $a$ and $b$.

Answer 1

$p\left(x\right)=(x+1)(x^2+ax+4)$ ........ $\left(i\right)$

and $q\left(x\right)=(x+4)(x^2+bx+2)$ ........ $\left(ii\right)$

$h\left(x\right)=x^2+5x+4$ is HCF of $p\left(x\right)$ and $q\left(x\right)$

$h\left(x\right)=x^2+5x+4$

$=x^2+4x+x+4$

$=x(x+4)+(x+4)$

$=(x+4)(x+1)$

$\therefore x=4,-1$ are the roots of $p\left(x\right),q\left(x\right)$ and $h\left(x\right)$

$\therefore p\left(x\right)=(x+1)(x^2+ax+4)=(x+1)(x+4)f\left(x\right)$, for some $f\left(x\right)$

$⟹(x^2+ax+4)=(x+4)f\left(x\right)$

Take $x=-4$

$⟹\left(\right(-4{)}^2+a(-4)+4)=0$

$⟹16-4a+4=0$

$⟹a=5$

Similarly, $q\left(x\right)=(x+1)(x+4)g\left(x\right)$, for some factor $g\left(x\right)$ of $q\left(x\right)$

$⟹(x^2+bx+2)=(x+1)g\left(x\right)$

Take $x=-1$

$⟹(-1{)}^2+b(-1)+2=0$

$⟹1-b+2=0$

$⟹b=3$

Hence, $a=5$ and $b=3$