Question

If $p\left(x\right)=(x+1)(3x^2+bx+2),q\left(x\right)=(x-2)(2x^2+ax+1)$ and their H.C.F. $h\left(x\right)=x^2-x-2$, then find the values of a and b.

• A. $2,-4$
• B. $3,-7$
• C. $4,-2$
• D. $5,-4$

Answer 1

Answer: (B)

$3,-7$

H.C.F. $h\left(x\right)=x^2-x-2=x^2-2x+x+2$
$=x(x-2)+1(x-2)=(x-2)(x+1)$
Now, the H.C.F. of $p\left(x\right)$ and $q\left(x\right)$ is $(x-2)(x+1)$
Hence, $(x-2)$ and $(x+1)$ are both factors of $p\left(x\right)$ as well as $q\left(x\right)$.
Now, $p\left(x\right)=(x+1)(3x^2+bx+2)$ and $(x-2)$ is a factor of $p\left(x\right)$.
$\therefore p\left(2\right)=0$
$\therefore (2+1)\left\{3\right(2{)}^2+b\left(2\right)+2\}=0$
$\therefore \left(3\right)(12+2b+2)=0$
$\therefore 2b+14=0$
$\therefore 2b=-14$ $\therefore b=-7$
Now, $q\left(x\right)=(x-2)(x^2+ax+1)$ and (x+1) is a factor of q(x).
$\therefore q(-1)=0$
$\therefore (-1,-2)\left\{2\right(-1{)}^2+a(-1)+1\}=0$
$\therefore (-3)(2-a+1)=0$
$\therefore 3-a=0\therefore a=3$
Thus $a=3$ and $b=-7$