Question

If $f\left(x\right)=(x-2)(x^2-x-a),g\left(x\right)=(x+2)(x^2+x-b)$ and their HCF is $x^2-4,$ then find $a-b$ ($a$ and $b$ are constants)

• A. 0
• B. 4
• C. 1
• D. 6

Answer 1

Answer: (A)

0

Given, $f\left(x\right)=(x-2)(x^2-x-a)$
and $g\left(x\right)=(x+2)(x^2+x-b)$
HCF of $f\left(x\right)$ & $g\left(x\right)$ is $x^2-4=(x-2)(x+2)$
i.e., both$f\left(x\right)$ & $g\left(x\right)$ are divisible by $(x-2)(x+2)$
It means $(x^2-x-a)$ is divisible by $x+2$
$(x^2-x-a)=(x+2)(x-3)+(-a+6)$
Remainder $=0$
Therefore, $-a+6=0\Rightarrow a=6$
Similarly, $(x^2+x-b)$ is divisible by $(x-2)$
$(x^2+x-b)=(x-2)(x+3)+(-b+6)$
remainder$=0$
Therefore, $-b+6=0\Rightarrow b=6$
$\Rightarrow a-b=6-6=0$