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 the value of $a-b$, ( a and b are constants ) from the following options.

• A. $2$
• B. $3$
• C. $-4$
• D. $0$

Answer 1

The correct option is D $0$

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

$g\left(x\right)=(x+2)(x^2+x-b)$

$HCF=$ $x^2-4=(x+2)(x-2)$

Therefore, $x+2$ is a factor of $(x^2-x-a)$

$\therefore$ $(-2{)}^2-(-2)-a=0$

$a=6$

$x-2$ is a factor of $x^2+x-b$

Therefore, $2^2+2-3=0$

$b=6$

Therefore, $a-b=0$