Question

If H.C.F of $(x-5)(x^2-x-a)$ and $(x-4)(x^2-2x-b)$ $(x-4)(x-5)$ is $(x–4)(x-5)$, find the values of a and b respectively

• A. $15,18$
• B. $10,7$
• C. $-8,10$
• D. $12,15$

Answer 1

Answer: (D)

$12,15$

It is given that $(x–4)(x-5)$ is the HCF of $(x-5)(x^2-x-a)$ and $(x-4)(x^2-2x-b)$, that is $(x–4)(x-5)$ is a factor of both the polynomials.

Let $f\left(x\right)=(x-5)(x^2-x-a)$ and $g\left(x\right)=(x-4)(x^2-2x-b)$

Now, $f\left(x\right)=g\left(x\right)$ at $x=4$ and $x=5$.

Let us first substitute $x=4$ as shown below:

$(4-5)(4^2-4-a)=(4-4)(4^2-2(4)-b)\Rightarrow -1(16-4-a)=0(16-8-b)\Rightarrow -1(12-a)=0\Rightarrow a-12=0\Rightarrow a=12$

Now, we substitute $x=5$ as follows:

$(5-5)(5^2-5-a)=(5-4)(5^2-2(5)-b)\Rightarrow 0(25-5-a)=1(25-10-b)\Rightarrow 0=25-10-b\Rightarrow 0=15-b\Rightarrow b=15$

Hence, $a=12,b=15$.