Question

If $p{q}^3+q{x}^2+rx+s$ is exactly divisible by $x^2-1$, then which of the following is/are necessarily true ?

(A) p=r

(B) q=s

(C) p=-r

(D) q=-s

• A. Both A and B
  
• B. Both C and D
  
• C. Both A and D
  
• D. Both B and C
  

Answer 1

The correct option is B Both C and D

Let $f\left(x\right)=p{x}^3+q{x}^2+rx+s$

Given, $x^2-1$ is a factor of $f\left(x\right)$.

Therefore, $(x+1)$ and $(x-1)$ are factors of $f\left(x\right)$.

Therefore, $f(-1)=0$ and $f\left(1\right)=0$

$-p+q-r+s=0$

$p+r=q+s$ ........(1)

$p+q+r+s=0$ ......(2)

From these equations, we get,

$p=-r$ and $q=-s$

Hence, option B is correct.