Find the values of p and q when $$px^3+ x^2-2x-q$$ is exactly divisible by (x-1) and (x+1).

Correct option is A

​Given:
$$px^3 + x^2 - 2x - q \ is \ divisible \ by \ (x-1) \ and \ (x+1).$$​

Concept:
If f(x) is divisible by (x -a), then f(a) = 0

Solution:

Substitute x = 1 in $$px^3 + x^2 - 2x - q = 0:$$​ ​

p(1)3+(1)2−2(1)−q=0p+1−2−q=0⟹p−q−1=0(Equation 1)p(1)^3 + (1)^2 - 2(1) - q = 0 \\p + 1 - 2 - q = 0 \implies p - q - 1 = 0 \quad \text{(Equation 1)}p(1)3+(1)2−2(1)−q=0p+1−2−q=0⟹p−q−1=0(Equation 1)​

Substitute x = -1 in $$px^3 + x^2 - 2x - q = 0:$$

p(−1)3+(−1)2−2(−1)−q=0−p+1+2−q=0⟹−p−q+3=0(Equation 2)p(-1)^3 + (-1)^2 - 2(-1) - q = 0 \\ -p + 1 + 2 - q = 0 \implies -p - q + 3 = 0 \quad \text{(Equation 2)}p(−1)3+(−1)2−2(−1)−q=0−p+1+2−q=0⟹−p−q+3=0(Equation 2)​​

Solve the two equations:

From Equation 1:  $p-q=1$

From Equation 2: $$-p - q = -3$$​​

Add the equations: $$(p - q) + (-p - q) = 1 + (-3) \implies -2q = -2 \implies q = 1$$​​
 Substitute q = 1 into p - q = 1:

​$$p - 1 = 1 \implies p = 2$$​​