If $x^3 + 2x^2 -ax - b$ is exactly divisible by $(x^2 - 1)$, then the values of a and b are :
Answer & explanation
Correct answer: option 3
If $x^3 + 2x^2 -ax - b$ is exactly divisible by $(x^2 - 1)$
So, (x2 - 1) = 0
= x = 1, -1
Put x = 1
= 13 + 2(1)2 - a - b = 0
= a + b = 3 ----(a)
Put x = -1
= (-1)3 + 2(-1)2 + a - b = 0
= a - b = -1 ----(b)
= from eq(a) and (b)
= a = 1, b = 2