Question

The polynomial ax^3 + 4x^2 + 3x – 4 and x^3 – 4x + a leave the same remainder when divided by (x-3), find the value of a.

Answer 1

Given ax^3 + 4x^2 + 3x - 4 and x^3 - 4x + a leave the same remainder when divided by x - 3.
Let p(x) = ax^3 + 4x^2 + 3x - 4 and g(x) = x^3 - 4x + a
By remainder theorem, if f(x) is divided by (x − a) then the remainder is f(a)
Here when p(x) and g(x) are divided by (x − 3) the remainders are p(3) and g(3) respectively.
Also given that p(3) = g(3)  → (1)
Put x = 3 in both p(x) and g(x)
Hence equation (1) becomes,
a(3)^3 + 4(3)^2 + 3(3) - 4 = (3)^3 - 4(3) + a
⇒ 27a + 36 + 9 − 4 = 27 − 12 + a
⇒ 27a + 41 = 15 + a
⇒ 26a  = 15 − 41 = − 26
∴ a = −1

Alternate Method :

According to remainder theorem, if f(x) is divided by (x-a)then remainder f(a)

f(x) = ax³+4x²+3x-4
g(x)= x³-4x +a
f(3)=A(27)+4(9)+3(3)-4
27a+41
g(3)=27-4(3)+a
15+a
f(3)=G(3)
27a+41=15+a
26a=15-41
a=15-41/26
a=-26/26
a=-1

Hope This Helps :)