Question

if the polynomials ax^3+4x^2+3x-4 and x^3-4x-a leaves the same remainder when divided by x-2, find the vlaue of a. ​

Answer 1

Answer:

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 remainderis 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 :)

Answer 2

Step-by-step explanation:

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answer is given