If polynomials $$4x^3+ax^2-3x+1$$​ and $$x^4+x^3-x^2+6$$​ leave the same remainder when each is divided by $$(x+1)$$​, then the value of a is:

Correct option is D

Given:​

Polynomial 1$$: 4x^3 + ax^2 - 3x + 1$$​​

Polynomial 2: $$x^4 + x^3 - x^2 + 6$$​​

Divisor: x + 1

Both polynomials leave the same remainder when divided by x + 1

Concept Used:

Remainder Theorem: If a polynomial p(x) is divided by (x - a), the remainder is p(a).

Solution:

Let p(x) =$$4x^3 + ax^2 - 3x + 1$$​ and q(x) $$= x^4 + x^3 - x^2 + 6.$$​​

According to the Remainder Theorem, if we divide p(x) and q(x) by x + 1, the remainders are p(-1) and q(-1), respectively.

Since both polynomials leave the same remainder when divided by x + 1,

we have: p(-1) = q(-1)

Now, we evaluate p(-1) and q(-1):

p(-1) $$= 4(-1)^3 + a(-1)^2 - 3(-1) + 1$$​​

p(-1) = -4 + a + 3 + 1

p(-1) = a

q(-1) =$$(-1)^4 + (-1)^3 - (-1)^2 + 6$$​​

q(-1) = 1 - 1 - 1 + 6

q(-1) = 5

Since p(-1) = q(-1)

a = 5

Therefore, the value of a is 5.