Finding the Value of 'a' Using the Remainder Theorem

The problem asks for the value of 'a' given two polynomials, $P(x) = 4x^3 + ax^2 - 3x + 1$ and $Q(x) = x^4 + x^3 - x^2 + 6$, which leave the same remainder when divided by $(x+1)$.

Applying the Remainder Theorem

The Remainder Theorem states that if a polynomial $f(x)$ is divided by $(x-c)$, the remainder is $f(c)$. Here, the divisor is $(x+1)$, so $c = -1$. We need to find the remainder for each polynomial when divided by $(x+1)$.

Remainder for the First Polynomial

Let $P(x) = 4x^3 + ax^2 - 3x + 1$. The remainder when $P(x)$ is divided by $(x+1)$ is $P(-1)$.

• Calculate $P(-1)$: $P(-1) = 4(-1)^3 + a(-1)^2 - 3(-1) + 1$ $P(-1) = 4(-1) + a(1) + 3 + 1$ $P(-1) = -4 + a + 3 + 1$ $P(-1) = a$

The remainder for the first polynomial is $a$.

Remainder for the Second Polynomial

Let $Q(x) = x^4 + x^3 - x^2 + 6$. The remainder when $Q(x)$ is divided by $(x+1)$ is $Q(-1)$.

• Calculate $Q(-1)$: $Q(-1) = (-1)^4 + (-1)^3 - (-1)^2 + 6$ $Q(-1) = 1 + (-1) - 1 + 6$ $Q(-1) = 1 - 1 - 1 + 6$ $Q(-1) = 5$

The remainder for the second polynomial is $5$.

Equating the Remainders

Since both polynomials leave the same remainder when divided by $(x+1)$, we set the calculated remainders equal to each other:

$P(-1) = Q(-1)$

$a = 5$

Conclusion

The value of 'a' is $5$. This corresponds to Option D.