#### Question

When the polynomial x^{3} + 2x^{2} – 5ax – 7 is divided by (x – 1), the remainder is A and when the polynomial x^{3} + ax^{2} – 12x + 16 is divided by (x + 2), the remainder is B. Find the value of ‘a’ if 2A + B = 0.

#### Solution

It is given that when the polynomial x^{3} + 2x^{2} – 5ax – 7 is divided by (x – 1), the remainder is A.

∴ (1)^{3} + 2(1)^{2} – 5a(1) – 7 = A

1 + 2 – 5a – 7 = A

– 5a – 4 = A …(i)

It is also given that when the polynomial x^{3} + ax^{2} – 12x + 16 is divided by (x + 2), the remainder is B.

∴ x^{3} + ax^{2} – 12x + 16 = B

(-2)^{3} + a(-2)^{2} – 12(-2) + 16 = B

-8 + 4a + 24 + 16 = B

4a + 32 = B …(ii)

It is also given that 2A + B = 0

Using (i) and (ii), we get,

2(-5a – 4) + 4a + 32 = 0

-10a – 8 + 4a + 32 = 0

-6a + 24 = 0

6a = 24

a = 4

Is there an error in this question or solution?

Solution When the Polynomial X3 + 2x2 – 5ax – 7 is Divided by (X – 1), the Remainder is a and When Polynomial X3 + Ax2 – 12x + 16 is Divided by (X + 2), the Remainder is B. Find the Value of ‘A’ If 2a + B = 0. Concept: Remainder Theorem.