If the polynomials 2x³ + ax² + 3x – 5 and x³ + x² – 4x + a leave the same remainder when divided by x – 2, find the value of a.

Let p(x) = 2x³ + ax² + 3x – 5 and q(x) = x³ + x² – 4x + a

As we know by Remainder Theorem,

If a polynomial p(x) is divided by a linear polynomial (x – a) then, the remainder is p(a)

⇒ Remainder of p(x) when divided by x – 2 is p(2). Similarly, Remainder of q(x) when divided by x – 2 is q(2)

⇒ p(2) = 2(2)³ +a(2)² + 3(2) – 5

⇒p(2) = 16 + 4a +6 – 5

⇒p(2) = 17 + 4a

Similarly, q(2) = (2)³ + (2)² + –4(2) + a

⇒ q(2) = 8 + 4 –8 + a

⇒ q(2) = 4 + a

Since they both leave the same remainder, so p(2) = q(2)

⇒ 17 + 4a = 4 + a

⇒ 13 = 3a

∴ The value of a is –13/3