If the polynomials 2x³ + ax² + 3x - 5 and x³ + x² - 2x + a leave the same remainder when divided by (x - 2), find the value of a. Also, find the remainder in each case.

By remainder theorem,

If f(x) is divided by (x - a), then remainder = f(a).

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

Divisor :

⇒ x - 2

⇒ x = 2

⇒ p(2) = 2(2)³ + a(2)² + 3(2) - 5

= 2(8) + 4a + 6 - 5

= 16 + 4a + 1

= 4a + 17.

⇒ q(2) = (2)³ + (2)² - 2(2) + a

= 8 + 4 - 4 + a

= 8 + a.

Given,

Polynomials 2x³ + ax² + 3x - 5 and x³ + x² - 2x + a leave the same remainder when divided by (x - 2).

∴ p(2) = q(2)

⇒ 4a + 17 = 8 + a

⇒ 4a - a = 8 - 17

⇒ 3a = -9

⇒ a = $\dfrac{-9}{3}$

⇒ a = -3.

Substituting value of a in p(2) :

⇒ p(2) = 4a + 17

= 4(-3) + 17

= -12 + 17

= 5.

Substituting value of a in q(2) :

⇒ q(2) = 8 + a

= 8 - 3

= 5.

Hence, the value of a = -3 and remainder in each case is 5.