Find a if the two polynomials ax³ + 3x² - 9 and 2x³ + 4x + a leave the same remainder when divided by (x + 3).

By remainder theorem,

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

Let p(x) = ax³ + 3x² - 9 and q(x) = 2x³ + 4x + a

Given,

Divisor :

⇒ x + 3 = 0

⇒ x = -3

On dividing ax³ + 3x² - 9 by x + 3, we get :

⇒ p(-3) = a(-3)³ + 3(-3)² - 9

= -27a + 27 - 9

= -27a + 18.

On dividing 2x³ + 4x + a by x + 3, we get :

⇒ q(-3) = 2(-3)³ + 4(-3) + a

= -54 - 12 + a

= -66 + a.

Given,

Polynomials ax³ + 3x² - 9 and 2x³ + 4x + a leave the same remainder when divided by (x + 3).

∴ p(-3) = q(-3)

⇒ -27a + 18 = -66 + a

⇒ -27a - a = -66 - 18

⇒ -28a = -84

⇒ a = $\dfrac{-84}{-28}$

⇒ a = 3.

Hence, the value of a = 3.