The polynomials ax³ + 3x² - 3 and 2x³ - 5x + a when divided by x - 4 leave the remainder r₁ and r₂ respectively. If 2r₁ = r₂, then find the value of a.

By remainder theorem, on dividing f(x) by (x - b), remainder = f(b)

∴ On dividing, f(x) = ax³ + 3x² - 3 by (x - 4)

Remainder = f(4) = a(4)³ + 3(4)² - 3 = 64a + 45

∴ On dividing, f(x) = 2x³ - 5x + a by (x - 4)

Remainder = f(4) = 2(4)³ - 5(4) + a = 128 - 20 + a = 108 + a

According to question,

r₁ = 64a + 45

r₂ = 108 + a

2r₁ = r₂

$\therefore 2(64a + 45) = 108 + a \\[0.5em] \Rightarrow 128a + 90 = 108 + a \\[0.5em] \Rightarrow 128a - a = 108 - 90 \\[0.5em] \Rightarrow 127a = 18 \\[0.5em] \Rightarrow a = \dfrac{18}{127}.$