If (2x³ + ax² + bx - 2) when divided by (2x - 3) and (x + 3) leaves remainders 7 and -20 respectively, find values of a and b.

By remainder theorem,

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

Let, f(x) = 2x³ + ax² + bx - 2.

Given,

Divisor :

⇒ 2x - 3 = 0

⇒ 2x = 3

⇒ x = $\dfrac{3}{2}$

Given,

On dividing 2x³ + ax² + bx - 2 by 2x - 3, remainder is 7.

$\Rightarrow f\Big(\dfrac{3}{2}\Big) = 7 \\[1em] \Rightarrow 2\Big(\dfrac{3}{2}\Big)^3 + a\Big(\dfrac{3}{2}\Big)^2 + b\Big(\dfrac{3}{2}\Big) - 2 = 7 \\[1em] \Rightarrow 2\Big(\dfrac{27}{8}\Big) + \Big(\dfrac{9a}{4}\Big) + \Big(\dfrac{3b}{2}\Big) - 2 = 7 \\[1em] \Rightarrow \dfrac{27}{4} + \dfrac{9a}{4} + \dfrac{3b \times 2}{4} = 7 + 2 \\[1em] \Rightarrow \Big(\dfrac{27 + 9a + 6b}{4}\Big) = 9 \\[1em] \Rightarrow 27 + 9a + 6b = 9 \times 4 \\[1em] \Rightarrow 27 + 9a + 6b = 36 \\[1em] \Rightarrow 9a + 6b = 36 - 27 \\[1em] \Rightarrow 9a + 6b = 9 \\[1em] \Rightarrow 3(3a + 2b) = 9 \\[1em] \Rightarrow 3a + 2b = \dfrac{9}{3} \\[1em] \Rightarrow 3a + 2b = 3…..(1)$

Divisor :

⇒ x + 3 = 0

⇒ x = -3

On dividing 2x³ + ax² + bx - 2 by x + 3, remainder is -20.

⇒ f(-3) = -20

⇒ 2(-3)³ + a(-3)² + b(-3) - 2 = -20

⇒ 2(-27) + 9a - 3b - 2 = -20

⇒ -54 + 9a - 3b - 2 = -20

⇒ 9a - 3b - 56 = -20

⇒ 9a - 3b = -20 + 56

⇒ 9a - 3b = 36

⇒ 3(3a - b) = 36

⇒ 3a - b = $\dfrac{36}{3}$

⇒ 3a - b = 12

⇒ b = 3a - 12 ….(2)

Substituting value of b from equation (2) in 3a + 2b = 3, we get :

⇒ 3a + 2(3a - 12) = 3

⇒ 3a + 6a - 24 = 3

⇒ 9a = 27

⇒ a = $\dfrac{27}{9}$

⇒ a = 3.

Substituting value of a in equation (2), we get :

⇒ b = 3(3) - 12

⇒ b = 9 - 12

⇒ b = -3.

Hence, the value of a = 3 and b = -3.