If (x³ + ax² + bx + 6) has (x - 2) as a factor and leaves a remainder 3 when divided by (x - 3), find the values of a and b.

Let f(x) = x³ + ax² + bx + 6

By factor theorem,

If, (x - 2) is a factor of f(x), then f(2) = 0.

⇒ (2)³ + a(2)² + b(2) + 6 = 0

⇒ 8 + 4a + 2b + 6 = 0

⇒ 4a + 2b + 14 = 0

⇒ 2(2a + b + 7) = 0

⇒ 2a + b + 7 = 0

⇒ 2a + b = -7 ….(1)

Given,

On dividing f(x) by (x − 3), the remainder is 3.

By remainder theorem,

∴ f(3) = 3

⇒ (3)³ + a(3)² + b(3) + 6 = 3

⇒ 27 + 9a + 3b + 6 = 3

⇒ 9a + 3b + 33 = 3

⇒ 9a + 3b = 3 - 33

⇒ 9a + 3b = -30

⇒ 3(3a + b) = -30

⇒ 3a + b = $-\dfrac{30}{3}$

⇒ 3a + b = -10 ….(2)

Subtracting equation (1) from equation (2),

⇒ 3a + b - (2a + b) = -10 -(-7)

⇒ 3a + b - 2a - b = -10 + 7

⇒ a = -3.

Substituting value of a in equation (1), we get :

⇒ 2(-3) + b = -7

⇒ -6 + b = -7

⇒ b = -7 + 6

⇒ b = -1.

Hence, the value of a = -3 and b = -1.