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.

x - 2 = 0 ⇒ x = 2.

Since, x - 2 is a factor of x³ + ax² + bx + 6,

∴ On substituting x = 2 in x³ + ax² + bx + 6, remainder = 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

⇒ b = -(7 + 2a) …….(i)

x - 3 = 0 ⇒ x = 3.

Given, when x³ + ax² + bx + 6 is divided by x - 3, the remainder is 3.

∴ On substituting x = 3 in x³ + ax² + bx + 6, remainder = 3.

⇒ (3)³ + a(3)² + b(3) + 6 = 3

⇒ 27 + 9a + 3b + 6 = 3

⇒ 9a + 3b + 33 = 3

⇒ 9a + 3b = -30

⇒ 3(3a + b) = -30

⇒ 3a + b = -10

⇒ b = -10 - 3a = -(10 + 3a) ……..(ii)

From (i) and (ii) we get,

⇒ -(7 + 2a) = -(10 + 3a)

⇒ 7 + 2a = 10 + 3a

⇒ 3a - 2a = 7 - 10

⇒ a = -3.

Substituting a = -3 in (i) we get,

⇒ b = -(7 + 2a) = -(7 + 2(-3)) = -(7 - 6) = -1.

Hence, a = -3 and b = -1.