(x - 2) is a factor of the expression x³ + ax² + bx + 6. When this expression is divided by (x - 3), it leaves the remainder 3. Find the values of a and b.

Let f(x) = x³ + ax² + bx + 6

Given, (x - 2) is factor of f(x), hence, f(2) = 0 by factor's theorem

$\therefore 2^3 + a(2)^2 + b(2) + 6 = 0 \\[0.5em] \Rightarrow 8 + 4a + 2b + 6 = 0 \\[0.5em] \Rightarrow 4a + 2b + 14 = 0 \\[0.5em] \Rightarrow 2a + b + 7 = 0 \\[0.5em] b = -7 - 2a \text{ (Equation 1) }$

Given, on dividing f(x) by (x - 3) remainder left is 3

By remainder theorem, remainder = f(3)

$\therefore 3^3 + a(3)^2 + b(3) + 6 = 3 \\[0.5em] \Rightarrow 27 + 9a + 3b + 6 = 3 \\[0.5em] \Rightarrow 33 + 9a + 3b = 3$

On dividing equation by 3,

$\Rightarrow 11 + 3a + b = 1$

Putting value of b from equation 1,

$\Rightarrow 11 + 3a - 7 - 2a = 1 \\[0.5em] \Rightarrow a + 4 = 1 \\[0.5em] \Rightarrow a = -3 \\[0.5em] \text{ and } b = -7 - 2a = -7 - 2(-3) = -7 + 6 = -1$

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