表达式 $2x^3 + ax^2 + bx + 10$（其中 $a$ 和 $b$ 为常数）含有因式 $x + 2$，且除以 $x - 3$ 的余数为 10。 求 $a$ 和 $b$ 的值。

1. Apply the Factor Theorem to form the first equation

Let $f(x) = 2x^3 + ax^2 + bx + 10$.
Since $x + 2$ is a factor, $f(-2) = 0$.
$2(-2)^3 + a(-2)^2 + b(-2) + 10 = 0$
$-16 + 4a - 2b + 10 = 0$
$4a - 2b = 6$
$2a - b = 3 \quad \text{--- (1)}$

2. Apply the Remainder Theorem to form the second equation

Since the remainder is 10 when divided by $x - 3$, $f(3) = 10$.
$2(3)^3 + a(3)^2 + b(3) + 10 = 10$
$54 + 9a + 3b + 10 = 10$
$9a + 3b = -54$
$3a + b = -18 \quad \text{--- (2)}$

3. Solve the simultaneous equations for $a$

Adding equation (1) and equation (2):
$(2a - b) + (3a + b) = 3 + (-18)$
$5a = -15$
$a = -3$

4. Solve for $b$

Substitute $a = -3$ into equation (1):
$2(-3) - b = 3$
$-6 - b = 3$
$b = -9$