当多项式 $x^{3}+2kx+2$ 除以 $x+2$ 时，所得余数比该多项式除以 $x-1$ 时所得余数小 3。
求 $k$ 的值。 解方程 $2x^{3}-5x^{2}-3x+10=0$，非整数根用精确值表示。

1. Find the remainder when the expression is divided by $x+2$.

Let $f(x) = x^{3}+2kx+2$.
By the Remainder Theorem, the remainder when divided by $x+2$ is $f(-2)$.
$f(-2) = (-2)^{3} + 2k(-2) + 2$
$= -8 - 4k + 2$
$= -6 - 4k$

2. Find the remainder when the expression is divided by $x-1$.

The remainder when divided by $x-1$ is $f(1)$.
$f(1) = (1)^{3} + 2k(1) + 2$
$= 1 + 2k + 2$
$= 3 + 2k$

3. Form an equation using the given condition and solve for $k$.

The remainder for $x+2$ is 3 less than the remainder for $x-1$:
$f(-2) = f(1) - 3$
$-6 - 4k = (3 + 2k) - 3$
$-6 - 4k = 2k$
$6k = -6$
$k = -1$

1. Use the Factor Theorem to find a linear factor of the cubic polynomial.

Let $f(x) = 2x^3 - 5x^2 - 3x + 10$.
$f(1) = 2(1)^3 - 5(1)^2 - 3(1) + 10 = 4 \neq 0$
$f(-1) = 2(-1)^3 - 5(-1)^2 - 3(-1) + 10 = 6 \neq 0$
$f(2) = 2(2)^3 - 5(2)^2 - 3(2) + 10 = 16 - 20 - 6 + 10 = 0$
Since $f(2) = 0$, $(x - 2)$ is a factor.

2. Find the quadratic factor by comparing coefficients or long division.

Let $2x^3 - 5x^2 - 3x + 10 = (x - 2)(ax^2 + bx + c)$.
By observation, $a = 2$ and $c = -5$.
So, $2x^3 - 5x^2 - 3x + 10 = (x - 2)(2x^2 + bx - 5)$.
Comparing the coefficients of $x^2$:
$-4 + b = -5 \implies b = -1$
The quadratic factor is $2x^2 - x - 5$.

3. Solve the quadratic equation to find the remaining roots.

$2x^2 - x - 5 = 0$
Using the quadratic formula:
$x = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(2)(-5)}}{2(2)}$
$x = \frac{1 \pm \sqrt{1 + 40}}{4}$
$x = \frac{1 \pm \sqrt{41}}{4}$