24. (本题 10 分)
先阅读第(1)题的解答过程,然后再解第(2)题.
(1)已知多项式$2x^{3} - x^{2} + m有一个因式是2x + 1$,求$m$的值.
解法一：设$2x^{3} - x^{2} + m = (2x + 1)(x^{2} + ax + b)$,
则$2x^{3} - x^{2} + m = 2x^{3} + (2a + 1)x^{2} + (a + 2b)x + b$.
比较系数得$\begin{cases}2a + 1 = - 1\\a + 2b = 0\\b = m\end{cases} $,解得$\begin{cases}a = - 1\\b = \frac{1}{2}\\m = \frac{1}{2}\end{cases} $,$\therefore m = \frac{1}{2}$.
解法二：设$2x^{3} - x^{2} + m = A \cdot (2x + 1)$,$A$为整式,
由于上式为恒等式,为方便计算,取$x = - \frac{1}{2}$,
$2 × ( - \frac{1}{2})^{3} - ( - \frac{1}{2})^{2} + m = 0$,故$m = \frac{1}{2}$.
(2)已知$x^{4} + mx^{3} + nx - 16有因式(x - 1)和(x - 2)$,求$m$,$n$的值.

设$f(x) = x^4 + mx^3 + nx - 16$。
因为$(x - 1)$是因式，所以$f(1) = 0$：
$1^4 + m \cdot 1^3 + n \cdot 1 - 16 = 0$
$1 + m + n - 16 = 0$
$m + n = 15$ ①
因为$(x - 2)$是因式，所以$f(2) = 0$：
$2^4 + m \cdot 2^3 + n \cdot 2 - 16 = 0$
$16 + 8m + 2n - 16 = 0$
$8m + 2n = 0$
化简得$4m + n = 0$ ②
联立①②：
由② - ①得：$3m = -15$，解得$m = -5$
将$m = -5$代入①：$-5 + n = 15$，解得$n = 20$
$m = -5$，$n = 20$