我觉得挺好的就来写个总结
若 \(f(x) = x ^ 4 + 4x ^ 3 + 6px ^ 2 + 4qx + r\) 能被 \(q(x) = x ^ 3 + 3x ^ 2 - x - 3\) 整除，求 \((p + q) ^ r\)。
余数定理：多项式 \(f(x)\) 除以 \(x - a\) 的余数为 \(f(a)\)。
先把 \(q(x)\) 因式分解：
\[q(x) = x ^ 3 + 2x ^ 2 + x + x ^ 2 - 2x - 3 \\ q(x) = x(x ^ 2 + 2x + 1) + (x - 3)(x + 1) \\ q(x) = x(x + 1)(x + 1) + (x - 3)(x + 1) \\ q(x) = (x + 1)(x ^ 2 + 2x -3) \\ q(x) = (x + 1)(x + 3)(x - 1) \]所以可以得到
\[(x + 1) \mid f(x) \\ (x + 3) \mid f(x) \\ (x - 1) \mid f(x) \]根据余数定理，得
\[\begin{cases} f(-1) = (-1) ^ 4 + 4 \times (-1) ^ 3 + 6p \cdot (-1) ^ 2 + 4q \cdot(-1) + r = 0 \\ f(-3) = (-3) ^ 4 + 4 \times (-3) ^ 3 + 6p \cdot (-3) ^ 2 + 4q \cdot(-3) + r = 0 \\ f(1) = 1 ^ 4 + 4 \times 1 ^ 3 + 6p \cdot 1 ^ 2 + 4q \cdot 1 + r = 0 \end{cases} \]所以
\[\begin{cases}6p - 4q + r = 3 \\ 54p - 12q + r = 27 \\ 6p + 4q + r = -5\end{cases} \]所以解得
\[\begin{cases}p = \dfrac{1}{3} \\ q = -1 \\ r = -3\end{cases} \]\[\therefore (p + q) ^ r \\ = (\dfrac{1}{3} -1) ^ {-3} \\ = -\dfrac{27}{8} \] 标签：cdot,dfrac,2x,4q,6p,一道,数学题,cases From： https://www.cnblogs.com/tmjyh09/p/16936538.html