2020-I-13

由於 $f(x)$ 可被 $x-1$ 整除，根據因式定理，可得

$\begin{array}{rcl}
f(1) & = & 0 \\
(1^2 – 1) (a(1) + b) + k(1) + 8 & = & 0 \\
k & = & -8
\end{array}$

$\begin{array}{rcl}
f(-3) & = & 0 \\
[(-3)^2 -1] [(a(-3) + b] -8(-3) + 8 & = & 0 \\
-24a + 8b +32 & = & 0 \ \ldots \unicode{x2460}
\end{array}$

由於當 $f(x)$ 除以 $x$ 時，其餘數為 $24$。根據餘式定理，可得

$\begin{array}{rcl}
f(0) & = & 24 \\
[(0)^2 – 1][a(0) + b] – 8(0) + 8 & = & 24 \\
-b & = & 16 \\
b & = & -16
\end{array}$

把 $b = -16$ 代入 $\unicode{x2460}$，可得

$\begin{array}{rcl}
-24a + 8(-16) + 32 & = & 0 \\
a & = & -4
\end{array}$

由此，可得

$\begin{array}{rcl}
f(x) & = & 0 \\
(x^2 – 1)(-4x – 16) -8x +8 & = & 0\\
(x + 1)(x – 1)(-4x – 16) -8(x – 1) & = & 0\\
(x – 1)([(x+1)(-4x -16) – 8] & = & 0 \\
(x – 1)(-4x^2 – 20x -24) & = & 0 \\
-4(x – 1)(x^2 + 5x + 6) & = & 0 \\
-4(x – 1)(x + 2)(x + 3) & = & 0 \\
(x – 1)(x + 2)(x + 3) & = & 0
\end{array}$

所以，其根為 $1$、$-2$ 及 $-3$。它們全都是整數。

由此，該宣稱正確。