Ans: (a) $h=2$, $k=-5$ (b) No
- By the remainder theorem, we have
$\begin{array}{rcl}
f(2) & = & -5 \\
(2-2)^2(2+h) + k & = & -5 \\
k & = & -5
\end{array}$By the factor theorem, we have
$\begin{array}{rcl}
f(3) & = & 0 \\
(3-2)(3+h) – 5 & = & 0 \\
3 + h & = & 5 \\
h & = & 2
\end{array}$ - By the result of (a), we have
$\begin{array}{rcl}
f(x) & = & (x-2)^2(x+2) -5 \\
& = & x^3 -2x^2 -4x + 3 \\
& = & (x-3)(x^2 + x -1) \\
\end{array}$For the equation $f(x)=0$,
$\begin{array}{rcl}
(x-3)(x^2 + x – 1) & = & 0 \\
\end{array}$Therefore $x-3=0$ or $x^2 + x -1=0$. Consider the solution of $x^2 + x -1 = 0 $,
$\begin{array}{rcl}
x & = & \dfrac{-(1) \pm \sqrt{(1)^2 – 4(1)(-1)}}{2(1)} \\
x & = & \dfrac{-1\pm\sqrt{5}}{2}\text{, which are not integers}
\end{array}$Hence, I don’t agree.
