2015-I-11

$\begin{array}{rcl}f(2)& =& -5\\ (2-2{)}^2(2+h)+k& =& -5\\ k& =& -5\end{array}$

By the factor theorem, we have

$\begin{array}{rcl}f(3)& =& 0\\ (3-2)(3+h)–5& =& 0\\ 3+h& =& 5\\ h& =& 2\end{array}$

$\begin{array}{rcl}f(x)& =& (x-2{)}^2(x+2)-5\\ & =& x^3-2x^2-4x+3\\ & =& (x-3)(x^2+x-1)\end{array}$

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 $x^2+x-1=0$,

$\begin{array}{rcl}x& =& {\displaystyle \frac{-(1)\pm \sqrt{(1{)}^2–4(1)(-1)}}{2(1)}}\\ x& =& {\displaystyle \frac{-1\pm \sqrt{5}}{2}}\text{, which are not integers}\end{array}$

Hence, I don’t agree.