\section*{Problem 18}

\(p(x)\) is the cubic \(x^{3}-3 x^{2}+5 x\). If \(h\) is a real root of \(p(x)=1\) and \(k\) is a real root of \(p(x)=5\), find \(h+k\)


\section*{Solution}

Put \(y=2-h\), where \(p(h)=1\), then \((2-y)^{3}-3(2-y)^{2}+5(2-y)-1=0\), so \(8-12 y+6 y^{2}-y^{3}-12+12 y-\) \(3 y^{2}+10-5 y-1=0\), or \(y^{3}-3 y^{2}+5 y=5\), or \(p(y)=5\). So if \(h\) is a root of \(p(h)=1\), then there is a root \(k\) of \(p(k)=5\) such that \(h+k=2\). To complete the proof we have to show that \(p(x)=5\) has only one real root.

But \(x^{3}-3 x^{2}+5 x=(x-1)^{3}+2(x-1)+3\) which is a strictly increasing function of \(x-1\) and hence of \(x\). So \(p(x)=k\) has only one real root.

