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A third degree polynomial $p(x)=0$ when $x=1$ and $x=3$. We also learn that $p(x) \geq 0 $ when $x \geq 1$ and $p(2) =2$. Determine $p(x)$. How should I proceed? I presume no calculus is needed.

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Hint: $p(x) = a(x-1)(x-3)(x-b)$ and only for one value of $b$ this polynomial doesn't change its sign around the point $x=3$.

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So, the answer should be $p(x)=2(x-1)(x-3)^{2}$. – user100561 Oct 13 '13 at 20:09
Yes. ${}{}{}{}$ – njguliyev Oct 13 '13 at 20:10

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