Problem
Conservative Polynomials
What happens to a polynomial when we take its absolute value?
Suppose that \(P\) is a polynomial that takes both positive and negative values. Prove that the function \(\lvert P\rvert\) is not a polynomial anymore.
Explore the solution
Suppose, for a contradiction, that \(Q(x)=\lvert P(x)\rvert\) is a polynomial.
At least one of the sets
\[\{x\in\mathbb R:P(x)\geq0\} \qquad\text{and}\qquad \{x\in\mathbb R:P(x)\leq0\}\]is infinite, because together they cover the real line.
If the first set is infinite, the polynomial \(P-Q\) has infinitely many roots. A nonzero polynomial has only finitely many roots, so \(P=Q=\lvert P\rvert\) everywhere. This contradicts the fact that \(P\) takes negative values.
If the second set is infinite, apply the same argument to \(P+Q\). It follows that \(P=-Q=-\lvert P\rvert\) everywhere, contradicting the fact that \(P\) takes positive values.
Either way, \(\lvert P\rvert\) cannot be a polynomial. The proof uses only the elementary fact about roots of a polynomial; no continuity argument is needed.