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Let $p$ be an even degree polynomial with real coefficients such that the product of the constant term and the leading coefficient is negative. Show that $p$ has at least two real roots.


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Isn't it necessary to specify that the polynomial has real coefficients for this theorem to be true? For example, $x^2+ix-1$ has no real roots. – Carl Love Mar 28 '14 at 18:07
It is indeed necessary. In my original post I wrote that p has real coeffeicients, but Git Gut deleted that. – user010010001 Mar 28 '14 at 22:09
up vote 10 down vote accepted

Hint: Take a look at $p(0)$ and the limits of $p$ as $x$ approaches $\pm\infty$.

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The limits at $\pm\infty$ are the same. $p(0)<0$ if $\lim_{x \rightarrow \pm\infty} =\infty $ and $p(0)>0$ if $\lim_{x \rightarrow \pm\infty}=-\infty $ So $p$ has to have a positive and a negative root. Is that right? – user010010001 Mar 28 '14 at 10:45
Correct. I would mention that this is because on $(0,\infty)$, $p$ takes both positive and negative values, therefore (because it is continuous), it must also take a zero value. Same for $(-\infty, 0)$. – 5xum Mar 28 '14 at 12:00

By scaling, the polynomial can be written in the form $p(x)=x^{2n}+...-1=0$.

Then $p(0)<0$ and $p(x) > 0$ for large negative and positive $x$, so $p(x) $ has at least one positive and negative root.

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