8
$\begingroup$

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.

Thanks!

$\endgroup$
2
  • $\begingroup$ 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. $\endgroup$
    – Carl Love
    Mar 28, 2014 at 18:07
  • $\begingroup$ It is indeed necessary. In my original post I wrote that p has real coeffeicients, but Git Gut deleted that. $\endgroup$ Mar 28, 2014 at 22:09

2 Answers 2

10
$\begingroup$

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

$\endgroup$
2
  • 2
    $\begingroup$ 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? $\endgroup$ Mar 28, 2014 at 10:45
  • $\begingroup$ 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 $(-\infty, 0)$. $\endgroup$
    – 5xum
    Mar 28, 2014 at 12:00
2
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.