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!
7
$\begingroup$
$\endgroup$
-
$\begingroup$ 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 '14 at 18:07
-
$\begingroup$ It is indeed necessary. In my original post I wrote that p has real coeffeicients, but Git Gut deleted that. $\endgroup$ – user010010001 Mar 28 '14 at 22:09
10
$\begingroup$
$\endgroup$
Hint: Take a look at $p(0)$ and the limits of $p$ as $x$ approaches $\pm\infty$.
-
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? $\endgroup$ – user010010001 Mar 28 '14 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 '14 at 12:00
2
$\begingroup$
$\endgroup$
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.
answered Mar 28 '14 at 12:00
