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Let $P(x)=x^n+a_1x^{n-1}+...+a_n$, where $n$ is an even positive integer, the $a$'s are real, and $a_n \lt 0$. show that the equation $P(x)=0$ has at least two real roots. what more can you say about them?
I know that a polynomial has the same number of real roots as its highest degree and the fact that $a_n \lt 0$ prevents the polynomial from shifting up or down, removing one of the roots. Because $n$ is an even positive integer, the smalles the highest degree of the polynomial can be is 2, so this polynomial will have at least 2 real roots.
I know this but the question wants me to prove it somehow, and I don't know what I am supposed to show in order to prove it?

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    $\begingroup$ As $n$ is even, what happens with $P(x)$ when $x \to \pm \infty$? And what is $P(0)$? $\endgroup$
    – Hetebrij
    Commented Nov 14, 2015 at 18:44
  • $\begingroup$ $P(0)=a_n$ where $a_n \lt 0$ and $P(x)$ goes to $\infty$ when $x \to \pm \infty$, is that really all I have to say to prove this because it makes sense graphically what you said $\endgroup$
    – idknuttin
    Commented Nov 14, 2015 at 18:52
  • $\begingroup$ That's all, since then you can use the intermediate value theorem. $\endgroup$
    – Hetebrij
    Commented Nov 14, 2015 at 19:09
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    $\begingroup$ You can also show that any complex or real $z$ with $P(z)=0$ satisfies $1+\max (|a_1|,...,|a_n|)>|z|>1/(1+\max (|a_0/a_n|,...,|a_{n-1}/a_n|))$ where $a_0=1$. $\endgroup$ Commented Nov 14, 2015 at 19:44

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Maybe it is not formal enough, but $P(0)$ is lower than $0$, $P(-\infty)$ and $P(\infty)$ is higher than $0$. And $P$ is continuous. Therefore is must cross $y = 0$ both on the positive and negative side.

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