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let $p(x)=a_0 +a_1x +......+a_nx^n \ $ is real polynomial of odd degree such that $a_0a_n <0$. show that $p$ has at least two real roots.

i know it will have at least one root by using IVP property.but how at least two roots???? i try to find $\lim _{x \to \infty} \ p(x)$ and $\lim _{x \to \ -\infty} \ p(x)$. and compare it with $p(0)=a_0$. but did not get exactly what i want .

any hint.

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  • $\begingroup$ Are you sure about the hypothesis? Because $p(x)=x^3-10$ is a polynomial of odd degree with $a_n a_0 = (1\times -3) < 0$ and has only one real root. $\endgroup$ – Batman Dec 19 '17 at 12:57
  • $\begingroup$ Pretty sure you meant even, rarher than odd. $\endgroup$ – Andrés E. Caicedo Dec 19 '17 at 13:06
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    $\begingroup$ (@Batman Rather, $1\times-10$.) $\endgroup$ – Andrés E. Caicedo Dec 19 '17 at 13:07
  • $\begingroup$ Can you clarify your question? It is false as stated. If you meant "even" instead of "odd" then it is true, as $\frac {1}{a_n}p(x)$ is less than $0$ at $x=0$ and has lead term $x^n$, (and so must have at least one positive and one negative root). $\endgroup$ – lulu Dec 19 '17 at 13:24
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That $p$ has at least two real roots is not true:

$p(x)=1-x^3$.

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The result is false for any odd degree $n$. For example, if $n$ is odd, the only real root of $p(x)=x^n-1$ is $x=1$. All other roots are complex non-real roots of unity.

On the other hand, the result is true if $n$ is even (so perhaps the "odd" is just a typo.) Indeed, if $p(x)=a_0 +a_1x +......+a_nx^n$ with $a_n\ne0$, the $a_nx^n$ term dominates, meaning that for $x$ large, $p(x)$ has the same sign as $a_nx^n$. If $n$ is odd, this sign is different for $x$ positive and for $x$ negative (and, via the intermediate value theorem, this is the reason why odd degree polynomials have at least one real root). On the other hand, for $n$ even, this sign is the same for $x$ positive and negative (and it is the sign of $a_n $). If in addition $a_0a_n<0$, then this sign is the opposite of the sign of $p(0)=a_0$. This means, via the intermediate value theorem, that $p$ must have at least one negative and one positive real root.

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