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Show that $z^6 + 5z^4 - z^3 + 3z$ has at least two real roots given that all roots are distinct. Also, show that |3z - z^3 + 5z^4| < |z^6| when $|z| > 3$.

I can see that 0 is a real root; however, I am having trouble starting this one. I couldn't seem to see a way to factorize this. Other than testing points in the derivative and maybe using the intermediate value theorem, I don't know what direction to take on this one.

6 Answers 6

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Since the coefficients are real, all roots have conjugate roots as well. This means that excluding the $z=0$ root, you have five other distinct roots, so at least one of them is real. So we have at least two real roots.

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Let $P(z) = z^6+5z^4-z^3+3z$, then $P(-1/2) = -67/64$, but the highest power of $P$ is even with positive coefficient, so $\lim_{x\to+\infty}P(x) = +\infty$ and $\lim_{x\to-\infty}P(x) = +\infty$, so by the intermediate value property of continuous functions $P$ possess at least two real roots.

Edit: And for the second part, looks like triangle inequality suffices.

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    This, or $P(0)=0$ and $P'(0)\ne0$.2012-04-22
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You have the root $z=0$, so any other root would also be a root of $z^5 + 5z^3 - z^2 + 3$. Call that expression $g(z)$.

$g(0)=+3$ while $g(-3)= -384$ so by continuity and the intermediate value theorem there is a real root of $g(z)$ in $(-3,0)$, and this will be a real root of your original expression.

Meanwhile for $|z| \gt 3$ you have $|3z - z^3 + 5z^4| \le 3|z| +|z^3| +5|z^4| \lt \left(\frac{3}{3^5}+\frac{1}{3^3}+\frac{5}{3^2}\right)|z^6| \lt |z^6|.$

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    @Henning Note that he does state that the converse holds in the second paragraph. It would be clearer to write: if $\rm\:r\ne 0\:$ then $\rm\:f(r) = r\:g(r) = 0\iff g(r)=0.\quad$2012-04-21
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$P(-\frac{1}{2})=- \frac{67}{64}$ and $P(-1)=4$. Therefore the other real root is in the interval $(-1,-\frac{1}{2})$. In any case, when you factor out zero you get a polynomial of fifth degree. A polynomial of odd degree has at least one real root. That root in our case is not zero therefore it has at least two distinct real roots.

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The derivative at zero is non-zero, so it must be strictly negative at some point $a$ close to $0$. Also, the polynomial tends to $+\infty$ in both directions. So it must have another root, with the same sign as $a$.

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For the first part of the question:

We do not need to know that the polynomial has distinct roots. $0$ is an obvious simple root, dividing by $z$ gives a polynomial of degree 5, which must have a (clearly different) real root by the intermediate value theorem -it tends to both $\pm\infty$.

nbubis' proof is very nice but uses algebraic closure of $\mathbb C$ -the fundamental theorem of algebra. The intermediate value theorem is simpler to prove.

EDIT: Well I should have read Henry's answer, mine overlaps alot with his. The main thing I add perhaps is that we do not need the distinct roots assumption.