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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.

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6 Answers 6

up vote 22 down vote accepted

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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An elegant answer. –  nbubis Apr 21 '12 at 11:31
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Actually I prefer yours ^^ it seems that it is what was intended in the question (to learn such technique, etc.). Still, sometimes it is more useful to use more basic tools, we shouldn't forget about them ;-) –  dtldarek Apr 21 '12 at 11:35
    
This, or $P(0)=0$ and $P'(0)\ne0$. –  Did Apr 22 '12 at 9:02

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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"Any other real root is also a root of $g$" is the wrong direction (it's true but not useful because we don't yet know that there is another real root). What you need is "Any real root of $g$ is also a real root of the original polynomial"" –  Henning Makholm Apr 21 '12 at 12:18
    
@Henning: I think that is pedantry: we "know" there is another real root because we trust the question to be honest. But I have changed the wording slightly to make it more hypothetical. –  Henry Apr 21 '12 at 12:24
    
The matter is not whether we trust the exercise, but that the reasoning you describe is no the reasoning the exercise asks us to do. It doesn't matter which properties "any other root" would have, because once we have a root to apply that knowledge to, we're already done. What does matter is the opposite implication, that roots of $g$ are also roots of the original polynomial. –  Henning Makholm Apr 21 '12 at 12:29
    
To quote Bill Clinton "It depends upon what the meaning of the word 'is' is." –  Henry Apr 21 '12 at 12:34
    
@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 $\rm\:f(r) = r\:g(r) = 0\iff g(r)=0.\quad$ –  Bill Dubuque Apr 21 '12 at 14:22

$P(-\frac{1}{2})=- \frac{67}{64}$ 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.

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