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Let $x^7-10x^5+15x+5$ a polynomial over $\mathbb Q$. I would like to know how many real roots this polynomial has, I know we have to use the intermediate value theorem, but I don't know how to use in this case in particular.

I need help please.

Thanks

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Are you wanting real roots or rational roots? When you say a polynomial over $\Bbb Q$, I think of rational roots but you ask for real roots, so I wanted to clarify. –  Clayton Jan 10 '13 at 4:49
    
@Clayton real roots Clayton –  user42912 Jan 10 '13 at 4:52
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A strong tool is Sturm Theorem –  Maesumi Jan 10 '13 at 5:26
    
@Maesumi yes, really powerful. –  user42912 Jan 10 '13 at 5:29
    
Sturm is really powerful, but requires a lot of polynomial division. Using synthetic division, it's not too bad. –  Henry B. Jan 10 '13 at 5:33
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4 Answers

up vote 2 down vote accepted

Maybe this will work for you

http://www.cs.iastate.edu/~cs577/handouts/polyroots.pdf

Uses Descartes’ rules of sign of the Sturm’s Theorem

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I'm going to see this tomorrow, it seems this Sturm's Theorem is really powerful, thank you very much for your answer –  user42912 Jan 10 '13 at 5:28
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The most direct method that I can currently think of is the following:

Differentiate the polynomial, and make a substitution $w=x^2$. This gives us $$7w^3-50w^2+15,$$ a cubic equation in $w$. There is a general formula for solving for the zeroes of a cubic equations, hence we can find the exact roots of this equation, and in turn, we can find the exact roots of $f'(x)$, which tells us where local max/min occur. Checking the values of the function at these points will tell us how many zeroes the polynomial has.

This is a little bit of a long method and probably not the cleanest; I'll see if I can think of a simpler method, but this is certainly one approach to the problem.

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How did you arrive at this cubic equation in $w$? –  andybenji Jan 10 '13 at 5:08
    
$f'(x)=7x^6-50x^4+15$. Substitute $w=x^2$ and we have a cubic polynomial. –  Clayton Jan 10 '13 at 5:16
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Using Descartes' rule of signs, we count the number of sign changes. There are two so there are either two positive root or no positive roots. Since, $f(0)=5$ and $f(2)=-157$, we know there is at least one positive root, so there must be two. Further, by examining the transformed polynomial $(-x)^7-10(-x)^5+15(-x)+5 = -x^7+10x^5-15x+5$, we can conclude there are three sign changes and thus either three negative roots or just one. Note that, $f(-1)=-1$ and $f(-2)=167$. So, we can conclude that there are at least two sign changes in $f$ meaning the number of negative roots must be three, not one. We can stop here. Although, we can note that the polynomial tends to negative infinity for large $x$ as the $x^7$ term comes to dominate. Therefore, we could have concluded there are three sign changes in the value of $f$ and three roots without using the information from Descartes.

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We use the rational roots theorem, which says a root of $$f(x) = x^7 -10x^5+15x+5$$ must be an integer that divides $5$, so immediately we have four possibilities: $\pm 1$ or $\pm 5$. Plugging these into the polynomial we see that $$f(-5) = -46495 \neq 0\\f(-1) = -1 \neq 0 \\f(1) = 11 \neq 0\\f(5) = 46955 \neq 0$$And from this, we see that $f$ has no roots over $\mathbb{Q}$.

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yes, thank you for your answer, but the roots are real. This Polynomial is irreducible by Eisenstein theorem, it can't have roots over $\mathbb Q$ as you said. –  user42912 Jan 10 '13 at 5:08
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