MHB Can the Polynomial $x^7-2x^5+10x^2-1$ Have a Root Greater Than 1?

AI Thread Summary
The polynomial \( x^7 - 2x^5 + 10x^2 - 1 \) has no roots greater than 1. By applying the Rational Root Theorem, the only possible rational roots are \( \pm 1 \). A calculus approach shows that the function \( f(x) = x^7 - 2x^5 + 10x^2 - 1 \) is positive at \( x = 1 \) and that its derivative \( f'(x) \) is positive for all \( x > 1 \), indicating that the function is increasing in that interval. Therefore, since \( f(x) \) does not cross the x-axis for \( x > 1 \), it confirms that there are no roots in that range. The conclusion is that \( x^7 - 2x^5 + 10x^2 - 1 \) has no root greater than 1.
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Prove that $x^7-2x^5+10x^2-1$ has no root greater than 1.

This is one of my all time favorite challenge problems! :o
 
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anemone said:
Prove that $x^7-2x^5+10x^2-1$ has no root greater than 1.

This is one of my all time favorite challenge problems! :o

If the polynomial has a root,then it will be of the form $\frac{k}{l}, \text{ with } (k,l)=1 \text{ and } k \mid -1 \Rightarrow k= \pm 1 \text{ and }l \mid 1 \Rightarrow l \pm 1$

Therefore,the unique possible roots are $\pm 1$.

So,we conclude that the polynomial $x^7-2x^5+10x^2-1$ has no root greater than 1. (Nerd)
 
I'm sorry evinda, I'm afraid I don't follow your reasoning...but, thanks for attempting to solve this challenge...:)
 
anemone said:
Prove that $x^7-2x^5+10x^2-1$ has no root greater than 1.

This is one of my all time favorite challenge problems! :o

Suppose it has got a raional root.
because the coefficeint of highest power that is x^7 is one then the root has to be a factor of -1 that is 1 or -1 .

so it has no root > 1

The above follows straight from Rational root theorem - Wikipedia, the free encyclopedia
 
evinda said:
If the polynomial has a root,then it will be of the form $\frac{k}{l}, \text{ with } (k,l)=1 \text{ and } k \mid -1 \Rightarrow k= \pm 1 \text{ and }l \mid 1 \Rightarrow l \pm 1$

Therefore,the unique possible roots are $\pm 1$.

So,we conclude that the polynomial $x^7-2x^5+10x^2-1$ has no root greater than 1. (Nerd)

Hi! ;)

It should be: If the polynomial has a rational root, then the polynomial has no rational root greater than 1.

That leaves the real roots... (Sweating)
 
anemone said:
Prove that $x^7-2x^5+10x^2-1$ has no root greater than 1.

This is one of my all time favorite challenge problems! :o
Solution using calculus:
[sp]Let $f(x) = x^7-2x^5+10x^2-1$. Then $f(1) = 8 > 0$. Also, $f'(x) = 7x^6 - 10x^4 + 20x$, so that $f'(1) = 17 > 0.$ If we can show that $f'(x) > 0$ for all $x>1$, then $f(x)$ will increase for all $x>1$ and so can have no zeros there.

Next, $f'(x) = x(7x^5 - 10x^3 + 20)$. If we can show that $g(x) = 7x^5 - 10x^3 + 20 >0$ for all $x>1$, then it will follow that $f'(x) > 0$ for all $x>1$, as required. But $g(1) = 17 > 0$, and $g'(x) = 35x^4 - 30x^2 = 5x^2(7x^2-6)$, which is certainly positive for all $x>1$. Hence $g(x)$ increases for all $x>1$ and is therefore always positive there.[/sp]
That does not strike me as the sort of argument that would qualify as an "all time favourite", so I suspect that there must be a more attractive method of proof.
 
kaliprasad said:
Suppose it has got a raional root.
because the coefficeint of highest power that is x^7 is one then the root has to be a factor of -1 that is 1 or -1 .

so it has no root > 1

The above follows straight from Rational root theorem - Wikipedia, the free encyclopedia

Thanks kaliprasad for your solution. :)

Opalg said:
Solution using calculus:
[sp]Let $f(x) = x^7-2x^5+10x^2-1$. Then $f(1) = 8 > 0$. Also, $f'(x) = 7x^6 - 10x^4 + 20x$, so that $f'(1) = 17 > 0.$ If we can show that $f'(x) > 0$ for all $x>1$, then $f(x)$ will increase for all $x>1$ and so can have no zeros there.

Next, $f'(x) = x(7x^5 - 10x^3 + 20)$. If we can show that $g(x) = 7x^5 - 10x^3 + 20 >0$ for all $x>1$, then it will follow that $f'(x) > 0$ for all $x>1$, as required. But $g(1) = 17 > 0$, and $g'(x) = 35x^4 - 30x^2 = 5x^2(7x^2-6)$, which is certainly positive for all $x>1$. Hence $g(x)$ increases for all $x>1$ and is therefore always positive there.[/sp]

Thanks, Opalg for posting your solution through calculus method, and thanks for participating!

Opalg said:
That does not strike me as the sort of argument that would qualify as an "all time favourite", so I suspect that there must be a more attractive method of proof.
Hehehe, I can't speak for others, but the solution that I'm about to show appeals to me a lot...without further ado, here is what I wanted to share with the community here:

The trick is to substitute $y+1$ for $x$ in the given function (which if we let it as $f(x)=x^7-2x^5+10x^2-1$), to end up with $f(y+1)=y^7+7y^6+19y^5+25y^4+15y^3+11y^2+17y+8$ and no sign change in $f(y+1)$ means the function has no positive roots, which in turn suggest that $f(x)$ has no root greater than 1 and we're done.
 
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