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

[anemone]

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!

 

[evinda]

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!

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)

 

[anemone]

I'm sorry evinda, I'm afraid I don't follow your reasoning...but, thanks for attempting to solve this challenge...:)

 

[kaliprasad]

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!

Spoiler

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

 

[I like Serena]

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)

 

[Opalg]

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!

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.

 

[anemone]

Spoiler

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. :)

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!

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:

Spoiler

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.