## Friday, August 21, 2015

### Prove that $x^7-2x^5+10x^2-1$ has no root greater than 1 (First Post)

Prove that $x^7-2x^5+10x^2-1$ has no root greater than 1.

This is one thought provoking problem, as it presents the golden opportunity for one to think hard and long so they understand the problem more well and that eventually will improve one's reasoning skill.

You might not believe that is the case, where $x^7-2x^5+10x^2-1$ has no root greater than 1. You might even wonder if that polynomial has other real roots. In case of doubt, investigate it!

Remember we have the top notch information technology online that could assist us in finding the truth and all that we need to do is to type the function into the program and let's us what curve does it represent.

graph of y=x^7-2x^5+10x^2-1

Asking wolfram to solve the polynomial function gives us the three real roots:

solutions of x^7-2x^5+10x^2-1=0

$x\approx -1.8681,\,-0.3152,\,0.3172$

So yeap! $x^7-2x^5+10x^2-1$ has no root greater than 1.

The question is, how do we prove that?

In order to begin to work with the problem, we need a plan. If you haven't the faintest idea of how to approaching the problem, that means you're not fully understand the situation and the problem.

Is the idea to show the polynomial $x^7-2x^5+10x^2-1$ has only three real roots, and work them out a great one?

Well, it is at least an idea. I recall when I was studying in the local university, a female Professor, who taught me the subject of "Crystallography" (Crystallography is the experimental science of determining the arrangement of atoms in the crystalline solids), she wanted us to speak up our ideas in her classes and she would throw us many questions in the classes and she would then pick us randomly to answer the questions offhand. During that time, there was nothing more scarier than attending her lectures but I must admit that she did it right, we would only focus if we are "pushed" in that direction.

Now, back to our problem, yes, you could try in light of what you have suggested, but whether it requires much more work than the original question is asking is another story, I will let you tell me that.

I will also post the solution to this problem in the next blog post, before reading my solution, I hope you could at least try to attempt at a solution.