Polynomial factorization methodology

[olspookishmagus]

Hello everyone and special greetings to freemathhelp.com regulars.

I would like to state that this post could fit in some of the other sub-forums but I don't know where should it be I posted it here. If any moderator is concerned and if this is an inappropriate sub-forum please move it where it should be.

I need to get some methodoly and tips on how to get polynomials factorized and especially help with:
x⁵+x+1

Thanks in advance.

 

[stapel]

In general, you start with the Rational Roots Test, to get a list of possible zeroes.

Then you run those "possibles" through synthetic division (perhaps narrowing your list by using Descartes' Rule of Signs).

With luck, you can find enough rational (fractional and/or integral) zeroes to get yourself down to a quadratic, to which you can apply the Quadratic Formula.

If the above methodology doesn't work, then either you'll need some sort of specialized-to-the-polynomial "trick", some astounding good luck, or else numerical methods.

Eliz.

Note: The above applies in general. I haven't checked whether it applies to the specific polynomial posited.

 

[Gene]

I don't see any magic to apply to it. Have you learned about Newtons method? It is an iterative equation

x₁ = x₀ - f(x₀)/f'(x₀)
You make a guess for x₀ then use that and do it over untill the answer is close enough.

x₁=x₀-(x₀^5+x₀+1)/(5x₀^4+1)
Starting with x₀=-1
x₁ = -1-(-1/6) = -5/6 = -.8333333
x₂ = -.764382
x₃ = -.755025
x₄ = -.754878
That is correct to 6 places.

 

[olspookishmagus]

The example I posted is solved to:
x⁵ + x + 1 = (x² + x + 1)(x³ - x² + 1)

My question is how we get to that solution, step by step.

 

[stapel]

I think this would be a "lucky" case or a "trick" case, because the usual methods don't apply.

Eliz.

 

[olspookishmagus]

There must be a way, a methodology that even when these exercises pop-up - one will be able to factorize them.

 

[pka]

What is the possible point in factoring x<SUP>5</SUP>+x+1?
Where in the world would “these exercises pop-up”?

 

[olspookishmagus]

In my exercises book?
You want that page scanned and uploaded somewhere?

 

[pka]

With a lifetime of experience is mathematics; I can tell you those excises are an absolute waste of your time. Spend you time learning why one factors in the first place. Do you know and fully understand the Factor/Root theorem?

 

[olspookishmagus]

I don't feel like continuing this thread any more.

One says there is no need to bother with this exercise, another says that where on earth would one find these exercises the other that there is no method to get this factorized. I feel like that if I keep asking will get me nowhere.

If someone however wishes to contribute something useful feel free to do so this is only what will make me post a reply to this thread again.

 

[soroban]

Hello,olspookishmagus!

There must be a way, a methodology that even when these exercises pop-up -
one will be able to factorize them.

I agree . . . yet there may be hundreds of methods that we're not aware of.
I just "invented" a method for this problem (but only because I already knew the answer).

We have: .\(\displaystyle x^5\,+\,x\,+\,1\)

Multiply by \(\displaystyle (x\,-\,1)\)*:

. . . \(\displaystyle (x\,-\,1)(x^5\,+\,x\,+\,1)\:=\:x^6\,-\,x^5\,+\,x^2\,-\,1\:=\:x^6\,-\,1\,-\,x^5\,+\,x^2\)

Factor: .\(\displaystyle (x^3\,-\,1)(x^3\,+\,1)\,-\,x^2(x^3\,-\,1)\)

Factor: .\(\displaystyle (x^3\,-\,1)(x^3\,+\,1\,-\,x^2)\)

Factor: .\(\displaystyle (x\,-\,1)(x^2\,+\,x\,+\,1)(x^3\,-\,x^2\,+\,1)\)

Divide by \(\displaystyle (x\,-\,1):\;\;(x^2\,+\,x\,+\,1)(x^3\,-\,x^2\,+\,1)\) . . . ta-DAA!

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

* . Why?

I recognized the quadratic factor from: .\(\displaystyle x^3\,-\,1\:=\x\,-\,1)(x^2\,+\,x\,+\,1)\)

I knew that multiplying the polynomial by \(\displaystyle (x-1)\)
. . would produce a factor of \(\displaystyle x^3\,-\,1\).

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

Don't be put off by the cynicism.

I too have had a "lifetime of experience in mathematics",
having retired from 38 years of teaching college-level mathematics.
Yet I found it to be a rewarding "learning experience".

 

[stapel]

There must be a way....one will be able to factorize them....I don't feel like continuing this thread any more.

You asked for the standard methodology. This was provided to you. You asked for the method for doing your specific exercise. It was explained that there is no standard method for such an exercise. You appear now to be upset that we won't just tell you what the method is.

I'm sorry that you didn't like hearing the answer to your question, but I'm afraid your distaste is unlikely to change the mathematical facts. It has been proven that there is, unfortunately, no general method for factoring (or solving) polynomials of the type you have posted.

Eliz.

 

[olspookishmagus]

I do not controverst that there is no general method or that there are some special exercises which cannot be solved by general methods. What I wanted to know is the way some more experienced mathematicians would try to work this out. Of course the best way to have this questions answered is to learn by experience but I needed a start.

Soroban:
Thanks a lot!

And thanks a lot to everyone else for their time.

 

[Denis]

The example I posted is solved to:
x⁵ + x + 1 = (x² + x + 1)(x³ - x² + 1)

Hey spooky, why don't you play around with something similar to that
(x^2 + x + 1)(x^3 - x^2 + 1), multiply it out, then ask your teacher
to factorize what the multiplication turned out to be; should be fun :twisted:

 

[galactus]

When I plugged this expression into my TI-92 it gave me the factorization.

It makes me wonder how the program in the calculator works to arrive at the

answer. Someone had to write it.