Factoring polynomials of higher degree

MAthBeg

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Hello,

I want to take advice of factoring polynomials. I got a homework, but I don't know how to factor ?:confused:
Homework: Derive the formula for factoring the following polynomials:
P2n(x)=x2n+1
P2n(x)=x2n-1
P2n+1(x)=x2n+1+1
P2n+1(x)=x2n+1-1.

Could someone help me ?

I try to finding roots of these polynomials and plot them with program MATLAB for 2n=16, 32, 64 and 128. I have these results now from polynomial y1=x16+1, y2=x32+1, y3=x64+1 and y4=x128+1 :
attachment.php

I don't know how to find general solution '' by hand '' ?
 

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Homework: Derive the formula for factoring the following polynomials:
P2n(x)=x2n+1
P2n(x)=x2n-1
P2n+1(x)=x2n+1+1
P2n+1(x)=x2n+1-1.
Are they really using the exact same notation for two different functions? :shock:

I try to finding roots of these polynomials and plot them with program MATLAB for 2n=16, 32, 64 and 128. I have these results now from polynomial y1=x16+1, y2=x32+1, y3=x64+1 and y4=x128+1...
I don't know how to find general solution '' by hand '' ?
What, in algebraic terms (rather than in pictures), are the factorizations (rather than the solutions)? Also, how are you being required to use calculus when doing this algebra? ;)
 
Thank you for reply.

The polynomials assign in my homework:
P2n(x)=x2n±1
P2n+1(x)=x2n+1±1
.

I plot the roots of polynomial P2n(x)=x2n+1. But I don't know how to get factorization process without mathematical software ?

For example the equation x2-x=12. The factor of this equation is: (x-4)*(x+3)=0. Roots are x=4 and x=-3. It is easy, but in this case (P2n(x)=x2n+1) ?


eq0015M.gif
 
The polynomials assign in my homework:
P2n(x)=x2n±1
P2n+1(x)=x2n+1±1
.
Thank you; that's very helpful information.

I plot the roots of polynomial P2n(x)=x2n+1. But I don't know how to get factorization process without mathematical software ?

For example the equation x2-x=12.
Solving an equation is quite different from factoring an expression. You are not "finding the solutions to P2n = m for some integer m"; you are factorizing P2n. So let's not confuse things by introducing different topics. :oops:

By the way, with respect to the point plots you provided earlier: Are those graphs of the solutions of P2n = 0 for various values of n?

Using software, what did you get as the factorizations of the following?

. . .\(\displaystyle P_{2_+}\, =\, x^2\, +\, 1\)

. . .\(\displaystyle P_{4_+}\, =\, x^4\, +\, 1\)

. . .\(\displaystyle P_{6_+}\, =\, x^6\, +\, 1\)

. . .\(\displaystyle P_{8_+}\, =\, x^8\, +\, 1\)

. . .\(\displaystyle P_{10_+}\, =\, x^{10}\, +\, 1\)

(Note that, for every n which is a multiple of 3, you'll be getting something which, at least in part, can be viewed as factoring a sum of cubes.)

What do you get for the factorizations of these?

. . .\(\displaystyle P_{2_-}\, =\, x^2\, -\, 1\)

. . .\(\displaystyle P_{4_-}\, =\, x^4\, -\, 1\)

. . .\(\displaystyle P_{6_-}\, =\, x^6\, -\, 1\)

. . .\(\displaystyle P_{8_-}\, =\, x^8\, -\, 1\)

. . .\(\displaystyle P_{10_-}\, =\, x^{10}\, -\, 1\)

Your answers for the above should be products of factors, not lists (or graphs) of points. Thank you! ;)
 
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