My Math Forum  

Go Back   My Math Forum > High School Math Forum > Algebra

Algebra Pre-Algebra and Basic Algebra Math Forum


Thanks Tree1Thanks
  • 1 Post By skipjack
Reply
 
LinkBack Thread Tools Display Modes
February 23rd, 2016, 12:22 PM   #1
Member
 
Joined: Dec 2015
From: Europe

Posts: 42
Thanks: 1

Hard factorization problem

How can you factor $\displaystyle k^{10}+k^5+1$ to this beautiful form:
$\displaystyle (k^2+k+1)(k^8-k^7+k^5-k^4+k^3-k+1)$

Last edited by greg1313; February 23rd, 2016 at 12:48 PM. Reason: Corrected latex for k^10
Inflekx12 is offline  
 
February 23rd, 2016, 04:36 PM   #2
Global Moderator
 
Joined: Dec 2006

Posts: 20,469
Thanks: 2038

$\displaystyle \begin{align*}k^{10} + k^5 + 1 &= \frac{k^{15} - 1}{k^3 - 1}\cdot \frac{k^3 - 1}{k^5 - 1} \\
&= (k^{12} + k^9 + k^6 + k^3 + 1)\cdot \frac{k^2 + k + 1}{k^4 + k^3 + k^2 + k + 1} \\
&= (k^8 - k^7 + k^5 - k^4 + k^3 - k + 1)(k^2 + k + 1)\end{align*}$

The longer factor = $\displaystyle (k^2 - 2\cos(24°)k + 1)(k^2 - 2\cos(48°)k + 1)(k^2 - 2\cos(96°)k + 1)(k^2 - 2\cos(192°)k + 1)$.
Thanks from topsquark
skipjack is offline  
Reply

  My Math Forum > High School Math Forum > Algebra

Tags
factorization, hard, problem



Search tags for this page
Click on a term to search for related topics.
Thread Tools
Display Modes


Similar Threads
Thread Thread Starter Forum Replies Last Post
Integer factorization problem Alexander I Number Theory 0 December 20th, 2015 08:18 PM
Hard problem Ionika Algebra 0 February 5th, 2014 11:18 AM
Interesting Factorization Problem mathbalarka Algebra 13 October 25th, 2013 08:24 PM
Factorization problem mathbalarka Number Theory 5 April 21st, 2012 07:59 AM
Very hard problem. tenki Calculus 1 February 6th, 2010 04:17 PM





Copyright © 2019 My Math Forum. All rights reserved.