1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Factoring this thing with 6 degree

  1. Feb 1, 2010 #1
    1. The problem statement, all variables and given/known data

    I need to factor this:

    u - u^6 -u^3 +1 = 0

    2. Relevant equations

    3. The attempt at a solution

    I know that (u +1) and (u-1) are roots.. but not sure what to do now without long division..
    Do I put them into multiples like (u^3 + 1)(u^3 -1) ??
  2. jcsd
  3. Feb 1, 2010 #2


    User Avatar
    Science Advisor
    Homework Helper

    (u^3 + 1)(u^3 -1) is wrong. What's the problem with long division?
  4. Feb 1, 2010 #3
    It takes too long.. I was just wondering if there was simply rule I could follow when it's in this kind of form?
  5. Feb 1, 2010 #4


    User Avatar
    Science Advisor
    Homework Helper

    No, no simple rule. You know (u-1) and (u+1) are factors, but you don't how many times they divide into your original polynomial. In this case, the answer for both is once. You just have to divide them out. If you want to minimize the work a bit you could divide by (u-1)*(u+1)=u^2-1. Saves you dividing by both separately.
  6. Feb 1, 2010 #5
    Ah okay.. so there is no shortcut :(
  7. Feb 1, 2010 #6


    Staff: Mentor

    There is a sort of shortcut, synthetic division. You have to actually do the division, but you don't have to write down all the x powers. You might have learned it already and forgotten it.
  8. Feb 2, 2010 #7
    So I got (u^2-1)(-u^4-u^2-u-1) = -(u+1)(u-1)(u^4+u^2+u+1)
    Does that mean there are roots only at u = -1 and u = 1?
  9. Feb 2, 2010 #8


    User Avatar
    Science Advisor
    Homework Helper

    The quartic doesn't have any rational roots. It has complex roots, of course. But as far as a factoring exercise, I think you can just leave it there.
  10. Feb 2, 2010 #9
    Oh okay, thanks!
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook