Factoring this thing with 6 degree

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The polynomial equation u - u^6 - u^3 + 1 = 0 can be factored using the known roots u = 1 and u = -1, leading to the factors (u + 1) and (u - 1). While there is no simple rule for factoring this polynomial, synthetic division can streamline the process. The polynomial can be expressed as (u^2 - 1)(-u^4 - u^2 - u - 1), indicating that the quartic component does not have rational roots but may have complex ones. Ultimately, the primary roots of the original equation are u = -1 and u = 1, while the remaining quartic can be left as is for factoring purposes.
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Homework Statement



I need to factor this:

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


Homework Equations





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

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.
 
Ah okay.. so there is no shortcut :(
 
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.
 
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?
 
zeion said:
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?

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.
 
Oh okay, thanks!
 
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