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How do you factor # x^4 + x^3 + x^2 + x + 1#?
1 Answer
Explanation:
This quartic has four zeros, which are the non-Real Complex #5# th roots of #1# , as we can see from:
So if we wanted to factor this polynomial as a product of linear factors with Complex coefficients then we could write:
Feeder 3 6 4 X 24
A cleaner algebraic approach is to notice that due to the symmetry of the coefficients, if #x=r# is a zero of #x^4+x^3+x^2+x+1# , then #x=1/r# is also a zero.
Hence there is a factorisation in the form:
So let's look for a factorisation:
Equating coefficients we find:
Substituting #b=-1/a# in #a+b=1# we get:
Omnigraffle pro 7 4 3 download free. #a-1/a = 1#
Hence:
Using the quadratic formula, we can deduce:
Feeder 3 6 4 X 2 3 4
Since our derivation was symmetric in #a# and #b# , one of these roots can be used for #a# and the other for #b# , to find:
Feeder 3 6 4 X 25
Feeder 3 6 4 X 26
If we want to factor further, use the quadratic formula on each of these quadratic factors to find the linear factors with Complex coefficients.