How do you factor # x^4 + x^3 + x^2 + x + 1#?

1 Answer
Feb 20, 2016

Answer:

#x^4+x^3+x^2+x+1#

#=(x^2+(1/2+sqrt(5)/2)x+1)(x^2+(1/2-sqrt(5)/2)x+1)#

Explanation:

This quartic has four zeros, which are the non-Real Complex #5#th roots of #1#, as we can see from:

#(x-1)(x^4+x^3+x^2+x+1) = x^5-1#

So if we wanted to factor this polynomial as a product of linear factors with Complex coefficients then we could write:

#x^4+x^3+x^2+x+1#

#=(x-(cos((2pi)/5) + i sin((2pi)/5))) * (x-(cos((4pi)/5) + i sin((4pi)/5))) * (x-(cos((6pi)/5) + i sin((6pi)/5))) * (x-(cos((8pi)/5) + i sin((8pi)/5)))#

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:

#x^4+x^3+x^2+x+1#

#=(x-r_1)(x-1/r_1)(x-r_2)(x-1/r_2)#

#=(x^2-(r_1+1/r_1)x+1)(x^2-(r_2+1/r_2)x+1)#

So let's look for a factorisation:

#x^4+x^3+x^2+x+1#

#=(x^2+ax+1)(x^2+bx+1)#

#=x^4+(a+b)x^3+(2+ab)x^2+(a+b)x+1#

Equating coefficients we find:

#a+b = 1#

#2+ab=1#, so #ab = -1# and #b=-1/a#

Substituting #b=-1/a# in #a+b=1# we get:

#a-1/a = 1#

Hence:

#a^2-a-1 = 0#

Using the quadratic formula, we can deduce:

#a = 1/2 +- sqrt(5)/2#

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:

#x^4+x^3+x^2+x+1#

#=(x^2+(1/2+sqrt(5)/2)x+1)(x^2+(1/2-sqrt(5)/2)x+1)#

If we want to factor further, use the quadratic formula on each of these quadratic factors to find the linear factors with Complex coefficients.