#f(y) = y^5-y^4+y^3-y^2+y-1#
Then #f(1) = 1-1+1-1+1-1 = 0#
So #(y-1)# is a factor.
#f(y)=(y-1)(y^4+y^2+1)#
#y^4+y^2+1# has no linear factors since #y^4+y^2+1 >= 1 > 0# for all #y in RR#
What about quadratic factors?
Suppose #y^4+y^2+1 = (y^2+ay+b)(y^2+cy+d)# for some constants
#a#, #b#, #c# and #d#.
#(y^2+ay+b)(y^2+cy+d)#
#=y^4+(a+c)y^3+(d+ac+b)y^2+(ad+bc)y+bd#
Comparing coefficients, we get:
[1] #a+c = 0#
[2] #d+ac+b = 1#
[3] #ad+bc = 0#
[4] #bd=1#
From [1] we get #c = -a#, so we can rewrite:
[2a] #b+d = a^2 + 1#
[3a] #a(d-b) = 0#
From [3a] we can deduce that #a=0# or #b=d#.
Suppose #a=0#. Then [2a] gives us #b+d = 1#. Together with [4], this gives us #b+1/b=1#, hence #b^2-b+1 = 0# which has no real solutions since its discriminant is #-3#,
The other option is #b=d#. Together with [4], this means #b=d=1# or #b=d=-1#. From [2a] we know #b+d >= 1# so we can discount the possibility #b=d=-1# and focus on #b=d=1#.
From [2a] we get #a^2 = 2 - 1 = 1#, so #a=+-1#
In a rather lengthy way, we have found:
#y^4+y^2+1 = (y^2+y+1)(y^2-y+1)#
So
#f(y) = (y-1)(y^2+y+1)(y^2-y+1)#
