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)
