2011/063) factor x^5 + x^4 + 1

This is a qunitic polynomial for which there is no direct method
(
one of the methods commonly used is given below at the end but a short cut can be used

(since using the Rational Root Theorem implies that the only possible rational roots are -1 and 1, and neither (x - 1) nor (x + 1) are factors).

as it is of the form x^(3n+2) + x^(3m+1) + 1 so
x = w and x = w^2 are zeros ( w is cube root of 1)
(x-w)(x-w^2) or (x^2 + x + 1) is a factor and by division

x^5 + x^4 + 1 = (x^2+x+1)(x³ - x + 1)


The common method
It's easy to check that x^5 + x^4 + 1 has no linear factors (since using the Rational Root Theorem implies that the only possible rational roots are -1 and 1, and neither (x - 1) nor (x + 1) are factors).

So, if x^5 + x^4 + 1 factors, then it must be the product of a quadratic factor and a cubic factor.
Hence, we assume that x^5 + x^4 + 1 = (x² + Ax + B)(x³ + Cx² + Dx + E).

Expanding yields
x^5 + x^4 + 1 = x^5 + (A + C)x^4 + (B + AC + D)x³ + (AD + BC + E)x² + (BD + AE)x + BE

Equate like coefficients:
A + C = 1
B + AC + D = 0
AD + BC + E = 0
BD + AE = 0
BE = 1.

Note that C = 1 - A and E = 1/B.
We can say more; since these variables should be integers, we know that B = E = ±1.

Thus, we obtain
±1 + A(1 - A) + D = 0
AD ± (1 - A) ± 1 = 0
±D ± A = 0

Let's look at the +1 case (it turns out that we don't need the -1 case...)
1 + A(1 - A) + D = 0
AD + (1 - A) + 1 = 0
D + A = 0

Since D = -A, we have
1 + A(1 - A) - A = 0 ==> A^2 - 1 = 0
-A^2 + (1 - A) + 1 = 0 ==> A^2 + A - 2 = 0.
These two equations are both true when A = 1.

Hence assuming that B = E = 1, we have A = 1, D = -1 ==> C = 0.
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So, we have the factorization x^5 + x^4 + 1 = (x² + x + 1)(x³ - x + 1).)