How do you factor #x^5 - 5x^4 - x^3 + x^2 + 4 = 0#?
There are no simpler factors with rational coefficients.
First note that the sum of the coefficients is zero.
Reversing the signs on the coefficients of odd degree we also find:
#-1-5+1+1+4 = 0#
#x^5-5x^4-x^3+x^2+4 = (x^2-1)(x^3-5x^2-4)#
By the rational root theorem, the only possible rational zeros of the remaining cubic factor
None of these work, so this cubic only has factors with irrational and/or Complex coefficients.
If this question was asked in Precalculus, I might offer a solution of the cubic, but it's probably beyond the scope of your Algebra course, so let's leave the factoring at this point.