Start by noticing that both #x = 1# and #x = -1# are solutions, so #(x - 1)# and #(x + 1)# are both factors.
Then solve:
#(x - 1)(x + 1)(ax^3 + bx^2 + cx + d) = x^5 - 5x^4 - x^3 + x^2 + 4#
for #a#, #b#, #c# and #d#:
Expanding:
#(x - 1)(x + 1)(ax^3 + bx^2 + cx + d) = (x^2 - 1)(ax^3 + bx^2 + cx + d)#
#= ax^5 + bx^4 + (c - a)x^3 + (d - b)x^2 - cx - d#
Comparing the coefficients of #x^5#, #x^4#, etc., we can quickly see that #a = 1#, #b = -5#, #c = 0# and #d = -4#.
So, thus far we have:
#x^5 - 5x^4 - x^3 + x^2 + 4 = (x - 1)(x + 1)(x^3 - 5x^2 - 4)#.
Differentiating the cubic #x^3 - 5x^2 - 4# gives #3x^2 - 10x#. This is zero for #x = 0# and #x = 10/3#. These are the #x# coordinates of the turning points of the cubic. Substituting each of these two values of #x# into the cubic give negative values, so the cubic has only 1 real root - the other two are complex numbers. Notice that #x^3 - 5x^2 - 4# is negative for #x = 5# and positive for #x = 6#. So the real root lies somewhere between 5 and 6, but it is not a rational number, let alone an integer. So the factorisation above is 'complete'.
