It can be. x^2 + x + 1 is a factor of 2x^2 + 2x + 2
(x2 - x + 1)(x2 + x + 1)
x2 + 7x + 6 = (x + 6) (x + 1)
(x + 1)(x + 1)
(x + 1)(x2 - x + 1)
(x + 2)(x - 1)
(x - 1)(x2 + x - 1)
x3 + x2 + 4x + 4 = (x2 + 4)(x + 1)
x3+x2+x First, factor out the x: x(x2+x+1) Could stop there, but the rest is factor-able using imaginary numbers: x(x+.5(1+i(3).5)(x+.5(1-i(3).5), where i = (-1).5, or sqrt(-1)
x2 + 6x + 5 can be factored into (x+1) (x+5)
x3 + 1 = (x + 1)(x2 - x + 1) The x + 1's cancel out, leaving x2 - x + 1
(x+1)(x+9)
(x-1)(x+4)