x3 + x = x2(x + 1)
x3 + 1 = (x + 1)(x2 - x + 1) The x + 1's cancel out, leaving x2 - x + 1
x3 - 2x2 + x - 2 =(x - 2)(x2 + 1)
x3 - 3x2 + x - 3 = (x - 3)(x2 + 1)
x3 + x2 + 4x + 4 = (x2 + 4)(x + 1)
If: x3+1 = 65 Then: x3 = 65-1 And: x3 = 64 So: x = 4 by means of the cube root function on the calculator
x3 + 4x2 + x + 4 = (x + 4)(x2 + 1)
x3 + 13x2 + 42x = x(x + 6)(x + 7).
No. Although it is increasing most of the time, it is decreasing between x=-1 and x=1.
The cubic function.
x3 + 2x2 - 35x = x(x + 7)(x - 5)
Anything plus 0 is itself, so x3 + 0 is x3.
4
(-x3 + 75x - 250) / (x + 10) = x2 - 10x - 25
x3 - x2 + x - 2 has no rational factors.
x(x + 8)(x + 5)
x(x+3)(x+5)