x3 + 2x2 - 8x + 5 = 0 x(2x - 8) + 5 = 0
x3 + 2x2 - 35x = x(x + 7)(x - 5)
It is x = -5
x(x+3)(x+5)
(x+1)(x+5)(x+2)
It has two complex roots.
10x + 5 unless you know what x & y are * * * * * Perhaps this answer will be of help: x3 + x2 + 5x + 5 = (x + 1)(x2 + 5). If you are willing to go to complex roots, then, x3 + x2 + 5x + 5 = (x + 1)(x2 + 5). = (x + 1)(x + i√5)(x - i√5); in which case, x = -1 or ±i√5.
x3 + 2x2 - 8x + 5 = 0 x(2x - 8) + 5 = 0
Factor out a 2. 2(x3 - 21x + 20) That factors to 2(x - 1)(x - 4)(x + 5) The roots are -5, 1 and 4
x3 + 2x2 - 35x = x(x + 7)(x - 5)
No, it is not.
It is x = -5
x(x + 8)(x + 5)
x(x+3)(x+5)
(x+1)(x+5)(x+2)
x3 + 5x2 - x - 5 = (x2 - 1)(x + 5) = (x + 1)(x - 1)(x + 5)
They are -1, 1 and 5.