x3 + 2x2 - 8x + 5 = 0 x(2x - 8) + 5 = 0
It is x = -5
Suppose x3-4x = 0. To solve, factor: x3-4x = x(x2-4) = x(x+2)(x-2) = 0 Now, a product equals 0 if and only one or more of the factors equals 0, so set each factor to 0 and solve. The roots are 0,-2 and +2.
x3 - 11x2 + 18x = 0x (x2 - 11x + 18) = 0 . . . . x = 0 is a root.x (x - 9) ( x - 2) = 0 . . . . x = 9 and x = 2 are roots.X = 0, 2, 9
3x3 -3x2 - 48x - 60 = 0 simply factor out a 3... 3(x3 - x2 - 16x - 20) = 0
x3 + 2x2 - 8x + 5 = 0 x(2x - 8) + 5 = 0
It is x = -5
x5 - 2x4 - 24x3 = 0 x3(x2 - 2x - 24) = 0 x3(x2 + 4x - 6x - 24) = 0 x3(x - 6)(x + 4) = 0 x = {-4, 0, 6}
Suppose x3-4x = 0. To solve, factor: x3-4x = x(x2-4) = x(x+2)(x-2) = 0 Now, a product equals 0 if and only one or more of the factors equals 0, so set each factor to 0 and solve. The roots are 0,-2 and +2.
x3 - 11x2 + 18x = 0x (x2 - 11x + 18) = 0 . . . . x = 0 is a root.x (x - 9) ( x - 2) = 0 . . . . x = 9 and x = 2 are roots.X = 0, 2, 9
x3=x2 x3-x2=0 Factor out x2 x2(x - 1) = 0 If x2=0 then x=0. If x - 1 = 0 then x = 1. Therefore the number is 0 or 1.
It has two complex roots.
x3 - 3x2 - 9x - 5 = 0 (x + 1)(x2 - 4x - 5) = 0 (x + 1)(x - 5)(x + 1) = 0 x ∈ {-1, 1, 5}
It could be x3 + 2x + 0It could be x3 + 2x + 0It could be x3 + 2x + 0It could be x3 + 2x + 0
Anything plus 0 is itself, so x3 + 0 is x3.
3x3 -3x2 - 48x - 60 = 0 simply factor out a 3... 3(x3 - x2 - 16x - 20) = 0
x3 + 8 = 0, then x3 = -8, therefore x = -2 -2 x -2 = 4 4 x -2 = -8