x3 + 2x2 - 8x + 5 = 0 x(2x - 8) + 5 = 0
It is x = -5
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
x2-x3+2x = 0 x(-x2+x+2) = 0 x(-x+2)(x+1) = 0 Points of intersection are: (0, 2), (2,10) and (-1, 1)
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 + 2x2 - 8x + 5 = 0 x(2x - 8) + 5 = 0
It is x = -5
Anything plus 0 is itself, so x3 + 0 is x3.
It has two complex roots.
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
It could be x3 + 2x + 0It could be x3 + 2x + 0It could be x3 + 2x + 0It could be x3 + 2x + 0
x3 + 8 = 0, then x3 = -8, therefore x = -2 -2 x -2 = 4 4 x -2 = -8
x2-x3+2x = 0 x(-x2+x+2) = 0 x(-x+2)(x+1) = 0 Points of intersection are: (0, 2), (2,10) and (-1, 1)
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.
Yes because if 1+0=1 than 0 plus b equals b
x3 + 3x2 - 9x + 5 = 0 has roots of -5,1 and 1. CHECK : x3 + 3x2 - 9x + 5 = (x + 5)(x - 1)(x - 1)