x3 + x = -2 => x3 + x + 2 = 0 => x3 + x2 - x2 - x + 2x - 2 = 0 => x2(x+1) - x(x+1) + 2(x+1) = 0 => (x+1)*(x2-x+2) = 0 Setting the first bracket equal to zero gives the only real solution, which is x = -1 The second bracket gives the complex roots, x = ½*[1 +or- i*sqrt(7)]
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)
x is approximately 0.7549
A number multiplied by itself thrice. 2 x 2 x 2 equals 2 cubed.
x^3 + y^3 = (x + y)(x^2 - xy + y^2)
(x + 6)(x^2 - 6x + 36)
x+2) by Algebraica Long Division. x^(3) + 3x^(2) + 3x + 2 ) = x^(2) + x + 1 -x^(3)- 2x^(2) ¬ 0 + x^(2) + 3x 0 - x^(2) - 2x ¬ 0 + 0 + x + 2 0 + 0 - x - 2 ¬ = 0 Hence answer is x^)2) + x + 1 ( Thos does NOT factor .
2x4
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)
x is approximately 0.7549
x3 + 8 = 0, then x3 = -8, therefore x = -2 -2 x -2 = 4 4 x -2 = -8
-2, 1.74 and 0.46
A number multiplied by itself thrice. 2 x 2 x 2 equals 2 cubed.
x^3 + y^3 = (x + y)(x^2 - xy + y^2)
x(x^2 + 1)
x 3-8
x = ¼x^4 + 4x + k where k can be any number
2(x^2 + 2)(x + 3)