What are the solutions to the equation -4x^3-4x=0 ?

The solutions to the equation -4x^3-4x=0 are x=0,x=i,x=-i

Find the zeros of -4x^3-4x=0

The zeros of -4x^3-4x=0 are x=0,x=i,x=-i

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- 4 x^{3} - 4 x = 0

x = 0, x = i, x = - i

Solution steps

- 4 x^{3} - 4 x = 0

Factor

- 4 x^{3} - 4 x : - 4 x (x^{2} + 1)

- 4 x (x^{2} + 1) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

x = 0 or x^{2} + 1 = 0

Solve

x^{2} + 1 = 0 : x = i, x = - i

The solutions are

x = 0, x = i, x = - i

2 y^{3} + y^{2} - 13 y - 6 y = \frac{2}{3}

(x + 1) (- 3 x^{2} + 9 x + 30) = 0

(x^{2} - x) + 1 = (x^{3} - 2)

x, 2 f - f (\frac{1}{x}) = ln (x^{3})

x^{102} - 5 = 2

What are the solutions to the equation -4x^3-4x=0 ?
The solutions to the equation -4x^3-4x=0 are x=0,x=i,x=-i
Find the zeros of -4x^3-4x=0
The zeros of -4x^3-4x=0 are x=0,x=i,x=-i