What are the solutions to the equation 0=x^3+2x^2-4x-8 ?

The solutions to the equation 0=x^3+2x^2-4x-8 are x=-2,x=2

Find the zeros of 0=x^3+2x^2-4x-8

The zeros of 0=x^3+2x^2-4x-8 are x=-2,x=2

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0 = x^{3} + 2 x^{2} - 4 x - 8

x = - 2, x = 2

Solution steps

0 = x^{3} + 2 x^{2} - 4 x - 8

Switch sides

x^{3} + 2 x^{2} - 4 x - 8 = 0

Factor

x^{3} + 2 x^{2} - 4 x - 8 : (x + 2)^{2} (x - 2)

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

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

x + 2 = 0 or x - 2 = 0

Solve

x + 2 = 0 : x = - 2

Solve

x - 2 = 0 : x = 2

The solutions are

x = - 2, x = 2

x, y + 3 = x^{3}

x, y^{4} = - x^{4}

x^{3} - 5 x^{2} - 50 x = 0

s^{3} + 132 s^{2} + 1540 s + 12000 = 0

3 x^{36} - 140 x + 137 = 0

What are the solutions to the equation 0=x^3+2x^2-4x-8 ?
The solutions to the equation 0=x^3+2x^2-4x-8 are x=-2,x=2
Find the zeros of 0=x^3+2x^2-4x-8
The zeros of 0=x^3+2x^2-4x-8 are x=-2,x=2