What are the solutions to the equation 0=x^3-4x^2-12x ?

The solutions to the equation 0=x^3-4x^2-12x are x=0,x=-2,x=6

Find the zeros of 0=x^3-4x^2-12x

The zeros of 0=x^3-4x^2-12x are x=0,x=-2,x=6

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0 = x^{3} - 4 x^{2} - 12 x

x = 0, x = - 2, x = 6

Solution steps

0 = x^{3} - 4 x^{2} - 12 x

Switch sides

x^{3} - 4 x^{2} - 12 x = 0

Factor

x^{3} - 4 x^{2} - 12 x : x (x + 2) (x - 6)

x (x + 2) (x - 6) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

x = 0 or x + 2 = 0 or x - 6 = 0

Solve

x + 2 = 0 : x = - 2

Solve

x - 6 = 0 : x = 6

The solutions are

x = 0, x = - 2, x = 6

2 a^{4} - 15 a^{2} + 28 = 0

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

5 x^{4} - 1 2^{3} = 0

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

π \cdot r^{4} = 4.86 \cdot 1 0^{- 7}

What are the solutions to the equation 0=x^3-4x^2-12x ?
The solutions to the equation 0=x^3-4x^2-12x are x=0,x=-2,x=6
Find the zeros of 0=x^3-4x^2-12x
The zeros of 0=x^3-4x^2-12x are x=0,x=-2,x=6