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

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

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

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

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

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

Solution steps

- 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

d^{3} \cdot 1 = 512 \cdot 1

y^{3} + 6 y^{2} - 7 y = 8

x^{4} - 2 x^{2} - 35 = 0

x^{3} - 2 x^{2} - 2 x + 1 = 0

x^{9} + 3 = 11

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