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

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

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

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

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

x = 3, x = 2 i, x = - 2 i

Solution steps

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

Switch sides

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

Factor

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

(x - 3) (x^{2} + 4) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

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

Solve

x - 3 = 0 : x = 3

Solve

x^{2} + 4 = 0 : x = 2 i, x = - 2 i

The solutions are

x = 3, x = 2 i, x = - 2 i

2 x^{3} + 6 x^{2} + 5 x + 15 = 0

- 9 m^{3} + 18 m^{2} - 8 m = 0

x, (x - 3)^{3} = 8 (y - 2)

x^{3} + 3 x^{2} + 3 x + 9 = 0

x^{4} - 9 x^{2} - 22 = 0

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