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

The solutions to the equation 0=3x^3-x^2-12x+4 are x= 1/3 ,x=-2,x=2

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

The zeros of 0=3x^3-x^2-12x+4 are x= 1/3 ,x=-2,x=2

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

x = \frac{1}{3}, x = - 2, x = 2

Decimal Notation

x = 0.33333 \dots, x = - 2, x = 2

Solution steps

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

Switch sides

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

Factor

3 x^{3} - x^{2} - 12 x + 4 : (3 x - 1) (x + 2) (x - 2)

(3 x - 1) (x + 2) (x - 2) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

3 x - 1 = 0 or x + 2 = 0 or x - 2 = 0

Solve

3 x - 1 = 0 : x = \frac{1}{3}

Solve

x + 2 = 0 : x = - 2

Solve

x - 2 = 0 : x = 2

The solutions are

x = \frac{1}{3}, x = - 2, x = 2

z^{5} = 163 + 16 i

15 = k^{3}

12 y^{2} - 13 y^{3} = 2 y^{2} - 2 y

x^{813} = 2

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

What are the solutions to the equation 0=3x^3-x^2-12x+4 ?
The solutions to the equation 0=3x^3-x^2-12x+4 are x= 1/3 ,x=-2,x=2
Find the zeros of 0=3x^3-x^2-12x+4
The zeros of 0=3x^3-x^2-12x+4 are x= 1/3 ,x=-2,x=2