What are the solutions to the equation x(3x^2-2x-12x+8)=0 ?

The solutions to the equation x(3x^2-2x-12x+8)=0 are x=0,x= 2/3 ,x=4

Find the zeros of x(3x^2-2x-12x+8)=0

The zeros of x(3x^2-2x-12x+8)=0 are x=0,x= 2/3 ,x=4

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x (3 x^{2} - 2 x - 12 x + 8) = 0

x = 0, x = \frac{2}{3}, x = 4

Decimal Notation

x = 0, x = 0.66666 \dots, x = 4

Solution steps

x (3 x^{2} - 2 x - 12 x + 8) = 0

Factor

x (3 x^{2} - 2 x - 12 x + 8) : x (- 2 + 3 x) (x - 4)

x (- 2 + 3 x) (x - 4) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

x = 0 or - 2 + 3 x = 0 or x - 4 = 0

Solve

- 2 + 3 x = 0 : x = \frac{2}{3}

Solve

x - 4 = 0 : x = 4

The solutions are

x = 0, x = \frac{2}{3}, x = 4

x, x^{3} + y^{3} + z^{3} = 1

2 x^{3} + 7 x^{2} - 4 x = 0

x^{3} + 5 x^{2} + 8 x + 3 = 0

x, 9 x^{4} + 6 y^{3} = 8

x^{2} (x - 1) (2 x + 7) = 0

What are the solutions to the equation x(3x^2-2x-12x+8)=0 ?
The solutions to the equation x(3x^2-2x-12x+8)=0 are x=0,x= 2/3 ,x=4
Find the zeros of x(3x^2-2x-12x+8)=0
The zeros of x(3x^2-2x-12x+8)=0 are x=0,x= 2/3 ,x=4