What are the solutions to the equation 6x^2+3x^3=-x(2x^3-4) ?

The solutions to the equation 6x^2+3x^3=-x(2x^3-4) are x=0,x= 1/2 ,x=-1+sqrt(3)i,x=-1-sqrt(3)i

Find the zeros of 6x^2+3x^3=-x(2x^3-4)

The zeros of 6x^2+3x^3=-x(2x^3-4) are x=0,x= 1/2 ,x=-1+sqrt(3)i,x=-1-sqrt(3)i

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

x = 0, x = \frac{1}{2}, x = - 1 + 3 i, x = - 1 - 3 i

Solution steps

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

Switch sides

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

Move

3 x^{3}

to the left side

- x (2 x^{3} - 4) - 3 x^{3} = 6 x^{2}

Move

6 x^{2}

to the left side

- x (2 x^{3} - 4) - 3 x^{3} - 6 x^{2} = 0

Factor

- x (2 x^{3} - 4) - 3 x^{3} - 6 x^{2} : - x (2 x - 1) (x^{2} + 2 x + 4)

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

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

x = 0 or 2 x - 1 = 0 or x^{2} + 2 x + 4 = 0

Solve

2 x - 1 = 0 : x = \frac{1}{2}

Solve

x^{2} + 2 x + 4 = 0 : x = - 1 + 3 i, x = - 1 - 3 i

The solutions are

x = 0, x = \frac{1}{2}, x = - 1 + 3 i, x = - 1 - 3 i

7 x^{3} + 12 x^{2} - 11 x + 2 = 0

s^{4} + 3 s^{3} + 70 s^{2} + 2 s + 10 = 0

\frac{27}{8} = x^{3}

x^{4} - 2 x^{3} - 4 x^{2} + 5 x - 6 = 0

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

What are the solutions to the equation 6x^2+3x^3=-x(2x^3-4) ?
The solutions to the equation 6x^2+3x^3=-x(2x^3-4) are x=0,x= 1/2 ,x=-1+sqrt(3)i,x=-1-sqrt(3)i
Find the zeros of 6x^2+3x^3=-x(2x^3-4)
The zeros of 6x^2+3x^3=-x(2x^3-4) are x=0,x= 1/2 ,x=-1+sqrt(3)i,x=-1-sqrt(3)i