What are the solutions to the equation 2x^3-3x^2+2x=-4x^3+10x-4 ?

The solutions to the equation 2x^3-3x^2+2x=-4x^3+10x-4 are x= 1/2 ,x=-2/(sqrt(3)),x= 2/(sqrt(3))

Find the zeros of 2x^3-3x^2+2x=-4x^3+10x-4

The zeros of 2x^3-3x^2+2x=-4x^3+10x-4 are x= 1/2 ,x=-2/(sqrt(3)),x= 2/(sqrt(3))

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2 x^{3} - 3 x^{2} + 2 x = - 4 x^{3} + 10 x - 4

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

Decimal Notation

x = 0.5, x = - 1.15470 \dots, x = 1.15470 \dots

Solution steps

2 x^{3} - 3 x^{2} + 2 x = - 4 x^{3} + 10 x - 4

Move

4

to the left side

2 x^{3} - 3 x^{2} + 2 x + 4 = - 4 x^{3} + 10 x

Move

10 x

to the left side

2 x^{3} - 3 x^{2} - 8 x + 4 = - 4 x^{3}

Move

4 x^{3}

to the left side

6 x^{3} - 3 x^{2} - 8 x + 4 = 0

Factor

6 x^{3} - 3 x^{2} - 8 x + 4 : (2 x - 1) (3 x + 2) (3 x - 2)

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

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

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

Solve

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

Solve

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

Solve

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

The solutions are

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

x^{4} - 6 x^{2} - 1 = 0

x^{5} = (x - 1)^{5}

0 = 5 x - x^{3}

x^{3} + x^{2} - 333 x - 1002 = 0

x^{4} - x^{3} - 11 x^{2} + 9 x + 18 = 0

What are the solutions to the equation 2x^3-3x^2+2x=-4x^3+10x-4 ?
The solutions to the equation 2x^3-3x^2+2x=-4x^3+10x-4 are x= 1/2 ,x=-2/(sqrt(3)),x= 2/(sqrt(3))
Find the zeros of 2x^3-3x^2+2x=-4x^3+10x-4
The zeros of 2x^3-3x^2+2x=-4x^3+10x-4 are x= 1/2 ,x=-2/(sqrt(3)),x= 2/(sqrt(3))