What are the solutions to the equation (x+2)3=2x(x^2-1) ?

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

Find the zeros of (x+2)3=2x(x^2-1)

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

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(x + 2) 3 = 2 x (x^{2} - 1)

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

Solution steps

(x + 2) \cdot 3 = 2 x (x^{2} - 1)

Expand

(x + 2) \cdot 3 : 3 x + 6

Expand

2 x (x^{2} - 1) : 2 x^{3} - 2 x

3 x + 6 = 2 x^{3} - 2 x

Switch sides

2 x^{3} - 2 x = 3 x + 6

Move

6

to the left side

2 x^{3} - 2 x - 6 = 3 x

Move

3 x

to the left side

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

Factor

2 x^{3} - 5 x - 6 : (x - 2) (2 x^{2} + 4 x + 3)

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

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

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

Solve

x - 2 = 0 : x = 2

Solve

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

The solutions are

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

(x - 2)^{3} = 8

(x - 2)^{3} = 0

5 x^{4} + 8 x^{3} = 0

4 x^{4} - 16 x^{3} + 7 x^{2} + 16 x + 4 = 0

4 x^{4} - 65 x^{2} + 16 = 0

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