# What is the solution for 2x^3+2x=0?

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To get the solution of *2x^3 + 2x, *solve for the greatest common factor ( GCF) of each terms. You can solve the GCF by prime factorization. List down the prime factors (when you say prime factors, ) of each term.

*2x^3 = *2 * x * x * x and

*2x *= 2 * x after listing the prime factors, bring down

GCF= 2 * x the factors that are common for both

terms. In this case only 2 and a single 'x' is present on both terms, so the GCF is '2x'.

Next step is to factor out the GCF. In factoring, you bring out GCF at the same time dividing the original expression by the GCF, thus:

2x (2x^3 + 2x) = 0 simplified as: 2x (2x^3 + 2x) = 0

2x 2x 2x

so : 2x (x^2 + 1) = 0

equate each factor to 0:

2x = 0 divide both sides by 2

2 2 to solve for x

**x = 0** (one solution)

then the other factor:

x^2 + 1 -1 = 0 - 1

x^2 = -1 get the square root of both

sides to eliminate the

exponent '2'

x = +/- sqrt (-1)

square root of a negative number is imaginary so you'll affix ** i **with

the square of 1 which is = 1. and since it is a square root, you should have a + and - answer . so **x = +/- i.**

**Therefore, you'll have x=0, x=+i and x=-i as the solution for the problem. Your solution must be equal to the highest exponent of the expression which is 3 in this case... :)**

`2x^3+2x=0`

Factor out the GCF of 2x from each term.

`2x(x^2) +2x(1)=0`

`2x(x^2+1)=0`

Set the single term on the left side of the equation to 0.

2x=0

Divide each side by 2 to get x alone.

`(2x)/2 = 0/2`

`x= 0/2`

`x=0`

Now set the other factor on the left side of the original factored equation to 0.

`x^2+1=0`

`x^2=-1`

Take the square root of both sides to eliminate the exponent.

`x=+-sqrt(-1)`

Pull out perfect square roots from beneath the radical. Here remove "i" because it is a perfect square.

`x=+-i`

The complete solution is the set of individual solutions.

`x=0,i,-i`

**2x^3+2x=0 **

**x= 0, i ,-i**

**Sources:**

`2x^3+2x=0`

factor out the greatest common factor

2x(x^2+1)=0 now set them equal to 0

`2x=0`

`(2x)/2 = 0/2`

`x=0`

`x^2+1=0`

`x^2=-1`

`sqrt(x^2)=+-sqrt(-1)`

the square root of -1 is an imaginary number so it becomes i

`x=+-i`

`x=-i ` `x=i`

the final answer is `0,i,-i`