8x^3+4x^2-4x solution sets

solution sets? Is there an equals sign anywhere?

If all this is equal to zero..

4x(2x^2+x-1)=4x(2x-1)(x+1)=0
x=0, x=1/2, x=-1
check those

thank you!

To find the solution sets for the equation 8x^3 + 4x^2 - 4x, we need to set the equation equal to zero and solve for x.

Step 1: Set the equation equal to zero:

8x^3 + 4x^2 - 4x = 0

Step 2: Factor out common terms, if possible:

We can see that there is a common factor of 4x, so we can factor that out:

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

Step 3: Solve for x by setting each factor equal to zero:

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

Solving the first equation, we get:

x = 0

To solve the second equation, we can use the quadratic formula:

x = [-b ± sqrt(b^2 - 4ac)] / (2a)

In this case, a = 2, b = 1, and c = -1. Plugging these values into the quadratic formula, we get:

x = [-1 ± sqrt(1 - 4(2)(-1))] / (2 * 2)

x = [-1 ± sqrt(1 + 8)] / 4

x = [-1 ± sqrt(9)] / 4

x = [-1 ± 3] / 4

So, the solutions for the second equation are:

x = (-1 + 3) / 4 = 1/2

x = (-1 - 3) / 4 = -1

Therefore, the solution set for the equation 8x^3 + 4x^2 - 4x = 0 is:

{x = 0, x = 1/2, x = -1}