How Do You Solve Complex Roots in a Cubic Equation?

[cue928]

I am feeling stupid over this one but I cannot remember how to solve it. I have the following cubic equation:
4x^3 - 32x = -12

I get the solution of x = -3 but not the other two.
I initially had it factored as: 4x(x^2 - 8) = -12
Did 4x = -12, got x = -3
But then I could not get the other two solutions.

Sorry for the simple question, just can't remember what you're supposed to do on these.

 

[Mark44]

I am feeling stupid over this one but I cannot remember how to solve it. I have the following cubic equation:
4x^3 - 32x = -12

I get the solution of x = -3 but not the other two.
I initially had it factored as: 4x(x^2 - 8) = -12
Did 4x = -12, got x = -3
But then I could not get the other two solutions.

Sorry for the simple question, just can't remember what you're supposed to do on these.

This is not a useful factorization: 4x(x^2 - 8) = -12
Write your equation as 4x^3 - 32x +12 = 0. Since x = -3 is a solution, then x + 3 must be a factor.

This means that you have (x + 3)(?) = 0, where ? is a quadratic expression, which is easier to factor than a cubic.

 

[cue928]

This is not a useful factorization: 4x(x^2 - 8) = -12
Write your equation as 4x^3 - 32x +12 = 0. Since x = -3 is a solution, then x + 3 must be a factor.

This means that you have (x + 3)(?) = 0, where ? is a quadratic expression, which is easier to factor than a cubic.

Thanks for your reply; I'm still not following where I need to go from there?

 

[Mark44]

Use polynomial division to divide 4x^3 - 32x +12 by x + 3.

 

[elect_eng]

I happen to have a cheat sheet for the general method to solve cubic equations, so I'll post it.

Don't take this to mean you shouldn't follow the advice given above. When you see a simple method, there is no good reason not to use it.

You'll see that your equation is already in the normal form (EDIT: that is, if you divide your equation by 4), and the solution is easier in that case.