# How to factor x^3 - 4x + 4

1. Dec 8, 2008

### temaire

1. The problem statement, all variables and given/known data
$$x^3-4x+4$$

2. Relevant equations
None

3. The attempt at a solution
I tried using the quadratic formula but it doesnt work. I started with this, but I don't know what to do now.$$x^{3}+0x^2-4x+4$$

2. Dec 8, 2008

### gabbagabbahey

Re: Factoring

What exactly are you trying to do?!

$x^3-4x+4$ is an expression, not a problem statement! Do you mean that you are trying to find the roots of the equation $x^3-4x+4=0$?

3. Dec 8, 2008

### temaire

Re: Factoring

yes I am trying to find the roots of the equation

4. Dec 8, 2008

### gabbagabbahey

Re: Factoring

Well, there are no simple solutions to this cubic equation. You can try using the cubic formula found here.

Is this the entire question, or is this part of a larger problem? Are you sure you have have written the equation correctly?

5. Dec 8, 2008

### temaire

Re: Factoring

This is the correct equation, and it is the entire question.

6. Dec 8, 2008

### gabbagabbahey

Re: Factoring

Are you asked to use any specific method? (such as approximating the real root using newton's method)
Are you allowed to use a calculator/computer?

7. Dec 8, 2008

### temaire

Re: Factoring

No specific method is being asked but you can use a simple scientific calculator.

8. Dec 8, 2008

### gabbagabbahey

Re: Factoring

What methods have you been taught in whichever course this is for? Use one of those.

9. Dec 8, 2008

### TayTayDatDude

Re: Factoring

Edit~

10. Dec 8, 2008

### gabbagabbahey

Re: Factoring

There are no rational roots to this equation, so that method won't work.

11. Dec 9, 2008

### Дьявол

Re: Factoring

Find the roots using Cardano's method.

$$D=(\frac{4}{2})^2 + (\frac{-4}{3})^3=4 - \frac{64}{27}=\frac{108-64}{27}=\frac{44}{27} > 0$$, so it got one rational and two complex roots.

The rational one is ~(-2.38)

12. Dec 9, 2008

### gabbagabbahey

Re: Factoring

The root (-2.38) is real not rational. The rational roots theorem tells you that the only possible candidates for rational roots of this equation are $\pm 1$, $\pm 2$, and $\pm 4$....and since none of these are roots; there are no rational roots.

13. Dec 10, 2008

### Дьявол

Re: Factoring

Rational numbers are also real numbers. So I suppose -2.38....(infinite number of decimals) is irrational number (also real number).

14. Dec 10, 2008

### gabbagabbahey

Re: Factoring

A rational number is a real number that can be expressed as a ratio of two integers....for example, 1/3, 0.25, and 3658793659/548709568017097 are rational numbers.

However, things like $\pi$, $\sqrt{2}$ and any non-repeating non-terminating decimal are irrational numbers.