# Root of f(x) = x^3-x-1 ?

1. Apr 16, 2012

### LearninDaMath

1. The problem statement, all variables and given/known data

How does one find the root of f(x) = x^3 - x - 1 ? Quadratic Equation only works on power of 2. I can't factor out an x to get a first term of x^2 because then Quadratic equation still won't work because the middle and last term would be messed up, I think.

What are the rules or the process?

2. Apr 16, 2012

### Staff: Mentor

There is a formula, like the one you've memorized for quadratics, though the one for cubics is more frightening. Try a google search for it.

Or you can sometimes see a solution by inspection, esp. integer solutions. Or you can use a numerical method to find an approximate solution, that approximation can be to whatever accuracy you desire. Or you could plot a graph and read off from there.

Are you content to settle for a solution correct to 4 sig figs?

3. Apr 16, 2012

### Staff: Mentor

It is always recommended that you sketch the curve so that you know what you are dealing with. There are plenty of online plotting resources, quickmath is a handy one. Here's a sample, but it's not of the exact function you have (I don't plan to do your work for you!) http://www.quickmath.com/webMathema...uation&v1=x^3+-+x+-+2=y&v2=-2&v3=2&v4=-4&v5=4

Bookmark that site, it offers a lot more than just curve plotting.

4. Apr 16, 2012

### e^(i Pi)+1=0

Use the rational root test, then test for each one using synthetic division.

5. Apr 17, 2012

### HallsofIvy

The rational root test won't give a root here because this polynomial has NO rational roots. The rational root test, for this polynomial, says that 1 and -1 are the only possible rational roots and they clearly are not roots.

If a and b are any two numbers then $(a- b)^3= a^3- 3a^2b+ 3ab^2- b^3$ and $3ab(a- b)= 3a^2b- 3ab^2$ so that $(a- b)^3+ 3ab(a- b)= a^3- b^3$. That means that if we let x= a- b, m= 3ab, and $n= a^3- b^3$, we have $x^3+ mx= n$.

Now, suppose we know m and n. Can we solve for a and b and so find x? Yes, we can!

From m= 3ab, we have b= m/3a. Putting that into $n= a^3- b^3$, we get $n= a^3- m^3/3^3a^3$. Multiplying through by $a^3$ we have $na^3= (a^3)^2- m^3/3^3$ which is a quadratic $(a^3)^2- na^3- m^3/3^3= 0$ for $a^3$.

Solving that with the the quadratic formula,
$$a^3= \frac{n\pm\sqrt{n^2+4\frac{m^3}{3^3}}}{2}= \frac{n}{2}\pm\sqrt{\left(\frac{n}{2}\right)^2+ \left(\frac{m}{3}\right)^3}$$

Since $a^3- b^3= n$,
$$b^3= a^3- n= -\frac{n}{2}\mp\sqrt{\left(\frac{n}{2}\right)^2+ \left(\frac{m}{3}\right)^3}$$

Last edited by a moderator: Apr 19, 2012
6. Apr 18, 2012

### e^(i Pi)+1=0

You are right that in this case it wouldn't help, but it is a good rule of thumb in general.

7. Apr 19, 2012

### Staff: Mentor

Very neat!

Is it too late to fix the typo?

8. Apr 19, 2012

### HallsofIvy

Thanks. I have edited it.