# How to solve x^3-x-1=0

1. Jun 21, 2009

2. Jun 21, 2009

### symbolipoint

Re: x^3-x-1=0

If you can divide the polynomial expression by a linear binomial, then you can determine what value of x of each binomial will make the binomial equal to zero. Try first divisors x+1, and x-1.

3. Jun 22, 2009

### HallsofIvy

Staff Emeritus
Re: x^3-x-1=0

By the "rational root theorem", which says that if m/n (m and n integers) is a rational root of $a_nx^n+ a_{n-1}x^{n-1}+\cdot\cdot\cdot+ a_1x+ a_0$ then n must divide the "leading coefficient", $a_n$, and m must divide the "constant term", a_0. That tells us that the only possible rational roots for this equation are the intgers 1 and -1. Unfortunately, trying them $1^3- 1- 1= -1$ and $(-1)^3- (-1)-1= -1$. Since neither of those is equal neither 1 nor -1 is a root and so this equation has no rational roots.

But if x= 2, $2^3- 2- 1= 5$ so there is clearly a root between 1 and 2. The only thing I could suggest is Cardano's "cubic formula": http://www.sosmath.com/algebra/factor/fac11/fac11.html.

Last edited: Jul 13, 2009
4. Jun 22, 2009

### Epic Jeff

Re: x^3-x-1=0

i have a similar problem

x^2 -x -20 < 0

this is what i have so far:

x^2 - x < 20

x^2 - x < 20
_______x to eliminate the power

x - 1 < 20
_______x

i'm pretty much stuck there, any help?
p.s. underscore is just to put the X where i want it

5. Jun 22, 2009

### rock.freak667

Re: x^3-x-1=0

What exactly are you to do with that inequality? Find the range of values for x?

x - 1 < 20
_______x

I think at one point you divided by x to get this, but you can't just divide by x in an inequality, as the sign of x can change the inequality.

But if you want to get $x^2-x-20$ to have x to a single power then try completing the square.

6. Jun 22, 2009

### HallsofIvy

Staff Emeritus
Re: x^3-x-1=0

Please do not "hijack" other people's threads for a new problem. And, I can see no similarity, except that they both involve polynomials.

I recommend you factor $x^2- x- 20$ which can be done relatively easily. Use the fact that the product of two numbers is negative if and only if one is positive and the other negative.

7. Jun 23, 2009

### Epic Jeff

Re: x^3-x-1=0

My bad, thanks for the help though.