# Homework Help: This would be really simple if I knew how, I know it

1. Mar 17, 2009

### hurliehoo

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

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

Find x

2. Relevant equations

This is for an eigenvalue problem :

[0 1 1,
1 0 1,
1 1 0]

3. The attempt at a solution

Well I know for a fact that the answers are -1, -1 and 2 but HOW from the above equation ... I would like to understand =/

2. Mar 17, 2009

### Staff: Mentor

Multiply this equation to put it in standard form: -x (x^2 - 1)+ 2x+ 2= 0.
The equation is a cubic, and you should be able to find one linear factor, and then the other two.

3. Mar 17, 2009

### hurliehoo

I tried to do that (using the quadratic formula) and it didn't work. :( The reason was that b=2/x which only gave one correct answer. I actually tried this in two different ways and each time only one solution was correct.

4. Mar 17, 2009

### matt grime

First, don't multiply anything out. There is an obvious factor of x+1 (you know how to factorize x^2 -1), and you can spot the others (that is allowed).

5. Mar 17, 2009

### hurliehoo

Am I missing something here ... when I factorise (x^2 -1), I get x(x -1/x) and not sure what that means ..?

6. Mar 17, 2009

### Staff: Mentor

x^2 - 1 factors into (x - 1)(x + 1)

7. Mar 17, 2009

### matt grime

I think we need to start from the beginning.

You must keep things in polynomials in x at all times. Do you understand what I mean by that? Do you understand what you're trying to do in factorising things? If I know that U*V=0, then I know one of U or V is zero, right? So stepping back a little, how do you solve x^2 -1 =0? This is not part of this question (though it will help).

You need to write x^2-1 as the product of two simpler things, the U and V above. What things multiply together to give x^2 - 1? If you don't know the rule of the difference of two squares, what are the roots of x^2 -1? I.e. what are the values for x that satisfy x^2 - 1? What relationship do the roots have with the factorisation?

8. Mar 17, 2009

### matt grime

Please, Mark, the point of the HW help is not to just give out answers. It is to help people work things through.

9. Mar 17, 2009

### hurliehoo

Yeah I get this, thanks for clearing it up though as it's good to have things confirmed ... it just seems a bit of a leap from having a possible solution of x=1, when it isn't correct, to producing a solution of x=2 from somewhere... this is what I'm struggling to understand. I suppose I'm just wondering if there's a better method than just randomly plugging numbers in when it gets to a certain point.

10. Mar 17, 2009

### matt grime

The best thing to do is to look and think. Remember, the point of the question is not to make you have to find strange and difficult expressions for the roots (which exist for quadratic, cubic and quartics only, in a precise sense). It is to make sure that you understand eigenvalues and eigenvectors.

Thus you can guess that the roots will always be integers close to 0. And remember an important rule:

if f(x) is a polynomial with integer coefficients, and the highest power of x has coefficient 1 (i.e. is x^n with n the largest power appearing) then any rational number that is a root of f is actually an integer.

Anyway, I don't understand what you mean by saying that you thought you had a possible solution of x=1.

11. Mar 17, 2009

### Staff: Mentor

I am well aware of these. I am also assuming that someone working on a problem about the eigenvalues of a matrix knows the basics of factoring polynomials.

12. Mar 17, 2009

### matt grime

But it is clear that the OP doesn't know the fundamentals of factorisation. (And if they did they wouldn't need that factorisation pointing out.)

13. Mar 19, 2009

### hurliehoo

Mmm... There are some steps missing here which have not been pointed out. I am aware that x=1 is not a factor ... if you read my OP you will see I already know the correct answers. Also, there is no need to fight guys! Please, lol.

Actually I do understand at least a few of the fundamentals although I admit I do have regular brief periods of exceptional stupidity.

You should see me trying to tie my shoelaces in the mornings - I tie myself to the nearest piece of furniture by accident half the time.

...What I would like, and what prompted me to ask for advice here, is more efficient methods of solving equations like the one above, more than anything else, though.

Anyway, both of you have been very helpful. My question has been answered and many thanks to both Matt and Mark.

Last edited: Mar 19, 2009
14. Mar 19, 2009

### matt grime

Of the original equation. That does not alter the fact that x^2-1 = (x-1)(x+1).

Look at the expression

-x(x^2-1) +2(x+1)

You can factor x^2-1 so you have

x(x+1)(x-1) + 2(x+1)

My view was informed by your writing x^2 -1 as x(x-1/x), and thus moving out of the realms of polynomials, and not seeing the link between roots and factors.

15. Mar 19, 2009

### hurliehoo

Hmm okay this makes it very clear. This is what I was missing earlier. In future similar problems I will try to use this approach. Many thanks for your kindness in taking the time for this.