# Hi guys, Need some help on something I didn't know I didn't

1. Jan 12, 2012

### Rob K

Hi guys,

Need some help on something I didn't know I didn't know.

How do you factorize this?
3+7λ+6=0

I assumed this?

-λ(λ2-7) + 6 = 0
λ = 6
λ = ±√7

But that's wrong as the answers in the book are -2, -1, and 3.

Can someone give me a little guidance on this, I have only ever done factorization where one of them is already know.

Kind Regards

Rob K

2. Jan 12, 2012

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

Re: Factorisation

First, I would just change the lambdas to X to make the problem look more familiar. To find one of the zeros, graph it if you're allowed, or use the rational root test to find one. Then, divide like normal, ect..

http://www.purplemath.com/modules/rtnlroot.htm

3. Jan 12, 2012

### Rob K

Re: Factorisation

Actually, I am working out an eigenvalue problem hence the Lamdas. I was kind of expecting to find a way to do this algebraically. If there were an x2 or λ2 term in there, I would be able to do it, but the lack of it has stumped me.

4. Jan 12, 2012

### Staff: Mentor

Re: Factorisation

I suppose you could solve it like you do for quadratics:

rewriting it to x^3 - 7x -6 = 0

and then looking at the factors of 6: 1, 2, 3, 6

then plug each one into the equation to find a root realizing you must try both +- versions.
From there you have one factor (x +- ??) that you can use to divide
the original polynomial by to get a quadratic that you can solve.

5. Jan 12, 2012

### micromass

Re: Factorisation

So if I write it as x3 + 0x2 - 7x -6 = 0, you'll be able to solve it?

6. Jan 12, 2012

### Staff: Mentor

Re: Factorisation

Makes me think of Cardano's method.

7. Jan 12, 2012

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

Re: Factorisation

How is the rational root test and algebraic/synthetic long division not algebraic?

8. Jan 12, 2012

### Staff: Mentor

Re: Factorisation

No offense meant, but that's not even remotely close. What you're saying here is something like, if a*b = 6, then a = 6 or b = 6.

If two numbers multiply to some other number, there's not much you can say about either of the two numbers, unless the third number happens to be 0.

In other words, writing the equation as
λ(λ2-7) = 6
is pretty much a nonstarter.

This is definitely the way to go.

9. Jan 12, 2012

### Rob K

Re: Factorisation

Apologies, I missed the comment about the rational root test. I don't know it, but now I know what it is called, I can learn it, thank you.

After some reading I didn't realize that this is actually a fairly complex problem to do manually. I assumed as quadratics are easily solved by either factorization or the equation method, that other polynomials are as easy.

Regards

Rob

10. Jan 12, 2012

### ehild

Re: Factorisation

I suppose you know that a polynomial p(x)=∑aixi of order n can be factorised in terms of the roots x1, x2, ...xn of the equation p(x)=0:
p(x)=an(x-x1)(x-x2)...(x-xn).

Multiplying the factors, the product has the constant term anx1 x2....xn which must be the same as a0, the constant term of the original polynomial. When all coefficients are integer rational roots can exist. They can be obtained in the form (divisor of a0)/(divisor of an)

You can write your polynomial as λ3-7λ -6, so an=1 and a0=-6 The integer divisors of 6 are ±1, ±2, ±3, ±6. Try to find a root among these numbers. What do you get when subbing -1 for λ, for example? If you find one root, x1, divide the polynomial by (x-x1) and you get a quadratics easy to factorise, or try other possible root.

ehild

11. Jan 12, 2012

### ehild

Re: Factorisation

It is certainly effective but I'd rather death ...

ehild