# Product of 3 linear factors

1. May 17, 2009

### hostergaard

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

http://img42.imageshack.us/img42/1231/lobgave.jpg [Broken]

2. Relevant equations

3. The attempt at a solution
i think that solution shuld look something like this:
(x-a)2*(x-k)
but im pretty lost, so i need some help...

Last edited by a moderator: May 4, 2017
2. May 17, 2009

### epenguin

I don't think your conjecture works.

Firstly there is one rather easy value of k you might notice that will then give you two equal roots.

If that were the highest one it would save work. So that is worth asking - think about the form of your cubic function.

If you can't find any shortcuts like that, there are several ways to solve a problem like this.
Do you know the relation between coefficients of a a polynomial equation and its roots?
Or do you know and understand the discriminant of a cubic - what does that have to do in order for the cubic to contain two equal roots?

I have not worked out out yet, have to leave, and there seems to be a fair amount of calculation in it unless you get lucky. I may have missed something clever - but in the end you have to know and sometimes use the things that will always work.

3. May 17, 2009

### gabbagabbahey

Choosing $(x-a)^2(x-k)$ to represent your polynomial causes two problems:

(1)You are effectively assuming that $k$ is a single root of your cubic. Since you are not told this, I would not make this assumption.

Instead, use a different letter to represent the single root: $(x-a)^2(x-b)$ represents a cubic with a double root $x=a$ and a single root $x=b$

(2) The second problem is that $(x-a)^2(x-b)$ has a leading coefficient of $1$, whereas the given cubic should have a leading coefficient of $2$. To solve this problem, just multiply your proposed solution by 2: $p(x)=2(x-a)^2(x-b)$

Try expanding this proposed solution, and then compare coefficients

4. May 19, 2009

### epenguin

Any progress?

I have done it by the discriminant method and promise you there is a simple answer. Even knowing the answer I cannot see any reasonable short cuts, so it seems to be a case of needing to know a general method as I said.

On the way you run into some pretty big numbers which is discouraging and often a warning of a mistake, but it works out in the end. So be careful to write out very clearly so you can pick up mistakes.

(If you can get that you can also easily get the least of the k's - it is a rational number but rather wierd for a student exercise leaving me wondering how these guys invented the problem.)

Last edited: May 20, 2009
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