# Product of 3 linear factors

• hostergaard
In summary: The Attempt at a Solutioni think that solution shuld look something like this: (x-a)2*(x-k)but I am pretty lost, so i need some help...:cry: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

## Homework Statement

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

## The Attempt at a Solution

i think that solution shuld look something like this:
(x-a)2*(x-k)
but I am pretty lost, so i need some help...

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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.

hostergaard said:

## The Attempt at a Solution

i think that solution shuld look something like this:
(x-a)2*(x-k)
but I am pretty lost, so i need some help...

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

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 weird for a student exercise leaving me wondering how these guys invented the problem.)

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## 1. What is a "product of 3 linear factors"?

A product of 3 linear factors is a mathematical expression that consists of three linear terms being multiplied together. Each linear term is composed of a variable raised to the first power and a constant.

## 2. How is a product of 3 linear factors written?

A product of 3 linear factors is typically written in the form (ax + b)(cx + d)(ex + f), where a, b, c, d, e, and f are constants and x is the variable.

## 3. What is the purpose of finding the product of 3 linear factors?

The purpose of finding the product of 3 linear factors is to factorize a polynomial expression and simplify it into its linear factors. This can help in solving equations and identifying important features of the polynomial, such as its zeros and end behavior.

## 4. How do you solve for the roots of a polynomial using the product of 3 linear factors?

To solve for the roots of a polynomial, you can set each linear factor equal to zero and solve for the variable. The values of the variable that satisfy all three equations will be the roots of the polynomial.

## 5. Can the product of 3 linear factors be used to solve any polynomial equation?

No, the product of 3 linear factors can only be used to solve polynomial equations that have three distinct roots. If a polynomial has repeated roots or complex roots, the product of 3 linear factors method will not work.