SOLUTION: I am having a hard time trying to factor this equation. (x²-x+1) Im starting to think it is prime but I'd like to make sure. This is how i worked it out. i know that each set

Algebra ->  Rational-functions -> SOLUTION: I am having a hard time trying to factor this equation. (x²-x+1) Im starting to think it is prime but I'd like to make sure. This is how i worked it out. i know that each set       Log On


   

Question 188497: I am having a hard time trying to factor this equation. (x²-x+1)
Im starting to think it is prime but I'd like to make sure.
This is how i worked it out.
i know that each set of parantheses will start out with "x". then i would figure out what times what will give you a positive one and subtract to give you a negative one. If you could please explain to me how to come about getting this answer, that would be greatly apprciated.
Found 2 solutions by Alan3354, solver91311:
Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
I am having a hard time trying to factor this equation. (x²-x+1)
Im starting to think it is prime but I'd like to make sure.
-------------------
It is not factorable, in the sense that it can't be factored in terms of integers.
Solved by pluggable solver: SOLVE quadratic equation (work shown, graph etc)
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case 1x%5E2%2B-1x%2B1+=+0) has the following solutons:

x%5B12%5D+=+%28b%2B-sqrt%28+b%5E2-4ac+%29%29%2F2%5Ca

For these solutions to exist, the discriminant b%5E2-4ac should not be a negative number.

First, we need to compute the discriminant b%5E2-4ac: b%5E2-4ac=%28-1%29%5E2-4%2A1%2A1=-3.

The discriminant -3 is less than zero. That means that there are no solutions among real numbers.

If you are a student of advanced school algebra and are aware about imaginary numbers, read on.


In the field of imaginary numbers, the square root of -3 is + or - sqrt%28+3%29+=+1.73205080756888.

The solution is , or
Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+1%2Ax%5E2%2B-1%2Ax%2B1+%29

Answer by solver91311(24713) About Me  (Show Source):
You can put this solution on YOUR website!


The short answer is that this trinomial is prime, meaning that it is not factorable over the integers. In fact, this one isn't factorable over the real numbers.

This might be getting a little ahead of you, given the way you phrased the question, but it is something that you will be learning in a week or two anyway. What I'm talking about is a little trick to determine the character of the roots of a quadratic equation of the form:



Take the coefficient on the x term and square it:

Multiply 4 times the lead coefficient times the constant term:

Find the difference of these two values:

If the result is zero, then you have a perfect square trinomial, and this would be factorable over the integers..

If the result is positive and a perfect square, then the trinomial is factorable over the integers.

If the result is positive and not a perfect square, the factors are real but irrational numbers, i.e. they involve the square root of a quantity that is not a perfect square. For the time being, you would consider such a trinomial as prime.

If the result is less than zero, then the factors are complex numbers of the form where i is the imaginary number defined by . You would consider this situation prime as well.

So, just to illustrate with your problem:







so



and your factors would contain a conjugate pair of complex numbers; they would look like:



Like I said, the trinomial is prime as far as you are concerned.

John