Factorizing polynomial of degree 4 - II

[Paul Belino]

This can be solved simply by plotting it out. Say on acell phone graphs APP or Excel. When I do this the roots are real but not integers. Two are negative and two are positive. As another clue plot between x = -3 and x = +3

 

[HallsofIvy]

Try this. It's a bit hairy but it can work. We know that a quartic can always be factored into two quadratic factors:
[math]m^4 - 8m^2 + m + 12 = (m^2 + am + b)(m^2 + cm + d)[/math]

Not if we expect to have integer or just rational coefficients! It might factor as an irreducible cubic and a linear function. If you are going to say that "a quartic can always be factored into two quadratic factors" (with possibly irrational or complex coefficients) why not assert that it can be factored into four linear factors?

Multiply things out and match coefficients. After some work you will find that [math]m^4 - 8m^2 + m + 12 = (m^2 - m - 3)(m^2 + m - 4)[/math].

-Dan

 

[JeffM]

Not if we expect to have integer or just rational coefficients! It might factor as an irreducible cubic and a linear function. If you are going to say that "a quartic can always be factored into two quadratic factors" (with possibly irrational or complex coefficients) why not assert that it can be factored into four linear factors?

This is wrong.

Every polynomial of degree n with real coefficients can be factored into n/2 quadratics with real coefficients if n is even and into one linear term with real coefficients and (n - 1)/2 quadratic terms with real coefficients if n is odd.

So Dan was quite right. A quartic with real coefficients can always be factored into two quadratics with real coefficients.

And what is meant by "irreducible cubic" when there is a formula that will give a real root for any cubic.

 

[HallsofIvy]

What was wrong about it? I said "if we expect to have integer or just rational coefficients" which is normally assumed in "factoring" problems. I also said that it could be factored into two quadratics if you are willing to use irrational coefficients. And noted that, with general complex coefficients, you can factor as four linear factors. There was nothing in the original post that said "real coefficients".

 

[JeffM]

What you said was that a quartic can always be factored into quadratics with possibly complex coefficients in disagreeing with the statement meaning that a quartic with real coefficients can always be factored into quadratics with real coefficients, a statement that is relevant for students in pre-algebra. So the fact that you can decompose a quartic into two quadratics, then decompose those into linear terms that may contain complex numbers, and then rearrange so that you get quadratics with complex coefficients does not contradict the fact that you can always decompose a quartic with real coefficients into two quadratics with real coefficients.

And why jump over real numbers like the square root of two to get into complex numbers? People in the thread had already talked about the rational root theorem and pointed that there was no rational root in this case.

And what do you mean by "irreducible cubic" unless you are restricting coefficients to real numbers.

Are you really saying that we should not expect a student in pre-algebra to factor x^2 - 2 = 0 but we should expect a solution to x^2 + 4 = 0.

 

[topsquark]

Well, either way I appreciate the comments. I had not considered non-rational coefficients.

-Dan