Is there a logic behind the factorization of polynomials?

2.0 years ago by

I need help with the concept of factoring. I keep trying to think of some logical way to deduce the ansers to the problems, but I can't. The whole concept is kind of fuzzy to me because, unlike so many other things in algebra that have definite formulas that are logical and clean-cut, factoring isn't this way. Just to give you an idea of exactly what I'm talking about, here are some example from the textbook

x4 - 13x2 + 36
a + b + 3a2 - 3b2

Is there no way to factor polynomials using some sort of logic?


1 Answer

24 months ago by

You ask whether it is possible to factor polynomials using some sort of logic. The answer is yes, and there are computer programs, for example, which can be used to factor polynomials. I'll try to give
you a general idea of one way this can be done. It probably isn't practical for your typical homework problem, at least not without a computer, because there's too much computation involved.

First of all, let's take a look at what makes factorization hard. You probably know that integers can be factored into primes. This is pretty easy to do in practice because to factor a given number, you can just try dividing it by all the smaller numbers. If you don't find a factor, you're done and the number is prime. If you do, you can repeat the process on the factors. This works pretty well because you have a definite set of numbers to try and you can go through all of them.

For polynomials it isn't so simple. We can still prove without too much trouble that any polynomial can be factored in essentially just one way. But the proof doesn't really tell us how to do it. We know that any factors have to have smaller degree than the given polynomial. In your problem x^4 - 13x^2 + 36, you could have factors of degree 1 or 2 or 3, but not of degree 5, for example. But that doesn't really narrow things down enough, because there are infinitely many polynomials of degree 2, and we can't try them all.

Now I'll give you a brief idea about one of the tricks that can be used to cut down the possibilities. One thing we can show about polynomials is that if you know enough values of a polynomial of a given degree, then you can find the coefficients. For example, if you know that a polynomial of degree 2 takes on the value 0 when x is 0, 1 when x is 1, and 6 when x is 2, you can figure out that the polynomial must be 2x^2 - x. And there are general, clear rules for doing this.

Now suppose you want to factor a polynomial, say x^4 + 13x^2 + 36 (I changed your example just a little). One of the ways you might try to factor it is into two polynomials of degree 2. Now to determine a polynomial of degree 2, it is enough to find its value for three values of x, as I mentioned in the previous paragraph. So let's substitute three values of x into the polynomial we're trying to factor. It doesn't matter which values we use as long as there are three of them. Let's try -1, 0, and 1. This gives us values of 50, 36, and 50. So if our factors are going to work, they have to yield these products when we substitute -1, 0, and 1 into both of them. So, for example, the first quadratic could have the values 5, 4, and 5 while the second had the values 10, 9, and 10; or the first could have the values 25, 2, and 5 while the second had the values 2, 18, and 10
when we substituted -1, 0, and 1 for x. There are quite a lot of possibilities depending on how you choose the factors for the values 50, 36, and 50, but given enough time, or the help of a computer, you can certainly try them all. In particular, you will eventually try the values 5, 4, 5 for the first factor and 10, 9, 10 for the second, and this will give the right factorization:

x^4 + 13x^2 + 36 = (x^2 + 4) (x^2 + 9)

Of course, you could have solved this one in a fraction of the time by other methods like trial and error. But the method I have sketched will work in principle on any polynomial.

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