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Factoring Polynomials

Date: 11/14/96 at 17:37:34
From: John Cairns
Subject: Factoring Polynomials

I am a freshman in high school in an Algebra II course.  I don't have 
a specific problem that I need help with.  Instead, 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 my textbook (but I don't need help 
with these specific ones, just the concept):
x^4 - 13x^2 + 36 
a + b + 3a^2 - 3b^2 
Is there no way to factor polynomials using some sort of logic?  Thank 
you very much for any light you can shed on this aspect of algebra.

Date: 11/17/96 at 23:04:59
From: Doctor Wilkinson
Subject: Re: Factoring Polynomials

I congratulate you for thinking seriously about problems and not being 
satisfied with only what you need to know in order to do your 
homework.  You have noticed that factoring polynomials is difficult, 
which is a very good observation.  You have also probably noticed that 
all you have been given by your teacher and textbook is a handful of 
special tricks, like how to factor the difference of two squares.  You 
haven't been given a general rule which will work for all polynomials.  

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.

I hope this helps.  You are obviously very smart so I have tried
to give a fairly complete answer even though it may be a little 
advanced for you right now.

-Doctor Wilkinson,  The Math Forum
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Associated Topics:
High School Basic Algebra
High School Polynomials

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