Factoring Polynomials

Date: 02/08/2003 at 13:27:40
From: Giovanni 
Subject: Factoring polynomials

9y^4 + 12y^2 + 4

The first term in the polynomial is not raised to the 2nd degree, and 
when I do change it so that it becomes a term to the 2nd degree, it's 
hard for me to find a number whose sum is 12 and product is 12. I 
don't know if I'm approaching this question in the right way.

Date: 02/08/2003 at 16:33:37
From: Doctor Kastner
Subject: Re: Factoring polynomials

Hi Giovanni -

When you are factoring polynomials, the first thing to think about is 
the pattern. Let me give you an example. We know how to factor the 
difference of squares:

   y^2 - 4 = (y-2)(y+2)

and whenever we see the form

   a^2 - b^2

we should factor it into

   (a-b)(a+b)

In actuality, a and b may be complicated expressions, but that doesn't 
matter. We just need to ask if it is something squared minus something 
squared:

   625y^4 - 16x^4

can be written as

   (25y^2)^2 - (4x^2)^2

which factors into

   (25y^2 - 4x^2)(25y^2+4x^2)

but notice that we have another difference of squares in the first 
term

   (25y^2 - 4x^2) = (5y)^2 - (2x)^2 = (5y-2x)(5y+2x)

so we have that

   625y^4 - 16x^4 = (5y-2x)(5y+2x)(25y^2+4x^2)

So even though it started off looking very complicated, it was just a 
couple of difference of squares. 

Now back to your problem. We are used to the form ay^2+by+c, but 
notice that ax^4+bx^2+c is pretty much the same thing. If we write it 
like

   a(x^2)^2 + b(x^2)^1 + c

we can let y = x^2, and now it looks like

   ay^2 + by + c

And you could factor this, making sure that you put the x^2 back in at 
the end. But we don't always need to do this explicit substitution. 
Instead, just remember that the factored form will have an x^2 in it 
instead of an x. Both of the signs are positive, so that means it will 
be of the form

   (__ x^2 + __)(__x^2+__)

if we can factor it at all. The numbers that can go into the first 
slots are 3 and 3 or 9 and 1, while the numbers that can go into the 
last slots are 4 and 1 or 2 and 2. Finally, we need the middle term to 
be 12, and looking at the combinations we could pick for the slots, 
one of them leaps out. Do you see which one it is?

Does this help?  Let me know if you'd like to talk about this some 
more, or if you have any other questions.

- Doctor Kastner, The Math Forum
  http://mathforum.org/dr.math/