Factoring Polynomials Using Substitution

Date: 09/26/2001 at 19:20:39
From: Scott
Subject: Factoring

I have a 25-question factoring assignment, and I can't get two of the 
questions on my own. Here they are:

x^2+10xy+25y^2-81z^2

a^4-6a^2b^2-27b^4

Date: 09/27/2001 at 11:17:52
From: Doctor Ian
Subject: Re: Factoring

Hi Scott,

In the first one, the z's are independent of the x's and y's, so the 
last term can be 'factored' by noting that 9^2 = 81, and by recalling 
that any expression like

  a^n * b^n

can be written more simply as 

  (a * b)^n

But what about the x and y terms? Well, if you've factored 23 other 
expressions, you've probably got the drill down, so you know that you 
need to find two numbers, u and v, such that

  x^2 + 10xy + 25y^2 = (x + uy)(x + vy)

                     = x^2 + (u+v)xy + (uv)y^2

In other words, you need to find u and v such that

  u + v = 10

  y * v = 25


In the second problem, you're looking for u and v such that

  a^4 - 6a^2b^2 - 27b^4 = (a^2 + ub^2)(a^2 + vb^2)

                        = a^2 + (u+v)b^2 + (uv)b^4

which is really just the same kind of problem with different 
exponents. (What a surprise!)

One nice trick in a case like this is to use substitution.  That is, 
define x and y such that

  x = a^2

  y = b^2

Now the problem becomes

  x^2 - 6xy + 27y^2

which should look familiar to you. When you factor this, you'll end up 
with something that looks like 

  (x + uy)(x + vy)

at which point you can undo the substitution, to get

  (a^2 + ub^2)(a^2 + vb^2)

which is what we got without doing the substitution.  

Whenever you see a problem with exponents that you don't know how to 
handle, you should consider whether you can use this kind of 
substitution to convert it into a problem that you already know how to 
solve.  

I hope this helps. Write back if you'd like to talk about this some 
more, or if you have any other questions. 

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