Factoring with 3 Variables

Date: 4/8/96 at 11:49:19
From: Anonymous
Subject: Factoring when either square is a square of a polynomial

This one has been frustrating.  Factor: 9a^2+4bc-4c^2-b^2

I re-arranged: 9a^2 -4c^2 +4bc-b^2

Then: (3a)^2+(-4c+b)(c-b), and that does not work, but it is the 
closest to being correct for the middle term.  I have tried 
several possible combinations for the last factors.  Surely you 
have seen this before and can help me over this hump.

Thank you.
LF

Date: 4/19/96 at 17:10:29
From: Doctor Cameron
Subject: Re: Factoring when either square is a square of a polynomial

Thanks for writing in.

 9a^2 + 4bc - 4c^2 - b^2

 I would rewrite it as

9a^2 - b^2 + 4bc -4c^2

then isolate a polynomial.

9a^2 - (b^2 - 4bc + 4c^2)  

The signs change when you subtract in the parentheses.

Then factor the polynomial

9a^2 - (b - 2c)^2.

Note that this polynomial is the difference of two squares, so it 
is in the form y^2 - z^2 where  y = 3a and z = b - 2c.  
We know that we can factor the difference of two squares as 
follows: 

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

Therefore we get
 
(3a - (b - 2c))(3a + (b - 2c))

as our finished answer.

-Doctor Cameron,  The Math Forum