Associated Topics || Dr. Math Home || Search Dr. Math

### 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

```
Associated Topics:
High School Basic Algebra

Search the Dr. Math Library:

 Find items containing (put spaces between keywords):   Click only once for faster results: [ Choose "whole words" when searching for a word like age.] all keywords, in any order at least one, that exact phrase parts of words whole words

Submit your own question to Dr. Math
Math Forum Home || Math Library || Quick Reference || Math Forum Search