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Factoring by Grouping


Date: 05/27/98 at 18:26:44
From: Ashley Bullard
Subject: Algebra 1(factoring by grouping)

How do you factor 2r^2 + rt - 6t to the second power?

So far I have multiplied the ends of the equation (2r^2 and -6t^2). 
What do I do now?


Date: 06/01/98 at 13:12:15
From: Doctor Peterson
Subject: Re: Algebra 1 (factoring by grouping)

Hi, Ashley. This is about the hardest type of problem you'll see for 
this sort of factoring, because both squared terms have coefficients 
(that is, multipliers, which in your problem are 2 and -6) that are 
not one. Let's go through a problem very much like yours (so you can 
try out what you learn on your own), and then if you need another 
example you can look on the Web at:

 http://mathforum.org/dr.math/problems/thorne11.14.95.html    

for a problem involving bigger and uglier numbers.

Here's my problem:

   Factor 3r^2 - 7rt - 6t^2

We want to get this in the form:

   (ar + bt)(cr + dt)

which multiplies out to:

   ac*r^2 + (ad + bc)*rt + bd*t^2

You don't have to memorize that. The important thing is that the 
coefficient of r^2 will be the product of a and c, and the coefficient 
of t^2 will be the product of b and d. So the first thing to do is to 
factor the 3 and the 6, and list (at least in your mind) all the 
products that can produce them:

        a   c
       -------
    3 = 1 * 3 or
        3 * 1

         b   d
        -------
   -6 = -1 * 6 or
        -2 * 3 or
        -3 * 2 or
        -6 * 1 or
        1 * -6 or
        2 * -3 or
        3 * -2 or
        6 * -1

(Maybe this is what you meant when you said you "multiplied the ends." 
There's no reason to multiply them, but factoring them is the first 
step.)

I don't usually make tables like this, and I don't think about a, b, 
c, and d; I just write down some parentheses and try filling them in:

   (r    t)( r    t)

Now we just have to search for a set of a, b, c, and d for which 
(ad + bc) is -7. The way I do this is to try writing the factored 
expression with some choice of values and see how it looks. In this 
case, it doesn't really matter whether I take a and c to be 1 and 3 or 
3 and 1, because that would just switch the order of the factors I end 
up with, so the only real decision is what to try for b and d. I 
usually start with something more or less balanced (2 * 3 rather than 
1 * 6) just to see what will happen:

   Goal: 3r^2 - 7rt - 6t^2

         3    6
        / \  / \
       /   \/   \
      /    /\    \
   (1r   2t)(3r   3t)
     \    \ /    /
      \    6    /
       \  +/-  /
        -- 3 --
          ===
           7 ?

The r terms give 3r^2 and the t terms give 6t^2. (We know we'll need a 
+ and a -, but we'll leave that for later.) Try multiplying the two rt 
pairs: 1r * 3t = 3rt, and 2t * 3r gives 6rt. Subtracting in either 
order won't give us -7, so this won't work. Let's try something else. 
Maybe switch the factors of 6:

         3    6
        / \  / \
       /   \/   \
      /    /\    \
   (1r   3t)(3r   2t)
     \    \ /    /
      \    9    /
       \  +/-  /
        -- 2 --
          ===
           7 ?

Now the rt terms are 1r * 2t = 2rt and 3t * 3r = 9rt. We can combine 
these to make -7: 2rt - 9rt. So we put a minus sign where it will 
multiply the 9rt term:

   (1r - 3t)(3r + 2t) = 3r^2 - 7rt - 6t^2

So we've factored it. As you can see, there's a lot of trial and 
error. If you don't feel like you know where you're going all the 
time, don't worry - we're all just feeling around when we do this. 
When you get enough experience, you get used to the process so it gets 
a little more orderly.

-Doctor Peterson,  The Math Forum
Check out our web site! http://mathforum.org/dr.math/   
    
Associated Topics:
High School Basic Algebra
High School Polynomials
Middle School Algebra
Middle School Factoring Expressions

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