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Factoring another Equation


Date: 11/14/95 at 19:26:13
From: Anonymous
Subject: Factoring

40y squared  +  yz  -  6z squared


Date: 11/14/95 at 21:47:12
From: Doctor Ethan
Subject: Re: Factoring

Good question.

For y squared I am going to use y^2

So your question can be written 40y^2 - zy - 6z^2

When you do these problems, you are looking for two binomials 
in the form:

(ay+bz)(cy+dz) = acy^2 + (ad+bc)zy + bdz^2 = 40y^2 - zy - 6z^2

This means that ac = 40 and bd = -6
 
Let us look at what a and c can be:

a  * c  = 40
1  * 40 = 40
2  * 20 = 40
4  * 10 = 40
5  * 8  = 40
8  * 5  = 40
10 * 4  = 40
20 * 2  = 40
40 * 1  = 40

and b and d can be:

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

Whew, that is a lot of choices.  But now we can narrow them down.

We need choices so that ad + bc = -1, so now we just go through by 
trial and error.  The idea is to pick values for a and c from our possible 
values list and then see if we can find a, b, and d that will work.

In order to speed this up I am going to choose 5 for a and 8 for c.

Then let us try values for b and d.

Let b = -1 and d = 6

Then ad + bc = 30 - 8 = 22.  Well, that isn't it. 
So try b = -2 and d = 3 - then ad + bc = 15 - 16 = -1 

Hey, that worked.  Yeah!

So a = 5, b = -2, c = 8, d = 3 is what we want.

Then (5y - 2z)(8y - 3z) = 40y^2 - zy -6z^2  

Just like we wanted.  Yeah!

Write back if this doesn't make sense to you.

-Doctor Ethan,  The Geometry Forum

    
Associated Topics:
High School Polynomials
Middle School Factoring Expressions

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