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Factoring Quadratic Trinomials


Date: 06/02/98 at 01:42:30
From: Anonymous
Subject: Factoring

I need to learn how to factor correctly. I sort of have it, but not 
all the way. Can you help me please?

I'm trying to figure out how to factor y^2 + 8y + 16. How do you know 
how to correctly find the greatest common factor and factor it?


Date: 06/02/98 at 09:13:05
From: Doctor Rob
Subject: Re: Factoring

In general, you will be trying to factor quadratic trinomials that 
look like this:  

   a*y^2 + b*y + c

where a, b, and c are expressions not involving y. In your case above, 
a, b, and c are constants, 1, 8, and 16, respectively.

If this polynomial factors, it will have two factors that look like 
this: 

   (r*y + s)(t*y + u)

Multiplying this out, and setting the coefficients of y^2, y^1 = y, 
and y^0 = 1, from the two polynomials equal, you get three equations:

   r*t = a
   r*u + s*t = b
   s*u = c

You want to find r, s, t, and u to make these equations true. Now:

   a*c = (r*t)*(s*u) = (r*u)*(s*t)
   b = (r*u) + (s*t)

This means that you are looking for two numbers d = r*u and s*t such 
that their product is a*c and their sum is b. In your case, you are 
looking for two numbers whose product is 1*16 = 16, and whose sum is 
8. Now run through all positive and negative factors d of a*c, and 
write down d and (a*c)/d. Pick out the one (if any) such that d and 
(a*c)/d add up to b. In your case, here is the table:

    d     16/d    sum
   -16     -1     -17
    -8     -2     -10
    -4     -4      -8
    -2     -8     -10
    -1    -16     -17
     1     16      17
     2      8      10
     4      4       8
     8      2      10
    16      1      17

There are some shortcuts available. First of all, you might as well
choose |d| <= |a*c/d|, so that d is the smaller in absolute value of 
the two numbers for which you are looking. Secondly, d must have the 
same sign as a*b*c. That will cut down the number of rows in the table 
by a factor of nearly 4. In your case, only the 6th, 7th, and 8th 
rows survive.

The one which works is 16 = 4*4, 4 + 4 = 8. If there are none which 
work, then (if you have done this correctly) the polynomial cannot be 
factored. Now write:

   a*y^2 + b*y + c = r*t*y^2 + (r*u + s*t)*y + s*u 
                                              (substituting a, b, c)
                   = r*t*y^2 + r*u*y + s*t*y + s*u       (expanding)
                   = (r*t*y^2 + r*u*y) + (s*t*y + s*u)    (grouping)
                   = r*y*(t*y + u) + s*(t*y + u) 
                                           (removing common factors)
                   = (r*y + s)*(t*y + u)  (removing a common factor)

In your case,

   y^2 + 8*y + 16 = y^2 + (4 + 4)*y + 16
                  = y^2 + 4*y + 4*y + 16
                  = (y^2 + 4*y) + (4*y + 16)
                  = y*(y + 4) + 4*(y + 4)
                  = (y + 4)*(y + 4)
                  = (y + 4)^2

You try to do these:

1. Factor y^2 + 10*y + 25.
2. Factor y^2 - 7*y + 12.
3. Factor y^2 - 7*y - 18.
4. Factor 2*y^2 + 7*y - 4.
5. Factor y^2 + y + 1.

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

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