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Factoring Quadratics

Date: 03/05/98 at 17:08:02
From: Jamie Stanfield
Subject: Adding and Subtracting Rational Expressions

  2            6y - 2
_____   +   ____________
y - 1       y^2 + 2y - 3

I'm totally lost I don't even know how to figure out how to factor to 
start off... also,

 2a^2 - a               6
___________    _   ____________
2a^2 + a - 3       2a^2 + a - 3

I really have no idea how to work either of these!

Date: 03/06/98 at 10:05:49
From: Doctor Rob
Subject: Re: Adding and Subtracting Rational Expressions

To factor a polynomial there are basically three steps:

   1. Identify and factor out common factors of all the coefficients.
   2. Identify and factor out variables appearing in all the terms.
   3. Factor the remaining piece, if possible.

The first step would be successful on 6*y-2, since both coefficients 
have a factor of 2: 6*y - 2 = 2*(3*y-1).

The second step would be successful on 2*a^2 - a, since the variable 
"a" appears in both terms: 2*a^2 - a = a*(2*a-1).

When the remaining piece has degree 1 (the variable appears only to 
the first power, not to any higher power), you are done. When the 
remaining piece has degree 2 or higher, it may factor into pieces of 
lower degree.

There are quite a large array of different ways to try to factor that 
kind of polynomial. I will only describe the one for factoring 
quadratic (or degree 2) polynomials.

You want to factor a quadratic polynomial of the form 

    a*x^2 + b*x + c,

where a, b, and c are some expressions not involving x, often whole 
number constants.

Multiply together a*c. Try to find a way of factoring that product 
into two numbers r and s, whose sum is b.  Thus, 

   r*s = a*c
   r + s = b.

Remember that r and s may be positive or negative! If satisfying these 
equations is impossible, then the polynomial cannot be factored.

Example:  20*x^2 + x - 12.  

          a*c = 20*(-12) = -240.  
          Can you factor -240 into two numbers whose sum is b = +1?  
          Yes, 16 and -15 will do fine.

How do we use these numbers? They are the r and s which fit into

   (a*x + r)*(a*x + s)/a = (a^2*x^2 + a*[r+s]*x + r*s)/a,
                         = a*x^2 + (r+s)*x + (a*c)/a,
                         = a*x^2 + b*x + c.

In the example, then,

   20*x^2 + x - 12 = (20*x + 16)*(20*x - 15)/20,
                   = (5*x + 4)*(4*x - 3).

In the last step we canceled factors in common between the denominator
and each factor in the numerator, which were a 4 from the first factor
and a 5 from the second.


Now you try this on your denominators y^2 + 2*y - 3 and 2*a^2 + a - 3.
In the first one, a*c = -3 and b = 2, so you want to find r and s with
r*s = -3 and r + s = 2.  In the second one, you want to find r and s 
with r*s = -6 and r + s = 1.

-Doctor Rob, The Math Forum
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Associated Topics:
High School Analysis
High School Polynomials

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