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Factoring Trinomials in Quadratic Equations

Date: 02/15/2004 at 01:06:16
From: Shoaib
Subject: Solving quadratic equations by factoring

I'm stuck on this question: 15x^2 - 16x - 7 = 0

My book teaches me two methods to factor quadratic equations: guess 
and check, and grouping.  I can't decide which method to use.



Date: 02/15/2004 at 14:53:29
From: Doctor Greenie
Subject: Re: Solving quadratic equations by factoring

Hello, Shoaib -

Factoring trinomials is a very interesting process because there are
many different ways that people approach it and think about it.  Many
students find a method that works for them and stick with it, but
finding that method may involve trying out lots of different ways.  
I'm going to show you a couple of ways you may not have seen before,
and hopefully one of them will help you. 

One way that ALWAYS works is to use the quadratic formula, as 
explained very nicely on the following page in the Dr. Math archives:

  Quadratic Trinomials
    http://mathforum.org/library/drmath/view/52902.html 

This page also shows an alternative method for performing the 
factorization.

Using the method involving the quadratic formula on your example, we 
have

  15x^2 - 16x - 7 = 0

  x = [16 + or - sqrt(16^2 - 4(15)(-7))]/30

  x = [16 + or - sqrt(676)]/30

  x = [16 + or - 26]/30

  x = 42/30 = 7/5  or  x = -10/30 = -1/3

Then we reconstruct the quadratic equation from which we obtained 
these roots:

  15(x - (7/5))(x + (1/3)) = 0

  5(x - (7/5)) * 3(x + (1/3)) = 0

  (5x - 7)(3x + 1) = 0

So the factorization of the original quadratic expression is

  15x^2 - 16x - 7 = (5x - 7)(3x + 1)

Below are explanations of a couple of other ways to factor quadratic 
expressions.  One of them may be the "grouping" method you say you are 
familiar with; the other one I suspect you haven't seen before.

A Grouping Method...

We have the quadratic expression

  15x^2 - 16x - 7

(1) We first multiply the leading coefficient and the constant to get 
a "target number":

  (15)(-7) = -105

(2) Next we find two numbers a and b whose product is this "target 
number" and whose sum is the coefficient of the middle term:

  a*b = -105
  a+b = -16

The numbers we want are -21 and 5.

(3) Now we "group" the middle term using these two numbers we have 
found:

    15x^2 - 16x - 7 = 15x^2 - 21x + 5x - 7

(4) And now we factor by grouping:

     15x^2 - 21x + 5x - 7
  = (15x^2 - 21x) + (5x - 7)
  = 3x(5x - 7) + 1(5x - 7)
  = (3x + 1)(5x - 7)

Here's another closely related method.  We start with the same 
quadratic expression

  15x^2 - 16x - 7

(1) First, we arbitrarily multiply the constant term by the leading 
coefficient and change the leading coefficient to 1; this gives us in 
this case

  x^2 - 16x - 105

  (So we multiplied the -7 by 15 and changed the "15" to "1"....)

(2) Next we factor in this form:

  x^2 - 16x - 105 = (x - 21)(x + 5)

(3) Now we multiply BOTH the leading coefficients by the original 
leading coefficient; this gives us the "factorization"

  (15x - 21)(15x + 5)

(4) Finally, remove the common factor from each term to get

  (5x - 7)(3x + 1)

Try any (or all) of these methods on a few more examples you have and 
find the one which you understand the best....

I hope this helps.  Please write back if you have any further 
questions about any of this.

- Doctor Greenie, The Math Forum
  http://mathforum.org/dr.math/ 
Associated Topics:
High School Polynomials
Middle School Factoring Expressions

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