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```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,

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

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

factorization.

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...

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)

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

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

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