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Are Binomial Factors Unique?Date: 04/22/2003 at 19:07:17 From: Joe Subject: Factoring Hello, I know how to factor a given polynomial into two binomials, starting like this, x^2 -5x+6 ... (x )(x ) then listing the factors of the third term 2*3, 6*1 and of course 2+3 is equal to 5 and I would place 2 and 3 into the binomials (x 2)(x 3) and then I would add the appropriate sign for each binomial (x-2)(x-3) However, I was wondering if there is any way that in the third term, where most of the time there is more than one pair of factors, more than one pair of factors could be used for the middle term.
Date: 04/23/2003 at 14:34:55
From: Doctor Dotty
Subject: Re: Factoring
Hi Joe,
One way of thinking about a question like this is to use a graph.
The graph of a quadratic where the coefficient of the x^2 term is
positive has this general shape:
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* | *
|
|
* | *
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---------*-|-----*-------
*| *
| *
|
|
|
When the coefficient of the x^2 term is negative, the graph is the
same shape, but the other way up. In both instances it crosses the
x-axis a maximum of two times.
The main reason we factorise a quadratic like x^2 - 5x + 6 is
to make it easy to find its roots, i.e., the points where its
graph crosses the x-axis.
Now, suppose we've factored x^2 - 5x + 6 as follows:
x^2 - 5x + 6 = 0
(x - 3)(x - 2) = 0
This tells us that the graph of the line crosses the x-axis when
x=3, and when x=2. (Do you see why?) Now suppose we factored it
a different way, to get
x^2 - 5x + 6 = 0
(x - b)(x - c) = 0 where b, c are not equal to 2 or 3
This would mean that the graph is crossing the x-axis at different
places! Which means it would have to be the graph of a different
quadratic, since each quadratic can have only one graph that
represents it. So if you've found a way to factor a quadratic,
you can be sure that it's the _only_ way.
Does that make sense?
- Doctor Dotty, The Math Forum
http://mathforum.org/dr.math/
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