Discussion:  All Topics in Algebra 
Topic:  Solving Equations Using Backtracking 
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Subject:  RE: Solving Equations Using Backtracking 
Author:  Mathman 
Date:  Nov 2 2006 
> On Nov 2 2006, Mathman wrote:
> I gave a method for factoring
>
> binomials with nonsingular leading coefficient to a colleague
Let's hear about that method...
No problem. It's based upon well founded theory, which you can develop if you
care to go that far. Much as I hate and detest doing math in ASCII, I'll give a
simple example that oculd be done otherwise as readily, but the method here is
universal. It takes much longer to explain in words than to show by a number of
examples.
Example: Factor 6x^2 + x  15
Start the process with two factors that begin with 6x the whole divided by 6,
like so: (6x )(6x )/6
Now, as is usual with a leading coefficient of 1, you need to find two numbers
whose sum ...or difference as the case may be... is 1 [middle coefficient], and
whose product is the same as 6 times 15; that is, two numbers whose product is
90. The methods of teaching that part vary, as they do also in the simpler
case, and it becomes easy only with practice. In any event, the student will
come up with 9 and 10 i nthis example.
So far: (6x 9)(6x 10)/6
Now for the signs. That rule is the same as in the simpler example. The outer
sign determines whether the signs are different, and the middle is that of the
larger. So it's 9 and +10, giving the result:
(6x  9)(6x + 10)/6
Now the two small factors are easily again factored.. and the product of those
will contain the 6 in the denominator:
3(2x  3)2(3x + 5)/6
The two required factors are then (2x  3)(3x + 5).
The two factors of 6, or whatever is the denominator, are usually readily seen,
and will always factor out the denominator.
Another example:
Factor 12x^2  47x + 40
Start:
(12x )(12x )/12
Signs: Both the same since it's +40, and the larger is negative since it's
47:
===> (12x  )(12x  )/12
Numbers: I tend to use prime factors if nothing jumps out at me as in the first
example: Two numbers whose product is 12 times 40 and whose sum is 47. Prime
factors of 12 and 40 are 4, 3, 2, 2, 2, 5.
===> (12x  32)(12x  15 )/12
Remove smaller factors:
===> (4x  5)(3x  8)
As I suggested, although it works ...every time... any such process needs
practice to be proficient for a test or exam [for which simple examples should
be given.] Note in the second example I did the signs first. That is not
absolutely necessary, but is better to determine at the same time if there is a
sum or a difference for the middle coefficient.
It is worth going back to basics to understand the reasoning for this process,
but that sort of thing is not done any more these days, so the method should
suffice as an alternative.
David.
 
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