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Topic:  Conceptual approach to factoring polynomials 
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Subject:  RE: Conceptual approach to factoring polynomials 
Author:  Mathman 
Date:  Dec 7 2006 
> If you would be willing to offer your
> email address,....
I realise that may be expecting a bit much so I'll do a single example:
Factor 6x^2  11x 35
Start with this setup using the leading coefficient, you have at least ...
(6x ...)(6x ...)/6
Next, find two numbers whose *algebraic* sum [i.e. sum or difference]is 11
[inside coefficient] and whose product is (6)(35) [outside coefficients]. That
will be 10 and 21.
So far... (6x .. 10)(6x .. 21)/6
The signs are found in the same way as with simpler quadratics: he last is
negative, so they are different. The middle is the sign of the larger. So
....
(6x + 10)(6x  21)/6
Take out common factors...
2(3x + 5)3(2x  7)/6
Leaving (3x + 5)(2x  7)
It might seem difficult at first glance, but, after a few examples explained in
full at each step, students catch onto it and prefer it where all else failed
for them. I've had some learn that and other methods and use either at will.
The explanation takes much longer than to actually do it. There is still a
certain number sense needed, of course, to find the numbers in the second phase
of the solution.
David.
 
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