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Q&A #104

Teachers' Lounge Discussion: Factoring trinomials

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From: Elizabeth Appelbaum <eappelbaum@kc.rr.com>
To: Teacher2Teacher Public Discussion
Date: 2005012621:30:56
Subject: Re: factoring trinomials

My previous message was truncated: I intended:
Thank you for posting

Here's what you posted: 

From: Elizabeth Appelbaum <eappelbaum@kc.rr.com>
To: Teacher2Teacher Public Discussion
Date: 2005012621:26:00
Subject: factoring trinomials

	
I read the interesting discussion on factoring trinomials. I suggest
that factoring is overemphasized in the United States. The main reason
for factoring is to solve equations. If the equation can be solved
more easily with the quadratic formula, then that method should be
used. In applied mathematics students encounter quadratic functions,
most of which do not factor. Students should recognize easy factoring,
like x^2 - 5x + 6 = (x  -2)(x - 3), zeros 2 and 3. Most of the time
they should use the quadratic formula.

Students should be encouraged to factor over the reals and complex
numbers ,rather than over the integers, so they learn the fundamental
theorem of algebra: a polynomial of degree n > 0 with complex
coefficients factors into n linear factors with complex coefficients.
Equivalently, if the polynomial has real coefficients, it factors into
linear and irreducible quadratic factors with real coefficients.

The quadratic formula can be used to factor a trinomial. For example,
the polynomial 6x^2 - x - 12 (Lloyd) has zeros 3/2 and -4/3, so it has
linear factors (x - 3/2) and (x + 4/3). Since the leading coefficient
is 6, the complete factorization is

     6(x - 3/2)(x + 4/3)

If you want integral coefficients, note the expression equals
          (2)(3)(x - 3/2)(x + 4/3) = (2x - 3)(3x + 4)






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