Q&A #18119

Teachers' Lounge Discussion: "Bottoms up" as a method to factor trinomials

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From: Jay Hinkelman

To: Teacher2Teacher Public Discussion
Date: 2009111001:22:50
Subject: Partial disproof of bottom up method in factoring

I'll do better - I'll *disprove* it.

Or, specifically, I'll demonstrate an important circumstance in which
it does not work.

First, let's create a trinomial by multiplying two binomials in one
variable: (ax + b)(cx + d), where a,b,c,&d are all integers. We will
assume, for the sake of argument, that neither binomial contains a
common monomial factor - in other words, a & b are coprime, as are c &
d.  Multiplying, we get acx² + (ad + bc)x + bd. Now let's use the
"bottoms up" method to factor this.

The method starts by multiplying the leading coefficient ac by the
trailing constant bd (yielding, of course, abcd in our case), and then
finding two integers whose product is abcd and whose sum is (ad + bc).
In our case, those numbers are - duh - ad & bc. Obviously, when actual
numbers are involved, finding these two numbers takes some effort.

Now, here's where the bottoms up method diverges from the commonly
taught "grouping method":
* We write our product of two binomial factors as (x + bc)(x + ad).
* We next "divide" the two constant terms by the leading coefficient
of the original trinomial, which in our case is ac. This gives us:
     (x + bc/ac)(x + ad/ac)
* Next we reduce the fractions; this produces (x + b/a)(x + d/c).
* Since d & c are coprime, and b & a are as well, the two fractions
can be reduced no further.
* Now, according to this method, we "move" the denominator of each
fraction in front of the variable within its binomial, where it
becomes a coefficient: (x + d/c) becomes (cx + d),  and (x + b/a)
becomes (ax + b). Our answer is (ax + b)(cx + d).

Now, this method appears to work just fine, and in practice it's much
smoother than the description above would indicate. For example:

1.      3x² + 2x - 16
2.    3(-16) = -48
3.    Two numbers with a product of -48 and a sum of 2: 8 and -6.
4.    Write (x + 8)(x - 6).
5.    Write the leading trinomial coefficient, 3, under the two
constant terms as a denominator:
         (x + 8/3)(x - 6/3)
6.    Reduce: (x + 8/3)(x - 2)
7.    Move any denominators in front of the respective x's:
         (3x + 8)(x - 2) is the correct answer.

Having gone through all that, there are two major problems with this
method. The first is that, in steps 5 and 7, it forgoes proper
algebraic logic for "magic steps." That's something to be considered
when trying to teach students to think algebraically.

The bigger problem is that if the trinomial has any common monomial
factors that aren't factored out first, the METHOD DOESN'T WORK!!! In
the "proof" I wrote above, we assumed that a&b and c&d were coprime
pairs, creating a trinomial with no common monomial factor. But in
reality, algebra students fail to correctly factor out the GCF all the
time. For example, let's factor 2x² + 4x + 2 using the bottoms-up
method, and let's pretend we don't see the common factor of 2. (The
correct answer should be 2(x+1)(x+1) or 2(x+1)² .)

2*2 = 4. We need two numbers with a product of 4 and a sum of 4.
Obviously, those numbers are 2 & 2.

We write (x + 2)(x + 2).

We put the leading trinomial coefficient 2 under the constant terms as
a denominator: (x + 2/2)(x + 2/2). This reduces to (x + 1)(x + 1).
Notice that we have the correct binomial factors, but the monomial GCF
has completely disappeared!

This is guaranteed to happen if the trinomial includes a GCF that
isn't factored out first:

kacx² + k(ad + bc)x + kbd = kacx² + (kad + kbc)x + kbd

Product = k²abcd

Find target numbers kad & kbc.

Write (x + kad)(x + kbc). Place denominators: (x + kad/kac)(x +
kbc/kac). The k's all cancel.

So, I do *NOT* recommend teaching this method to beginning algebra
students - they commonly (rim shot) miss GCFs, and an unnoticed GCF
spoils the method.

-- Jay Hinkelman
Tutoring Coordinator, Ivy Tech Community College of Indiana (East
Central Region)

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