Teacher2Teacher 
Q&A #18119 
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From: Jay Hinkelman <jhinkelm@ivytech.edu> To: Teacher2Teacher Public Discussion Date: 2009111000:22:50 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 bottomsup 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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