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I was watching a first-year high-school-algebra student struggle with factoring quadratics last night. Given a quadratic $ax^2+bx+c$ (I'll give you the exact example in a moment), her method — presumably her teacher's — was as follows: find the factors of $ac$, and see which pair add up to $b$.

It seems to me that multiplying $a$ by $c$ is needless work. True, it's $ac$ whose factors sum to $b$. But when writing out the factors as (say) $(a_1x+c_1)(a_2x+c_2)$, one's actually working with not the factors of $ac$ but rather the $c_i$ and the $a_i$, factors of, respectively, $c$ and $a$.

So my question is: Is there any advantage to working with $ac$ — finding its factors, seeing which ones sum to $b$ — and, if so, what is that advantage?

Here's the example she was working, so you get a better under standing of what I mean. The problem was (or amounted to) $9x^2-47x+60=0$. This poor girl found $9\cdot60$ and started examining its factors to see which sum to $47$. Eventually, she hit upon the answer, $20\cdot27$, and put them in her parentheses as $(9x-20)(x-3)$ (somehow divining that the $27$ was to be split up as $9\cdot3$, and the $9$ as $9\cdot1$; I'm not sure how she hit upon that).

My method would have been instead to consider $(9x-c_1)(x-c_2)$ or $(3x-c_1)(3x-c_2)$. (I'd reject the latter because $3\nmid47$, but I wouldn't expect that of my high schooler. So consider both possibilities.) Then find factors of $60$ that possibly fit in one of those pairs of parentheses, and hit upon $3\cdot20$.

Again, what if anything is the advantage to factoring $ac$? (The advantage, if any, may be pedagogic.)

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    Did she multiply and then find factors of $540$, or did she continue straight to $3^2\cdot 6\cdot 10 = 2^2\cdot 3^3\cdot 5$?2012-04-30
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    @Neal, I wish she'd have done the latter (if she had to deal with $ac$ at all)! No, she found factors of $540$.2012-04-30
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    Maybe the same advantage as the supposed $1+2+3+\cdots +100$ assignment to Gauss's class. Keeps them busy.2012-04-30
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    I think that pedagogically, it creates uniformity with the method for factoring $1x^2 + bx + c$ (that is, find factors of $c$ that add to $1$).2012-04-30
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    @Neal: But factoring $a$ and $c$ also creates pedagogical uniformity. The special case $a=1$ is factor $1$, factor $c$.2012-04-30

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