(Please read my last blog entry either before or after this one. They go together.)\[\begin{array}{c}\left( {nx + m} \right)\left( {px + s} \right)\\nx\left( {px + s} \right) + m\left( {px + s} \right)\\np{x^2} + nsx + mpx + ms\\np{x^2} + (ns + mp)x + ms\end{array}\]
The above should be recognized by all secondary teachers of mathematics as a generic example of using the distributive property (of multiplication over addition, as it is frequently phrased) to multiply a pair of binomials.
Simple and to the point. Anyone can do it.
What I wish to point out to those who do not notice is that \[(np)(ms) = (ns)(mp)\]
This might seem like a bit of obvious but but seemingly irrelevant trivia, except for the fact that this little fact is the key that unlocks what I believe to be the way that quadratic factoring should be done (and taught).
Take a close look at an actual example (using numbers).
\[\begin{array}{c}(2x + 3)(5x + 7)\\2x(5x + 7) + 3(5x + 7)\\10{x^2} + 14x + 15x + 21\\10{x^2} + 29x + 21\end{array}\]
This is how the multiplication of binomials should look. (Forget that mnemonic FOIL. Forget it now and forget it forever.)
Now we will see the same steps displayed in reverse order.
\[\begin{array}{c}10{x^2} + 29x + 21\\10{x^2} + 14x + 15x + 21\\2x(5x + 7) + 5(3x + 7)\\(2x + 5)(3x + 7)\end{array}\]
Take note the first step involves separating \(29x\) into \(14x + 15x\). Why choose 14 and 15? Because they add to 29 AND multiply to 210, the product of 10 and 21. The rest is just applying that same old distributive law.
Take a look at one from scratch, say \(4{x^2} + 43x + 63\)
Our first step will be to find two numbers that add to 43 and multiply to 252, which is the product of 4 and 63.
\(\begin{array}{c}(1)(252)\,\,\,add\,\,to\,\,253\\(2)(126)\,\,\,add\,\,to\,\,128\\(3)(84)\,\,\,add\,\,to\,\,87\\(4)(63)\,\,\,add\,\,to\,\,67\\(6)(42)\,\,\,add\,\,to\,\,48\\(7)(36)\,\,\,add\,\,to\,\,43\end{array}\)
Take note: the numbers on the left are no more than counting, 5 was skipped because it is not a divisor of 252, and we stop at 7 because we found the numbers we need. With these numbers we can continue:
\(4{x^2} + 7x + 36x + 63\)
\(1x(4x + 7) + 9(4x + 7)\)
\((1x + 9)(4x + 7)\)
Please please recognize that this process is highly programmable. It is actually a system based on action, not a fallback "guess and check" that is pushed on kids far far too often. Also, take note that I do accept "1" as a meaningful numeral, and do not always jump at the chance to avoid writing it.
Here is a sample with negatives: \[6{x^2} - 7x - 5\]
We start by taking pairs of factors of \[ - 30\]. Since the two numbers must add to a negative, we will make the larger number in each pair negative and keep the smaller one positive.
\[\begin{array}{c}(1)( - 30)\,\,\,add\,\,\,to\,\, - 29\\(2)( - 15)\,\,\,add\,\,\,to\,\, - 13\\(3)( - 10)\,\,\,add\,\,\,to\,\, - 7\end{array}\]
We got our two numbers pretty quickly this time, and we can continue.
\[\begin{array}{c}6{x^2} - 7x - 5\\6{x^2} + 3x - 10x - 5\\3x(2x + 1) - 5(2x + 1)\\(3x - 5)(2x + 1)\end{array}\]
Please please remember that the special cases with leading coefficient 1 are special only to someone who knows the whole story. They are not special to a student seeing them for the first time. What may be seen by those in the know as a shortcut cannot be seen by beginners as a shortcut. A shortcut is never meaningful unless and until a longer route is known.
If you would like a seemingly endless list of practice problems (with check-ability), check this out.
No comments:
Post a Comment