Marveling At The Historical

Math Oldies But Goodies

  • About This Blog

    This blog is mostly about math procedures in textbooks dated from about 1825-1900. I’m writing about them because some of the procedures are exquisite and much more powerful, and simpler, than some of the procedures in current text books. Really!

    I update this blog as frequently as possible ... every 2-3 days. And, if you are a lover of old texts and unique procedures, you might want to talk to me about them, at markdotmath@gmail.com. I’m not an antiquarian; the books I have are dusty, musty, brown-paged scribbled-in texts written by authors with insights into how math works. Unfortunately, most of their procedures have vanished. They’ve been overcome by more traditional perspectives, but you have to realize that at that time, they were teaching the traditional methods.

Are We Lying About Factoring?

Posted by mark schwartz on May 9, 2016

This is a short piece. I didn’t mean to do this but I was playing with factoring and here’s the thought I had.

Using a simple example, the factors of x2 + 2x + 1 are (x + 1)(x + 1). However, what is also a set of factors of this trinomial is ( ̶  x  ̶ 1) ( ̶  x  ̶ 1). Although this latter is technically correct, it is not traditionally correct. If a student in class gave you this as an answer, would you accept it? Technically, yes; traditionally, no. You know what a computer-based delivery system would say … and that’s because it’s been programmed by traditionalists.

So, why  won’t we accept  ( ̶  x  ̶ 1) ( ̶  x  ̶ 1) as a pair of factors? We teach factoring as a prelude to solving an equation. The expression x2 + 2x + 1 becomes an equation by simply stating that the expression equals something, like 5 or y. In solving the equation using the factoring method, would it make a difference?

Given the equation x2 + 4x + 1 = – 3, adding 3 to both sides gives x2 + 4x + 4 = 0. Solving this by factoring we get (x + 2)(x + 2) = 0 and setting both to zero, x equals – 2. If we now set the factors to be (-x – 2)(-x – 2) and set both to zero, x again equals – 2. So far, it seems to work.

When wouldn’t it work? I don’t know. Just a thought to share with you …

 

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