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.

An 1851 Quadratic Factoring Method

Posted by mark schwartz on April 7, 2016

I’ve never seen anything like this. In A Theoretical and Practical Treatise on Algebra (1851), Mr. Robinson demonstrates a factoring method that may have been common at that time, but in the set of old texts I have, no other author has presented it. I can see why it may have gone out of use, since there are more steps than current procedures. But nonetheless, it’s worth getting a sense of how some students were taught to factor in 1851.

This procedure applies to quadratics of the form ax2 + bx + c, so be sure to remove all common factors first and if a ≠1, divide all terms by the coefficient of x2.

Take the quadratic x2 + 7x + 12.

Step 1. The sum of two numbers gives the coefficient of the “x” term, so, a + b = 7.

Step 2. The product of two numbers gives the constant term, so ab = +12.

Step 3. Square both sides of the equation a + b = 7, giving a2 + 2ab + b2 = 49.

Step 4. Multiply “ab = +12” by 4, giving 4ab = 48.

Step 5. Subtract 4ab = 48 from a2 + 2ab + b2 = 49, giving a2 ̶ 2ab + b2 = 1.

Step 6. a2 ̶ 2ab + b2 = 1; therefore (a ̶ b)2 = 1.

Step 7. Take the square root of both sides of (a ̶ b)2 = 1, giving a ̶ b = ± 1.

Step 8. Add a + b = 7 and a ̶ b = + 1. The result is a = 4 and b = 3.

Step 9.   Subtract  a ̶ b = ̶ 1 from a + b = 7. The result is a = 3 and b = 4.

Mr. Robinson simply states that given these values for “a” and “b”, these are also the values for x. He doesn’t make it clear why this is true … but it is. Basically, his method has captured the quadratic formula but in a somewhat round-about way.

You’re done. See? It is interesting but as I say, there are too many steps compared to the number of steps in today’s procedures.

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