16-year-old Helena Muffly wrote exactly 100 years ago today:
Thursday, October 26, 1911: Have such difficult algebra problems. So much work it is to find the H.C. F. and L.C.M. Good bye for me if we happen to get one of these in examination.
Her middle-aged granddaughter’s comments 100 years later:
Important: If you aren’t into math—skip my comments today and come back tomorrow. Suffice it to say that Grandma was doing some fairly difficult algebra.
But, if you enjoy math here’s my take on what this diary entry is talking about–
First I’ll give an example of the L.C.M. (lowest common multiple) and H.C.F. (highest common factor) of two whole numbers (integers); then I’ll explain how it’s done for algebraic expressions.
The L.C.M. is the smallest integer that two whole numbers can be divided by. 1 would always be the L.C.M.
For example, for 8 and 12 the L.C.M. would be 1.
The H.C.F.(highest common factor) is the largest integer that two whole numbers can be divided by.
For the same two numbers (8 and 12), the H.C.F. would be 4.
If the H.C. F. is 1, it is a prime number.
The basic idea is the same as for algebraic expressions. For example, for H.C.F. of 2ab and 4a2b is 2a.
But it quickly gets complicated. I’m going to give you directions and examples from a 1911 algebra textbook below for H.C.F. [An aside: If you really want to understand this concept you might find the information on the algebrahelp.com website helpful.]
Now, here are the directions for finding the H.C.F. in Durrell’s School Algebra (1912):
The method of finding the H.C.F. is to:
Factor the given expressions, if necessary:
Take the H.C.F. of the numerical coefficients:
Annex the literal factors common to all of the expressions, giving to each factor the lowest exponent which it has in any expression.
Ex. 1: Find the H.C. F. of 6x2y – 12xy2 + 6y3 and 3x2y2 + 9xy3 – 12y4
6x2y – 12xy2 + 6y3 = 6y(x – y)2
3x2y2 + 9xy3 – 12y4 = 3y2(x2 + 3xy – 4y2) = 3y2(x + 4y)(x – y)
H.C.F. = 3y(x – y)
Whew, I’m getting a headache just typing these expressions. But if you’re still with me, here’s a couple problems you could try from the 1911 textbook:
Find the H.C.F.
1. 4a2b , 6ab2
2. x2 – 3x , x2 – 9
3. x2 + x , x2 – 1 , x2 – x – 2
4. 4a3x – 4ax3 , 8a2x3 – 8ax4 , 4a2x2(a – x)2
5. 3a2 – 10a + 3 , 9a – a3 , (3 – a)3