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.

Archive for the ‘factoring’ Category

Blended Factoring

Posted by mark schwartz on May 23, 2016

Introduction:

Factoring trinomials is a core operation in Algebra and is more than a stand-alone procedure. Consider how it is used in solving quadratic equations. Consider the basic algorithm for factoring. I’ve seen about 7 different ones, one of which I developed, which I call the “blended” method. This method is called blended because the core of it is the traditional procedure but it is blended with a set of operations which many students have found more comfortable. For this method, it is expected that students have studied and can perform identifying and factoring out a common factor from a binomial. So, a review of it, if not covered recently, should be done. Being able to do this is an essential part of the Blended Method. This became a student handout and was given to them at the time of practicing the procedure.

The Story:

When factoring a trinomial of the form ax2 + bx + c, traditionally one always looks for and factors out any common factor, referred to as a greatest common factor (GCF). Further, the method for factoring is typically presented in two steps: first, address a trinomial of the form ax2 + bx + c where a = 1, and then later, when a >1. Then there are special cases – the perfect square trinomial and the difference of two squares, which are handy but not essential. All factoring of these can be done with what is described as the blended method.  It is not necessary to find the GCF, nor to pause to recognize and recall the procedure for the special cases.

It starts with a trinomial of the form ax2 + bx + c, with a = 1 but is not limited to this. This allows for exploring why the blended method works when a = 1 and also when a > 1.

For the first demonstration of how this method works, factor x2 + 5x + 6. The key thing to notice is that the coefficient of x2 is 1. So, as in the traditional method, set up two binomials as (1x     )(1x     ) but include the coefficient of x2.  It may seem odd to include the coefficient of 1 in both binomials, but it provides for a clear transition to factoring trinomials when a>1. Have students write it in cases where the coefficient of the x2 term is 1.

Step two is to factor both ‘a’ and ‘c’, combining these factors into one set. Typically, only pairs of factors of both ‘a’ and ‘c’ are needed but using prime factoring is very helpful. In this case the factors would be 1•2•3 (again, it seems odd but include the 1). This set then has to be played with until two products add to ‘b’, the coefficient of x. For example, 1 +  2•3 = 7; 1•2 + 3 = 5, as does 1•3 + 2 = 5. Either of these latter two is acceptable. Again, this may seem trivial because ‘1’ is one of the factors but it establishes and reinforces a pattern which isn’t quite so obvious. More examples will be provided later with different sets of factors.

Place the ‘2’ and ‘3’ as the second term in each binomial, and it makes no difference which binomial. So, the factors of x2 + 5x + 6 are (1x + 2)(1x + 3). Again, the coefficient of 1 for the x term need not be written once students are comfortable with the procedure, but for now, they are to use it to demonstrate and emphasize the procedure. At this point, the next step is to examine each of the binomials for a common factor, and if any are found, factor them out. In this case, there are none, but in later examples this step will demonstrate why it is to be done. Try another one.

Factor x2 – x – 12. Step one gives (1x     )(1x     ). Step two gives several possibilities because ‘c’ is negative and one of the factors has to be negative. However, the factors don’t have to be handled as a positive and a negative; they could both be treated as positive, with the awareness that one of the products is subtracted from the other to give the value of  ─1, the coefficient of ‘b’.

Twelve has 3 sets of factors: 1•12, 2•6, and 3•4. Again, notice that prime factoring wasn’t the beginning point, but in some cases, prime factoring is recommended. Which pair can give a difference of −1? The only pair with a difference of 1 is 3 and 4, and since we’re looking for −1, the two factors are 3 and −4. Again, multiplying either of these by 1 seems trivial (and perhaps it is in this case), so the two factors are still 3 and −4¸ and the factors of x2 – x – 12 are (1x +3 )(1x −4 ). Again, look for a common factor in each binomial, factoring it out if found. In this case, there are none. More practice with a = 1 is recommended before moving to cases where a>1. This is to establish that the coefficient of ‘x’ in both binomials is the coefficient of the x2 term in the trinomial. It also provides practice in handling trinomials where ‘c’ is negative.

Now try factoring 2x2 – 11x – 6. In this case, the two initial binomials are (2x     )(2x     ). Notice that the coefficient of ‘x’ in both binomials is the coefficient of the x2 term in the trinomial. The factors of ‘ac’ are 1•2•2•3. These factors have to be played with until two numbers result such that their difference is ‘b’, or – 11. The outcome is 1 and 2•2•3, or 1 and 12. The 12 has to be negative because the sum has to be – 11.  So, the two binomial factors of 2x2 – 11x – 6 are (2x + 1)(2x −12). The next step is examining each binomial for a common factor, and in this case, (2x −12) gives 2(x – 6). So, we now have (2x + 1)•2•(x – 6). The question is whether or not the factor of 2 is part of the final factoring or not. To determine this, divide this answer by the coefficient of x2 in the original trinomial. The outcome is (2x + 1)(x – 6), and these are the factors of 2x2 – 11x – 6.

Here’s a summary of the steps for the blended method. Given a trinomial of the form ax2 + bx + c, where a ≥ 1:

  1. Write 2 binomials with only a first term in each of (ax    )(ax    ), where ‘a’ is the coefficient of the x2 term. Indicate that this expression will be divided by ‘a’, giving the expression.
  2. Factor ‘a’ and ‘c’, producing a set of factors containing those of ‘a’ and ‘c’. Prime factors are not required but typically make the work easier.
  3. Use this set of factors to determine two values – “m” and “n” – which added together give the coefficient of “b”.
  4. Write the two binomials (ax & m)(ax & n), where the “&” could be either a plus or a minus.
  5. Extract any common factor in each binomial, identifying them as c1 and c2.
  6. Cancel all common factors in c1, c2 and “a”.
  7. This gives the factors of ax2 + bx + c, two binomials and in some cases a constant.

 

 

More practice is to be done with this procedure using examples from the text, but a demonstration of how to use it for the two special cases – perfect square trinomial or difference of two squares – follows.

Factor 4x2  ̶   12x + 9. First, there is no need to identify if this is a perfect square trinomial. It seems that it’s likely to be since a = 4 and c = 9, both of which are perfect squares. But for the moment, ignore that.

Step 1 then is to write (4x  )(4x   )/4 , based on step 1 of the procedure.

Step 2 gives the set of prime factors 2⦁2•3•3.

Step 3 has to give two products whose sum is 12, and that is 2⦁3 and 2⦁3. But notice that it’s a minus twelve in the trinomial so, the two products have to be   ̶ 6 and    ̶ 6.

Step 4 gives the expression (4x   ̶ 6)( 4x   ̶ 6)/4

Step 5 gives 2(2x  ̶ 3)⦁2⦁(2x  ̶ 3)/4

Applying step 6, gives the answer of (2x  ̶  3)(2x  ̶  3).

Now factor 8 x2   ̶  18. At first, it appears as though this might be the difference of two squares but it isn’t. There is a greatest common factor, and if that isn’t noticed, the question is where to start. Try the blended method.

Step 1 then is to write (8x   )(8x   )/8, based on step 1 of the procedure.

Step 2 gives 2•4•2•9.

Step 3 using 2•4•2•9 won’t give zero, the coefficient of ‘b’, so moving toward prime factoring gives 2•4•2•3•3, and this can simplify to 2•2•3 and 4•3, which is 12 and 12. Making one of them negative and adding gives zero.

Step 4 gives the expression (8x   ̶ 12) (8x   ̶ 12)/8

Step 5 gives 4(2x   ̶  3)⦁4⦁(2x   ̶  3)/8

Applying step 6, gives the answer of 2(2x   ̶  3) (2x  + 3).

Using this process seems like a rather long way to factor as opposed to learning and recognizing the special cases of a perfect square trinomial and difference of two squares …  and it may well be. However, consider learning one procedure – the blended method – or 5 procedures (a trinomial with ‘a’ equal 1, a trinomial with ‘a’ > 1, using the grouping method, a perfect square trinomial or the difference of two squares). The choice is obviously yours, but give the blended method a try. Once practiced and understood, it’s as automatic as using 4 different procedures.

One caution. Some texts include problems in the form ax3 + bx2 + cx. Notice that this is not the form

ax2 + bx + c and since it isn’t, the blended method won’t work. Given this kind of problem, the first step has to be to factor out the common term “x”, bringing the trinomial to a form that can be factored with the Blended Method.

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What is the Question to This Answer?

Posted by mark schwartz on May 11, 2016

Introduction:

In Greenleaf’s 1862 New Elementary Algebra, he devotes considerable time to fractions, including exercises which require considerable mastery of the operations with fractions. He introduces the section on fractions with a discussion of unit fractions, which are fractions with a 1 in the numerator. He establishes a very handy algorithm for adding them, and then posits the question “Doesn’t it seem that if given the answer one could identify the fractions that added together gave that answer?” He doesn’t answer his own question, but I found it intriguing enough to play with it and it led to an interesting relationship with another standard Algebraic operation.

The Story:

Given the answer to the addition of two unit fractions, can one find those two fractions?

For example, given the answer 5/6, what two fractions added together give this answer? For those familiar with adding fractions, the answer of 1/2 and 1/3 might be found after a little trial and error. On the other hand, finding the “question” to the answer 14/24 might take a little longer.

But there is one caveat of caution to be considered throughout this discussion, particularly if you are the one creating the answers to be used. In all cases, if the answer is one that can be reduced, use the non-reduced form first. If not, it is possible that the reduced form will give no “question” or a “question” that is correct but different from the one used to generate the answer. For example, 14/24 can be reduced to 7/12, and the answer 14/24 gives the question 1/2 + 1/12, whereas the answer 7/12 gives the question 1/3 + 1/4. So, in presenting this kind of problem, students may be asked to find both sets of answers. But continuing this caveat of caution, 20/96 can be reduced to 5/24 but 5/24 gives no answer. So, when generating the answer, check out the possible questions.

Let’s start by identifying some useful algebraic strategies that already exist and can be used to find the question to the answer.

There is a method presented in Algebra books for finding the sum of two fractions without finding a lowest common denominator. Yes, there is such a procedure and it is:

m/a + n/b = (mb + an)/ab

For purposes of this discussion, remember the caveat provided earlier. Further, since we are discussing this premised on unit fractions, the value of ‘m’ and ‘n’ in this expression would both be equal to 1, and the expression is then

Expression 1.               (a + b)/ab

This expression is linked to the quadratic equation! Take a look.

Using ax2 + bx + c = 0, where a = 1, the solution is to find two factors of the constant ‘c’ which when added together are equal to ‘b’, the coefficient of the x-term, which is exactly what is seen in Expression 1. In the quadratic, the value “b” is the numerator of Expression 1, while the value “c” is the denominator of Expression 1. Here’s how to use this.

We were given the answer of 5/6 and asked to find the two fractions that added together gave this answer. If the numerator 5 is seen as the coefficient of the x-term ‘b’ in the quadratic (the sum of two factors) and the denominator 6 is seen as the constant ‘c’ in the quadratic (the product of two factors), the quadratic equation can be stated as

x2  – 5x + 6 = 0

The x-term is negative because the values in the two fractions are all positive, and when solving the quadratic equation for the values of x, it becomes apparent that the factors of the constant term both have to be negative and therefore when added together give a negative value for the coefficient of the x-term.

Factoring this equation gives (x – 2)(x – 3) = 0 and setting both to zero and solving for x gives 2 and 3, which are the denominators of the unit fractions 1/2  and 1/3, which when added together give the answer 5/6.

So, try another one. Given the answer 8/15, what two fractions added together give this answer? Using the above concept, we have

x2  – 8x + 15 = 0

Factoring, we get (x – 3)(x – 5) = 0 and setting both to zero and solving for x gives 3 and 5, which are the denominators of the unit fractions 1/3  and 1/5, which when added together give the answer 8/15.

This type of problem may never appear in today’s texts but if it ever does, thank Mr. Greenleaf for providing an algorithm for solving it.

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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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