To go "Beyond the Basics", I suppose it is necessary to illustrate a quick review of "the basics" in the beginning. The image above is taken from a website that was active in January of 2020. This is how synthetic division is taught in thousands of classrooms across America, (millions more ignore the topic because they see it as without value). And the disclaimer about "we will see and example...." they say , "We followed the synthetic division process, but arrived at a wrong answer." but they didn't show any method of adapting to this problem....
I grew up being told that this was the extent of what synthetic division could do, and like any reasonable student, decided that it wasn't really a useful tool to worry about. But when I went in to teaching, it was still in textbooks occasionally, and it came with the same set of warnings. One day in a boring staff meeting, I decided to test that hypothesis. The results of that exploration are in the sections below.
The most basic form of synthetic division is to have a quadratic or higher polynomial divided by a linear term with a unit coefficient of the variable; something like x+3, or x-5. While to many people this is "synthetic division:, it is better known as Ruffini's rule, and was first described by Paolo Ruffini in 1804 according to Florian Cajori. This work by Ruffini actually anticipated the method of Horner. More on this historical note at the at the bottom. So let's create a simple problem, and start with the most basic approach, polynomial long division, show what the algebra really means, and then show why synthetic division was created as "an advantage" Then we will break all the rules and divide by a quadratic, and a term with a leading coefficient different than one.
IF you oppose synthetic division, then thoroughly teach the polynomial long division in a way that develops understanding.
THE BASE, OF THE "BASIC" METHOD:
Let's start with a simple problem of dividing a quadratic expression by a linear term, and then explain what it really says. We will divide 2x^2 - 2x + 5 by the linear term x+3
_______________________
x+3 )2x^2 - 2x + 5
We ask what we multiply the x in the divisor by, to get 2x^2. The answer of course is 2x, so we write that in over the bar, and multiply by all terms in the divisor.
2x
_______________________
x+3 )2x^2 - 2x + 5
2x^2 + 6x
------------------------
At this point, like any long division problem, we subtract the bottom line from the original dividend. (Subtraction is harder, and more prone to trivial mistakes than addition, and this step will explain a major difference in synthetic division.
2x
_______________________
x+3 )2x^2 - 2x + 5
2x^2 + 6x
------------------------
-8x + 5
So now, we divide x into -8x, with the obvious result of -8, and we multiply by the divisor again.
2x -8
_______________________
x+3 )2x^2 - 2x + 5
2x^2 + 6x
------------------------
-8x + 5
-8x -24
------------------- and again we subtract again to get a remainder 29 (yep,but 5 minus -24 is -17... and teachers with experience KNOW that kids mess this up incredibly often).
At this point the algebra teacher often has the student resort to polynomial multiplication of (x+3) (2x-8) and then add 29(the remainder) and sure enough they get x^2 -2x -24 +29 which simplifies to x^2 - 2x + 5, and hurrah it all checks. The student, not really sure of either the new division process, or the meaning of the multiplication process he may have learned only a few weeks ago, silently accepts the result, but really doesn't understand any thing about the operation he just preformed.
BUT, suppose we tell the students, "Let's pretend that x = 30. (we want to pick a first number greater than our remainder) Then this problem would say divide 2(30^2) -2(30) + 5, by 30 + 3.... or 1740 divided by 33. Go on do it, you can even use your calculator. They will tell you that the answer is 52 with remainder of 29 or (2*30-8) with a remainder of 29. Now do it with another number, 50 or 100 or more more more..... this is the long division they "sort of" knew from middle school.. it's just a more useful way of finding an answer whatever value of x you choose.
NOW, they may be ready to approach a shortcut to something the are closer to understanding, synthetic division by a polynomial.
Important Instruction number 1, the approach of synthetic division simply avoids writing out all the variables by use of positioning, and reduces the common mistakes of subtraction by changing the steps to an addition process.
We write the same problem of divide 2x^2 - 2x + 5 by the linear term x+3 in synthetic division:
To change the subtraction to addition, we will use only the opposite of the constant term, change the 3, to -3
Now we write the problem in the same way as the long division, but without listing the exponents (but if any terms are missing, we need to put in the missing zeros)
For this problem we begin
________________
-3 ) 2 -2 5
Yeah, lots less writing. Now we begin by bringing down the first number in the divisor, the 2, to the third row, leave a space under the 2, -2, 5. We'll use that later.
________________
-3 ) 2 -2 5
------------------
2
From here on out, it goes quick, we multiply the bottom number by -3, place that in the second row of the next column, and then we ADD (hurrah, no subtraction of a negative from a negative) Since 2 & (-3) = -6, we put that below the -3, and add putting the answer in the bottom row.
-3 ) 2 -2 5
-6 -
------------------
2 -8
And now we just keep repeating those two steps, multiply the last number on the bottom row( -8) by the divisor (-3) and add to 5
-3 ) 2 -2 5
-6 +24
------------------
2 -8 29
Now look at the bottom line, 29 is the remainder, so the -8 is a constant term, and the 2 is the coefficient of x. Yep, 2x-8 with a remainder of 29. (and we stress this a lot at early levels.... NO MATTER WHAT WE DESIGNATE THE x TO BE. (After they are comfortable with this, try showing them that even if x is less than the remainder, the arithmetic still works out. For example, if we said x was 5 in this problem, then 2x^2 - 2x + 5 = 2(25)- 10 + 5 = 45 divided by x+3=8. 45 divided by 8 is 5 remainder of five... (that means that 5 x 8 plus the remainder of 5 is the dividend, 45. .. but if we look at our answer, 2x-8 R29 the 2(5) -8 is 2. When we multiply 2 (the quotient) by 8, the divisor, we get 16. But the divisor is 45... oh wait, we add back the remainder of 29 and indeed, we get 45.
Maybe a table of values stresses this aspect
x dividend = quotient * divisor + remainder
2x^2 - 2x + 5 = 2x-8 * x+3 + 29
5 45 = 2 * 8 +29
10 185 = 12 * 13 +29
15 425 = 22 * 18 +29
And now on to the fun stuff for the teachers
Synthetic Division by A Quadratic
I was sitting in a school improvement meeting paying less attention than I ought to, and started trying to figure out how to figure out the topic above. What is below is my way.. I tried to find something about this on the internet, but only found sites that deal with linear factors, and some of them said that what I have done was not possible (I hope they are wrong)... Then when I came home I found a post on the AP Calculus Discussion site that asked how to do this very thing, with a citation for an article in the Mathematics Teacher, (March, 1980 journal (hopefully the archives go back that far), there is an article called "Synthetic Division for Nonlinear Factors" ...thanks to Lisa Lewis) that describes the process. Unfortunatly I can't get that on-line.
So here is how I did it.. I made up a problem involving a fourth power polynomial, and a quadratic divisor(A clever student could break this divsor into two linear terms and do the same in two divisions... IF he is careful about remainders, but pretend I had been clever enough to use a irreducible quadratic for the divisor)
Divide x4 + 8x3+ 15x2 + 4x + 1 by x2 + 3x + 2, using synthetic division:
The dividend is written out in the usual order with the coefficients (including any missing zeros)... and to the left two columns with the negative of the B term (3 and the C term (2) as shown below... notice that I use the -B term above the -C term.. and as usual, the bottom row will be the quotient.
The first step is still to bring down the leading coefficient.. from there there are two multiplications: the product of the -B term and the number in the bottom row go in the row with -B and the column with the last bottom number... The product of the -C term and the last bottom number will go in the column one to the right of the last bottom number. Note the locations of the -3 under the 8, and the -2 under the 15...Then we add up all the values in the second row to get a new bottom value. 
We continue this process, always keeping the product of the -C term entered one to the right of the last bottom number. 
Since we are dividing a fourth degree polynomial by a second degree polynomial, the answer will be of the second degree, and the last two cells on the bottom represent a linear remainder. In this case the quotient is x2 + 5x -2 and the remainder is 0x + 5... You can check the solution by multiplication.
Big Idea..... Gauss proved that all polynomials with real valued coefficients can be factored into a product of linear factors (like x-a) and irreducible quadratic factors (qudratics that don't have real roots, and hence can't be factored over the reals). So we can factor any real valued polynomial with nothing bigger than quadratic division....
OK, I don't know, but if I get a free moment in the next 24 hours, I think I will try extending this method to a third term and see if I get lucky...
Ok, so the remark about "breaking the cubic down" was not clear, so I will try to cover that in a day or two..... HEY, I have five classes to prep..give a guy a break...
A Little More on Synthetic Division
I wanted to follow up a few things I didn't make as clear as I might have yesterday; first the question about the leading coefficient, then the extension to dividing by a polynomial of higher than the second degree, and the remark about dividing by "breaking up a cubic"
First the question about the leading coefficient.... It doesn't matter what the leading coefficient of the dividend is, but the leading term of the divisor must have a coefficient of one...That is the same as with dividing by a linear factor...... but you can reduce the problem to an equivalent division by dividing out that coefficient...(wow, that was a mouthful.. so here is an example, using the easier linear factor )
Supppose we wanted to divide 6x2 + 16x + 10 by 3x+5.. We need to turn the 3 in 3x+5 into a one, so we divide to get x+5/3.. but this is like simplifying fractions(think of 6x2 + 16x + 10 as the numerator and 3x+5) as the denomiantor) we have to divide both the top and bottom by three to keep the equality, so we also divide 6x2 + 16x + 10 by three, and our new problem is to divide 2x2 + 16/3 x + 10/3 by x + 5/3
Keep in mind that if there is a remainder, it will not be the same in the simplified problem as in the original.. If we add 6 to the constant term of the previous problem we get 6x2 + 16x + 16 by 3x+5. Eliminating the leading term of the divisor we get 2x2 + 16/3 x + 16/3 by x + 5/3 and now when we do the synthetic division
we get a remainder of 2, instead of six.... ahhh, but we divided both terms by three, so our true remainder is 3x2=6...Keep in mind that one of the things synthetic division does is give us f(2) for a function (the dividend) of x.. If we divide f(x) by three, the value at any point (and hence the remainder on division) is going to be 1/3 of what it would have been otherwise.
And now... the higher degree issue, which seems to work out quite nicely... If we wanted to divide by a cubic we just add three lines, and continue to move the products over one column.. Here is an example dividing x5+ 6x4+4x3-39x2-122x-120 by the cubic x3+3x2-10x-24... note that we still take the negative of each coefficient of the divisor.
As we continue we see that the quotient is x2+3x+5 with no remainder...
Finally for the cryptic remark about "breaking up a cubic"... In the last problem, if we had wanted to, and if we recognized that x3+3x2-10x-24 could be factored into (x-3)(x2+6x+8), we could have done the last problem in steps. If we divide by one factor and then by the other, we will stil get the same result (although there is still need to pay special attention to remainders). In arithmetic we can divide 24 by 6, or we can divide it by 3, then divide the answer by 2... In the same way we can do the synthetic division in two steps.. I decided (no special reason.. ) to divide by the linear term first.. then the quadratic, as shown here.
Synthetic Division when the Leading Coefficient is NOT One
I was hoping this would be the last one on synthetic division, I have a problem I promised Al Harmon from Misawa, Japan I would answer, and I wanted to talk about the Law of Sines (and how much I hate when people write it upside down)... So I will defer the responses to those of you who asked about the history of synthetic divison for a later day. Today I wanted to answer the challenge to post the way to divide by ax+b without factoring anything out first.
Apparently my ex-students never throw away their notes hoping to catch me if i ever do something a different way, as I did a couple of days ago with dividing both divisor and dividend before I did the division so as to eliminate the leading coefficient of the divisor..... so in case there are other teachers who teach this out there...here is a modified synthetic division approach that will work for 3x+5 and such and still give the right remainder...
I will use the same example 6x2 + 16x + 16 by 3x+5 . To accomodate the 3 in the leading coefficent, a new line is added below the usual bottom line. After each column is summed, (except for the remainder) the result is divided by the linear coefficient, in this case 3.
Notice that after we add the first two numbers in a column, we divide that sum by three to get the actual value of the quotient for that term.... but we do NOT divide the remainder.I also wondered (I had never tried it) if you could do the same, or something similar with the higher degree divisors. I had previously divide x5+ 6x4+4x3-39x2-122x-120 by the cubic x3+3x2-10x-24, so I thought I would try 6x5+ 6x4+4x3-39x2-122x-120 by the cubic 2x3+3x2-10x-24, which Should give a quotient of 3x2 - 3x/2 +77/4 with a remainder of 159/4 x2 + 69x/2 + 342 and see what happens.....
Once more we create an extra line to divide by the coefficient of the term with the highest degree (in this case 2) and proceed as we do in the non-linear cases by shifting over one column as we go down the list of coefficients (or the negatives of them, actually).
And once more, the result emerges rather effortlessly... Ok, we have some way ugly fractions in there... but you would get them even with all the mess you create with long division...
Call that a wrap...
Synthetic Division to find the Derivative of a Polynomial
Questions, questions, questions... It seems like synthetic division has struck a cord..and I get responses asking or telling me that
a) No one should teach synthetic division;
b) I didn't show you the way to divide by linear terms like 3x-5 directly, and I had shown someone in my class a few years ago (and they remembered,... "thank you, math Spirits)...
c) I should have shown you how to find the derivative like I showed their calculus class (not sure how far back that was..I've done it a couple of times)
My answer to a) is, "If you don't want to teach synthetic division, don't; but please give me the freedom to make that choice for myself."
That leaves c) so today I will illustrate that it is almost as easy to find the value of the derivative at a given value using synthetic division as it is to find the value of a function.
I will use the simple f(x) = x3 - 3x2 + 6x - 7 and we wish to evaluate f(2) and f '(2) ...
From the factor-remainder theorem we know that if a polynomial f(x) is divided by x - b, the remainder is f(b)... so we can simply find the value by synthetic division as below.
From this we see that f(2) = 1... but how does that help us find the derivative... Calculus teachers know that f '(x) = 3x 2 - 6x + 6 , and nothing like that seems to pop out of what we see above... and if we evaluate f"(2) we get 6... but where is it in the synthetic division?
Patience,... one more run... now evaluate the quotient of the above problem at 2 again....and
Behold, as Brahmagupta supposedly wrote.. the remainder of this second division is the value of the first derivative at that point...... and I know your heart is thumping in your chest, wondering, wishing.... what if we did it again, would it be,........ could it be?
Oh, Yeah... and now the descent through the derivatives in similar form is probably clear...
SO WHY does that work?
Let's walk through the second part using some pronumeral value x=a instead of a number
Notice that in the place where we expect the remainder, we see that we get the evaluation f(a)... and when we continue down, you can see the derivative terms accumulate as the division works across to finally reveal, that f '(a) does indeed become 3a2 -6a + 6...
I love math.... You just don't have clever things like that pop out in any other discipline...
For those who would like a slightly more technical explanation, her is one from William Rose:
Write P(x) = (x-a)Q(x) + r
By Remainder Theorem, this is:
P(x) = (x-a)Q(x) + P(a)
P'(a) = lim as x--> a of [ P(x) - P(a) ]/(x-a)
= lim as x--> a of [(x-a)Q(x) + P(a) - P(a)]/(x-a)
=lim as x--> a of Q(x)
=Q(a)
So the theorem says that the quotient after dividing P by a, evaluated at a, is the derivative of P at a. This seems totally transparent when you look at P(x) = (x-a)Q(x) + P(a) in that form. Q(a) is the rate at which P is changing near x =a.
Some Notes on the History
Ruffini seems to have been the first person in Western mathematics to introduce this shortcut for finding the roots of a polynomial. The Italian Scientific Society offered a prize in 1802 for the best method for determining the roots of a polynomial equation of any degree. Ruffini was awarded the gold medal in 1804. He also wrote simpler explanations of his method in 1807 and 1813. Horner would publish his paper in the 1819 in the Philosophical Transactions. Fortunately for Horner, he had two well known ambassadors for his method in Augustus De Morgan and J R Young. Through their influence, the work by Horner swept through European Mathematical circles. Almost a century expired before Florian Cajori credited Ruffini in 1910.
Of course there was a very similar approached used by the Chinese as early as the clasic "Nine Chapters on the Mathematical Arts" before the 1st Century BC. Both D. E. Smith (1925) and Cajori (1924) recognize this as the first "synthetic division" . (For student's we can point out that this ancient text also had the Pythagorean Theorem (but not by that name) and Gaussian Elimination (also not by that name).
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