Synthetic Division When the Linear Coefficent 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 for Derivatives of Polynomials

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."

I will try to answer b in my next blog (soon...)

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 that for x=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 first derivative...... 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.

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 Divison 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....

But what if we have a seventh power polynomial and want to divide it by a cubic and we don't want to do it without breaking the cubic down... can it be done...?

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...

**** This blog has been expanded to include this method and some other little known uses of synthetic division.  

Synthetic Division, Beyond the Basics