Note: In the following I use the words power, and order somewhat interchangeably, in relation to the exponent of x.

Please also forgive the rather bad formatting for some of the expressions in this text. WordPress and LaTeX are somewhat unforgiving.

One of the topics which seems to have caused the most problems in the assignments for MAM1000W so far this year is that of polynomial division. Thus I want to go through an example here to show you exactly what we’re doing when we think about performing such a calculation.

If you think you already know what to do, but want to have some more practice, plug in some random polynomials into this page and make sure that you get the same steps as them.

In fact, the first question might be: what’s the point? Often the expression looks more complicated in the end than it does at the beginning. Well, part ofthe answer is that sometimes there’s nothing to do at all, but sometime’s there is. It’s important to ask what we are trying to achieve. A polynomial division question can be thought of very much like a normal fraction. If I give you the fraction:

You can think about ‘simplifying’ this to be

In what way is this simpler? Well, when you see the fraction, it’s hard to get an idea of how big it is when it’s written in the original form. What we’ve done in writing it out the second way is to say that is a little more than 3 and a little less than 4.

What has this got to do with polynomial division? Well we can think of the integer part in the above final result (ie. the 3) as being like having a polynomial sitting on its own – it is complete, and not being divided by another polynomial. Very importantly, the second term – ie. the remainder part in the fraction calculation, is equivalent to a one polynomial divided by another where the order of the polynomial in the numerator is strictly less than the order of the denominator. Just in the same way that the integer in the numerator above (2) is strictly less than the integer in the denominator (7). There’s really a direct correspondence, but rather than thinking about the magnitude of the integers in the fraction, with think about the order of the polynomial.

If you were given the fraction and asked to perform a similar calculation as above, the answer would be simply . If you take 2 and divide it by 5, you don’t get any integer plus a remainder term. It’s just the remainder term.Â So, if you are given a polynomial division where the order of the numerator is less than the order of the denominator, then you don’t have to do anything at all. This is known in fact as being in proper rational form. ie:

is already in proper rational form and you don’t have to do anything, but:

is not in proper rational form as the order of the numerator is 4 and the order of the denominator is 2, and so we can think about what we have to do to get it into the form of a polynomial plus an expression in proper rational form.

What we are trying to do is, given an expression of the form:

where a(x) and b(x) are polynomials in x, write it in the form:

where c(x) and d(x) are polynomials in x. Note importantly that just as in the integer fraction example at the top, the denominator of the final term (7 in the integer fraction case), we want to have the same denominator in the remainder term as we had in the original expression.

So, let’s see what we can do with the above calculation.

We want to calculate:

We are going to essentially do long division by polynomials in a very long-hand way, but I want to do this to show you that every step of the process is really perfectly acceptable to your mathematics.Â When we do it in the normal way of long division we will be taking short-cuts, but those short-cuts are simply short-hand for the long way of doing it that we show here.

We start by seeing what theÂ term with the highest power of x is in the denominator. We see that it’s the term. Let’s see which terms in the numerator are of higher power than . In fact here there’s just one term: . Now we are looking at a much simpler question: What’s divided by . Well, that answer is easy, it’s .

You can see that this might be thought of as a first approximation of the above calculation.

Now we’re saying:

What is that something? Well, let’s multiply everything in the above expression by the denominator:

and expand out:

Now we see that there is a on both sides, so we can simplify the above equation. Let’s manipulate it to give:

and therefore:

So, what we seem to have found so far is:

Notice that we started with a fourth order polynomial divided by a second order polynomial. Now we have a polynomial plus a third order divided by a second order. The orders in the fraction on the right hand side are closer than when we started.

We’re still not in proper rational form on the right hand side because the order is higher in the numerator than in the denominator. Let’s look at this term and repeat exactly what we’ve just done for this term:

We see that the highest power in the numerator is the term, again (it had to be as the denominator hasn’t changed from our original denominator. Let’s see which terms in the numerator we can divide by to give polynomials. Well, we can clearly divide the , but we can also happily divide the . These first two terms, when divided by give: . So, let’s write:

where again we need to figure out what is by multiplying through by the denominator again:

Now we multiply out the brackets, which gives:

Again, the important thing to note here is that the highest powers of x cancel on both sides. We can now simplify this to give:

So:

Now we see that the numerator and denominator are of the same power, so we are really almost there. We just need to do this last fraction. Let’s remind ourselves though what we’ve got so far. We found that:

but we just showed that:

and:

so, putting everything together, we have:

Now let’s look at this last expression. Again, the highest power in the denominator is the term. We see what we can divide by this in the numerator (to give a non-negative power of x), and it’s only the term. So:

Again, multiply everything by the denominator:

which gives:

Again, see that the highest powers cancel. This is by design. Finally we then get:

Aha! We see that now this new term has a higher power in the denominator than it does in the denominator – ie. it’s in proper rational form. Let’s put it all together:

which gives:

I know that this looks like a lot of work, but that’s because we’ve taken the long route. Make sure that you agree with every expression in here. Now let’s do the short route. Note that at one step we will make a slightly different choice.

Step 0: This is the expression we want to calculate (I’ve put 0’s in as placeholders):

Step 1: Spot that is the highest power in the denominator and divide through the terms in the numerator which have the highest power, at least as high as :

Step 2: Now we multiply the term at the top by the denominator, just as you would for regular long division:

Step 3: And now subtract this from the original numerator:

Step 4: Now we use this new term as the expression that we are dividing the denominator by. Now let’s just divide the term by the and we get a contribution :

Step 5: Now multiply this new contribution by the denominator to get:

Step 6: Now subtract again, to get:

Step 7: Now we have a term with $22x^2$ which can be divided by the term in the denominator to give a contribution of , so write this again at the top:

Step 8: Multiply the by the denominator to give:

Step 9: and subtract:

Now this last line has a maximum power which is less than that of the denominator, therefore this is our remainder.

Note that there were a couple of places in this calculation where we had a choice. In particular, in step 4 we could also have divided the term by the but we chose here the highest power of . Prove to yourself that this would have given the same result.

So, the answer to the question:

is

MikeMarch 6, 2016 at 2:13 pmNice post for those who have been struggling with this. Any chance you would consider doing a post of this sort of step by step nature with regards to mathematical induction? I know that at least a few people have been finding it difficult to properly grasp the concept, and would likely find it helpful.

Jonathan ShockMarch 6, 2016 at 2:25 pmHi Mike, I plan on doing precisely that.

FranceApril 18, 2017 at 12:57 amThis really was a big deal of help to me, I only realised now that I have been making the same mistake over and over when it came to polynomial division. But I was wondering, when will you do proof by Induction and proof by contradiction?, if you have already done it can I you please post the link no the comments.

FranceApril 18, 2017 at 12:58 amI mean on the comments section

Jonathan ShockApril 18, 2017 at 6:58 amHi France, here are four posts on proof by contradiction:

http://www.mathemafrica.org/?p=12493

http://www.mathemafrica.org/?p=12520

http://www.mathemafrica.org/?p=11735

http://www.mathemafrica.org/?p=13399

India BabbMarch 3, 2018 at 8:08 pmThank you thank you thank you!

Jonathan ShockMarch 3, 2018 at 8:09 pmNo problem at all.

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