In this section we’re going to take a brief look at dividing
polynomials. This is something that
we’ll be doing off and on throughout the rest of this chapter and so we’ll need
to be able to do this.
Let’s do a quick example to remind us how long division of
Example 1 Divide
Let’s first get the problem set up.
Recall that we need to have the terms written down with
the exponents in decreasing order and to make sure we don’t make any mistakes
we add in any missing terms with a zero coefficient.
Now we ask ourselves what we need to multiply to get the first term in first
polynomial. In this case that is . So multiply by and subtract the results from the first
The new polynomial is called the remainder. We continue the
process until the degree of the remainder is less than the degree of the divisor, which is in this case. So, we need to continue until the degree of
the remainder is less than 1.
Recall that the degree
of a polynomial is the highest exponent in the polynomial. Also, recall that a constant is thought of
as a polynomial of degree zero.
Therefore, we’ll need to continue until we get a constant in this
Here is the rest of the work for this example.
Okay, now that we’ve gotten this done, let’s remember how
we write the actual answer down. The
There is actually another way to write the answer from the
previous example that we’re going to find much more useful, if for no other
reason that it’s easier to write down.
If we multiply both sides of the answer by we get,
In this example we divided the polynomial by a linear
polynomial in the form of and we will be restricting ourselves to only
these kinds of problems. Long division
works for much more general division, but these are the kinds of problems we
are going to seeing the later sections.
In fact we will be seeing these kinds of divisions so often
that we’d like a quicker and more efficient way of doing them. Luckily there is something out there called synthetic division that works
wonderfully for these kinds of problems.
In order to use synthetic division we must be dividing a polynomial by a
linear term in the form . If we aren’t then it won’t work.
Let’s redo the previous problem with synthetic division to
see how it works.
Example 2 Use
synthetic division to divide by .
Okay with synthetic division we pretty much ignore all the
x’s and just work with the numbers
in the polynomials.
First, let’s notice that in this case r=4.
Now we need to set up the process. There are many different notations for
doing this. We’ll be using the
The numbers to the right of the vertical bar are the
coefficients of the terms in the polynomial written in order of decreasing
exponent. Also notice that any missing
terms are acknowledged with a coefficient of zero.
Now, it will probably be easier to write down the process
and then explain it so here it is.
The first thing we do is drop the first number in the top
line straight down as shown. Then
along each diagonal we multiply the starting number by r (which is 4 in this case) and put this number in the second
row. Finally, add the numbers in the
first and second row putting the results in the third row. We continue this until we get reach the
final number in the first row.
Now, notice that the numbers in the bottom row are the
coefficients of the quadratic polynomial from our answer written in order of
decreasing exponent and the final number in the third row is the remainder.
The answer is then the same as the first example.
We’ll do some more examples of synthetic division in a
bit. However, we really should
generalize things out a little first with the following fact.
Given a polynomial P(x)
with degree at least 1 and any number r
there is another polynomial Q(x),
called the quotient, with degree
one less than the degree of P(x)
and a number R, called the remainder, such that,
Note as well that Q(x)
and R are unique, or in other words,
there is only one Q(x) and R that will work for a given P(x) and r.
So, with the one example we’ve done to this point we can see
Now, let’s work a couple more synthetic division problems.
Example 3 Use
synthetic division to do each of the following divisions.
(a) Divide by
Okay in this case we need to be a little careful
here. We MUST divide by a term in the
form in order for this to work and that minus
sign is absolutely required. So, we’re
first going to need to write as,
and in doing so we can see that .
We can now do synthetic division and this time we’ll just
put up the results and leave it to you to check all the actual numbers.
So, in this case we have,
[Return to Problems]
(b) Divide by
In this case we’ve got r=6. Here is the work.
In this case we then have.
[Return to Problems]
So, just why are we doing this? That’s a natural question at this point. One answer is that, down the road in a later
section, we are going to want to get our hands on the Q(x). Just why we might want
to do that will have to wait for an explanation until we get to that point.
There is also another reason for this that we are going to
make heavy usage of later on. Let’s
first start out with the division algorithm.
Now, let’s evaluate the polynomial P(x) at r. If we had an actual polynomial here we could
evaluate P(x) directly of course, but
let’s use the division algorithm and see what we get,
Now, that’s convenient.
The remainder of the division algorithm is also the value of the
polynomial evaluated at r. So, from our previous examples we now know
the following function evaluations.
This is a very quick method for evaluating polynomials. For polynomials with only a few terms and/or
polynomials with “small” degree this may not be much quicker that evaluating
them directly. However, if there are
many terms in the polynomial and they have large degrees this can be much
quicker and much less prone to mistakes than computing them directly.
As noted, we will be using this fact in a later section to
greatly reduce the amount of work we’ll need to do in those problems.