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Polynomial Functions

Division Algorithm
In this page we show how to divide one polynomial by another. We first explain what it means to divide one
polynomial by another.

Let and be two polynomials . To divide by means to find two additional
polynomials and such that



where the degree of is less than the degree of . The polynomial is called the quotient
and the polynomial is called the remainder.

The polynomial doing the dividing, , is called the divisor, and the polynomial being divided, , is called
the dividend.
Example 1: Before looking at an example involving polynomials, let’s look at an example involving
integers. Divide 27 by 4.

Solution : Remember, to divide 27 by 4, means to find two integers. The quotient, which tells us the
largest number of times 4 goes into 27, and the remainder which tells us how much is left over. To find the
quotient we list multiples of 4.

4, 8, 12, 16, 20, 24, 28

Looking at the list we see that the largest number of times 4 goes into 27 is six. The remainder is
. Thus, we have . The quotient is 6 and the remainder is 3.
Example 2: Divide the polynomial by the polynomial .
Solution: The first step is to write each polynomial so that the exponents are decreasing in size. is
already in that order, but we need to write as . The next step is to display the two
polynomials as follows

What do we have to multiply x by to get 3x2? The answer is 3x. The entire procedure is shown below.

In a) we write 3x above the horizontal line, because 3x times x equals 3x2; 3x2 is then subtracted from the
dividend . The next term x is then brought down, and (see b) is written above the horizontal
line. Note 1 times x equals x. The x is then subtracted from and the 2 is brought down. The degree of the
constant polynomial 2 is zero, which is less than the degree of the divisor x. We now know that



Thus, the quotient is and the remainder is 2. This can also be written

Historical Comment: The use the word algorithm means a step by step breakdown of a complicated
mathematical procedure . The word comes from the name of a famous Islamic astronomer/mathematician,
, who is considered the founder of algebra. In fact the word algebra comes from the title of a
famous book, Al-Jabrwa-al-Muqabilah, which was written by .

Example 3: Divide the polynomial by the polynomial .

Solution: First a few observation about what the quotient and remainder have to look like .

• The remainder is a polynomial of degree one or less because the divisor, , is a polynomial of
degree 2. The remainder is a polynomial of degree less than the degree of the divisor. Hence its degree
is 1 or less.

• The quotient is a polynomial of degree 1 because the leading term of the divisor is 3x2 and the leading
term of the dividend is 4x3. Thus, we need to multiply 3x2 by to get 4x3, and the degree of is
1. Another way to get this is to subtract the degree of the divisor from the degree of the dividend.

• The computations are shown below:

• The row is obtained by multiplying (the divisor) by .
• The row is obtained by subtracting from the dividend


• The row is obtained by multiplying by -1, and the last row is obtained by
subtracting from , the row immediately above.

• We stop at this point since the degree of the last row is less than the degree of the divisor.
• The remainder is and the quotient is . Other ways to write this are:

and

Example 4: Divide the polynomial by the polynomial .

Solution: Before we show the details of the division process. Let’s see what we can surmise about the
remainder and quotient.

• The degree of the quotient is 3, because . The leading coefficient of the quotient is 3
because 6 equals 3 times 2. That is, .

• The degree of the remainder is no more than 1, as the degree of the divisor is 2. It is possible that the
degree is 0, but we can’t know for sure without doing the actual computations.

• In the details below, notice that we included in the dividend the missing terms. That is the missing
term x4 is included as 0x4, etc. This is done to help eliminate computational errors.

The quotient is and the remainder is .

Example 5: Divide by .

Solution: Notice that the degree of the quotient is 2 and its leading term is 1. The remainder must be a
constant, since it will be a polynomial of degree less than 1.

The remainder is 0 and the quotient is . This can also be written as

or

Notice that since the remainder is zero, the dividend is a multiple of the divisor.

Here we state the division algorithm for polynomials, and discuss synthetic division . Synthetic division is a
shorthand way to divide a polynomial by a polynomial of the form x - c. Note this last polynomial has degree
1 and its leading coefficient is also 1. Synthetic division is only used when the divisor is a polynomial of
degree 1 with leading coefficient 1 also.

The Division Algorithm
Let and be any two polynomials. Then there exist unique polynomials and ,
where the degree of is less than the degree of , such that



is called the quotient, the remainder, the divisor, and is called the dividend.

In the previous pages we showed how to calculate the quotient and remainder. So the only thing new in the
above statement is the uniqueness of the quotient and remainder.

If the divisor is a polynomial of degree 1, with leading coefficient also 1, (the divisor is of the form x - c,) then
there is a process called synthetic division which enables us to find the quotient and remainder fairly easily.

In the example below we divide by the polynomial . In fact we do the division twice.
The first time using long division and the second time using synthetic division.
Example 6: Divide by .

Solution:

The long division of by should be clear by now. However the process of synthetic
division needs some explanation.

There are three rows shown and the procedure to obtain rows two and three is as indicated. The first row is
obtained from the coefficients of the dividend, this must include zeros for missing terms. These coefficients are
placed to the right of the vertical line and above the horizontal line. Be sure to leave space for the second row.
The negative of the divisor ’s constant term (in this example the constant term is - 2) is placed just to the left of
the vertical line. The first entry of row 3 is the leading coefficient of the divident. The remaining entries in
rows two and three are obtained as indicated. Notice the last term in the third row is the remainder and the
remaining terms of the third row are the coefficients of the quotient.

The last row in the synthetic division is 1, 4, 3, 8. Thus, we have

and


One question which remains is how do we know that the quotient is a polynomial of degree 2? Well, the
dividend is a polynomial of degree 3 and the divisor is a polynomial of degree 1, and 3 -1 = 2.

One last remark is in order before you look at some of the examples below.

Suppose we divide by x - c, then the remainder is a constant k, which may be zero, and we have the
equality



If we evaluate this equality at , we get .

Ohhh, the remainder when is divided by is . This is important, make sure you understand and
remember it.
Question: If is divided by x - 1, what is the remainder?
Answer: If is divided by x - 1, the remainder is



Question: If is divided by x + 1, what is the remainder?
Answer: If is divided by x + 1, the remainder is

For complicated polynomials if the value of is desired it is sometimes easier to use synthetic division
to calculate the remainder, and then use the fact that the remainder is . Although, since we now have ready
access to computers and calculators, evaluating polynomials is an easy task.

Example 7: Using synthetic division, divide by .

Solution: We take notice that the degree of the quotient is 3 and that the x3 term is missing in , but
we’ve included it below by placing a zero in the top row.

Thus, the quotient equals and the remainder equals 74.

Example 8: Using synthetic division divide by .
Solution: We have to be careful here. Synthetic division assumes that we are dividing by x - c. So in
this particular example .

The quotient equals , the remainder equals - 745, and .

Example 9: Divide , by using synthetic division.
Solution: Note that

The quotient equals , the remainder equals 0, and

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