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 ![]() ![]() ![]() ![]() polynomials ![]() ![]() ![]() where the degree of ![]() ![]() ![]() and the polynomial ![]() The polynomial doing the dividing, ![]() ![]() 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 ![]() ![]() |
Example 2: Divide the polynomial
![]() ![]() Solution: The first step is to write each polynomial so that the exponents are decreasing in size. ![]() already in that order, but we need to write ![]() ![]() 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 |
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, ![]() famous book, Al-Jabrwa-al-Muqabilah, which was written by ![]() Example 3: Divide 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, ![]() 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 ![]() ![]() 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
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Example 4: Divide 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 ![]() 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 |
Example 5: Divide
![]() ![]() 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
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 ![]() ![]() ![]() ![]() where the degree of ![]() ![]() ![]() ![]() ![]() ![]() ![]() 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 ![]() ![]() The first time using long division and the second time using synthetic division. |
Example 6: Divide
![]() ![]() Solution:
The long division of
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Suppose we divide ![]() equality ![]() If we evaluate this equality at ![]() ![]() Ohhh, the remainder when ![]() ![]() ![]() remember it. |
Question: If ![]() Answer: If ![]() ![]() Question: If ![]() Answer: If ![]() ![]() |
For complicated polynomials ![]() ![]() to calculate the remainder, and then use the fact that the remainder is ![]() access to computers and calculators, evaluating polynomials is an easy task. Example 7: Using synthetic division, divide ![]() ![]() Solution: We take notice that the degree of the quotient is 3 and that the x3 term is missing in ![]() we’ve included it below by placing a zero in the top row.
Thus, the quotient equals
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Example 8: Using synthetic division divide
![]() ![]() 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
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Example 9: Divide
![]() ![]() Solution: Note that ![]()
The quotient equals
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