Polynomial Long Division
Polynomial Long Division
In this section you will learn how to rewrite a rational function such as
in the form
The expression
is called the quotient, the expression
is called the divisor and the term
is called the remainder.
What is special about the way the expression above is written?
The remainder 28x+30 has degree 1, and is thus less than the degree of the divisor 
.
It is always possible to rewrite a rational function in this manner:
DIVISION ALGORITHM: If f(x) and  are polynomials, and the degree of d(x) is less than or equal to the degree of f(x), then there exist unique polynomials q(x) and r(x), so that
and so that the degree of r(x) is less than the degree of d(x). In the special case where r(x)=0, we say that d(x) divides evenly into f(x). |
How do you do this? Let's look at our example
in more detail. Write the expression in a form reminiscent of long division:
First divide the leading term 
of the numerator polynomial by the leading term 
of the divisor, and write the answer 3x on the top line:
Now multiply this term 3x by the divisor , and write the answer
under the numerator polynomial, lining up terms of equal degree:
Next subtract the last line from the line above it:
Now repeat the procedure:
Divide the leading term 
of the polynomial on the last line by the leading term
of the divisor to obtain -11, and add this term to the 3x on the top line:
Then multiply "back": 
and write the answer
under the last line polynomial, lining up terms of equal degree:
Subtract the last line from the line above it:
You are done! (In the next step, you would divide 28x by , not yielding a polynomial expression!)
The remainder is the last line: 28x+30, and the quotient is the expression on the very top: 3x-11. Consequently,
The easiest way to check your answer algebraically is to multiply both sides by the divisor:
then to multiply out:
and then to simplify the right side:
Indeed, both sides are equal! Other ways of checking include graphing both sides (if you have a graphing calculator), or plugging in a few numbers on both sides (this is not always 100% foolproof).
Let's use polynomial long division to rewrite
Write the expression in a form reminiscent of long division:
First divide the leading term 
of the numerator polynomial by the leading term x of the divisor, and write the answer on the top line:
Now multiply this term by the divisor x+2, and write the answer
under the numerator polynomial, carefully lining up terms of equal degree:
Next subtract the last line from the line above it:
Now repeat the procedure:
Divide the leading term 
of the polynomial on the last line by the leading term x
of the divisor to obtain -2x, and add this term to the on the top line:
Then multiply "back": 
and write the answer
under the last line polynomial, lining up terms of equal degree:
Subtract the last line from the line above it:
You have to repeat the procedure one more time.
Divide:
Multiply "back":
and subtract:
You are done! (In the next step, you would divide -9 by x, not yielding a polynomial expression!)
The remainder is the last line: -9 (of degree 0), and the quotient is the expression on the very top: 
. Consequently,
Let's use polynomial long division to rewrite
Write the expression in a form reminiscent of long division:
First divide the leading term of the numerator polynomial by the leading term of the divisor, and write the answer x on the top line:
Now multiply this term x by the divisor 
, and write the answer
under the numerator polynomial, carefully lining up terms of equal degree:
Next subtract the last line from the line above it:
Now repeat the procedure:
Divide the leading term 
of the polynomial on the last line by the leading term
of the divisor to obtain -5, and add this term to the x on the top line:
Then multiply "back": 
and write the answer
under the last line polynomial, lining up terms of equal degree:
Subtract the last line from the line above it:
You are done! In this case, the remainder is 0, so divides evenly into 
.
Consequently,
Multiplying both sides by the divisor yields:
In this case, we have factored the polynomial , i.e., we have written it as a product of two "easier" (=lower degree) polynomials.
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Helmut Knaust
