2.6: Finding Real Zeros of Polynomial Functions
Learning Objectives
 Use polynomial long division and synthetic division to find roots
 Apply synthetic division to find the rational zeros of an unfactored polynomial (of degree >2)
Synthetic Division
As you know from years of algebra, any linear equation , can be easily solved. You have also learned how to find a solution to a quadratic equation using the quadratic formula. In this section, we will learn how to solve polynomial equations of higher degrees. We begin with a process called the synthetic division. It is a special case of the polynomial long division method. To develop the method, we will present the following example in which the divisor has the form . We will first use the method of long division and then redo the division by the synthetic division method.
Keep in mind that if we wish to divide
where is the dividend and is the divisor, then
where is the quotient and is the remainder. We can also rewrite this equation in the following form
as a solution.
Let's say we are interested in dividing by .
By long division, we get
As you can see, all the work involved is to manipulate the coefficients of the variables and the constants. Thus we could just as easily complete the division by omitting the variables, as long as we write the coefficients in the proper places. So the division problem will look like this:
Since the underlined numerals are repetitions of those immediately above them, we can shorten the process by simply deleting them. Further, since these underlined numbers are products of the numbers in the quotient by the 1 in the divisor, we eliminate this 1. Thus we get the following
It is also unnecessary to bring down the 10 and 13:
Moving the numerals upward, we get
When the numeral 1 (the first number in the quotient) is brought down to the last line, it will contain the remainder and the quotient,
Therefore, the quotient is and the remainder is . This is the method of the synthetic division.
Example 1
Use synthetic division to find the quotient and the remainder of
Solution
First we write the divisor in the form , as . Then use 2 as a “divisor” in the synthetic division as follows:
Notice that 0 is used as the coefficient of the “missing” term. Also, we wrote the coefficients of the dividend in descending order. The process of synthetic division is as follows: Bring down the first coefficient 3 and multiply by 2 (the divisor) to get 6, and then add 0 to 6 to get 6. Next, multiply 6 by 2 (the divisor) to get 12, and then add 8 to 12 to get 4. Finally, multiply 4 by 2 to get 8, and then add 1 to 8 to get the remainder, 7. As a result of this process, the quotient is
and the remainder is
In other words, since
Remember, this method will only work when the divisor is in the form , that is, when the coefficient of in the divisor is 1.
The Remainder Theorem and The Factor Theorem
This example leads us to the Remainder Theorem.
The Remainder Theorem
If a polynomial of degree is divided by , then the remainder is a constant and it is equal to the value of the polynomial when is substituted for . That is
Proof
The division algorithm states
since the degree of the remainder must be less than the degree of the division, , then must be a constant. We will denote it by . In addition, if , we will find,
The Factor Theorem
If is a polynomial of degree and , then is a factor of the polynomial . Further, if is a factor, then is a zero of .
Proof
If , then by the Remainder Theorem,
and the division algorithm becomes,
This shows that is a factor of .
In addition, since is a factor of , then
and is thus a zero of .
Example 2
Use synthetic division and the remainder and factor theorems to find the quotient and the remainder if is divided by .
Solution
Hence
Notice that the remainder is 181 and it can also be obtained if we simply substituted into ,
Example 3
If , evaluate
Solution
We can simply use the synthetic division to evaluate at the given values. By the remainder theorem, the remainder is equal to .
 Using synthetic division,
Hence, .
 By synthetic division,
Hence, .
 By synthetic division,
Hence, .
Example 4
Show that is a factor of . Find the quotient and express is factored form.
Solution
By the factor theorem, if , then is a factor of the polynomial. In other words, if the synthetic division produces a remainder equal to zero, then is a factor of the polynomial. Using the synthetic division with ,
Hence, , and the quotient is
so that can be written as
Rational Zero Theorem
Recall that the zeros of a polynomial represent the intercepts on the graph. It is easy to find the zeros of first or second degree polynomials but as the degree increases, the problem of determining the zeros becomes more difficult. There is a general method however for finding the zeros of polynomials (if they exist) for degrees . Before we state the method it is important to recall one of the fundamental properties of integers: Every positive integer can be expressed as a product of prime numbers. For example, , and so on. Here 5, 2, and 11 are prime numbers.
The Rational Zero Theorem
Given the polynomial
and is a positive integer. If the coefficients are integers and is a rational zero in lowest terms, then is a divisor of and is a divisor of .
Example 5
Use the theorem above and synthetic division to find all the possible rational zeros of the polynomial
Solution
From the rational zero theorem, is a rational zero of the polynomial . So is a divisor of 2 and is a divisor of 1. Hence, can take the following values: 1, 1, 2, 2 and can be either 1 or 1. Therefore, the possible values of are
So there are four possible zeros. Of these four, not more than three can be zeros of because is a polynomial with degree 3. To test which of the four possible candidates are zeros of , we use the synthetic division. Recall from the remainder theorem if , then is a zero of . We have
Hence, 2 is a zero of . Further, by the division algorithm,
The remaining zeros of are simply the zeros of which is easier to manipulate,
and thus the remaining zeros are 1 and 1. Thus the rational zeros of are 1, 1, and 2.
Example 6
Use the theorem above and synthetic division to find all the possible rational zeros of the polynomial
Solution
Assume is a rational zero of . By the rational zero theorem, is a divisor of 6 and is a divisor of 1. Thus and can assume the following respective values
and
Therefore, the possible rational zeros will be
Notice that with these choices for and there could be rational zeros. But, eight of them are duplicates. For example . The next step is to test all these values by the synthetic division (we'll let you do this on your own for practice) and we finally find that
are zeros of . That is
Keep in mind that the rational zero theorem has only a limited ability in finding the zeros of a polynomial equation. For example, the equation can not be solved by using the rational zero theorem because the roots of this equation are irrational numbers.
Graphing Polynomials Using Rational Zero Theorem
Example 7
Graph the polynomial function .
Solution
Notice that the leading term is , where odd and . This tells us that the end behavior will take the shape of a power function with an odd exponent.
Here, as you can see, there is no straightforward way to find the zeros of . However, with the use of the factor theorem and the synthetic division, we can find the rational roots of . First, we use the rational zero theorem and find that the possible rational zeros are
testing all these numbers by the synthetic division,
1 is not a root. Now let's test .
we find that 1 is a zero of and so we can rewrite ,
Looking at quadratic part,
and so
Thus 1 and are the intercepts of . The intercept is
Further, the synthetic division can be also used to form a table of values for the graph of :
We choose test points from each interval and find .
Interval  Test Value  Sign of  Location of points on the graph  

1  28    below the axis  
  below the axis  
3  4  +  above the axis 
From this information, the graph of is shown in the two graphs below. Notice that the second graph is a magnification of in the vicinity of the axis.
Descartes’ Rule of Signs
One challenge for finding the zeros of polynomials is that you may not know how many roots a polynomial actually has. Descartes Rule of Signs helps with this problem. Descartes' Rule of Signs is given as a procedure for finding the number of possible positive and negative roots of a polynomial.
Descartes Rule of Signs
Given any polynomial, ,
 Write it with the terms in descending order, i.e. from the highest degree term to the lowest degree term.
 Count the number of sign changes of the terms in . Call the number of sign changes .
 Then the number of positive roots of is less than or equal to .
 Further, the possible number of positive roots is
 To find the number of negative roots of , write in descending order as above (i.e. change the sign of all terms in with odd powers), and repeat the process above. Then the maximum number of negative roots is .
Example 8
Use Descartes Rule of Signs to identify the possible number of positive and negative roots of .
Solution
First, rewrite in descending order
The number of sign changes of is 2, so the number of positive roots is either 2 or 0.
For the negative roots, write
The number of sign changes of is 2, so the maximum number of negative roots is 2.
The graph of below shows that there is one negative root and two positive roots.
Exercises
 Use (i) long division and (ii) synthetic division to perform the divisions. Express each result in the form .
 by
 by
 by
 Use synthetic division to find and so that if and
 If , use synthetic division to determine the followings:
 What are the factors of ?
 Find so that is a factor of
 Use synthetic division to determine all the zeros of the polynomials.
 Graph the polynomial function by using synthetic division to find the intercepts and locate the intercepts.
 Graph the polynomial function by using synthetic division to find the intercepts and locate the intercepts.
 Let . Find the solution of the following equations:
 Refer back to Problem 6 above. Use the graph of to find the solution set of the inequality .
 Use the graph of to find the solution set of the inequality .
Answers
 Answers:

 21
 31
 10
 66
 Answers:
 intercepts: 2, 1, 3; intercepts: 6
 intercepts: 1, 2; intercepts: 4
 Answers: