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Polynomial Functions


This lesson covers polynomial functions. The graphing, writing, and use of the functions are


After completing this lesson, you should be able to

• identify a polynomial function;

• evaluate a polynomial function using synthetic division and determine its zeros;

• use synthetic division to apply the remainder and factor theorems;

• graph a polynomial function and determine an equation for a polynomial graph;

• write a polynomial function for a given situation and find the maximum or minimum value of
the function;

• use technology to approximate the real roots of a polynomial equation;

• solve polynomial equations by various methods of factoring and the RATIONAL ROOT

• apply several theorems about polynomial functions.


Chapter 2, Sections 2-1 through 2-7 (pages 53–93)


Section 2-1: Zeros and Factors of Polynomial Functions, pp. 53–58

A polynomial function involves an equation that can be written in the following form:

The powers of x can be any number, and they should be written in decreasing order. Examples of
polynomials follow with key terms for each.


The roots of a polynomial function are x-values that give the function a value of zero. The roots of
P (x) = 3 x − 7 and P (x) = 3 x2 + 2 x − 8 are found below:

is a root and a zero for P (x) = 3 x − 7.

and −2 are roots and zeros for P (x) = 3 x2 + 2 x − 8.

Other values of the function can be found by placing desired values into the function.

Example 1: Find P (−3) given P (x) = 2 x2 − 4 x + 6:

Example 2: Let's try problem 18b on page 56 of your text.

Find k (1 + i) given k(x) = x2 (x2 + 16)

Synthetic substitution is another method that can be used to find values of functions. I will use
synthetic substitution to determine the value of f (−5) when f (x) = 2 x3 − 3 x2 + 4 x + 5.

List all of the coefficients. Drop the leading coefficient down, multiply 2 by −5, and place in the

Add −3 and −10. Multiply −13 by −5 and place in the box:

Add 4 and 65. Multiply 69 by −5 and place in the box:

Add 5 and −345:

The result is −340, so f (−5) = − 340.

Let's check it out:

For this problem, synthetic substitution would have been the quicker method. The synthetic
substitution can be done in one step as follows

Example 3: Use synthetic substitution to find P (−2) when P (x) = x5 − 1.

There are several missing terms in x5 − 1. The complete polynomial equation is
x5 + 0 x4 + 0 x3 + 0 x2 + 0 x − 1. All terms must be used in synthetic substitution:

In this problem P (−2) = (−2)5 − 1, but −32 − 1 = − 33 is quicker. Which method would be
quicker for P (−1) = x15 − 2? Think of all the missing terms! This would be quicker using
substitution. Use the method best suited for the situation.

Example 4: Let's try problem 24 on page 57 of your text.

If 2i is a zero, then f (2i) = 0:

Example 5: Let's try problem 30b on page 57 of your text.

Given g (x) = 3 − 8 x, find g (x + 2) − g (x):

Study Exercises

Complete odd-numbered problems 1–27 in the Written Exercises section on pages 56–57 of your
text. Then check your answers in the back of the text.

Section 2-2: Synthetic Division: The Remainder and Factor Theorems, pp. 58–61

Synthetic substitution is versatile. Used not only to find the value of a function, it can also be used to
find the quotient and remainder in a polynomial division problem. When it is used in this manner, it
is referred to as synthetic division. I will work through a polynomial division problem and compare
the results to those in a corresponding synthetic division.

  subtract Multiply

Since I divided by x − 2, I will use 2 in my synthetic division. If I had divided by x + 3, I would
have used −3 in my synthetic division:

The remainder is 4 and 3 x2 + 1 x + 3 is the quotient.

If the binomial x − 2 would have been a factor of 3 x3 − 5 x2 + x − 2, the remainder would have
been zero. So if my only desire is to determine if x − 2 is a factor of 3 x3 − 5 x2 + x − 2, I don't need
to divide it out. I could use synthetic division and just look at the remainder. Or, I could use
substitution. In this case, I would find P(2) given P(x) = 3 x3 − 5 x2 + x − 2:

P(2) = 3 (2)3 − 5 (2)2 + 2 − 2.

P(2) = 4 is the same as the remainder.

This demonstrates the REMAINDER THEOREM.


When a polynomial P (x) is divided by x − a, the remainder is P (a)

So, depending on the problem, I could choose synthetic division or substitution to determine if x − a
is a factor of P(x). If I wish to divide a polynomial by a polynomial of a degree higher than 1, I need
to use long division.