Download 5 Locating Zeros of a Polynomial Function

Advanced Mathematical Concepts
Chapter 4
Lesson 4-5
Example 1
Determine between which consecutive integers the real zeros of f(x) = x3 + 3x2 - 4x + 6 are located.
There are three complex zeros for this function. According to Descartes’ Rule of Signs, there are two or zero
positive real roots and one negative real root. You can use substitution, synthetic division, or the TABLE
feature on a graphing calculator to evaluate the function for consecutive integral values of x.
Method 1: Synthetic Division
r
-6
-5
-4
-3
-2
-1
0
1
2
3
1
1
1
1
1
1
1
1
1
1
1
3
-3
-2
-1
0
1
2
3
4
5
6
-4
14
6
0
-4
-6
-6
-4
0
6
14
6
-78
-24
6
18
18
12
6
6
18
48
Method 2: Graphing Calculator
Use the TABLE feature.
The change in sign between -5 and -4 indicates that a zero exists between -5 and -4. This result is consistent
with Descartes’ Rule of Signs.
Advanced Mathematical Concepts
Chapter 4
Example 2
Approximate the real zeros of f(x) = 5x3 - 2x2 - 4x + 1 to the nearest tenth.
There are three complex zeros for this function. According to Descartes’ Rule of Signs, there are two or zero
positive real roots and one negative real root.
Use the TABLE feature of a graphing calculator. To find the zeros to the nearest tenth, use the TBLSET feature
changing Tbl to 0.1. There are zeros between -0.9 and -0.8, between 0.2 and 0.3, and at 1.
Since 0.36 is closer to zero than -0.665, the zero
is about -0.8.
Since 0.16 is closer to zero than -0.245, the zero
is about 0.2.
The third zero occurs at 1.
Advanced Mathematical Concepts
Chapter 4
Example 3
Use the Upper Bound Theorem to find an integral upper bound and the Lower Bound Theorem to find
an integral lower bound of the zeros of f(x) = x3 + 5x2 - 2x - 8.
The Rational Root Theorem tells us that 1, 2, 4, and 8 might be roots of the polynomial equation x3 +
5x2 - 2x - 8 = 0. These possible zeros of the function are good starting places for finding an upper bound.
f(x) = x3 + 5x2 - 2x - 8
1
5
-2
-8
r
1
1
6
4
-4
2
1
7 12
16
f(-x) = -x3 + 5x2 + 2x - 8
-1
5
2
-8
r
1
-1
4
6
-2
2
-1
3
8
8
3
-1
2
8 16
4
-1
1
6 16
5
-1
0
2
2
6
-1
-1
-4 -32
An upper bound is 2. Since 6 is an upper bound of f(-x), -6 is a lower bound of f(x). This means that all real
zeros of f(x) can be found in the interval -6  x  2.
Example 4
HOUSING The homeless rate in a large urban city is modeled by the function
f(x) = -0.0025x4 + 0.0082x3 - 0.0317x2 - 0.1624x + 3.05, where x represents the number of months since
January, 2002 and f(x) represents the homeless rate as a percent. Use this model to predict when the
homeless rate will be 1.6%.
You need to know when f(x) has a value of 1.6.
1.6 = -0.0025x4 + 0.0082x3 - 0.0317x2 - 0.1624x + 3.05
0 = -0.0025x4 + 0.0082x3 - 0.0317x2 - 0.1624x + 1.45
Now search for the zero of the related function,
g(x) = -0.0025x4 + 0.0082x3 - 0.0317x2 - 0.1624x + 1.45
r
4
5
-0.0025
-0.0025
-0.0025
0.0082
-0.0018
-0.0043
-0.0317
-0.0389
-0.0532
There is a zero between 4 and 5 months.
Confirm this zero using a graphing calculator.
The zero is about 4.3 months. So, about
4 months after January, 2002 or May, 2002,
the homeless rate would be 1.6%.
-0.1624
-0.318
-0.4248
1.45
0.178
1.0216