Chapter 4: Polynomial and
Rational Functions
Section 4-5: Locating Zeros of a
polynomial function
The Location Principle
• Suppose y=f (x) represents a polynomial function with
real coefficients. If a and b are two numbers with f (a)
negative and f (b) positive, the function has at least one
real zero between a and b.
• When we set up a table of synthetic division, a change in
signs in the remainder indicates that there is a zero
between the two integers.
Example
• Determine between which consecutive integers the real zeros of
are located.
F(x)=12x3-20x2-x+6
• 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 root.
Using synthetic division:
So the zeros according to the location
Principle, will be between -1 and 0, 0 and 1,
1 and 2
r
12 -20 -1
6
-3
12 -56
-2
12 -44
-1
12 -32
-
0
12 -20 -1
+
1
12 -8 -9
-3
2
12 4 7
20
3
12 16
4
12
+
+
Upper and Lower Bound
• The upper and lower bound theorem helps
you narrow down the possibilities of zeros
from p /q without graphing, and helps you
make sure you found all of them.
Upper Bound Theorem
• Suppose c is a positive real number and P (x) is divided by x - c. If
the resulting quotient and remainder have no change is sign,
then P (x) has no real zero greater than c. Thus c is an upper bound
of the zeros of P (x). (Zero coefficients are ignored when counting
sign changes).
• What you are looking for: In the table using synthetic division all of
the signs across the row are the same.
Lower Bound Theorem
• If c is an upper bound of the zeros of
P (-x), then –c is a lower bound of the
zeros of P (x).
Example
• 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)=6x3-7x2-14x+15
• The rational root theorem tells us that
±15/2,
±1, ±3, ±5, ± 15, ±1/2, ±1/3,
±1/6, ±3/2, ±5/2, ±5/3, ±5/6,
r
6
1/2
6 -4
5/6
6
1
6 -1
3
6
-7 -14
15
±15/6
-2
11 19
72
No change in signs
3 is the upper bound
Finding the Lower Bound
r
• Let’s2 look at f (-x)=6(-x)3- 1
7(-x) -14(-x)+15=
-6x3-7x2+14x+15
2
• I do not have any r>0
3
(negative) because a
negative r x -6 would give 4
me a positive
5
• 2 is an upper bound of
• f (-x), so -2 is a lower
6
bound of f (x)
7
8
9
-6
-6
-6
-6
-6
-6
-6
-6
-6
-6
-7 14
15
-13 +1
-19 -24 -33
HW#29
• Section 4-5
• Pp 240-242
• #13-25 odd, 41-44 all