Transcript Document

Zeros of Polynomials
2.5
Properties of Polynomial Equations
 A polynomial of degree n, has n roots
(counting multiple roots separately)
 If a + bi is a root, then a – bi is also a root.
Complex roots occur in conjugate pairs!
Descartes’ Rule of Signs
Find Numbers of Positive and Negative
Zeros
State the possible number of positive real zeros, negative real
zeros, and imaginary zeros of
p(x) = –x6 + 4x3 – 2x2 – x – 1.
Since p(x) has degree 6, it has 6 zeros. However, some
of them may be imaginary. Use Descartes’ Rule of
Signs to determine the number and type of real zeros.
Count the number of changes in sign for the coefficients
of p(x).
p(x) =
–x6
+
yes
– to +
2 or 0 positive real
zeros
4x3
–
yes
+ to –
2x2
–
no
– to –
x
–
no
– to –
1
Find Numbers of Positive and Negative
Zeros
Since there are two sign changes, there are 2 or 0 positive real
zeros. Find p(–x) and count the number of sign changes for its
coefficients.
p(–x) = –(–x)6 +
no
– to –
4(–x)3 – 2(–x)2 –
no
– to –
yes
– to +
(–x) –
yes
+ to –
Since there are two sign changes, there are 2 or 0 negative real
zeros. Make a chart of possible combinations.
1
Find Numbers of Positive and Negative
Zeros
Answer:
Find all of the zeros of f(x) = x3 – x2 + 2x + 4.
Since f(x) has degree of 3, the function has three zeros.
To determine the possible number and type of real
zeros, examine the number of sign changes in f(x) and
f(–x).
f(x) =
x3
–
x2
yes
f(–x) = –x3
–
no
+
2x
yes
x2
–
no
+
4
2 or 0 positive real
zeros
no
2x
+
yes
4
1 negative real
zero
Rational Zero Theorem
Factors of a 0
Possible Zeros 
Factors of a n
Identify Possible Zeros
A. List all of the possible rational zeros of
f(x) = 3x4 – x3 + 4.
Answer:
B. List all of the possible rational zeros of
f(x) = x3 + 3x + 24. <YU TRY>
A.
B.
C.
D.
Use the Factor Theorem
Show that x – 3 is a factor of x3 + 4x2 – 15x – 18. Then find the
remaining factors of the polynomial.
The binomial x – 3 is a factor of the polynomial if 3 is a
zero of the related polynomial function. Use the factor
theorem and synthetic division.
1
4 –15
3
21
1
7
6
–18
18
0
Use the Factor Theorem
Since the remainder is 0, (x – 3) is a factor of the polynomial. The
polynomial x3 + 4x2 – 15x –18 can be factored as (x – 3)(x2 + 7x + 6).
The polynomial x2 + 7x + 6 is the depressed polynomial. Check to
see if this polynomial can be factored.
x2 + 7x + 6 = (x + 6)(x + 1)
Answer:
Factor the trinomial.
So, x3 + 4x2 – 15x – 18 = (x – 3)(x + 6)(x + 1).
Example
0  x 4  6 x 2  8 x  24
Solve
Possible Zeros :  1,  2,  3,  4,  6,  8,  12,  24
2 1 0 -6
2 4
-8
24
2 1 2 -2
- 12
2
8
12
1 4
6
0
- 4 - 24
1 2 - 2 - 12
0


x 4  6 x 3  8 x  24  x 2  4 x  6 x  2
Repeated zero at x = 2
2
Upper and Lower Bounds



Suppose f(x) is divided by x – c using
synthetic division
If c > 0 and each number in the last row is
either positive or zero, then c is an upper
bound for real zeros
If c < 0 and each number in the last row are
alternatively positive or negative (zero counts
as both), then c is a lower bound
Solve 0  x 4  6 x 3  22 x 2  30 x  13
Possible
roots are
±1, ±13
The degree is 4,
so there are 4
roots!
1 1 - 6 22
Use
Synthetic
Division to
find the roots
1
-5
- 30
13
17 - 13
1 - 5 17 - 13
0
4  16  4(1)(13) 4   36 4  6i


 2  3i
2(1)
2
2
Multiplicity
of 2
x  1, 1, 2  3i
1 1 - 5 17
- 13
1
13
-4
1 - 4 13
0
x 2  4 x  13
Use Quadratic
Formula to find
remaining
solutions
f(x) = x – x + 2x + 4.
3
2
x  1  i 3,  1
Possible
roots are
±1, ±2, ±4
There are 3 zeros, we
know there are 2 or 0
positive and 1 negative
-4, -2, -1, 1, 2, 4
Use
Synthetic
Division to
find the roots
Use Quadratic
Formula to find
remaining
solutions
211111- 1- 1- 1222 444
We need either 2
or 0 positive, so 4
cannot be a zero
21- 1202 82- 4
1 111 0- 242 4126 0
-1 works, so it is our 1
negative zero. The
other two have to be
imaginary.
2  4  4(1)( 4) 2   12 2  2i 3


 1 i 3
2(1)
2
2
x  1  2i,  1, 3
Solve 0  x 4  2 x 2  16 x  15
There are 4 zeros, We
know there are 1 positive
and 3 or 1 negative
(-3x)-1,
 1,
(3,x)5, 152( x)
-15,f-5,
4
Possible
 16(are
 x±1,
)  15
±3, ±5, ±15
2 roots
35 11 00 --22 --16
16 --15
15
Now try
negatives
35 25
9
115
21
15
495
11 35 23
7
99
5
480
0
 13 11 33 77 55
- 13 -02 -- 21
5
1 02 75 |-016
3 is
Allapositive
solution.soThis
5 isisanour
1upper
positive
bound
zero
-1 Alternates
is our 1 negative.
between
Use
positive
the quadratic
and negative
to find the
so 3remaining
is a lower2 bound
zeros
 2  4  4(1)(5)  2   16  2  4i


 1  2i
2(1)
2
2
Linear Factorization Theorem
 Find a fourth degree polynomial with real
coefficients that has 2, -2 and i as zeros and
such that f(3) = -150
f ( x)  an x  2x  2x  i x  i 
  
f ( x)  a x  3x  4
 150  a (3)  3(3)  4
f ( x )  an x 2  4 x 2  1
4
2
n
4
2
n
 3  an

Foil
Foil
Substitute f(3) = -150
Solve for a
Substitute back in equation

f ( x)  3 x 4  3x 2  4  3x 4  9 x 2  12
Linear Factorization Theorem
 An nth – degree polynomial can be
expressed as the product of a nonzero
constant and n linear factors
Completely factor x 4  3x 2  28
x
2


7 x
 7 x2  4
factored over the rationals
x  7 x 
 4 factored over the reals
x  7 x  7 x  2i x - 2i completely factored
2