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8
Complex
Numbers,
Polar
Equations,
8.1-1
8
Complex Numbers, Polar Equations,
and Parametric Equations
8.1 Complex Numbers
8.2 Trigonometric (Polar) Form of Complex
Numbers
8.3 The Product and Quotient Theorems
8.4 De Moivre’s Theorem; Powers and Roots of
Complex Numbers
8.5 Polar Equations and Graphs
8.1-2
8.1 Complex Numbers
Basic Concepts of Complex Numbers ▪ Complex Solutions of
Equations ▪ Operations on Complex Numbers
8.1-3
Basic Concepts of Complex
Numbers
 i is called the imaginary unit.
 Numbers of the form a + bi are called complex
numbers.
a is the real part.
b is the imaginary part.
 a + bi = c + di if and only if a = c and b = d.
8.1-4
8.1-5
Basic Concepts of Complex
Numbers
 If a = 0 and b ≠ 0, the complex number is a
pure imaginary number.
Example: 3i
 If a ≠ 0 and b ≠ 0, the complex number is a
nonreal complex number.
Example: 7 + 2i
 A complex number written in the form a + bi or a
+ ib is written in standard form.
8.1-6
The Expression
8.1-7
Example 1
WRITING
AS
Write as the product of real number and i, using the
definition of
8.1-8
Example 2
SOLVING QUADRATIC EQUATIONS FOR
COMPLEX SOLUTIONS
Solve each equation.
Solution set:
Solution set:
8.1-9
Example 3
SOLVING QUADRATIC EQUATIONS FOR
COMPLEX SOLUTIONS
Write the equation in standard form,
then solve using the quadratic formula with a = 9,
b = –6, and c = 5.
8.1-10
Caution
When working with negative radicands,
use the definition
before
using any of the other rules for radicals.
In particular, the rule
is
valid only when c and d are not both
negative.
For example
while
so
,
,
8.1-11
Example 4
FINDING PRODUCTS AND QUOTIENTS
INVOLVING NEGATIVE RADICANDS
Multiply or divide as indicated. Simplify each answer.
8.1-12
Example 5
Write
SIMPLIFYING A QUOTIENT INVOLVING
A NEGATIVE RADICAND
in standard form a + bi.
Factor.
8.1-13
Addition and Subtraction of
Complex Numbers
For complex numbers a + bi and c + di,
8.1-14
Example 6
ADDING AND SUBTRACTING COMPLEX
NUMBERS
Find each sum or difference.
(a) (3 – 4i) + (–2 + 6i) = [3 + (–2)] + (–4i + 6i)
= 1 + 2i
(b) (–9 + 7i) + (3 – 15i) = –6 – 8i
(c) (–4 + 3i) – (6 – 7i)
= (–4 – 6) + [3 – (–7)]i
= –10 + 10i
(d) (12 – 5i) – (8 – 3i) + (–4 + 2i)
= (12 – 8 – 4) + (–5 +3 + 2)i
= 0 + 0i = 0
8.1-15
Multiplication of Complex
Numbers
8.1-16
Example 7
MULTIPLYING COMPLEX NUMBERS
Find each product.
(a) (2 – 3i)(3 + 4i) = 2(3) + (2)(4i) + (–3i)(3) + (–3i)(4i)
= 6 + 8i – 9i – 12i2
= 6 –i – 12(–1)
= 18 – i
(b) (4 + 3i)2 = 42 + 2(4)(3i) + (3i)2
= 16 + 24i + 9i2
= 16 + 24i + 9(–1)
= 7 + 24i
8.1-17
Example 7
(c) (2 + i)(–2 – i)
MULTIPLYING COMPLEX NUMBERS
= –4 – 2i – 2i – i2
= –4 – 4i – (–1)
= –4 – 4i + 1
= –3 – 4i
(d) (6 + 5i)(6 – 5i) = 62 – (5i)2
= 36 – 25i2
= 36 – 25(–1)
= 36 + 25
= 61 or 61 + 0i
8.1-18
Example 7
MULTIPLYING COMPLEX NUMBERS
(cont.)
This screen shows how the TI–83/84 Plus displays
the results found in parts (a), (b), and (d) in this
example.
8.1-19
Example 8
SIMPLIFYING POWERS OF i
Simplify each power of i.
(a)
(b)
(c)
Write the given power as a product involving
or
.
(a)
(b)
(c)
8.1-20
Powers of i
i1 = i
i5 = i
i9 = i
i2 = 1
i6 = 1
i10 = 1
i 3 = i
i 7 = i
i11 = i
i4 = 1
i8 = 1i12 = 1, and so on.
8.1-21
Property of Complex Conjugates
For real numbers a and b,
8.1-22
Example 9(a) DIVIDING COMPLEX NUMBERS
Write the quotient in standard form a + bi.
Multiply the numerator and
denominator by the complex
conjugate of the denominator.
Multiply.
i2 = –1
Factor.
Lowest terms; standard form
8.1-23
Example 9(b) DIVIDING COMPLEX NUMBERS
Write the quotient in standard form a + bi.
Multiply the numerator and
denominator by the complex
conjugate of the denominator.
Multiply.
i2 = –1
Standard form
This screen shows how
the TI–83/84 Plus
displays the results in
this example.
8.1-24
8.2 Trigonometric (Polar) Form
of Complex Numbers
The Complex Plane and Vector Representation ▪ Trigonometric
(Polar) Form ▪ Converting Between Rectangular and
Trigonometric (Polar) Forms ▪ An Application of Complex
Numbers to Fractals
8.1-25
The Complex Plane and Vector
Representation
 Horizontal axis: real axis
 Vertical axis: imaginary axis
Each complex number a + bi
determines a unique position
vector with initial point (0, 0)
and terminal point (a, b).
8.1-26
The Complex Plane and Vector
Representation
The sum of two complex numbers is represented
by the vector that is the resultant of the vectors
corresponding to the two numbers.
(4 + i) + (1 + 3i) = 5 + 4i
8.1-27
Example 1
EXPRESSING THE SUM OF COMPLEX
NUMBERS GRAPHICALLY
Find the sum of 6 – 2i and –4 – 3i. Graph both
complex numbers and their resultant.
(6 – 2i) + (–4 – 3i) = 2 – 5i
8.1-28
Relationships Among x, y, r, and θ.
8.1-29
Trigonometric (Polar) Form
of a Complex Number
The expression r(cos θ + i sin θ) is called
the trigonometric form (or polar form) of
the complex number x + yi.
The expression cos θ + i sin θ is
sometimes abbreviated cis θ.
Using this notation, r(cos θ + i sin θ) is
written r cis θ.
The number r is the absolute value (or modulus)
of x + yi, and θ is the argument of x + yi.
8.1-30
Example 2
CONVERTING FROM TRIGONOMETRIC
FORM TO RECTANGULAR FORM
Express 2(cos 300° + i sin 300°) in rectangular form.
The graphing calculator
screen confirms the
algebraic solution. The
imaginary part is an
approximation for
8.1-31
Converting From Rectangular Form to
Trigonometric Form
Step 1 Sketch a graph of the number x + yi
in the complex plane.
Step 2 Find r by using the equation
Step 3 Find θ by using the equation
choosing the
quadrant indicated in Step 1.
8.1-32
Caution
Be sure to choose the correct
quadrant for θ by referring to the
graph sketched in Step 1.
8.1-33
Example 3(a) CONVERTING FROM RECTANGULAR
FORM TO TRIGONOMETRIC FORM
Write
measure.)
in trigonometric form. (Use radian
Step 1:
Sketch the graph of
in the complex plane.
Step 2:
8.1-34
Example 3(a) CONVERTING FROM RECTANGULAR
FORM TO TRIGONOMETRIC FORM
(continued)
Step 3:
The reference angle for θ is
The graph shows that θ is in
quadrant II, so θ =
8.1-35
Example 3(b) CONVERTING FROM RECTANGULAR
FORM TO TRIGONOMETRIC FORM
Write –3i in trigonometric form. (Use degree measure.)
Sketch the graph of –3i in
the complex plane.
We cannot find θ by using
because x = 0.
From the graph, a value for θ is 270°.
8.1-36
Example 4
CONVERTING BETWEEN
TRIGONOMETRIC AND RECTANGULAR
FORMS USING CALCULATOR
APPROXIMATIONS
Write each complex number in its alternative form,
using calculator approximations as necessary.
(a) 6(cos 115° + i sin 115°)
≈ –2.5357 + 5.4378i
8.1-37
Example 4
CONVERTING BETWEEN
TRIGONOMETRIC AND RECTANGULAR
FORMS USING CALCULATOR
APPROXIMATIONS (continued)
(b) 5 – 4i
A sketch of 5 – 4i shows that θ
must be in quadrant IV.
The reference angle for θ is approximately –38.66°.
The graph shows that θ is in quadrant IV, so
θ = 360° – 38.66° = 321.34°.
8.1-38
Example 5
DECIDING WHETHER A COMPLEX
NUMBER IS IN THE JULIA SET (cont.)
Determine whether each number belongs to the Julia
set.
The calculations
repeat as 0, –1, 0, –1,
and so on.
The absolute values are either 0 or 1, which do not
exceed 2, so 0 + 0i is in the Julia set, and the point
(0, 0) is part of the graph.
8.1-39
Example 5
DECIDING WHETHER A COMPLEX
NUMBER IS IN THE JULIA SET (cont.)
The absolute value is
so 1 + 1i is not in the Julia set and (1, 1) is
not part of the graph.
8.1-40
8.3 The Product and Quotient
Theorems
Products of Complex Numbers in Trigonometric Form ▪
Quotients of Complex Numbers in Trigonometric Form
Copyright © 2009 Pearson Addison-Wesley
1.1-41
8.1-41
Let's try multiplying two complex numbers in polar form together.
z1  r1 cos1  i sin 1 
z2  r2 cos2  i sin 2 
z1 z2   r1  cos 1  i sin 1    r2  cos  2  i sin  2  



Look at where we started and where we ended up and see if
 r1r2a cos
1  i as
sinto1what
coshappens
2  i sin

you can make
statement
to the2 r 's
Must FOIL
and the  's when you multiply
twothese
complex numbers.

 r1r2 cos 1 cos  2  i sin  2 cos 1  i sin 1 cos  2  i 2 sin 1 sin  2
Replace i 2 with -1 and group real terms and then imaginary terms
Multiply the Moduli and Add the Arguments
 r1r2 cos1 cos2  sin 1 sin 2   sin 1 cos2  cos1 sin 2 i
use sum formula for cos
use sum formula for sin
 r1r2 cos1  2   i sin 1  2 

Product Theorem
are any two complex numbers, then
In compact form, this is written
Copyright © 2009 Pearson Addison-Wesley
1.1-43
8.1-43
Example 1
USING THE PRODUCT THEOREM
Find the product of 3(cos 45° + i sin 45°) and
2(cos 135° + i sin 135°). Write the result in
rectangular form.
Copyright © 2009 Pearson Addison-Wesley
1.1-44
8.1-44




If z  4 cos 40  i sin 40 and w  6 cos 120  i sin 120 ,
find : (a) zw
(b) z w





 

zw   4 cos 40  i sin 40  6 cos120  i sin120 

 


 4  6  cos 40  120  i sin 40  120
multiply the moduli





add the arguments
(the i sine term will have same argument)
 24  cos160  i sin160 
 24 0.93969  0.34202i 
 22.55  8.21i
If you want the answer in
rectangular coordinates simply
compute the trig functions and
multiply the 24 through.
Let z1  r1 cos 1  i sin 1  and z 2  r2 cos  2  i sin  2 
be two complex numbers. Then
z1 z2  r1r2 cos1  2   i sin 1  2 
(This says to multiply two complex numbers in polar form, multiply the moduli
and add the arguments)
If z 2  0, then
z1 r1
 cos1  2   i sin1  2 
z2 r2
(This says to divide two complex numbers in polar form, divide the moduli and
subtract the arguments)
Quotient Theorem
are any two complex numbers, where
In compact form, this is written
Copyright © 2009 Pearson Addison-Wesley
1.1-47
8.1-47
Example 2
USING THE QUOTIENT THEOREM
Find the quotient
rectangular form.
Copyright © 2009 Pearson Addison-Wesley
Write the result in
1.1-48
8.1-48
Example 3
USING THE PRODUCT AND QUOTIENT
THEOREMS WITH A CALCULATOR
Use a calculator to find the following. Write the results
in rectangular form.
Copyright © 2009 Pearson Addison-Wesley
1.1-49
8.1-49
Example 3
Copyright © 2009 Pearson Addison-Wesley
USING THE PRODUCT AND QUOTIENT
THEOREMS WITH A CALCULATOR
(continued)
1.1-50
8.1-50
8.4
De Moivre’s Theorem; Powers
and Roots of Complex Numbers
Powers of Complex Numbers (De Moivre’s Theorem) ▪ Roots of
Complex Numbers
Copyright © 2009 Pearson Addison-Wesley
1.1-51
8.1-51
You can repeat this process raising complex numbers to
powers. Abraham DeMoivre did this and proved the
following theorem:
DeMoivre’s
Theorem
If z  rcos  i sin is a complex number,
Abraham de Moivre
(1667 - 1754)
then
z  r cos n  i sin n 
n
n
where n  1 is a positive integer.
This says to raise a complex number to a power, raise the modulus to that power
and multiply the argument by that power.
De Moivre’s Theorem
is a complex number,
then
In compact form, this is written
Copyright © 2009 Pearson Addison-Wesley
1.1-53
8.1-53
Example 1
Find
form.
First write
FINDING A POWER OF A COMPLEX
NUMBER
and express the result in rectangular
in trigonometric form.
Because x and y are both positive, θ is in quadrant I,
so θ = 60°.
Copyright © 2009 Pearson Addison-Wesley
1.1-54
8.1-54
Example 1
FINDING A POWER OF A COMPLEX
NUMBER (continued)
Now apply De Moivre’s theorem.
480° and 120° are coterminal.
Rectangular form
Copyright © 2009 Pearson Addison-Wesley
1.1-55
8.1-55
This theorem is used to raise complex numbers to powers. It would
be a lot of work to find




  3i  3i  3i  3i

r

 3
1  4  2
2
 1  but in Quad II

 3
  tan 1 

3 i

you would need to FOIL and
multiply all of these together
and simplify powers of i --UGH!
Instead let's convert to polar form and use
DeMoivre's Theorem.
2

4
5

6
4
 
5
5  
 3  i   2  cos
 i sin
   2 4 cos 4  5   i sin  4  5 
6
6 
6 
6 

 
 
4
  10 
 16  cos 
  3

 10 
  i sin 

 3



 1 
 
3
i 
 16     
 
 2
2

 

  8  8 3i
nth Root
For a positive integer n, the complex
number a + bi is an nth root of the
complex number x + yi if
Copyright © 2009 Pearson Addison-Wesley
1.1-57
8.1-57
Solve the following over the set of complex numbers:
z 1
3
We know that if we cube root both sides we could
get 1 but we know that there are 3 roots. So we
want the complex cube roots of 1.
Using DeMoivre's Theorem with the power being a rational exponent (and therefore
meaning a root), we can develop a method for finding complex roots. This leads to
the following formula:
   2 k 
  2 k  
z k  r  cos  
  i sin  

n 
n 
n
 n
n
where k  0, 1, 2,  , n  1
nth Root Theorem
If n is any positive integer, r is a positive
real number, and θ is in degrees, then the
nonzero complex number r(cos θ + i sin θ)
has exactly n distinct nth roots, given by
where
Copyright © 2009 Pearson Addison-Wesley
1.1-59
8.1-59
Note
In the statement of the nth root theorem,
if θ is in radians, then
Copyright © 2009 Pearson Addison-Wesley
1.1-60
8.1-60
Let's try this on our problem. We want the cube roots of 1.
We want cube root so our n = 3. Can you convert 1 to polar form? (hint: 1 = 1 +
0i)
r
1  0
2
2
1
0
  tan    0
1
1
  0 2 k 
 0 2 k
z k  1 cos 
  i sin  
3 
3
3
 3
3

, for k  0, 1, 2

Once we build the formula, we use it first with k = 0 and get
one root, then with k = 1 to get the second root and finally with
k = 2 for last root.
We want cube root
so use 3 numbers
here
   2 k 
  2 k  
z k  r  cos  
  i sin  

n 
n 
n
 n
n
  0 2k 
 0 2k
zk  1cos 
  i sin  
3 
3
3
 3
3
  0 20
z0  1cos 
3
 3
3

 0 20
  i sin  
3

3

, for k  0, 1, 2


 1cos 0   i sin 0   1
 Here's the root we
already knew.
  0 21 
 0 21 
z1  1 cos 
  i sin  

3 
3 
3
 3
  2 
1
3
 2  
 1cos

i
sin



i



2 2
 3 
  3 
 0 22  
 0 22   
3 
z 2  1 cos 
  i sin  

3 
3 
3
 3
3
  4 
 4
 1cos
  i sin 
 3
  3 
1
3

i
   
2 2

If you cube any of these
numbers you get 1.
(Try it and see!)
We found the cube roots of 1 were:
Let's plot these on the complex plane
1
3
1
3
1,  
i,  
i
2 2
2 2
about 0.9
each line is 1/2 unit
Notice each of the
complex roots has the
same magnitude (1).
Also the three points are
evenly spaced on a
circle. This will always
be true of complex
roots.
Example 2
FINDING COMPLEX ROOTS
Find the two square roots of 4i. Write the roots in
rectangular form.
Write 4i in trigonometric form:
The square roots have absolute value
and argument
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1.1-64
8.1-64
Example 2
FINDING COMPLEX ROOTS (continued)
Since there are two square roots, let k = 0 and 1.
Using these values for , the square roots are
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1.1-65
8.1-65
Example 2
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FINDING COMPLEX ROOTS (continued)
1.1-66
8.1-66
Example 3
FINDING COMPLEX ROOTS
Find all fourth roots of
rectangular form.
Write
Write the roots in
in trigonometric form:
The fourth roots have absolute value
and argument
Copyright © 2009 Pearson Addison-Wesley
1.1-67
8.1-67
Example 3
FINDING COMPLEX ROOTS (continued)
Since there are four roots, let k = 0, 1, 2, and 3.
Using these values for α, the fourth roots are
2 cis 30°, 2 cis 120°, 2 cis 210°, and 2 cis 300°.
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1.1-68
8.1-68
Example 3
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FINDING COMPLEX ROOTS (continued)
1.1-69
8.1-69
Example 3
FINDING COMPLEX ROOTS (continued)
The graphs of the roots lie on a circle with center at
the origin and radius 2. The roots are equally spaced
about the circle, 90° apart.
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1.1-70
8.1-70
Example 4
SOLVING AN EQUATION BY FINDING
COMPLEX ROOTS
Find all complex number solutions of x5 – i = 0. Graph
them as vectors in the complex plane.
There is one real solution, 1, while there are five
complex solutions.
Write 1 in trigonometric form:
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1.1-71
8.1-71
Example 4
SOLVING AN EQUATION BY FINDING
COMPLEX ROOTS (continued)
The fifth roots have absolute value
argument
and
Since there are five roots, let k = 0, 1, 2, 3, and 4.
Solution set: {cis 0°, cis 72°, cis 144°, cis 216°, cis 288°}
Copyright © 2009 Pearson Addison-Wesley
1.1-72
8.1-72
Example 4
SOLVING AN EQUATION BY FINDING
COMPLEX ROOTS (continued)
The graphs of the roots lie on a unit circle. The roots
are equally spaced about the circle, 72° apart.
Copyright © 2009 Pearson Addison-Wesley
1.1-73
8.1-73
4
Polynomial and
Rational Functions
4.3
Dividing Polynomials
Long Division of Polynomials
Long Division of Polynomials
Dividing polynomials is much like the
familiar process of dividing numbers.
When we divide 38 by 7, the quotient is 5
and the remainder is 3.
We write: 38
3
5
7
7
To divide polynomials, we use long division—as
in the next example.
E.g. 1—Long Division of
Polynomials
Divide 6x2 – 26x + 12 by x – 4.
The dividend is 6x2 – 26x + 12 and
the divisor is x – 4.
We begin by arranging them as follows:
x  4 6x 2  26x  12
E.g. 1—Long Division of Polynomials
Next, we divide the leading term in the dividend by the
leading term in the divisor to get the first term of the quotient:
6x2/x = 6x
Then, we multiply the divisor by 6x and
subtract the result from the dividend.
6x
2
x  4 6 x  26 x  12
6 x  24 x
 2x  12
2
E.g. 1—Long Division of Polynomials
We repeat the process using the last
line –2x + 12 as the dividend.
6x  2
2
x  4 6 x  26 x  12
6 x  24 x
 2 x  12
2 x  8
4
2
E.g. 1—Long Division of Polynomials
The division process ends when the last
line is of lesser degree than the divisor.
The last line then contains the remainder.
The top line contains the quotient.
Long Division of Polynomials
We summarize the long
division process in the following
theorem.
Division Algorithm
If P(x) and D(x) are polynomials, with
D(x) ≠ 0, then there exist unique polynomials Q(x)
and R(x), where R(x) is either 0 or of degree less
than the degree of D(x),
such that:
P(x) = D(x) . Q(x) + R(x)
Henc, Dividend= (divisor)(quotient)+ remainder.
The polynomials P(x) and D(x) are called
the dividend and divisor, respectively.
Q(x) is the quotient and R(x) is the remainder.
E.g. 2—Long Division of
Polynomials
Let
and
P(x) = 8x4 + 6x2 – 3x + 1
D(x) = 2x2 – x + 2
Find polynomials Q(x) and R(x)
such that:
P(x) = D(x) · Q(x) + R(x)
E.g. 2—Long Division of Polynomials
We use long division after first inserting
the term 0x3 into the dividend to ensure that the
columns line up correctly.
4 x 2  2x
2x 2  x  2 8 x 4  0 x 3  6 x 2  3 x  1
8x 4  4x 3  8x 2
4 x 3  2x 2  3 x
4 x 3  2x 2  4 x
7 x–7x
 1+ 1 is
The process is complete at this point 
as
of lesser degree than the divisor 2x2 – x + 2.
E.g. 2—Long Division of
Polynomials
From the long division, we see that:
Q(x) = 4x2 + 2x and R(x) = –7x + 1
Thus,
8x4 + 6x2 – 3x + 1
= (2x2 – x + 2)(4x2 + 2x)
+ (–7x + 1)
Synthetic Division
Synthetic Division
Synthetic division is a quick method
of dividing polynomials.
It can be used when the divisor is of the form
x – c.
In synthetic division, we write only the essential
parts of the long division.
Long Division vs. Synthetic
Division
Compare the following long and synthetic
divisions, in which we divide 2x3 – 7x2 + 5
by x – 3.
We’ll explain how to perform
the synthetic division in Example 3.
Long Division vs. Synthetic
Division
2x 2  x  3
x  3 2x  7 x  0 x  5
3
2
32
2x 3  6 x 2
 x 2  0x
x  3x
2
2
7
0
5
6
3
9
1  3
4
Quotient
3 x  5
3 x  9
4
Remainder
Synthetic Division
Note that, in synthetic division, we:
Abbreviate 2x3 – 7x2 + 5 by writing only
the coefficients: 2, –7, 0, and 5.
Simply write 3 instead of x – 3.
(Writing 3 instead of –3 allows us to add
instead
of subtract. However, this changes the sign of
all the numbers that appear in the gold boxes.)
E.g. 3—Synthetic Division
Use synthetic division to divide
2x3 – 7x2 + 5 by x – 3.
We begin by writing the appropriate coefficients
to represent the divisor and the dividend.
32
7
0
5
E.g. 3—Synthetic Division
We repeat this process of
multiplying and then adding until
the table is complete.
32
2
7
0
6
3
1
3
5
E.g. 3—Synthetic Division
32
2
7
0
5
6
3
9
1
3
4
From the last line, we see that the quotient
is 2x2 – x – 3 and the remainder is –4.
Thus,
2x3 – 7x2 + 5 = (x – 3)(2x2 – x – 3) – 4.
The Remainder
and Factor Theorems
Remainder Theorem
The Remainder Theorem shows how synthetic
division can be used to evaluate polynomials easily.
If the polynomial P(x) is divided by x – c,
then the remainder is the value P(c).
Remainder Theorem—Proof
If the divisor in the Division Algorithm is
of the form x – c for some real number c,
then the remainder must be a constant.
This is because the degree of the remainder
is less than the degree of the divisor.
Remainder Theorem—Proof
If we call this constant r, then
P(x) = (x – c) · Q(x) + r
Setting x = c in this equation, we get:
P(c) = (c – c) · Q(x) + r
=0+r
=r
That is, P(c) is the remainder r.
E.g. 4—Using the Remainder
Theorem
Let
P(x) = 3x5 + 5x4 – 4x3 + 7x + 3
(a) Find the quotient and remainder when
P(x) is divided by x + 2.
(b) Use the Remainder Theorem to find P(–2).
E.g. 4—Using Remainder
Example (a)
Theorem
As x + 2 = x – (–2) , the synthetic division
for the problem takes the following form.
2 3
3
5
4
0
7
3
6
2
4
8
2
1
2
4
1
5
The quotient is 3x4 – x3 – 2x2 + 4x – 1.
The remainder is 5.
E.g. 4—Using Remainder
Example (b)
Theorem
By the Remainder Theorem,
P(–2) is the remainder when P(x)
is divided by x – (–2) = x + 2.
From part (a), the remainder is 5.
Hence, P(–2) = 5.
Factor Theorem
The Factor Theorem says that zeros
of polynomials correspond to factors.
We used this fact in Section 4.2 to graph
polynomials.
c is a zero of P if and only if x – c
is a factor of P(x).
The Rational Zero Theorem
The Rational Zero Theorem gives a list of possible rational zeros of a
polynomial function. Equivalently, the theorem gives all possible rational roots
of a polynomial equation. Not every number in the list will be a zero of the
function, but every rational zero of the polynomial function will appear
somewhere in the list.
The Rational Zero Theorem
p
If f(x)  anxn  an-1xn-1 … a1x  a0 has integer coefficients and
q
p
(where is reduced) is a rational zero, then p is a factor of the constant
q
term a0 and q is a factor of the leading coefficient an.
EXAMPLE: Using the Rational Zero Theorem
List all possible rational zeros of f(x)  15x3  14x2  3x – 2.
Solution
The constant term is –2 and the leading coefficient is 15.
Factors of the constant term,  2
Factors of the leading coefficient, 15
1,  2

1,  3,  5,  15
Possible rational zeros 
 1,  2,
 13 ,  23 ,
Divide 1
and 2
by 1.
Divide 1
and 2
by 3.
 15 ,  52 ,
Divide 1
and 2
by 5.
1 ,  2
 15
15
Divide 1
and 2
by 15.
There are 16 possible rational zeros. The actual solution set to f(x)  15x3 
14x2  3x – 2 = 0 is {-1, 1/3, 2/5}, which contains 3 of the 16 possible solutions.
EXAMPLE:
Solve:
Solving a Polynomial Equation
x4  6x2  8x + 24  0.
Solution Because we are given an equation, we will use the word "roots,"
rather than "zeros," in the solution process. We begin by listing all possible
rational roots.
Factors of the constant term, 24
Factors of the leading coefficient, 1
1,  2  3,  4,  6,  8,  12,  24

1
 1,  2  3,  4,  6,  8,  12,  24
Possible rational zeros 
EXAMPLE:
Solve:
Solving a Polynomial Equation
x4  6x2  8x + 24  0.
Solution The graph of f(x)  x4  6x2  8x + 24 is shown the figure below.
Because the x-intercept is 2, we will test 2 by synthetic division and show that
it is a root of the given equation.
2 1 0 6 8 24
The zero remainder
2
4 4 24
indicates that 2 is a root
of x4  6x2  8x + 24  0.
1
2
2
12
0
2
2 8 12
1 4 6
0
The zero remainder
indicates that 2 is a root of
x3  2x2  2x  12 = 0.
EXAMPLE:
Solve:
Solution
Solving a Polynomial Equation
x4  6x2  8x + 24  0.
Now we can solve the original equation as follows.
x4  6x2  8x + 24  0
(x – 2)(x – 2)(x2  4x  6)  0
x – 2  0 or x – 2  0 or x2  4x  6  0
x2
x2
x2  4x  6  0
This is the given equation.
This was obtained from the first
,second synthetic division.
Set each factor equal to zero.
Solve.
EXAMPLE:
Solve:
Solution
Solving a Polynomial Equation
x4  6x2  8x + 24  0.
We can use the quadratic formula to solve x2  4x  6  0.
b  b2  4ac
x
2a
4  42  4 1  6 

2 1
4   8

2
4  2i 2

2
 2  i 2
We use the quadratic formula because x2  4x  6  0
cannot be factored.
Let a  1, b  4, and c  6.
Multiply and subtract under the radical.
8 
4(2)(1)  2i 2
Simplify.
The solution set of the original equation is {2, 2,2 i i2, 2 i i2 }.
EXAMPLE: Using the Rational Zero Theorem
2 x 4  3x 3  2 x 2  1  0
Solution
The constant term is –1 and the leading coefficient is 2.
possible rational zeros
Factors.of .the.cons tan t.term,1
Factors.of .the.leading.coefficient ,2
1
Divide 1
Divide 1

by 1
by 2
,1  2

possible rational zeros
 1,
1
2
There are 4 possible rational zeros. The actual solution set to 2x4  3x3  2x2 – 1 = 0
is {-1,1 1/2, 1/2}, which contains 2 of the 4 possible solutions.
EXAMPLE:
Solving a Polynomial Equation
2 x 4  3x 3  2 x 2  1  0
Solution The graph of f(x)  2 x  3x  2 x  1  0 is shown the figure below.
Because the x-intercept is 2, we will test 2 by synthetic division and show that it
is a root of the given equation.
4
-1
1
2
3
2
3 2 0 1
-2 -1 -1 1
21 1 -1 0
1 1 1
2 2 2 0
2
The zero remainder indicates
that -1 is a root of 2x4 3x3 
2x2 -1 0.
The zero remainder indicates that1/2 is
a root of 2x3 x2  x -1 0
EXAMPLE:
Solve:
Solution
Solving a Polynomial Equation
2 x 4  3x 3  2 x 2  1  0
Now we can solve the original equation as follows.
2 x 4  3x 3  2 x 2  1  0
This is the given equation.
1
2
(x +1)(x – )(2x2  2x  2)  0
x +1  0
x  -1
or x
1
– 2
x
0 or 2x2  2x  2  0
1
2
x2  x  1  0
Solve.
This was obtained from the first
,second synthetic division.
Set each factor equal to zero.
EXAMPLE:
Solution
Solving a Polynomial Equation
We can use the quadratic formula to solve x2  x  1  0.
b  b2  4ac
x
2a

 1  12  4(1)(1)
We use the quadratic formula because x2  x  1  0
cannot be factored.
Let a  1, b  1, and c  1.
2(1)

1  3
2

1 i 3
2
Multiply and subtract under the radical.
1
2
The solution set of the original equation is {-1, ,
 1  3i
2
}
Properties of Polynomial Equations
1. If a polynomial equation is of degree n, then counting multiple roots
separately, the equation has n roots.
2.
If a  bi is a root of a polynomial equation (b  0), then the non-real
complex number a  bi is also a root. Non-real complex roots, if
they exist, occur in conjugate pairs.
Descartes' Rule of Signs
If f(x)  anxn  an1xn1  …  a2x2  a1x  a0 be a polynomial with real
coefficients.
1. The number of positive real zeros of f is either equal to the number
of sign changes of f(x) or is less than that number by an even integer.
If there is only one variation in sign, there is exactly one positive real
zero.
2. The number of negative real zeros of f is either equal to the number
of sign changes of f(x) or is less than that number by an even
integer. If f(x) has only one variation in sign, then f has exactly one
negative real zero.
EXAMPLE:
Using Descartes’ Rule of Signs
Determine the possible number of positive and negative real zeros of
f(x)  x3  2x2  5x + 4.
Solution
1. To find possibilities for positive real zeros, count the number of sign
changes in the equation for f(x). Because all the terms are positive, there
are no variations in sign. Thus, there are no positive real zeros.
2. To find possibilities for negative real zeros, count the number of sign
changes in the equation for f(x). We obtain this equation by replacing x
with x in the given function.
f(x)  x3  2x2  5x + 4
Replace x with x.
f(x)  (x)3  2(x)2  5x  4
 x3  2x2  5x + 4
This is the given polynomial function.
EXAMPLE:
Using Descartes’ Rule of Signs
Determine the possible number of positive and negative real zeros of
f(x)  x3  2x2  5x + 4.
Solution
Now count the sign changes.
f(x)  x3  2x2  5x + 4
1
2
3
There are three variations in sign.
# of negative real zeros of f is either equal to 3, or is less than this number by
an even integer.
This means that there are either 3 negative real zeros
or 3  2  1 negative real zero.
Properties of Polynomial Equations
1. If a polynomial equation is of degree n, then counting multiple roots
separately, the equation has n roots.
2.
If a  bi is a root of a polynomial equation (b  0), then the non-real
complex number a  bi is also a root. Non-real complex roots, if
they exist, occur in conjugate pairs.
Complex Conjugates Theorem
Roots/Zeros that are not Real are Complex with an
Imaginary component. Complex roots with
Imaginary components always exist in Conjugate
Pairs.
If a + bi (b ≠ 0) is a zero of a polynomial function,
then its Conjugate, a - bi, is also a zero of the
function.
Find Roots/Zeros of a Polynomial
If the known root is imaginary, we can use the Complex
Conjugates Theorem.
Ex: Find all the roots of f (x)  x 3  5x 2  7x  51
If one root is 4 - i.
Because of the Complex Conjugate Theorem, we know that
another root must be 4 + i.
Can the third root also be imaginary?
Example (con’t)
Ex: Find all the roots of f (x)  x 3  5x 2  7x  51
If one root is 4 - i.
If one root is 4 - i, then one factor is [x - (4 - i)], and
Another root is 4 + i, & another factor is [x - (4 + i)].
Multiply these factors:
 x   4  i    x   4  i    x 2  x  4  i   x  4  i    4  i  4  i 
 x 2  4 x  xi  4 x  xi  16  i 2
 x 2  8 x  16  (1)
 x 2  8 x  17
Example (con’t)
Ex: Find all the roots of f (x)  x 3  5x 2  7x  51
If one root is 4 - i.
2
If the product of the two non-real factors is x  8x 17
then the third factor (that gives us the neg. real root) is
the quotient of P(x) divided by x2  8x 17 :
x 3
x 2  8x  17 x 3  5x 2  7x  51

x 3  5x 2  7x  51
0
The third root
is x = -3
So if asked to find a polynomial that has zeros, 2 and
1 – 3i, you would know another root would be 1 + 3i.
Let’s find such a polynomial by putting the roots in
factor form and multiplying them together.
If x = the root then
x - the root is the factor form.
x  2x  1  3i x  1  3i 
x  2x 1  3i x 1  3i 
x  2 x
2
Multiply the last two factors
together. All i terms should
disappear when simplified.
 x  3xi  x  1  3i  3xi  3i  9i

 x  2 x 2  2 x  10


-1
Now multiply the x – 2 through
 x  4 x  14 x  20
3
2
2
Here is a 3rd degree polynomial with roots 2, 1 - 3i and 1 + 3i
Conjugate Pairs
Complex Zeros Occur in Conjugate Pairs = If a + bi is a
zero of the function, the conjugate a – bi is also a zero of
the function
(the polynomial function must have
real coefficients)
EXAMPLES: Find the polynomial equation with the
given zeros
-1, -1, 3i, -3i
2, 4 + i, 4 – i
Now write a polynomial equation of least degree that has
real coefficients, a leading coeff. of 1 and 1, -2+i, -2-i
as zeros.
0= (x-1)(x-(-2+i))(x-(-2-i))
0= (x-1)(x+2 - i)(x+2+ i)
0= (x-1)[(x+2) - i] [(x+2)+i]
0= (x-1)[(x+2)2 - i2]
Foil
0=(x-1)(x2 + 4x + 4 – (-1))
Take care of i2
0= (x-1)(x2 + 4x + 4 + 1)
0= (x-1)(x2 + 4x + 5)
Multiply
0= x3 + 4x2 + 5x – x2 – 4x – 5
0= x3 + 3x2 + x - 5
Now write a polynomial equation of least degree that has
real coefficients, a leading coeff. of 1 and 4, 4, 2+i as
zeros.
Note: 2+i means 2 – i is also a zero
0= (x-4)(x-4)(x-(2+i))(x-(2-i))
0= (x-4)(x-4)(x-2-i)(x-2+i)
0= (x2 – 8x +16)[(x-2) – i][(x-2)+i]
0= (x2 – 8x +16)[(x-2)2 – i2]
0= (x2 – 8x +16)(x2 – 4x + 4 – (– 1))
0= (x2 – 8x +16)(x2 – 4x + 5)
0= x4– 4x3+5x2 – 8x3+32x2 – 40x+16x2 – 64x+80
0= x4-12x3+53x2-104x+80