# Some Polynomial Theorems

Document Sample

Some Polynomial Theorems

by

John Kennedy
Mathematics Department
Santa Monica College
1900 Pico Blvd.
Santa Monica, CA 90405

rkennedy@ix.netcom.com
Except for this comment explaining that it is blank for
This paper contains a collection of 31 theorems, lemmas, and corollaries that help explain some
fundamental properties of polynomials. The statements of all these theorems can be understood by
students at the precalculus level, even though a few of these theorems do not appear in any precalculus
text. However, to understand the proofs requires a much more substantial and more mature
mathematical background, including proof by mathematical induction and some simple calculus. Of
significance are the Division Algorithm and theorems about the sum and product of the roots, two
integer roots, a theorem about the equality of two polynomials, theorems related to the Euclidean
Algorithm for finding the KGH of two polynomials, and theorems about the Partial Fraction
Decomposition of a rational function and Descartes's Rule of Signs. It is rare to find proofs of either of
these last two major theorems in any precalculus text.

1. The Division Algorithm
If :ÐBÑ and .ÐBÑ ´ ! are any two polynomials then there exist unique polynomials ;ÐBÑ and <ÐBÑ
/
such that :ÐBÑ œ .ÐBÑ † ;ÐBÑ  <ÐBÑ where the degree of <ÐBÑ is strictly less than the degree of .ÐBÑ
when the degree of .ÐBÑ   " or else <ÐBÑ ´ !.
Division Algorithm Proof:
We apply induction on the degree 8 of :ÐBÑÞ We let 7 denote the degree of the divisor .ÐBÑÞ
We will establish uniqueness after we establish the existence of ;ÐBÑ and <ÐBÑÞ

If 8 œ ! then :ÐBÑ œ - where - is a constantÞ
Case ": 7 œ !Þ
Î
.ÐBÑ œ 5 where 5 is a constant and since .ÐBÑ ´ ! we know 5 Á 0.
-
In this case choose ;ÐBÑ œ and choose <ÐBÑ ´ !Þ
5
-
Then .ÐBÑ † ;ÐBÑ  <ÐBÑ œ 5 †  ! œ - œ :ÐBÑÞ In this case <ÐBÑ ´ !Þ
5
Case #: 7  !Þ
In this case let ;ÐBÑ ´ ! and let <ÐBÑ œ -Þ Then clearly .ÐBÑ † ;ÐBÑ  <ÐBÑ œ
.ÐBÑ † !  - œ !  - œ - œ :ÐBÑÞ In this case the degree of <ÐBÑ is strictly less
than the degree of .ÐBÑÞ

Now assume there exist polynomials ;" ÐBÑ and <" ÐBÑ such that :" ÐBÑ œ .ÐBÑ † ;" ÐBÑ  <" ÐBÑ
whenever :" ÐBÑ is any polynomial that has a degree less than or equal to 5 .
Let :ÐBÑ be a polynomial of degree 5  "Þ We assume :ÐBÑ œ +5 +" B5"  +5 B5  â  +" B  +!
where +5" Á !Þ We must show the theorem statement holds for :ÐBÑÞ
Case ": 7 œ !Þ
.ÐBÑ œ - where - is a constant and since .ÐBÑ ´ ! we know - Á !Þ
/
"
Let ;ÐBÑ œ :ÐBÑ and let <ÐBÑ ´ !Þ
-
1
Then .ÐBÑ † ;ÐBÑ  <ÐBÑ œ - † :ÐBÑ  ! œ :ÐBÑ  ! œ :ÐBÑÞ In this case <ÐBÑ ´ !Þ
-

proof continued on the next page

page 1
Case #: 7  !Þ
+5"
Let .ÐBÑ œ .7 B7  â  ." B  .! where .7 Á !Þ Note that               Á ! since both
.7
+5" 5"7
constants are nonzero. Let :" ÐBÑ œ :ÐBÑ           B       † .ÐBÑÞ Then the subtraction on
.7
the right cancels the leading term of :ÐBÑ so :" ÐBÑ is a polynomial of degree 5 or less and
we can apply the induction assumption to :" ÐBÑ to conclude there exist polynomials
;" ÐBÑ and <" ÐBÑ such that :" ÐBÑ œ .ÐBÑ † ;" ÐBÑ  <" ÐBÑ where the degree of <" ÐBÑ is
strictly less than that of .ÐBÑÞ
+5" 5"7
:" ÐBÑ œ .ÐBÑ † ;" ÐBÑ  <" ÐBÑ œ :ÐBÑ    B      † .ÐBÑ
.7
Now we solve the 2nd equation for :ÐBÑÞ
+5" 5"7
:ÐBÑ œ       B      † .ÐBÑ  .ÐBÑ † ;" ÐBÑ  <" ÐBÑ
.7

:ÐBÑ œ .ÐBÑ † ”               ;" ÐBÑ •  <" ÐBÑÞ
+5" 5"7
B
.7

So we may let ;ÐBÑ œ ”               ;" ÐBÑ • and let <ÐBÑ œ <" ÐBÑ and
+5" 5"7
B
.7
we have established the theorem holds for :ÐBÑ of degree 5  "Þ

The induction proof that establishes the existence part of the theorem is now complete.

Then we have .ÐBÑ † c;" ÐBÑ  ;# ÐBÑd œ <# ÐBÑ  <" ÐBÑÞ Call this equation (*).
To establish uniqueness, suppose :ÐBÑ œ .ÐBÑ † ;" ÐBÑ  <" ÐBÑ œ .ÐBÑ † ;# ÐBÑ  <# ÐBÑÞ

Case 1: 7 œ !Þ
In this case both remainders must be identically zero and this means <" ÐBÑ ´ <# ÐBÑÞ
Î
In turn, this means .ÐBÑ † Ò ;" ÐBÑ  ;# ÐBÑ Ó ´ !, and since .ÐBÑ ´ ! we must have
;" ÐBÑ  ;# ÐBÑ ´ ! which of course implies ;" ÐBÑ ´ ;# ÐBÑÞ

If c ;" ÐBÑ  ;# ÐBÑd ´ ! then we can compute the degrees of the polynomials on both sides of
Case #: 7  !Þ
Î
the (*) equation. The degree on the left side is greater than or equal to the degree of .ÐBÑÞ But
on the right side, both remainders have degrees less than .ÐBÑ so their difference has a degree

contradiction. So we must have c ;" ÐBÑ  ;# ÐBÑd ´ ! and when this is the case the entire left
that is less than or equal to that of either which is less than the degree of .ÐBÑÞ This is a

side of the (*) equation is identically ! and we may add back <" ÐBÑ from the right side to
conclude that the two remainders are also identically equal.
Q.E.D.

page 2
2. The Division Check for a Linear Divisor
Consider dividing the polynomial :ÐBÑ by the linear term ÐB  +ÑÞ Then, the Division Check states
that: :ÐBÑ œ ÐB  +Ñ † ;ÐBÑ  <
Division Check Proof:
This is just a special case of the Division Algorithm where the divisor is linear.
Q.E.D.

3. Remainder Theorem
When any polynomial :ÐBÑ is divided by ÐB  +Ñ the remainder is :Ð+ÑÞ
Remainder Theorem Proof:
By the Division Check we have :ÐBÑ œ ÐB  +Ñ † ;ÐBÑ  <Þ
Now let B œ +Þ This last equation says :Ð+Ñ œ Ð+  +Ñ † ;Ð+Ñ  <
:Ð+Ñ œ ! † ;Ð+Ñ  < œ !  < œ <Þ
Q.E.D.

4. Factor Theorem
ÐB  +Ñ is a factor of the polynomial :ÐBÑ if and only if :Ð+Ñ œ !Þ
Factor Theorem Proof:
Assume ÐB  +Ñ is a factor of :ÐBÑ. Then we know ÐB  +Ñ divides evenly into :ÐBÑÞ
The remainder when :ÐBÑ is divided by ÐB  +Ñ must be 0. By the Remainder Theorem
this says ! œ < œ :Ð+ÑÞ Next, assume :Ð+Ñ œ !Þ Divide :ÐBÑ by ÐB  +ÑÞ By the Remainder
Theorem, the remainder is :Ð+Ñ œ !Þ Since the remainder is 0, the division comes out even so that
ÐB  +Ñ is a factor of :ÐBÑÞ
Q.E.D.

5. Maximum Number of Zeros Theorem
A polynomial cannot have more real zeros than its degree.
Maximum Number of Zeros Theorem Proof:
By contradiction. Suppose :ÐBÑ has degree 8   1, and suppose +" ß +# ß á ß +8 ß +8" are 8+1 roots of

less than :ÐBÑ such that :ÐBÑ œ ÐB  +1 Ñ † ;" ÐBÑÞ Now since :a+# b œ ! and since +# Á +" ß we must
:ÐBÑÞ By the Factor Theorem, since :Ð+1 Ñ œ ! then there exists a polynomial ;" ÐBÑ of degree one

have ;" Ð+# Ñ œ ! and again by the Factor Theorem we can write :ÐBÑ œ ÐB  +" Ñ † ÐB  +# Ñ † ;# ÐBÑ
where ;# ÐBÑ is of degree # less than :ÐBÑÞ Now since +\$ is distinct from +" and +# we must
have ;# Ð+\$ Ñ œ ! and we can continue to factor :ÐBÑ œ ÐB  +" Ñ † ÐB  +# Ñ † ÐB  +\$ Ñ † ;\$ ÐBÑ where
the degree of ;\$ ÐBÑ is of degree \$ less than :ÐBÑÞ Clearly this argument can be repeated until we
reach the stage where :ÐBÑ œ ÐB  +" ÑâÐB  +8 Ñ † ;8 ÐBÑ and ;8 ÐBÑ is of degree 8 less than :ÐBÑÞ
Since :ÐBÑ only had degree 8 in the first place, ;8 ÐBÑ must be of degree 0 making ;8 ÐBÑ some
constant, say ;8 ÐBÑ œ -Þ Now +8" is still a zero of :ÐBÑß and since +8" is distinct from all the
other +3 ß we must have ;8 Ð+8" Ñ œ !Þ The only way this can happen is if - œ ! and this would
imply :ÐBÑ ´ !ß a contradiction since we are assuming 8   "Þ
Q.E.D.

page 3
6. Fundamental Theorem of Algebra
a) Every polynomial of degree 8   " has at least one zero among the complex numbers.
b) If :ÐBÑ denotes a polynomial of degree 8ß then :ÐBÑ has exactly 8 roots, some of
which may be either irrational numbers or complex numbers.
Fundamental Theorem of Algebra Proof:
This is not proved here. Gauss proved this in 1799 as his Ph.D. doctoral dissertation topic.

7. Product and Sum of the Roots Theorem
Let :ÐBÑ œ 1B8  +8" B8"  â  +\$ B\$  +# B#  +" B  +! be any polynomial with real
coefficients with a leading coefficient of 1 where 8   ". Then +! is Ð"Ñ8 times the product of all
the roots of :ÐBÑ œ ! and +8" is the opposite of the sum of all the roots of :ÐBÑ œ !.
Product and Sum of the Roots Theorem Proof:
By the Fundamental Theorem of Algebra we know :ÐBÑ has 8 roots which may be denoted by
<" ß <# ß <\$ ß á ß <8 Þ Now form the product of the 8 factors associated with these roots. Let ;ÐBÑ œ
ÐB  <" ÑÐB  <# ÑÐB  <\$ ÑâÐB  <8 Ñ and multiply out all these terms. Then inspect the coefficient
on B8" and inspect the constant term.

This can also be formally proved by using induction on 8. When 8 œ " we have :ÐBÑ œ B  +! and
in this case the only root of :ÐBÑ is <" œ +! Þ Since <" is the only root, <" is itself the product of all
the roots. But then Ð"Ñ8 +! œ Ð"Ñ" +! œ +! œ <" Þ So this establishes the part about the constant
term. Note again that since <" is the only root, <" is itself the sum of all the roots and the 2nd
leading coefficient is the opposite of the sum of all the roots since +! œ Ð <" ÑÞ

It is probably more instructive to manually look at the case when 8 œ # before setting up the
induction step. Note that ÐB  <" ÑÐB  <# Ñ œ B#  Ð<"  <# ÑB  <" <# Þ In this case it is
immediately apparent that the 2nd leading coefficient is the opposite of the sum of all the roots and
the constant term is product of all the roots. Because :ÐBÑ is quadratic, in this case 8 œ # so
(1)8 œ (1)# œ "Þ

Now lets assume the result is true whenever we have 5 roots and let :ÐBÑ be a polynomial with

:ÐBÑ œ eÐB  <" ÑÐB  <# ÑâÐB  <5 ÑfÐB  <5" Ñ. Let ;ÐBÑ œ eÐB  <" ÑÐB  <# ÑâÐB  <5 ÑfÞ
5  " roots, say :ÐBÑ œ ÐB  <" ÑÐB  <# ÑâÐB  <5" ÑÞ Now consider that we may write

Then ;ÐBÑ has degree 5 and we may apply the induction hypothesis to ;ÐBÑ. If we write
;ÐBÑ œ B5  +5" B5"  â  +" B  +! then we know +5" œ Œ ! <3  and we know
5

3œ"

+! œ Ð"Ñ5 † Œ # <3 .
5

3œ"

Now :ÐBÑ œ ;ÐBÑ † ÐB  <5" Ñ œ ÐB  <5" Ñ † ˜B5  +5" B5"  â  +" B  +! ™

proof continued on the next page

page 4
œ ˜B5"  +5" B5  â  +" B#  +! B™ 
˜Ð<5" ÑB5  Ð<5" Ñ+5" B5"  â  Ð<5" Ñ+" B  Ð<5" Ñ+! ™

œ B5"  Ö+5"  Ð<5" Ñ×B5  â  Ð+!  <5" † +" ÑB  Ð<5" Ñ † +! .

Clearly +5"  Ð<5" Ñ œ Œ ! <5   Ð<5" Ñ œ Œ ! <3  and Ð<5" Ñ † +! œ
5                        5"

3œ"                       3œ"

Ð<5" Ñ † Ð"Ñ5 † Œ # <3  œ Ð"Ñ5" † Œ # <3  which are what we needed to establish.
5                     5"

3œ"                   3œ"
Q.E.D.

8. Rational Roots Theorem
Let :ÐBÑ œ +8 B8  +8" B8"  â  +\$ B\$  +# B#  +" B  +! be any polynomial
-
with integer coefficients. If the rational number is a reduced rational root of :ÐBÑ œ ! then - must
.
be a factor of +! and . must be a factor of +8 Þ
Rational Roots Theorem Proof.
+8" 8" +8# 8#               +\$     +#       +"     +!
Let ;ÐBÑ œ B8          B             B     â  B\$  B#  B  Þ
+8            +8               +8     +8       +8     +8
By the Product of the Roots Theorem, we know the product of the roots of this
+!         -
polynomial is the fraction Ð"Ñ8 † Þ Thus if is a root, - must be a factor of +! and . must
+8         .
be a factor of +8 Þ
Q.E.D.

9. Integer Roots Theorem
Let :ÐBÑ œ B8  +8" B8"  â  +\$ B\$  +# B#  +" B  +! be any polynomial
with integer coefficients and with a leading coefficient of 1. If :ÐBÑ has any rational zeros,
then those zeros must all be integers.
Integer Roots Theorem Proof:
By the Rational Roots Theorem we know the denominator of any rational zero must divide into the
leading coefficient which in this case is 1. Thus any denominator must be „ 1 making the rational
zero into a pure integer.
Q.E.D.

page 5
10. Upper and Lower Bounds Theorem
Let :ÐBÑ be any polynomial with real coefficients and a positive leading coefficient.
(Upper Bound) If +  ! and :Ð+Ñ  ! and if in applying synthetic substitution to compute :Ð+Ñ all
numbers in the 3rd row are positive, then + is an upper bound for all the roots of :ÐBÑ œ !.
(Lower Bound) If +  ! and :Ð+Ñ Á ! and if in applying synthetic substitution to compute :Ð+Ñ all
the numbers in the 3rd row alternate in sign then + is a lower bound for all the roots of :ÐBÑ œ !.
[ In either bound case, we can allow any number of zeros in any positions in the 3rd row except in
the first and last positions. The first number is assumed to be positive and the last number is
:Ð+Ñ Á !. For upper bounds, we can state alternatively and more precisely that no negatives are
allowed in the 3rd row. In the lower bound case the alternating sign requirement is not strict either,
as any 0 value can assume either sign as required. In practice you may rarely see any zeros in the
3rd row. However, a slightly stronger and more precise statement is that the bounds still hold even
when zeros are present anywhere as interior entries in the 3rd row.]

Upper and Lower Bounds Theorem Proof:
(Upper Bound). Let , be any root of the equation :ÐBÑ œ !Þ Must show ,  +Þ
If , œ !, then clearly ,  + since + is positive in this case. So we assume , Á !.
If the constant term of :ÐBÑ is !, then we could factor B or a pure power of B from :ÐBÑ and just
operate on the resulting polynomial that is then guaranteed to have a nonzero constant term. So we
can implicitly assume :Ð!Ñ Á !Þ The last number in the third row of the synthetic substitution
process is positive and it is :Ð+ÑÞ Since , is a root, we know by the Factor Theorem that
:ÐBÑ œ ÐB  ,Ñ † ;ÐBÑ where ;ÐBÑ is the quotient polynomial. The leading coefficient of :ÐBÑ is also
the leading coefficient of ;ÐBÑ and since all of ;ÐBÑ's remaining coefficients are positive, and
since +  !, we must have ;Ð+Ñ  !Þ Finally, :Ð+Ñ œ Ð+  ,Ñ † ;Ð+ÑÞ Since ;Ð+Ñ  !, we may
:Ð+Ñ
divide by ;Ð+Ñ and get Ð+  ,Ñ œ         Þ Now since :Ð+Ñ and ;Ð+Ñ are both positive, Ð+  ,Ñ  !
;Ð+Ñ
which implies ,  +Þ Note that since the leading coefficient of ;ÐBÑ is positive and since +  !, we
don't really need all positive numbers in the last row. As long as ;ÐBÑ's remaining coefficients are
nonnegative we can guarantee that ;Ð+Ñ  !Þ

(Lower Bound). Let , be any root of the equation :ÐBÑ œ !Þ Must show +  ,Þ
As in the above Upper Bound proof, we can easily dispense with the case when , œ !Þ Clearly
+  , when , œ ! because + is negative. We can further implicitly assume no pure power of B is a
factor of :ÐBÑ and this also allows us to assume :Ð!Ñ Á !Þ Since :Ð,Ñ œ ! by the Factor Theorem
we may write :ÐBÑ œ ÐB  ,Ñ † ;ÐBÑ. Substituting B œ + we have :Ð+Ñ œ Ð+  ,Ñ † ;Ð+ÑÞ Since
:Ð+Ñ
:Ð+Ñ Á ! we know ;Ð+Ñ Á !Þ So we can divide by ;Ð+Ñ to get Ð+  ,Ñ œ          Þ
;Ð+Ñ

Now ;Ð+Ñ is either positive or negative. Because +  ! and the leading term in ;ÐBÑ has a positive
coefficient, the constant term in ;ÐBÑ has the same sign as ;Ð+Ñ. This fact can be established by
considering the two cases of the even or odd degrees that ;ÐBÑ must have.

proof continued on the next page

page 6
For examples:
;ÐBÑ œ "B&  #B%  \$B\$  %B#  &B  'Þ
With +  !ß ;Ð+Ñ  ! and ;Ð+Ñ and ;ÐBÑw s constant term agree in sign.
or
;ÐBÑ œ "B%  #B\$  \$B#  %B  &.
With +  !, ;Ð+Ñ  ! and again ;Ð+Ñ and ;ÐBÑw s constant term agree in sign.

We might note that in these examples, it would make no difference if any of the interior coefficients
were 0. This is because the first term has a positive coefficient, and all the remaining terms just add
fuel to the fire with the same sign as the first term. The presence of an interior zero just means you
might not get as big a fire, but the first term guarantees there is a flame!

Another note is that :Ð!Ñ œ Ð,Ñ † ;Ð!Ñß and since we are assuming , Á !ß we can divide this
equation by , to conclude that ;Ð!Ñ Á ! when :Ð!Ñ Á !Þ So assuming neither , nor the constant
term in :ÐBÑ are zero guarantees that the constant term in ;ÐBÑ must be strictly positive or strictly
negative. Since the numbers in the third row alternate in sign, :Ð+Ñ differs in sign from the constant
term in ;ÐBÑÞ But since the constant term in ;ÐBÑ has the same sign as ;Ð+Ñ we know :Ð+Ñ and ;Ð+Ñ
:Ð+Ñ
differ in sign. So Ð+  ,Ñ œ        !Þ +  ,Þ
;Ð+Ñ
Q.E.D.

11. Intermediate Value Theorem
If :ÐBÑ is any polynomial with real coefficients, and if :Ð+Ñ  ! and :Ð,Ñ  ! then
there is at least one real number - between + and , such that :Ð-Ñ œ !Þ
Intermediate Value Theorem Proof:
This result depends on the continuity of all polynomials and is a special case of the Intermediate
Value Theorem that normally appears in a calculus class.

12. Single Bound Theorem
Let :ÐBÑ œ B8  +8" B8"  +8# B8#  â  +\$ B\$  +# B#  +" B  +! be any polynomial with
real coefficients and a leading coefficient of 1. Let Q" œ "  7+BÖk+! kß k+" kß k+# kß á ß k+8" k× and
let Q# œ 7+BÖ"ß k+! k  k+" k  k+# k  á  k+8" k×. Finally let Q œ 738ÖQ" ß Q# ×. Then every
zero of :ÐBÑ lies between Q and Q .
Single Bound Theorem Proof:
We need to show Q is an Upper Bound and we need to show Q is a Lower Bound.

Case 1: Q œ Q" Þ

Then we know for ! Ÿ 3 Ÿ 8  " that Q   "  k+3 kÞ This implies two things. First, Q   " and
second, Q  k+3 k   "Þ These two inequalities are crucial and further imply that Q Ÿ " and
Q  k+3 k Ÿ "Þ

proof continued on the next page

page 7
To show Q is an Upper Bound, consider the synthetic substitution calculation of :ÐQ ÑÞ
We will label the second and remaining coefficients in the second row as ,3 values.
We will label the second and remaining coefficients in the third row as -3 values.

Q
"   +8"   +8#   +8\$   â    +"   +!
Q    ,8#   ,8\$   â    ,"   ,!
"   -8"   -8#   -8\$   â    -"   -!

We claim that each -3 value is not only positive, we claim each -3   "Þ Similarly we claim each
,3   Q Þ We will establish these two claims by working from left to right across the columns in the
synthetic substitution table, one column at a time.

First note that -8" œ +8"  Q   k+8" k  Q œ Q  k+8" k   "Þ
We are done with the 2nd column.

Now we will argue about the 3rd column in the above table.
Having established in the 2nd column that -8"   "ß multiply both sides of this inequality by Q to
obtain: ,8# œ -8" † Q   Q Þ

Here we use the fact that ,8#   Q   "  k+8# k. So ,8#  k+8# k   "Þ
Now we basically repeat the above argument to establish the size of -8# œ +8#  ,8# Þ

So -8# œ +8#  ,8#   k+8# k  ,8# œ ,8#  k+8# k   "Þ
We are now done with the 3rd column in the table. Each next column is handled like the 3rd
column.

Just to make sure you get the idea we will establish our claims for the 4th column.
Since -8#   ", we can multiply across this inequality by Q to get -8# † Q   Q Þ

-8\$ œ +8\$  ,8\$   k+8\$ k  ,8\$ œ ,8\$  k+8\$ k   Q  k+8\$ k   "Þ
,8\$ œ -8# † Q   Q Þ

Clearly we can continue working across the columns of the above table, one column at a time.
Since all the -3 coefficients are positive we know Q is an Upper Bound for the zeros of :ÐBÑÞ

Next, to show Q is a Lower Bound, consider the synthetic substitution calculation of :ÐQ ÑÞ

Q
"   +8"   +8#   a8\$   â    +"   +!
Q     ,8#   ,8\$   â    ,"   ,!
"   -8"   -8#   -8\$   â    -"   -!

proof continued on the next page

page 8
We claim that the coefficients in the 3rd row alternate in sign. Obviously the first coefficient is
"  !Þ
We claim not only that the -3 alternate in sign, we claim that when -3  ! then -3 Ÿ "Þ We also
claim that when -3  ! then -3   "Þ

In the 2nd column of the table we have -8" œ +8"  Q Ÿ k+8" k  Q œ Q  k+8" k Ÿ "Þ
So we have established our claim within the 2nd column.

Moving over to the 3rd column we note that ,8# œ -8" † aQ b and since both of the numbers in

-8" Ÿ " and multiply across by the negative number Q we get -8# † aQ b   Q Þ But
this product are negative, we have ,8#  !Þ In fact when we start with the inequality that

,8# œ -8# † aQ b so we know that ,8#   Q Þ Now lets compute -8# Þ
-8# œ +8#  ,8#   +8#  Q   k+8# k  Q œ Q  k+8# k   "Þ

Next, consider what happens in the 4th column of the above table. We just established that

Multiplying across this inequality by Q we get -8# † aQ b Ÿ Q Þ But ,8\$ œ -8# † aQ b so
-8#   "Þ

-8\$ œ +8\$  ,8\$ Ÿ +8\$  Q Ÿ k+8\$ k  Q œ Q  k+8\$ k Ÿ "Þ
we know ,8\$ Ÿ Q Þ Now lets compute -8\$ Þ

Clearly the above arguments may be repeated as we move across the columns of the above table.
Each time we multiply by Q to compute the next ,3 value we have a sign change. This is
primarily why the -3 values alternate in sign.

In any case, the values in the 3rd row alternate in sign and since Q  ! we know Q is a lower
bound for any zero of the equation :ÐBÑ œ !Þ

Case 2: Q œ Q# Þ

Subcase 1: Q œ " and k+! k  k+" k  k+# k  â  k+8" k  "Þ

In particular, for each 3 where ! Ÿ 3 Ÿ 8  " we know ! Ÿ k+3 k  " œ Q Þ
Then k+3 k  Q œ "  k+3 k  !Þ
Since Q œ ", the Synthetic Substitution table takes on a particularly simple form.
Note how the 2nd row elements are the same as the 3rd row elements shifted over
one column.
"
" +8" +8# a8\$ +8% â +" +!
" -8" -8# -8\$ â -# -"
" -8" -8# -8\$ -8% â -" -!

Starting in the 2nd column, -8" œ +8"  "   k+8" k  " œ "  k+8" k  !Þ
We claim that all the -3 values are positive.

proof continued on the next page

page 9
k+8# k  k+8" k  " œ ak+8# k  k+8" kb  "  !Þ
Now consider the 3rd column. -8# œ +8#  -8" œ +8#  +8"  "

k+8\$ k  k+8# k  k+8" k  " œ ak+8\$ k  k+8# k  k+8" kb  "  !Þ
Next consider the 4th column. c8\$ œ +8\$  -8# œ +8\$  +8#  +8"  "

Clearly we can continue to accumulate the sums of more and more terms and still apply
the main inequality that appears in the Subcase 1: statement. So all the elements in the
last row are positive and " œ Q is an upper bound for all the roots of :ÐBÑ œ ! by
applying the Upper/Lower Bounds Theorem.

To establish that 1 is a lower bound we compute synthetic substitution with Q œ  "Þ

"
"   +8"    +8#     a8\$      +8%        â    +"    +!
"    -8"    -8#     -8\$        â   -#   -"
"   -8"    -8#     -8\$      -8%        â    -"    -!

Starting in the 2nd column, -8" œ +8"  Ð"Ñ Ÿ k+8" k  a"b  !Þ
Now we must establish that the -3 values in the last row alternate in sign.

In the 3rd column, -8# œ +8#  a  -8" b œ +8#  +8"  "
k+8# k  k+8" k  " œ ak+8# k  k+8" kb  "  !Þ

In the 4th column, -8\$ œ +8\$  a-8# b œ +8\$  +8#  +8"  " Ÿ
k+8\$ k  k+8# k  k+8" k  "  !Þ

Clearly this argument may be repeated to establish that the coefficients in the 3rd
row really do alternate in sign. So Q œ " is a lower bound by the Upper/Lower Bounds
Theorem.

Subcase #: Q œ k+! k  k+" k  k+# k  â  k+8" k and Q   "Þ

To establish that Q is an upper bound, we consider the synthetic substitution table for
computing :ÐQ Ñ and we will show that all the values in the last row are nonnegative.

Q
"   +8"   +8#    +8\$   â    +"     +!
Q    ,8#    ,8\$   â    ,"     ,!
"   -8"   -8#    -8\$   â    -"     -!

proof continued on the next page

page 10
-8" œ +8"  Q   k+8" k  Q œ k+! k  k+" k  k+# k  â  k+8# k   !Þ
Now consider the 2nd column in the above table.

Next consider the 3rd column in the above table.
Since Q   "ß -8" † Q   -8" Þ But ,8# œ -8" † Q so ,8#   -8" Þ

k+8# k  k+8" k  Q œ k+! k  k+" k  k+# k  â  k+8\$ k   !Þ
Next, -8# œ +8#  ,8#   +8#  -8" œ +8#  +8"  Q

We continue to argue in the same manner for the 4th column.
Since Q   "ß -82 † Q   -82 Þ But ,83 œ -82 † Q so ,83   -82 Þ

k+8\$ k  k+8# k  k+8" k  Q œ k+! k  k+" k  k+# k  â  k+8% k   !Þ
Next, -83 œ +8\$  ,8\$   +8\$  -8#   +8\$  +8#  +8"  Q

This argument may be repeated across the columns in the above table to establish that all the
numbers in the last row are nonnegative. So by the Upper/Lower Bounds Theorem, Q is an
upper bound for all the roots of :ÐBÑ œ !Þ

Finally, consider the synthetic substitution table for computing :ÐQ ÑÞ

Q
"   +8"   +8#    +8\$   +8%    â    +"   +!
Q     ,8#    ,8\$   ,8%    â    ,"   ,!
"   -8"   -8#    -8\$   -8%    â    -"   -!

We claim the nonnegative numbers in the 3rd row of this table alternate in sign. In the first
column the number is 1 so we know we are starting with a positive value.

-8" œ +8"  aQ b œ +8"  ek+! k  k+" k  k+# k  â  k+8" kf œ
Now look at -8" in the 2nd columnÞ

+8"  k+8" k  ek+! k  k+" k  k+# k  â  k+8# k fÞ

and the first two terms either make ! or make # † k+8" k so the whole expression
Now the last term in the above expression is obviously less than or equal to zero,

is less than or equal to 0.

Now consider the 3rd column. We must show -8#   !.

-8" œ ek+! k  k+" k  k+# k  â  k+8# k fÞ
We take the worst case from -8" assuming this has the smallest absolute value where

Then -8# œ +8#  ek+! k  k+" k  k+# k  â  k+8# k f † aQ b œ
+8#  Q † k+8# k  ek+! k  k+" k  k+# k  â  k+8\$ k f † aQ b
+8#  k+8# k  ek+! k  k+" k  k+# k  â  k+8\$ k f † aQ b   !
since the third term is nonnegative and the first two terms make either 0 or # † k+8# kÞ

proof continued on the next page

page 11
Now consider the 4th column. We must show -8\$ Ÿ !Þ

-8# œ ek+! k  k+" k  k+# k  â  k+8\$ k f † aQ bÞ
We take the worst case from -8# assuming -8# has the smallest absolute value where

Then -8\$ œ +8\$  ek+! k  k+" k  k+# k  â  k+8\$ k f † aQ # b œ
+8\$  Q # † k+8\$ k  ek+! k  k+" k  k+# k  â  k+8% k f † aQ # b Ÿ
+8\$  k+8\$ k  ek+! k  k+" k  k+# k  â  k+8% k f † aQ # b Ÿ !
since the third term is negative and the first two terms make either ! or # † k+8\$ kÞ

Just to make sure you get the idea we will continue with the 5th column. We must show

value where -8\$ œ ek+! k  k+" k  k+# k  â  k+8% k f † aQ # bÞ
-8%   !Þ We take the worst case from -8\$ assuming -8\$ has the smallest absolute

Then -8% œ +8%  ek+! k  k+" k  k+# k  â  k+8% k f † aQ \$ b œ
+8%  Q \$ † k+8% k  ek+! k  k+" k  k+# k  â  k+8& k f † aQ \$ b
+8%  k+8% k  ek+! k  k+" k  k+# k  â  k+8& k f † aQ \$ b   !
since the third term is nonnegative and the first two terms make either 0 or # † k+8% kÞ

This argument may be repeated across the columns of the above table to conclude that the
nonnegative terms in the last row alternate in sign. By the Upper/Lower Bounds Theorem
we know Q is a lower bound for all the zeros of :ÐBÑ œ !Þ
Q.E.D.

13. Odd Degree Real Root Theorem
If :ÐBÑ has real coefficients and has a degree that is odd then it has at least one real root.
Odd Degree Real Root Theorem Proof:
Without loss of generality we assume the leading coefficient of :ÐBÑ is positive. Otherwise we can
factor " from :ÐBÑ apply the theorem to the polynomial that is the other factor.
By choosing Q  ! sufficiently large we can establish that :ÐQ Ñ  ! and :ÐQ Ñ  !Þ
For example, see the above Single Bound Theorem. Now apply the Intermediate Value Theorem.
There exists a number + such that Q  +  Q and :Ð+Ñ œ !Þ + is real and is a root of :ÐBÑÞ
Q.E.D.

page 12
14. Complex Conjugate Roots Theorem
If :ÐBÑ is any polynomial with real coefficients, and if +  ,3 is a complex root of
the equation :ÐBÑ œ !, then another complex root is its conjugate +  ,3Þ
(Complex number roots appear in conjugate pairs)
Complex Conjugate Roots Theorem.
This proof just depends on properties of the conjugate operator denoted by bars below.
If :ÐBÑ œ +8 B8  +8" B8"  â  +\$ B\$  +# B#  +" B  +! œ ! then

+8 B8  +8" B8"  â  +\$ B\$  +# B#  +" B  +! œ !.

+8 B8  +8" B8"  â  +\$ B\$  +# B#  +" B  +! œ !.

+8 B8  +8" B8"  â  +\$ B\$  +# B  +" B  +! œ !.

+8 B8  +8" B8"  â  +\$ B\$  +# B  +" B  +! œ !. This shows :ÐBÑ œ !Þ
Q.E.D.

15. Linear and Irreducible Quadratic Factors Theorem
Any polynomial :ÐBÑ with real coefficients may be written as a product of linear factors and
irreducible quadratic factors. The sum of all the degrees of these component factors is the degree of
:ÐBÑÞ
Linear and Irreducible Quadratic Factors Theorem Proof:
By the Fundamental Theorem of Algebra, :ÐBÑ œ -ÐB  <" ÑÐB  <# ÑÐB  <\$ ÑâÐB  <8 Ñ
where the <5 denote the 8 roots of :ÐBÑÞ The constant - is simply :ÐBÑ's leading coefficient.
If all the <5 roots are real then :ÐBÑ is a product of linear real factors only. However, if any <5 value
is a complex number then it can be paired with some other <4 value which is its complex conjugate.
If we assume <5 œ +  ,3 then <4 œ +  ,3Þ Note that since <5 is complex, we must have , Á !Þ
Next we note that <5  <4 œ #+ and <5 † <4 œ +#  ,# and we compute
(B  <5 ÑÐB  <4 Ñ œ B#  Ð<5  <4 ÑB  <5 <4 œ B#  Ð#+ÑB  Ð+#  ,# ÑÞ That this
last expression is an irreducible quadratic factor follows by computing its discriminant that is
Ð#+Ñ#  % † " † Ð+#  ,# Ñ œ %+#  %+#  %,# œ %,#  ! since , Á !Þ
If we re-order or rename the indices of the roots so that <5 and <4 were the last two roots then we can

:ÐBÑ œ -ÐB  <" ÑÐB  <# ÑâÐB  <8\$ ÑeÐB#  #+B  +#  ,# ÑfÞ
assume that :ÐBÑ now takes the form in which we put the irreducible quadratic just found at the end.

Now if any two of the preceding linear factors form complex <-value conjugate roots, we treat them
just like we did with the pair <5 and <4 . This produces another irreducible quadratic factor which we
also place as the last rightmost factor. Clearly this process can be continued until only real linear
factors remain at the beginning and only irreducible quadratic factors are at the end. There may be
no linear factors at the beginning and only irreducible quadratic factors, or there may be no
irreducible quadratic factors at the end and only linear factors at the beginning. It all depends on the
nature of the roots <3 and how many of these roots are real and how many are complex.
Q.E.D.

page 13
Let :ÐBÑ be any polynomial with rational real coefficients. If +  ,È- is a root of the
16. Irrational Conjugate Roots Theorem

equation :ÐBÑ œ ! where È- is irrational and + and , are rational, then another root is +  ,È- Þ
(Like complex roots, irrational real roots appear in conjugate pairs, but only when the polynomial
has rational coefficients.)

Assume +  ,È- is one root. Must show +  ,È- is also a root. If , œ ! we are done, so assume
Irrational Conjugate Roots Theorem Proof:

, Á !Þ Let >ÐBÑ œ ˆB  ˆ+  ,È- ‰‰ † ˆB  ˆ+  ,È- ‰‰ œ aB  +b#  ,# -Þ Then >ÐBÑ is a
quadratic polynomial with rational coefficients. Next, consider dividing :ÐBÑ by >ÐBÑÞ By the
Division Algorithm, there is a quotient polynomial ;ÐBÑ and there exists a remainder polynomial
<ÐBÑ such that :ÐBÑ œ >ÐBÑ † ;ÐBÑ  <ÐBÑ where the degree of <ÐBÑ is 1 or 0.
If we assume <ÐBÑ œ GB  H then G and H must be rational. In fact, since :ÐBÑ and >ÐBÑ have only

So we may write :ÐBÑ œ >ÐBÑ † ;ÐBÑ  GB  H and when we substitute B œ +  ,È- we conclude
rational coefficients, both ;ÐBÑ and <ÐBÑ must have only rational coefficients.

that ! œ G ˆ+  ,È- ‰  H from which we can further conclude that G œ H œ !Þ
So we really have :ÐBÑ œ >ÐBÑ † ;ÐBÑÞ Finally we substitute B œ +  ,È- in this last equation to
conclude that :ˆ+  ,È- ‰ œ ! which is what we needed to show.
Q.E.D.

17. Descartes's Rule of Signs Lemma 1.
If :ÐBÑ has real coefficients, and if :Ð+Ñ œ ! where +  !ß then :ÐBÑ has at least one more sign
variation than the quotient polynomial ;ÐBÑ has sign variations where :ÐBÑ œ ÐB  +Ñ;ÐBÑÞ
[When the difference in the number of sign variations is greater than 1, the difference is always an
odd number.]
Descartes's Rule of Signs Lemma 1 Proof:
The following particular example shows that ;ÐBÑ may indeed have fewer sign variations
than :ÐBÑ has. In this example, :ÐBÑ has three variations in sign while ;ÐBÑ has only two
variations in sign. Had the number 152 in the top row been 102 instead, then ;ÐBÑ
would have had even fewer sign variations as is shown in the next example.
#
"# ((         "&# (( \$!
#% "!'          *#      \$!
"# &\$          %'      "&       !

In the next example, :ÐBÑ has four sign variations while ;ÐBÑ has only one sign variation. So
the difference of the sign variation counts in the example below is the odd number \$.
#
"# ((         "!# ((         "(!
#% "!'        ) "(!
"# &\$         % )&             !

proof continued on the next page

page 14
Finally we show one more example before starting the formal proof. In this example, :ÐBÑ has
four sign variations and ;ÐBÑ has only three sign variations. Moreover, the coefficients in :ÐBÑ
and ;ÐBÑ match signs column by column from left to right through the constant column in ;ÐBÑÞ
\$
" ' "" ""              "&
\$ *         ' "&
" \$        #    &       !

Assume the leading coefficient of :ÐBÑ is positive and consider the synthetic substitution
form used to compute :Ð+ÑÞ Consider the constant term in :ÐBÑÞ If this constant term is
negative as in the first example above, then in the previous column the constant term in ;ÐBÑ
must have been positive in order for the final column numbers to add to make 0. If the
constant term in :ÐBÑ were positive as in the second example above, then in the previous
column the constant term in ;ÐBÑ must have been negative in order for the final column
numbers to add to make 0. So the constant terms in :ÐBÑ and ;ÐBÑ must have opposite signs.
This argument has depended on the facts that +  ! and that :Ð+Ñ œ !Þ

right, we claim ;ÐBÑ cannot change signs until :ÐBÑ changes signs. Whenever ;ÐBÑ changes
signs from one column to the next, :ÐBÑ must also change signs between those same two
columns. But as in the second example above (columns 2 & 3 and columns 3 & 4), ;ÐBÑ can
keep the same sign even when :ÐBÑ does change sign. But ;ÐBÑ can never change signs unless
:ÐBÑ changes signs first.

Now suppose in counting sign changes that at some point :ÐBÑ changes signs when ;ÐBÑ does
not as in the first two examples above. Then :ÐBÑ has one more sign variation, and from that
point forward, :ÐBÑ will continue to lead in the sign variation count because each
further time ;ÐBÑ changes signs, so does :ÐBÑÞ We can rest our case in this case.

The other case that needs to be considered is when counting sign changes, if we reach the end
of ;ÐBÑ, and if at that point :ÐBÑ and ;ÐBÑ have the same sign variation count, as in the third
example above. Then :ÐBÑ and ;ÐBÑ will have the same sign in the next to the last column, but
then :ÐBÑ will change signs one more time in its last column.

No matter how you look at it, :ÐBÑ has at least one more sign variation count than ;ÐBÑÞ
If the leading coefficient of :ÐBÑ is not positive, then factor out 1 from :ÐBÑ and then
apply the above argument to the resulting polynomial.

We have already proved :ÐBÑ has at least one more sign variation than the quotient polynomial
;ÐBÑ. To prove that the difference is always an odd number, we reiterate that the constant
terms in both polynomials always differ in sign while the first terms always agree in sign. So
when more than one sign variation occurs, it occurs at some interior coefficient. But
changing the sign of any interior coefficient either raises or lowers the sign variation count by
2 because such a change applies to the term before it and to the term after it. So when the
difference in the number of sign variations is more than ", it must be an odd difference.
Q.E.D.

page 15
18. Descartes's Rule of Signs Lemma 2.
If :ÐBÑ has real coefficients, the number of positive zeros of :ÐBÑ is not greater than the
number of variations in sign of the coefficients of :ÐBÑÞ

Descartes's Rule of Signs Lemma 2 Proof:
Let <" ß <# ß <\$ ß á ß <5 denote all the positive roots of the equation :ÐBÑ œ !Þ
Then we may write :ÐBÑ œ ÐB  <" ÑÐB  <# ÑÐB  <\$ ÑâÐB  <5 Ñ † UÐBÑÞ
Now consider the following regrouping of these factors:

:ÐBÑ œ ÐB  <" ÑeÐB  <# ÑÐB  <\$ ÑâÐB  <5 Ñ † UÐBÑfÞ

Let ;" ÐBÑ œ eÐB  <# ÑÐB  <\$ ÑâÐB  <5 Ñ † UÐBÑfÞ Then :ÐBÑ has at least one more sign
By Lemma 1 we know :ÐBÑ has at least one more sign variation than the rightmost factor.

;" ÐBÑ œ ÐB  <# ÑeÐB  <\$ ÑâÐB  <5 Ñ † UÐBÑf
variation than ;" ÐBÑ has. Moreover, since we may write

we can again apply Lemma 1 to conclude that ;" ÐBÑ has one at least more sign variation than
the polynomial ÖÐB  <\$ ÑâÐB  <5 Ñ † UÐBÑ×. Let ;# ÐBÑ œ ÖÐB  <\$ ÑâÐB  <5 Ñ † UÐBÑ×Þ

Now :ÐBÑ has at least one more sign variation than ;" ÐBÑ and ;" ÐBÑ has at least one more sign
variation than ;# ÐBÑß so :ÐBÑ has at least two more sign variations than ;# ÐBÑÞ

Clearly we may continue to regroup the rightmost factors and reduce the number of factors.
;# ÐBÑ œ ÖÐB  <\$ ÑâÐB  <5 Ñ † UÐBÑ× œ ÐB  <\$ Ñ † ÖÐB  <% ÑâÐB  <5 Ñ † UÐBÑ×Þ
So ;# ÐBÑ has one or more sign variations than ;\$ ÐBÑ œ ÖÐB  <% ÑâÐB  <5 Ñ † UÐBÑ×Þ

We argue that for each factor we drop, :ÐBÑ has yet at least another sign variation more than
the resulting rightmost factor. After dropping all 5 factors we conclude that :ÐBÑ has 5 or
more          sign variations than does UÐBÑ, since after dropping 5 factors, UÐBÑ is all that
remains.

Now assume UÐBÑ has 7 sign variations in its coefficients and assume :ÐBÑ has 8 sign
variations in its coefficients. Then because 7   !, 7  5 Ÿ 8 implies 5 Ÿ 8. The number of
positive zeros of :ÐBÑ is less than or equal to the number of sign variations in the coefficients
of :ÐBÑÞ
Q.E.D.

page 16
19. Descartes's Rule of Signs Lemma 3.
Let <" ß <# ß <\$ ß á ß <5 denote 5 positive numbers and let :ÐBÑ œ # ÐB  <3 ÑÞ
5

3œ"
Then the coefficients of :ÐBÑ are all alternating in sign and this polynomial has exactly 5 sign
variations in its coefficients.
Descartes's Rule of Signs Lemma 3 Proof:
Either use induction on 5 or else apply Lemma 2 to conclude that UÐBÑ œ " has 5 fewer
variations in sign than :ÐBÑÞ But since " has no variations in sign we know :ÐBÑ must have 5

As a simple example: ÐB  <" ÑÐB  <# Ñ œ B#  Ð<"  <# ÑB  a"b# a<" † <# bÞ
variations in sign which means all the coefficients alternate in sign.

Using induction, if 5 œ ", then we note ÐB  <" Ñ has exactly " sign variation. Next, assume
the theorem is true for any polynomial with 5 factors or less, and let
:ÐBÑ œ # ÐB  <3 Ñ œ ÐB  <5" Ñ † ” # ÐB  <3 Ñ•
5"                             5

3œ"                           3œ"

If we consider the second factor to be the quotient polynomial then we can apply the induction
assumption to conclude this quotient has exactly 5 sign variations. Next we apply Lemma " to
:ÐBÑ to conclude that :ÐBÑ has at least one more sign variation than the quotient. This means
:ÐBÑ has exactly 5  " sign variations. Being a Ð5  "Ñ=> degree polynomial, :ÐBÑ cannot have
more than 5  " sign variations because it has only 5  # coefficients.
Q.E.D.

20. Descartes's Rule of Signs Lemma 4.
The number of variations in sign of a polynomial with real coefficients is even if the first and last
coefficients         have the same sign, and is odd if the first and last coefficients have opposite signs.
Descartes's Rule of Signs Lemma 4 Proof:
Before giving the proof we look at one example.
:ÐBÑ œ B%  'B\$  ""B#  "#B  "&Þ
In this case, the first and last coefficients have the same sign and we can see that
:ÐBÑ has an even number of sign changes in its coefficients; it has 4 sign changes.
The degree of :ÐBÑ is %, it has & coefficients, and thus it has an a priori possibility of
having at most 4 sign variations. If we were to change the sign of either the first or the
last coefficient, we would have one less sign change, or an odd number of sign changes.
If we were to increase the degree of :ÐBÑ by adding just one term, then we would not
add a sign change unless the sign of that new term differed from the existing leading
term's sign.

We prove this theorem by strong induction on the degree 8 of the polynomial :ÐBÑÞ
When 8 œ ", we assume :ÐBÑ œ +B  ,. If + and , have the same sign then we have
0 or an even number of sign changes. If + and , have opposite signs then we have 1 or
an odd number of sign changes. So the theorem is true when 8 œ "Þ

proof continued on the next page

page 17
Next, assume the theorem is true whenever 8 Ÿ 5 , and let :ÐBÑ be a polynomial of degree
5  "Þ Must show the theorem is true for :ÐBÑÞ

Consider the polynomial of degree 5 , obtained by dropping the leading term from :ÐBÑÞ
Call this polynomial ;ÐBÑÞ The theorem is assumed true for ;ÐBÑ since its degree can be
assumed to be either 5 , or even better, less than 5 .

There are two cases.

Case 1: ;ÐBÑ's leading and trailing terms have the same sign.
Then we know by the induction assumption that ;ÐBÑ has an even number of sign
changes.
There are only two possibilities for the sign of the leading term that was dropped.
If the dropped leading term has the same sign as the leading term in ;ÐBÑ, then there is no
sign change when this term is added back. So :ÐBÑ would still have an even number of
sign changes and the leading and trailing terms of :ÐBÑ would still agree in sign.
If the dropped leading term has a different sign from the leading term in ;ÐBÑß then there
is one additional sign change that gets added when this term is put back. So in this case
:ÐBÑ would have an odd number of sign changes. But also in this case, the leading and
trailing terms of :ÐBÑ would have opposite signs.

Case 2: ;ÐBÑ's leading and trailing terms have opposite signs.
Then we know by the induction assumption that ;ÐBÑ has an odd number of sign changes.
There are only two possibilities for the sign of the leading term that was dropped.
If the dropped leading term has the same sign as the leading term in ;ÐBÑ, then there is no
sign change when this term is added back. So :ÐBÑ would still have an odd number of
sign changes and the leading and trailing terms of :ÐBÑ would still have opposite signs.
If the dropped leading term has a different sign from the leading term in ;ÐBÑß then there
is one additional sign change that gets added when this term is put back. So in this case
:ÐBÑ would have an even number of sign changes. But also in this case, the leading and
trailing terms of :ÐBÑ would have the same signs.

In either case, the theorem is true for :ÐBÑ with degree 5  "Þ This completes the proof by
induction on 8.
Q.E.D.

page 18
21. Descartes's Rule of Signs Lemma 5.
If the number of positive zeros of :ÐBÑ with real coefficients is less than the number of sign
variations in :ÐBÑ, it is less by an even number.
Descartes's Rule of Signs Lemma 5 Proof:

eÐB  <" ÑâÐB  <5 Ñf † eÐB  8" ÑâÐB  84 Ñf † eÐB#  ," B  -" ÑâÐB#  ,6 B  -6 Ñf
If the leading coefficient of :ÐBÑ isn't ", we can factor it out and just assume :ÐBÑ œ

where the <3 denote all the positive zeros of :ÐBÑ, the 83 denote the all the negative zeros of
:ÐBÑ, and the remaining factors are quadratics corresponding to all the complex-conjugate
paired complex zeros of :ÐBÑ. Let = be the number of sign changes in the coefficients of :ÐBÑÞ
We assume 5  = and we must show there exists an even integer / such that /  ! and

Let 0 ÐBÑ œ eeÐB  8" ÑâÐB  84 Ñf † eÐB#  ," B  -" ÑâÐB#  ,6 B  -6 ÑffÞ
5  / œ =Þ

By Lemma 2, we know the 0 ÐBÑ polynomial has at least 5 fewer sign variations than :ÐBÑÞ
We let > be the number of sign changes in the 0 ÐBÑ polynomial. Then we know >   =  5  !Þ
Next we note a special property of each of the irreducible quadratic factors. The discriminant
#                           #
of each quadratic must be negative, so we know ,3  % † " † - 3  !Þ So ! Ÿ ,3  %-3 and we
conclude that all the -3 coefficients are strictly positive.

we may write 0 ÐBÑ œ eeÐB  :" ÑâÐB  :4 Ñf † eÐB#  ," B  -" ÑâÐB#  ,6 B  -6 ÑffÞ
Also, each ÐB  83 Ñ factor of 0 ÐBÑ may be written as ÐB  :3 Ñ where :3 œ 83 is positive. So

Now it is clear that the leading coefficient of 0 ÐBÑ is ", and the constant term of 0 ÐBÑ is also
positive since it is the product of all positive numbers.

The constant term of 0 ÐBÑ œ – # :3 —” # -3 •. By Lemma 4, the number of sign variations
4       6

3œ"     3œ"
in the coefficients of 0 ÐBÑ is even. > is even. From above we have >   =  5  !Þ
Therefore >  5   =Þ Now if it happens that >  5 œ = then we let / œ > and we are done.

Otherwise, if >  5  = then we have to argue about the first 5 factors in :ÐBÑÞ Having just
established that the constant term in 0 ÐBÑ is positive, the sign of the constant term in :ÐBÑ is
the sign of a"b5 † ” # <5 • œ the sign of a"b5 since all the <3 values are positive. If 5 is even
5

3œ"
then by Lemma %, :ÐBÑ has an even number of sign variations in its coefficients which means
= is even. If 5 is odd then again by Lemma 4 we conclude = is odd.
So 5 and = are even together or are odd together. Now consider that >  5  =  !Þ What kind
of a positive number is this? Well > is even, and if 5 and = are both even then >  5  = must
be even. By the same token, if 5 and = are both odd, then 5  = is still even, and since > is
always even, >  Ð5  =Ñ must be even. Therefore >  5  = œ #@ for some positive integer @Þ
Then Ð>  #@Ñ  5 œ =Þ Let / œ Ð>  #@ÑÞ All that remains is to show that Ð>  #@Ñ is a
positive even         integer. First, Ð>  #@Ñ œ =  5 and since =  5  !, we know Ð>  #@Ñ is
a positive integer. Second, both > and #@ are even, so their difference Ð>  #@Ñ is even. In any
case, there exists a positive even integer / such that /  5 œ =Þ So when = is larger than 5 , it is
larger by a positive even integer.
Q.E.D.

page 19
22. Descartes's Rule of Signs Lemma 6.
Each negative root of :ÐBÑ corresponds to a positive root of :ÐBÑÞ That is, if +  !
and + is a zero of :ÐBÑ, then + is a positive zero of :ÐBÑÞ
Descartes's Rule of Signs Lemma 6 Proof:
The graph of the function C œ :ÐBÑ is just the graph of C œ :ÐBÑ reflected over the C-axis.

:ÐBÑ œ :aa+bb œ :Ð+Ñ œ !Þ So + is a positive zero of :aBbÞ
So, if +  ! and :Ð+Ñ œ !, then +  !ß and when B œ +ß then

Q.E.D.

23. Descartes's Rule of Signs Theorem
Let :ÐBÑ be any polynomial with real coefficients.
(Positive Roots) The number of positive roots of :ÐBÑ œ ! is either equal to the
number of sign variations in the coefficients of :ÐBÑ or else is less than this
number by an even integer.
(Negative Roots) The number of negative roots of :ÐBÑ œ ! is either equal to the
number of sign variations in the coefficients of :ÐBÑ or else is less than
this number by an even integer.
Note that when determining sign variations we can ignore terms with zero coefficients.

Proof of Descartes's Rule of Signs Theorem:
The statement about the number of positive roots of :ÐBÑ œ ! is exactly the statement of
Lemma 5 that has already been proved.
To prove the statement about the number of negative roots of :ÐBÑ we need only apply
Lemma 6. Each negative root of :ÐBÑ corresponds to a positive root of :ÐBÑ and by
Lemma 5, the number of positive roots of any polynomial Ölike :ÐBÑ × is either equal
to the number of sign variations in that polynomial Ö :ÐBÑ ×, or is less than the number
of sign variations in that polynomial Ö :ÐBÑ × by a positive even integer.
Q.E.D.

page 20
24. Lemma On Continuous Functions.
Let 0 ÐBÑ and 1ÐBÑ be two continuous real-valued functions with a common domain that is
an open interval Ð+ß ,Ñ. Furthermore let - − Ð+ß ,Ñ and assume that except when B œ - we have
0 ÐBÑ œ 1ÐBÑ for all B − Ð+ß ,ÑÞ Then we must also have 0 Ð-Ñ œ 1Ð-ÑÞ
Proof of Lemma On Continuous Functions:
By contradiction. Assume 0 Ð-Ñ Á 1Ð-ÑÞ Without loss of generality we may assume 0 Ð-Ñ  1Ð-Ñ
1Ð-Ñ  0 Ð-Ñ
and choose % œ                 Þ Note that %  !Þ By the continuity of both 0 ÐBÑ and 1ÐBÑ at
#
B œ - , there exists a \$0  0 and there exists a \$1  ! such that for all B − Ð+ß ,Ñ

1) if !  kB  - k  \$0 then k0 ÐBÑ  0 Ð-Ñk  %

and          2) if !  kB  - k  \$1 then k1ÐBÑ  1Ð-Ñk  %

Let \$ œ 738Ö\$0 ß \$1 × and choose B" − Ð+ß ,Ñ such that !  kB"  - k  \$ Þ

Note that since B" is chosen so that B" Á - , we must have 1ÐB" Ñ œ 0 ÐB" ÑÞ
Also, by our choice of \$ , parts 1) and 2) above apply so we can conclude that 0 ÐB" Ñ  0 Ð-Ñ  %
and with a little bit of thought, we can see that we must also have 1Ð-Ñ  1ÐB" Ñ  %Þ

So adding both inequalities we must have 0 ÐB" Ñ  0 Ð-Ñ  1Ð-Ñ  1ÐB" Ñ  #% œ 1Ð-Ñ  0 Ð-ÑÞ

Now since 1ÐB" Ñ œ 0 ÐB" Ñ the left expression simplifies and may be rearranged so that we have
1Ð-Ñ  0 Ð-Ñ  1Ð-Ñ  0 Ð-Ñ, a contradiction.
Q.E.D.

25. Theorem On the Equality of Polynomials
Let :ÐBÑ œ +8 B8  +8" B8"  â  +\$ B\$  +# B#  +" B  +! and let
;ÐBÑ œ ,7 B7  ,7" B7"  â  ,\$ B\$  ,# B#  ," B  ,! be any two real polynomials of
degrees 8 and 7 respectively. If for all real numbers B, :ÐBÑ œ ;ÐBÑ then
1) 7 œ 8
and 2) for all 3, if 0 Ÿ 3 Ÿ 8 then +3 œ ,3 Þ
Proof of Theorem On the Equality of Polynomials:
The following informal argument can be formalized using Mathematical Induction. However,
we prefer a more relaxed discussion that emphasizes technique over formality.
First note that if +8 B8  +8" B8"  â  +\$ B\$  +# B#  +" B  +! œ
,7 B7  ,7" B7"  â  ,\$ B\$  ,# B#  ," B  ,! for all B, we may let
B œ ! to conclude that +! œ ,! Þ

proof continued on the next page

page 21
Next, we subtract the common constant term from both sides of the equation to
conclude that for all B

+8 B8  +8" B8"  â  +\$ B\$  +# B#  +" B œ
,7 B7  ,7" B7"  â  ,\$ B\$  ,# B#  ," BÞ

Now divide both sides of this last equation by B, assuming B Á !. Then we have:

+8 B8"  +8" B8#  â  +\$ B#  +# B  +" œ
,7 B7"  ,7" B7#  â  ,\$ B#  ,# B  ,"

for all nonzero B. However, both of these last polynomials are defined and are continuous in a
neighborhood about B œ !, so we may apply the above Lemma with - œ ! to conclude this new
equation is true for all Bß including when B œ !Þ

Now we can repeat the above argument and let B œ ! to conclude that +" œ ," ß and we again
subtract this common constant term from both sides of the last equation to obtain the statement
that for all B

+8 B8"  +8" B8#  â  +\$ B#  +# B œ ,7 B7"  ,7" B7#  â  ,\$ B#  ,# BÞ

Again we divide both sides by B to obtain the simpler equation that

+8 B82  +8" B83  â  +\$ B  +# œ ,7 B72  ,7" B73  â  ,\$ B  ,# Þ

Even though this equation is only true for nonzero B because we just divided by B, we can
apply the above Lemma to conclude this equation must also be true when B œ !Þ So again we
may let B œ ! to conclude that +# œ ,# Þ

Clearly this argument may be continued to repeatedly pick off each of the coefficients one by one
in order until we run out of both coefficients. So every coefficient of :ÐBÑ matches the same
degree term coefficient of ;ÐBÑ. That we must run out of both coefficients at the same time is
because otherwise, if :ÐBÑ and ;ÐBÑ had different degrees, we could find a ! coefficient in one of
these polynomials that would match a nonzero coefficient in the other and that would be a

A final note about this theorem and its lemma is that the lemma is very easy for a non-calculus
student to understand when continuity is presented in an intuitive way (no epsilons or deltas!).
This theorem can also be proved assuming the Fundamental Theorem of Algebra, but the advantage
of this alternative approach is that we don't have to assume the Fundamental Theorem of Algebra
and we can introduce the fundamental property of continuity of polynomials.
Q.E.D.

page 22
26. Theorem Euclidean Algorithm for Polynomials
Let :ÐBÑ and ;ÐBÑ be any two polynomials with degrees   1. Then there exists a polynomial .ÐBÑ
such that .ÐBÑ divides evenly into both :ÐBÑ and ;ÐBÑ. Moreover, .ÐBÑ is such that if +ÐBÑ is any
other common divisor of :ÐBÑ and ;ÐBÑ, then +ÐBÑ divides evenly into .ÐBÑ. The polynomial .ÐBÑ is
called the Greatest Common Divisor of :ÐBÑ and ;ÐBÑ is sometimes denoted by KGHÐ:ÐBÑß ;ÐBÑÑÞ
Except for constant multiples, .ÐBÑ is unique.
Proof of the Euclidean Algorithm for Polynomials
Without loss of generality we assume the degree of :ÐBÑ is larger than or equal to the degree of ;ÐBÑÞ
By the Division Algorithm we may write

:ÐBÑ œ ;ÐBÑ † ;" ÐBÑ  <" ÐBÑ                                     (1)

where ;" ÐBÑ is the quotient polynomial and <" ÐBÑ is the remainder. If <" ÐBÑ ´ ! then we stop.
Î
Otherwise, if <" ÐBÑ ´ ! we note from the above equation that any common divisor of both ;ÐBÑ and
<" ÐBÑ must be a divisor of the right side of the above equation and therefore a divisor of the left side.
Any common divisor of ;ÐBÑ and <" ÐBÑ must be a divisor of :ÐBÑÞ Next, by writing
:ÐBÑ  ;ÐBÑ † ;" ÐBÑ œ <" ÐBÑ
we can see that every common divisor of :ÐBÑ and ;ÐBÑ must be a divisor of <" ÐBÑ and thus a
common divisor of ;ÐBÑ and <" ÐBÑÞ So KGHÐ:ÐBÑß ;ÐBÑÑ œ KGHÐ;ÐBÑß <" ÐBÑÑÞ

We continue by applying the Division Algorithm again to write

;ÐBÑ œ < " ÐBÑ † ;# ÐBÑ  <# ÐBÑ                                  (2)

If <# ÐBÑ ´ ! we stop. Otherwise, repeating the above reasoning,
KGHÐ;ÐBÑß <" ÐBÑÑ œ KGHÐ<" ÐBÑß <# ÐBÑÑ. Now apply the Division Algorithm again.

<" ÐBÑ œ <# ÐBÑ † ;\$ ÐBÑ  <\$ ÐBÑ                                 (3)

If <\$ ÐBÑ ´ ! we stop. Otherwise we note KGHÐ<" ÐBÑß <# ÐBÑÑ œ KGHÐ<# ÐBÑß <\$ ÐBÑÑ and we
continue to apply the Division Algorithm to get

<2 ÐBÑ œ <3 ÐBÑ † ;4 ÐBÑ  <4 ÐBÑ                                 (4)

proof continued on the nex page

page 23
If <% ÐBÑ ´ ! we stop. Otherwise we continue this process. However, we cannot this process forever
because the degrees of the remainders <3 ÐBÑ keep decreasing by 1.

degreeÐ<% ÐBÑÑ  degreeÐ<\$ ÐBÑÑ  degreeÐ<# ÐBÑÑ  degreeÐ<" ÐBÑÑ Ÿ degreeÐ;ÐBÑÑ

So after applying the Division Algorithm at most the number of times that is the degree of ;ÐBÑ we
must have some remainder become the identically zero polynomial.

We claim KGHÐ:ÐBÑß ;ÐBÑÑ is the last nonzero remainder. For example, suppose

<8# ÐBÑ œ <8" ÐBÑ † ;8 ÐBÑ  <8 ÐBÑ                               Ð8Ñ

and

<8" ÐBÑ œ <8 ÐBÑ † ;8" ÐBÑ                             Ð8  "Ñ

where <8" ÐBÑ ´ ! and is not written. The last equation shows <8 ÐBÑ is a divisor of <8" ÐBÑ so
KGHÐ<8" ÐBÑß <8 ÐBÑÑ œ <8 ÐBÑ.

Now KGHÐ:ÐBÑß ;ÐBÑÑ œ KGHÐ;ÐBÑß <" ÐBÑÑ œ KGHÐ<" ÐBÑß <# ÐBÑÑ œ KGHÐ<# ÐBÑß <\$ ÐBÑÑ
œ â œ KGHÐ<8" ÐBÑß <8 ÐBÑÑ œ <8 ÐBÑ, the last nonzero remainder.
Q.E.D.

page 24
27. Corollary to the Euclidean Algorithm for Polynomials
The KGH of any two polynomials :ÐBÑ and ;ÐBÑ may be expressed as a linear combination of :ÐBÑ
and ;ÐBÑÞ
Proof of the Corollary to the Euclidean Algorithm for Polynomials.
The following were the series of equations that led up to the creation of the KGH polynomial.

:ÐBÑ œ ;ÐBÑ † ;" ÐBÑ  <" ÐBÑ                                                                       Ð"Ñ
;ÐBÑ œ < " ÐBÑ † ;# ÐBÑ  <# ÐBÑ                                                                    Ð#Ñ
<" ÐBÑ œ <# ÐBÑ † ;\$ ÐBÑ  <\$ ÐBÑ                                                                   Ð\$Ñ
<2 ÐBÑ œ <3 ÐBÑ † ;4 ÐBÑ  <4 ÐBÑ                                                                   Ð%Ñ
ã
<8\$ ÐBÑ œ <8# ÐBÑ † ;8" ÐBÑ  <8" ÐBÑ                                                     Ð8  "Ñ
<8# ÐBÑ œ <8" ÐBÑ † ;8 ÐBÑ  <8 ÐBÑ                                                               Ð8Ñ

Now starting with the last equation, we solve for the KGH.

<8 ÐBÑ œ <8# ÐBÑ  <8" ÐBÑ † ;8 ÐBÑ                                                               Ð‡Ñ

Note this shows how to write the KGH as a linear combination of <8# ÐBÑ and <8" ÐBÑÞ
But in the next to the last equation we can solve for <8" ÐBÑ and substitute into Ð‡Ñ.

<8 ÐBÑ œ <8# ÐBÑ  Ò <8\$ ÐBÑ  <8# ÐBÑ † ;8" ÐBÑ Ó † ;8 ÐBÑ
œ <8# ÐBÑ  <8\$ ÐBÑ † ;8 ÐBÑ  <8# ÐBÑ † ;8" ÐBÑ † ;8 ÐBÑ
œ Ò"  ;8" ÐBÑ † ;8 ÐBÑÓ † <8# ÐBÑ  Ò  ;8 ÐBÑÓ † <8\$ ÐBÑ

We have now shown how to write <8 ÐBÑ as a linear combination of <8# ÐBÑ and <8\$ ÐBÑÞ
Clearly we can continue to work backwards, and solve each next equation for the previous
remainder, and then substitute that remainder (which is a linear combination of its two previous
remainders) into our equation to continually write <8 ÐBÑ as a linear combination of the two most
recent remainders.

proof continued on the next page

page 25
As we work our way up the list, we will eventually have

<8 ÐBÑ œ 0 ÐBÑ † <1 ÐBÑ  1ÐBÑ † <2 ÐBÑ

and when we solve the second equation for <# ÐBÑ and substitute we get

<8 ÐBÑ œ 0 ÐBÑ † <" ÐBÑ  1ÐBÑÒ ;ÐBÑ  <" ÐBÑ † ;# ÐBÑ Ó
œ 0 ÐBÑ † <" ÐBÑ  1ÐBÑ;ÐBÑ  1ÐBÑ † <" ÐBÑ † ;# ÐBÑ
œ Ò 0 ÐBÑ  1ÐBÑ † ;# ÐBÑ Ó † <" ÐBÑ  Ò 1ÐBÑ Ó † ;ÐBÑ

Lastly we solve the first equation for <" ÐBÑ and substitute and we get

<8 ÐBÑ œ Ò 0 ÐBÑ  1ÐBÑ † ;# ÐBÑ Ó † Ò :ÐBÑ  ;ÐBÑ † ;" ÐBÑ Ó  Ò 1ÐBÑ Ó † ;ÐBÑ
œ Ò 0 ÐBÑ  1ÐBÑ † ;# ÐBÑ Ó † :ÐBÑ  Ò 0 ÐBÑ  1ÐBÑ † ;# ÐBÑ Ó † ;ÐBÑ † ;" ÐBÑ  1ÐBÑ † ;ÐBÑ
œ Ò 0 ÐBÑ  1ÐBÑ † ;# ÐBÑ Ó † :ÐBÑ  eÒ 1ÐBÑ † ;# ÐBÑ  0 ÐBÑ Ó † ;" ÐBÑ  1ÐBÑf † ;ÐBÑ

This shows that the KGH can be written as a linear combination of :ÐBÑ and ;ÐBÑÞ
Q.E.D.

page 26
28. Lemma 1 for Partial Fractions
+ÐBÑ
If 0 ÐBÑ œ            where KGHÐ,ÐBÑß -ÐBÑÑ œ " then there exist polynomials .ÐBÑ and /ÐBÑ such
,ÐBÑ-ÐBÑ
that
.ÐBÑ /ÐBÑ
0 ÐBÑ œ        
,ÐBÑ   -ÐBÑ
Proof of Lemma 1 for Partial Fractions
The two polynomials ,ÐBÑ and -ÐBÑ are called relatively prime when their KGH is ". Of course this
means that ,ÐBÑ and -ÐBÑ have no common factor. Apply the Corollary to the Euclidean Algorithm
for polynomials to construct polynomials =ÐBÑ and >ÐBÑ such that

" œ =ÐBÑ † ,ÐBÑ  >ÐBÑ † -ÐBÑ

Then multiply both sides of this equation by +ÐBÑ to get

+ÐBÑ œ +ÐBÑ † =ÐBÑ † ,ÐBÑ  +ÐBÑ † >ÐBÑ † -ÐBÑ

and finally divide both sides of this last equation by the product ,ÐBÑ † -ÐBÑ

+ÐBÑ       +ÐBÑ † =ÐBÑ † ,ÐBÑ   +ÐBÑ † >ÐBÑ † -ÐBÑ
œ                    
,ÐBÑ † -ÐBÑ      ,ÐBÑ † -ÐBÑ         ,ÐBÑ † -ÐBÑ

+ÐBÑ       +ÐBÑ † =ÐBÑ   +ÐBÑ † >ÐBÑ
0 ÐBÑ œ               œ             
,ÐBÑ † -ÐBÑ     -ÐBÑ          ,ÐBÑ

Now let .ÐBÑ œ +ÐBÑ † >ÐBÑ and let /ÐBÑ œ +ÐBÑ † =ÐBÑÞ
Q.E.D.

page 27
29. Lemma 2 for Partial Fractions
:ÐBÑ
c ;ÐBÑ d7
If 0 ÐBÑ œ             then there exists a polynomial 1ÐBÑ and for " Ÿ 3 Ÿ 7 there exist polynomials

=3 ÐBÑ each with degree less than ;ÐBÑ such that

:ÐBÑ             =" ÐBÑ     =# ÐBÑ       =\$ ÐBÑ           =7 ÐBÑ
c ;ÐBÑ d                     c ;ÐBÑ d     c ;ÐBÑ d \$  â  c ;ÐBÑ d7
0 ÐBÑ œ            7 œ 1ÐBÑ                   # 
;ÐBÑ

Proof of Lemma 2 for Partial Fractions
Apply the Division Algorithm for the first time to write À
:ÐBÑ œ ;ÐBÑ † c ;" ÐBÑ d  <" ÐBÑÞ Note the degree of <" ÐBÑ is less than the degree of ;ÐBÑÞ
Now divide ;" ÐBÑ by ;ÐBÑ to get a second quotient and a second remainder so we may write
:ÐBÑ œ ;ÐBÑ † c ;ÐBÑ † ;# ÐBÑ  <# ÐBÑ d  <" ÐBÑ
:ÐBÑ œ c ;ÐBÑ d# † ;# ÐBÑ  ;ÐBÑ † <# ÐBÑ  <" ÐBÑ
Note that the degree of <# ÐBÑ is less than the degree of ;ÐBÑÞ
Now divide ;# ÐBÑ by ;ÐBÑ to get a third quotient and a third remainder and write
:ÐBÑ œ c ;ÐBÑ d# † c ;ÐBÑ † ;\$ ÐBÑ  <\$ ÐBÑ d  ;ÐBÑ † <# ÐBÑ  <" ÐBÑ
:ÐBÑ œ c ;ÐBÑ d\$ † ;\$ ÐBÑ  c ;ÐBÑ d# † <\$ ÐBÑ  ;ÐBÑ † <# ÐBÑ  <" ÐBÑ
We continue to divide each newest quotient ;3 ÐBÑ by ;ÐBÑ to get a newer quotient and a newer
remainder and substitute for the ;3 ÐBÑ quotient. Each remainder has a degree smaller than the degree
of ;ÐBÑÞ
:ÐBÑ œ c ;ÐBÑ d\$ † c ;ÐBÑ † ;% ÐBÑ  <% ÐBÑ d  c ;ÐBÑ d# † <\$ ÐBÑ  ;ÐBÑ † <# ÐBÑ  <" ÐBÑ
:ÐBÑ œ c ;ÐBÑ d% † ;% ÐBÑ  c ;ÐBÑ d\$ † <% ÐBÑ  c ;ÐBÑ d# † <\$ ÐBÑ  ;ÐBÑ † <# ÐBÑ  <" ÐBÑ
We may continue breaking down and substituting for each ;3 ÐBÑ quotient until we have
:ÐBÑ œ c ;ÐBÑ d7 † ;7 ÐBÑ  c ; aBb d7" † <7 ÐBÑ  â  c ;ÐBÑ d# † <\$ ÐBÑ  ;ÐBÑ † <# ÐBÑ  <" ÐBÑ

Finally we divide both sides of this last equation by c ;ÐBÑ d7 to get
:ÐBÑ               <7 ÐBÑ         <\$ ÐBÑ         <# ÐBÑ            <" ÐBÑ
c ;ÐBÑ d                         c ;ÐBÑ d       c ;ÐBÑ d 7"  â  c ;ÐBÑ d7
0 ÐBÑ œ            7 œ ;7 ÐBÑ         â          7# 
;ÐBÑ
Now we may let 1ÐBÑ œ ;7 ÐBÑ and let =3 ÐBÑ œ <73" ÐBÑÞ
Q.E.D.

page 28
30. Partial Fraction Decomposition Theorem
:ÐBÑ
Let        be a rational function where :ÐBÑ and ;ÐBÑ are polynomials such that the degree of :ÐBÑ
;ÐBÑ
is less than the degree of ;ÐBÑÞ Then there exist algebraic fractions J" ß J# ß á ß J< such that
:ÐBÑ
œ J"  J#  â  J< and where each J3 fraction is one of two forms:
;ÐBÑ
E3                  E5 B  F5
83
or      #  , B  - Ñ75
where E3 , +3 ß ,3 ß E5 ß F5 , +5 ß ,5 ß -5 are all real numbers and
Ð+3 B  ,3 Ñ       Ð+5 B     5     5

the 83 and the 75 are positive integers and each quadratic expression +5 B#  ,5 B  -5 has a
negative discriminant.

Proof of the Partial Fraction Decomposition Theorem
Since ;ÐBÑ is a polynomial, by the Linear and Irreducible Quadratic Factors Theorem we may write

;ÐBÑ œ – # Ð+3 B  ,3 Ñ:3 — † ” # Ð+5 B#  ,5 B  -5 Ñ;5 •
4                    6

3œ"                 5œ"

where for each 3, Ð+3 B  ,3 Ñ is a real linear factor of ;ÐBÑ of multiplicity :3 and for each 5 ,
Ð+5 B#  ,5 B  -5 Ñ is an irreducible quadratic factor of ;ÐBÑ of multiplicity ;5 Þ The +3 ß ,3 are
different from the +5 ß ,5 Þ Since the real linear and irreducible quadratic factors have no factors in
common their KGH is " and we may apply Lemma 1 for Partial Fractions to write:
:ÐBÑ               +ÐBÑ                            ,ÐBÑ

” # Ð+5 B#  ,5 B  -5 Ñ;5 •
œ                              
# Ð+3 B  ,3 Ñ:3
–                      —
;ÐBÑ          4                           6

3œ"                         5œ"

Now for each different 3, each factor of the form Ð+3 B  ,3 Ñ:3 is different from the next so we may
again apply Lemma 1 for Partial Fractions 4  " times to split the first fraction above into a sum of 4
other fractions. For each of those fractions that have an exponent of # or higher in the denominator
we apply Lemma 2 for Partial Fractions :3  " more times to split each denominator with
Ð+3 B  ,3 Ñ:3 into a sum of :3 more fractions.
For each different 5 , each factor of the form Ð+5 B#  ,5 B  -5 Ñ;5 is different from the next so we
may again apply Lemma 1 for Partial Fractions 5  " times to split the second fraction above into a
sum of 5 other fractions. For each of those fractions that have an exponent of # or higher in the
denominator we apply Lemma 2 for Partial Fractions ;5  " more times to split each denominator
with Ð+5 B#  ,5 B  -5 Ñ;5 into a sum of ;5 more fractions.
As a final note, the 1ÐBÑ term that appears in Lemma 2 for Partial Fractions will be the ! polynomial
because we are assuming the degree of :ÐBÑ is strictly less than that of ;ÐBÑÞ So our partial fraction
decomposition really does break down into a sum of pure algebraic fractions.
Q.E.D.

page 29
31. Partial Fraction Decomposition Coefficient Theorem
:ÐBÑ
Let        be a rational function where :ÐBÑ and ;ÐBÑ are polynomials such that the degree of :ÐBÑ
;ÐBÑ
is less than the degree of ;ÐBÑ. If B œ + is a root of ;ÐBÑ œ ! of multiplicity ", then in the partial
:ÐBÑ                                      E                       :Ð+Ñ
fraction decomposition of        which contains a term of the form         , the constant E œ w Þ
;ÐBÑ                                   ÐB  +Ñ                   ; Ð+Ñ

Proof of the Partial Fraction Decomposition Coefficient Theorem:
:ÐBÑ      E                                                     :ÐBÑ
Assume         œ          0 ÐBÑ is the partial fraction decomposition of      where 0 ÐBÑ is itself
;ÐBÑ   ÐB  +Ñ                                                  ;ÐBÑ
a rational function, but is such that 0 Ð+Ñ is well-defined. In fact, 0 ÐBÑ will be continuous at B œ +Þ
ÐB  +Ñ † :ÐBÑ
Then                  œ E  0 ÐBÑÐB  +Ñ
;ÐBÑ
ÐB  +Ñ † :ÐBÑ                         !
Since B œ + is a simple zero of ;ÐBÑ,    lim                is an indeterminate form and we may
BÄ+     ;ÐBÑ                               !
thus apply L'Hopital's Rule when evaluating the limit. Taking the limit on both sides of the above
equation we have

œ lim cE  0 ÐBÑÐB  +Ñd
ÐB  +Ñ:ÐBÑ
lim
BÄ+    ;ÐBÑ       BÄ+

" † :ÐBÑ  ÐB  +Ñ † :w ÐBÑ
lim                             œ lim E  lim 0 ÐBÑ † lim ÐB  +Ñ
BÄ+           ; w ÐBÑ             BÄ+     BÄ+         BÄ+

:ÐBÑ
lim         œ E  lim 0 ÐBÑ † !
BÄ+ ; w ÐBÑ       BÄ+

:Ð+Ñ
œ E  0 Ð+Ñ † ! œ E
; w Ð+Ñ
Q.E.D.

page 30

DOCUMENT INFO
Shared By:
Categories:
Tags:
Stats:
 views: 21 posted: 8/7/2011 language: English pages: 32