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Title: College Algebra
Description: The note is for 1st year students. College Algebra lecture note by Attila Berczes.
Description: The note is for 1st year students. College Algebra lecture note by Attila Berczes.
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College Algebra
´
Attila B´rczes and Akos Pint´r
e
e
TMOP-4
...
2
...
1
Introduction of real numbers
The most important notion in mathematics is belonging to numbers
...
The most important set of numbers is the set of real numbers
...
Decimals might be
finite decimals and infinite decimals, which are either infinite repeating decimals
or infinite non-repeating decimals
...
Example
...
04
25
3
= 1
...
6666
...
142857142857
...
25142857142857
...
REAL NUMBERS
and infinite non-repeating decimals are like
√
2 = 1
...
;
π = 3
...
On the set of real numbers there are defined several operations
...
In the
sequel we assume that addition and multiplication of real numbers is well known
for the reader
...
The set of real numbers is denoted by R
...
2
Axiomatic definition of real numbers*
Definition 1
...
Let F be a non-empty set with two operations + and · with the
following properties
(A1) For all a, b, c ∈ F , (a + b) + c = a + (b + c)
...
(+ commutative)
(A3) There exists 0 ∈ F such that for all a ∈ F , a + 0 = a
...
(Negatives)
(M1) For all a, b, c ∈ F , (ab)c = a(bc)
...
(· commutative)
(M3) There exists 1 ∈ F , 1 ̸= 0, s
...
for all a ∈ F , a1 = a
...
(Reciprocals)
(DL) For all a, b, c ∈ F , a(b + c) = ab + ac
...
(O2) For all a, b ∈ F , if a ≤ b and b ≤ a, then a = b
...
1
...
PROPERTIES OF REAL NUMBERS
5
(O4) For all a, b, c ∈ F , if a ≤ b, then a + c ≤ b + c
...
Finally, every nonempty subset of F that has an upper bound also has a least
upper bound (supremum)
...
More precisely it can be shown that the axioms above determine R completely,
that is, any other mathematical object with the same properties must be essentially
the same as R
...
3
Properties of real numbers
The below basic properties of addition and multiplication of real numbers are socalled axioms, so they are assumed to be self-evidently true and there is no need
to prove them
...
2
...
Then the following properties are true:
6
CHAPTER 1
...
The additive inverse of a is also called the negative of a or the opposite
of a, and the multiplicative inverse of a is also called the reciprocal of a
...
Do not confuse ”the negative of a number” with ”a negative number”!
Notation
...
Theorem 1
...
(Properties of real numbers) Let a, b and c be arbitrary real
numbers
...
3
...
Addition property
If a = b then
a + c = b + c
...
Theorem 1
...
(Properties of zero)
For all real numbers a and b we have
1)
...
ab = 0 if and only if a = 0 or b = 0
...
5
...
−(a + b) = (−a) + (−b),
2)
...
(−a)b = −(ab),
4)
...
(−a)(−b) = ab
...
The subtraction of two real numbers is defined by the
addition of the additive inverse of the second to the first
...
8
CHAPTER 1
...
6
...
Similarly, for all real numbers a and b ̸= 0 we define the quotient
a
b
by
a
= a · b−1
...
The quotient
a
b
is also denoted by a : b
...
Clearly, you may divide zero by any nonzero number, and the result is
zero:
0
= 0,
3
0
= 0,
−π
0
√ = 0
...
Theorem 1
...
(Properties of subtraction and division of real numbers)
Let a, b, c, d be arbitrary real numbers and suppose that all the denominators in the
formulas below are non-zero:
1)
...
a − 0 = a,
3)
...
−(a − b) = b − a,
5)
...
−a
b
= −a =
b
7)
...
4
...
ac
bc
9)
...
a
b
:
c
d
=
11)
...
a
b
+
c
d
=
9
= a,
b
ad+bc
...
8
...
If a + c = b + c, then a = b
...
9
...
If ac = bc, then a = b
...
1
...
To change the order of the operations we use grouping symbols: parentheses
( ), square brackets [ ], and braces { }
...
Order of operations:
10
CHAPTER 1
...
If no fraction lines and grouping symbols are present:
(1) First do all exponentiations and taking roots in the order they appear,
working from left to right,
(2) Then do all multiplications and divisions in the order they appear,
working from left to right
...
II
...
(2) Use the rules of point I
...
Example
...
√
√
√
7+ 3+2·3
7+ 3+6
7+ 9
7+3
10
√
=
=
=
=
=1
(3 + 2) · 2
5·2
10
10
(3 + 4) · 2
Remark
...
However, if used, then it means just a sign of division (:), and not a fraction line
...
5 · 4 = 2
...
5
...
5
11
Special subsets of the set of Real Numbers
During the history of mathematics the real numbers were not the first set of
numbers which appeared
...
)
developed together with the human race, being with us from the beginning
...
Partly, they are results of the ”wish” to be able to subtract any two natural
numbers
...
However, these ”do not fill completely” the coordinate line, so mathematicians
introduced the set of real numbers, containing the set of rational numbers and the
set of irrational numbers, which is the set of all non-rational real numbers
...
REAL NUMBERS
The set of natural numbers contains all numbers which can be obtained
by successively adding several copies of 1:
N := {1, 2, 3,
...
, −3, −2, −1, 0, 1, 2, 3,
...
10
...
Real numbers which have their decimal form starting
with a natural number (e
...
3
...
) or which start with zero (e
...
0
...
)
are called positive numbers, and their additive inverses are called negative
numbers
...
(Properties of the sign of real numbers)
• Zero is neither positive nor negative
...
6
...
• The additive inverse of a natural number is negative
...
• A rational number is negative if
– the numerator is negative and the denominator is positive,
– the numerator is positive and the denominator is negative
...
6
The real number line and ordering of the real
numbers
The real number line is a geometric representation of the set of real numbers
...
This is the case especially with the ordering of real
numbers
...
11
...
This point will
also be called the origin
...
This way using the distance of these two points we have fixed a unit
measure
...
to the right of the origin and
using central symmetry through the origin we also fix the negatives of the natural
14
CHAPTER 1
...
e
...
) on the left of the origin
...
4
31
Irrational numbers can be located by computing
their decimal representation to any desired accuracy
...
Now we define a ”natural” ordering among the real numbers
...
The easiest way is to say that the real number a is
larger then b if a is to the right of b on the coordinate line
...
Definition 1
...
Let a, b be real numbers
...
Further, we say that a is smaller than b and write
a < b, if a − b is negative
...
We use several variations of the relations > and <
...
For negation of statements involving such symbols we use the notations ̸>, ̸<, ̸≥, ̸≤
...
13
...
Then the following properties of the ordering < (”strictly smaller”)
are true:
1
...
THE REAL NUMBER LINE AND ORDERING OF THE REAL NUMBERS15
Irreflexive property
The statement a < a is always false
...
The ordering > (”strictly greater”) has completely similar properties
...
14
...
Then the following properties of the ordering ≤ (”smaller
or equal”) are true:
Reflexive property
The statement a ≤ a is always true
...
The ordering ≥ (”greater or equal”) has completely similar properties
...
REAL NUMBERS
1
...
An interval is a set containing all real numbers between the two endpoints
of the interval
...
Definition 1
...
Let a ≤ b be real numbers
...
If a = b then we have the following conventions:
[a, b] = [a, a] = {a}
and
]a, b[=]a, a[= [a, a[=]a, a] = ∅
...
7
...
This can
be useful when writing down proofs including intervals where the endpoints are
unknowns (i
...
letters) so a priory we do not know which of them is smaller or
larger, but we will never write down an interval with given numbers as endpoints
so that the left endpoint is larger than the right one
...
If the square bracket ”is looking toward the center” of the interval then
the endpoint is included, otherwise it is not included in the set
...
This
can be explained in the following way:
• the sign ] at the left endpoint of an interval means that
the endpoint is not included in the set,
• the sign ] at the right endpoint of an interval means that
the endpoint is included in the set,
• the sign [ at the left endpoint of an interval means that
the endpoint is included in the set,
• the sign [ at the right endpoint of an interval means that
the endpoint is not included in the set
...
1
...
3, −5, −4
...
99, 4, 4
...
REAL NUMBERS
Draw the graph representing the above intervals on a real number line
...
2
...
8
b) x ≥ 3
h) x > −1
f) 1 ≤ x ≤ 3
The Absolute Value of a Real Number
Definition 1
...
Let x be a real number
...
1)
−x if x < 0
Remark
...
Equivalent definitions for the absolute value:
All formulas below are equivalent reformulations of (1
...
17
...
Then we have
|x| :=
x
if x > 0
if x = 0
0
−x if x < 0
1
...
EXPONENTIATION
19
• |a| ≥ 0,
• | − a| = |a|,
• |a · b| = |a| · |b|,
•
a
b
=
|a|
,
|b|
• |a + b| ≤ |a| + |b|,
• |a| = b if and only if a = b or a = −b,
• |a| < b if and only if −b < a < b,
• |a| > b if and only if a < −b or a > b
...
8
...
Let f (x) = 2|x − 5| + 4 and give the graph of this
function
...
1)
...
2
shows the function |x − 5|
...
3
...
4
...
1
...
9
...
To shorten the notation for
the repeated multiplication a · a ·
...
REAL NUMBERS
Figure 1
...
9
...
2: Graph of the function |x − 5|
21
22
CHAPTER 1
...
3: Graph of the function 2|x − 5|
1
...
EXPONENTIATION
Figure 1
...
REAL NUMBERS
exponential notation an , i
...
an := a · a ·
...
n times
Here we call a the base and n the exponent or the power
...
This way we have defined
the exponentiation for any non-zero real base and any integer exponent
...
18
...
Further, if a or b is zero then suppose
that m and n are positive
...
Simplify the following expression containing exponentiations:
(a3 )2 · a4 · (a2 )5
;
a7 · (a2 )4
Solution:
a ̸= 0
...
a7+8
a
Example
...
1
...
EXPONENTIATION
25
Solution:
(a4 b−2 )−2 ·
a6 b
= (a4 )−2 ·(b−2 )−2 ·a6−(−2) b1−5 = a−8 ·b4 ·a8 b−4 = a−8+8 ·b4+(−4) = a0 b0 = 1
...
3
...
9
...
19
...
Suppose that if
√
n is even, then a and b are positive
...
√
√
2
a we use the notation a
...
√
4 = 2,
√
√
√
3
125 = 5, 5 −32 = −2, 4 −16 has no sense (it is not a real number)
Theorem 1
...
(Properties of radicals) If n, k are positive integers and a, b
are positive real numbers, then we have
26
CHAPTER 1
...
2)
...
4)
...
6)
...
√
√ n
n
an = ( n a) = a
√
√ n
k
an = ( k a)
√
√ √
nk
n
aka=
an+k
√
√
na
nk
√ =
an−k
k a
√√
√√
√
k n
a = n k a = nk a
√
√ √
n
a n b = n ab
√
√
na
√ = n a
n
b
b
Remark
...
In one hand, for even n some of the expressions above have no
sense in the set of real numbers
...
In the case of statement 1) of Theorem 1
...
Example
...
Solution:
√ √
√√
√
√√
√
√
√
√
√
3
3
6
4
4
2 a3 =
3 · a2 a3 =
5 4 a3 = 6 4 (a5 )4 · a3 = 24 a20 · a3 = 24 a23
...
Write the following expression using only one root sign:
√ √ 2 √ 5
3
4
a· a · a ;
a ≥ 0
...
1
...
EXPONENTIATION
27
Exercise 1
...
Simplify the following expression containing exponentiations, where
all the indeterminates are supposed to be positive:
√
√
√
a)
2
√
3 √
4
d)
a2 b
√ √ √
x y x
g)
y x y
√
√
x 10 y
√
j) 3 xy · 5 ·
y
x
√
√ √
m) a 3 a 4 a
√ √
√
3
p) a2 a3 a
3
1
...
3
b)
√
5
√
3
√ √
√
e) a a a
h)
√ √ √
4
3
a · b · ab
√ √
x 3 y√
4
k)
xy
y x
√
√
√
3
6
n) a3 · a2 · a11
q)
√ √ √
7
a· 5a· 3a
√
3
a2
√ √
√
4
f) a a2 a3
√ √
a 3 b √
i)
·
· 6a
b
a
√ √
√
5
4 1 3
l) x
x
x
√
√
5
o) a 4 a
√ √
√
4
3
r) a5 a2 a
c)
Rational exponents
To extend the exponentiation for rational exponents for integers m, n (n > 1) we
put
am/n = (a1/n )m =
( √ )m √
n
a = n am
...
2)
However, this has no sense among the real numbers when a < 0 and n is even
...
For example, ((−1)2 )1/2 = 1, but this is not
equal to (−1)2·1/2 = −1
...
Theorem 1
...
(Properties of exponentiation with rational exponents)
Let r, s be rational numbers, and let a, b be positive real numbers
...
REAL NUMBERS
1)
...
ar
as
= ar−s
3)
...
(ar )s = ars
5)
...
a = ar
b
b
Example
...
Solution:
(
( 3 )10 ( 1 )10 √
√
√
√
3
3 )−3 · 3 b2 = a 4
a b
·( a
· b3
· (a3 )−3 · b2
√
√
−9
3
1
15
10
2
3
= a 4 ·10 · b 3 ·10 · a−9 · b2 = a 2 · b 3 · a 2 · b 3
3
4
1
3
15
)10
=a2+
Exercise 1
...
( 2 1 )6
a3 · b2
a)
a 3 · b2
( 4
)9
−1
3 · b 9
a
d)
a10 · b2
( 4 3 )5
a3 · b2
g)
a 3 · b2
( 2 1 )6
a− 3 · b 4
j)
3
a−4 · b 2
−9
2
10
2
6
12
· b 3 + 3 = a 2 · b 3 = a 3 b4
...
1
Introduction to algebraic expressions
In algebra it is common to use letters to represent numbers
...
141592
...
In mathematics variables are used in two ways
...
(An example for this use is the
case of equations
...
A second use of the
variables is to describe general relationship between numbers, operations and other
mathematical objects
...
)
Definition 2
...
(Algebraic expression) An algebraic expression is the result
of performing a finite number of the basic operations addition, subtraction, multiplication, division (except by zero), extraction of roots on a finite set of variables
29
30
CHAPTER 2
...
By equivalent
expressions we mean expressions which represent the same real number for all
valid replacements of the variables
...
In many cases our goal is to simplify a given algebraic expression such
that the result is an equivalent expression to the original one, which is much
simpler in form
...
These are mainly the transformations described in our theorems
...
a2/3 − ab2
√ ,
a−1/3 · 5 b
x2 − 3,
x3 − 1
...
2
...
The above-mentioned real number is called the coefficient
...
Like terms which are added or
subtracted may be ”collected” (using the distributive law) to get again a
”like term” to the original ones whose coefficient can be computed by adding
or subtracting the coefficients of the original ”like terms”
...
• The degree of a monomial is the sum of the exponents of all unknowns
in the monomial
...
1
...
If the polynomial consists of only one term, then it is a monomial, if it consist of 2, 3, 4 terms, then we call it a binomial, trinomial,
quadrinomial, respectively
...
• A non-constant polynomial is called univariate if it contains a single variable, and multivariate otherwise
...
The quotient of two polynomials is called a
rational expression
...
The coefficient of the leading term of a univariate polynomial is called the
leading coefficient of the polynomial
...
ALGEBRAIC EXPRESSIONS
2
...
2
...
Subtracting polynomials: Subtraction of polynomials is performed in
the same way as addition, except that first we change the sign of all the
monomials of the subtrahend polynomial
...
Example
...
Then
we have
(P + Q)(x, y) =(x2 − 3xy + 2y 3 ) + (2x2 − 3xy 2 + 3y 3 ) =
= x2 − 3xy + 2y 3 + 2x2 − 3xy 2 + 3y 3 = 3x2 − 3xy + 5y 3 − 3xy 2 ,
(P − Q)(x, y) =(x2 − 3xy + 2y 3 ) − (2x2 − 3xy 2 + 3y 3 ) =
= x2 − 3xy + 2y 3 − 2x2 + 3xy 2 − 3y 3 = −x2 − 3xy − y 3 + 3xy 2 ,
(P · Q)(x, y) =(x2 − 3xy + 2y 3 ) · (2x2 − 3xy 2 + 3y 3 ) =
= 2x4 − 6x3 y + 4x2 y 3 −3x3 y 2 + 9x2 y 3 − 6xy 5 + 3x2 y 3 − 9xy 4 + 6y 6 =
(x2 −3xy+2y 3 )·2x2
(x2 −3xy+2y 3 )·(−3xy 2 )
(x2 −3xy+2y 3 )·3y 3
= 2x4 − 3x3 y 2 − 6x3 y + 16x2 y 3 − 6xy 5 − 9xy 4 + 6y 6
2
...
POLYNOMIALS
33
Exercise 2
...
Let P (x, y), Q(x, y) and H(x, y) be polynomials defined by
P (x, y) := x2 − 3xy + y 2 ,
Q(x, y) := 2x3 − x2 y + 3xy 2 + 5y 3 ,
H(x, y) := x2 + 5xy + 3x − 2xy 3 + y
...
2
...
Compute the following expressions:
a) P (x) + Q(x)
b) xP (x) + Q(x)
c) P (x) · Q(x)
d) P (x) + 2H(x)
e) P (x) · H(x)
f) (P (x) + Q(x)) · H(x)
g) (x · P (x) + 3 · Q(x)) · H(x)
h) P (x) + Q(x) + H(x)
i) P (x) · (Q(x) + H(x))
j) P (x) · Q(x) · H(x)
34
CHAPTER 2
...
2
...
Divide the sign of the two monomials
...
Divide the coefficients of the two monomials
...
Divide the like variables by subtracting their exponents
...
12x4 y 7 z
12
= − · x4−3 y 7−1 z 1−0 = −4xy 6 z
...
3
...
Divide each term (monomial) of the dividend polynomial
by the divisor monomial
...
Add the results to get the resulting polynomial
...
x2 y 3 − 3x5 y 2
x2 y 3 −3x5 y 2
= 2 +
= y 2 − 3x3 y
...
Although the formal division of two monomials can be executed always,
the result of this division is a monomial only when the dividend monomial is
divisible by the divisor monomial
...
The same is true for division of polynomials by monomials,
and even for division of polynomials by polynomials
...
2
...
2
...
This procedure is a straightforward generalization of the long division of integers
...
Write the dividend and the divisor polynomials in the following
scheme
dividend polynomial
divisor polynomial
quotient polynomial
2)
...
Multiply the above resulting monomial by the divisor, change the sign
of every monomial of the result, and write the resulting polynomial
below the dividend
...
Draw a horizontal line, add the dividend to the above resulting
polynomial and write the result of the addition below the line
...
Let the polynomial below the last horizontal line take the role of the
dividend and repeat steps 2)-4) until the polynomial below the last
horizontal line is zero or has degree strictly smaller than the degree
of the divisor
...
Divide the polynomial f (x) = x4 −3x3 +5x2 +x−5 by g(x) = x2 −2x+2
...
ALGEBRAIC EXPRESSIONS
x4
−3x3
+5x2
−x4
+2x3
−2x2
−x3
+3x2
+x
x3
−2x2
+2x
x2
+3x
−5
−x2
+2x
−2
5x
−7
+x
−5
x2 − 2x + 2
x2 − x + 1
−5
Remark
...
Example
...
−5
x4
+0x3
+5x2
−x4
+2x3
−2x2
2x3
+3x2
+x
−2x3
+4x2
7x2
−3x
−5
−7x2
+14x
−14
11x
x2 − 2x + 2
−4x
−19
+x
x2 + 2x + 7
−5
Exercise 2
...
Divide the polynomial f (x) by the polynomial g(x) using the pro-
2
...
POLYNOMIALS
37
cedure of Euclidean division:
a) f (x) := x3 + 2x2 − 4x + 2,
g(x) := x2 − x + 1
b) f (x) := x5 − 3x4 + 4x3 + 2x2 − 4x + 2,
g(x) := x2 − x + 1
c) f (x) := x5 − 3x4 + 2x2 − 4x + 2,
g(x) := x2 − 3x + 2
d) f (x) := x6 − 3x4 + 2x2 − 4x + 2,
g := x3 − 2x + 1
e) f (x) := x5 − 3x4 + 4x3 − 5x2 + x + 2 g(x) := x2 − 3x + 2
f) f (x) := x6 − 64,
g(x) := x2 − 2x + 4
g) f (x) := x5 − 2x4 − 3x + 2,
g(x) := x2 − 3x + 4
h) f (x) := x5 + 2x4 − 5x3 + 2,
g := x2 − 5x + 2
i f (x) := x5 − 2x4 − 5x3 + 2x2 − 3x + 2,
g(x) := x2 − 4x + 3
38
CHAPTER 2
...
2
...
Let us consider a concrete example:
2
...
POLYNOMIALS
1x4
+2x3
−x4
+5x2
39
−5
x−2
+2x3
+x
1x3 + 4x2 + 13x + 27
+x
−5
13x2
+x
−5
−13x2
+26x
4x3
+5x2
−4x3
+8x2
27x
−27x
−5
+54
49
It is clear that in fact we only need to compute the numbers typeset by red,
since they are exactly the coefficients of the quotient, and of course the remainder
typeset in blue
...
3
...
, a0 ∈ R, and g(x) = x − c with c ∈ R be two polynomials
...
ai+1
...
bi
...
, 0
r := c · b0 + a0
...
1)
40
CHAPTER 2
...
Further, we also have
f (c) = r
...
2)
Remark
...
Example
...
The first place in the first line is empty,
then we list the coefficients of f
...
Then we compute the consecutive elements of the second line using (2
...
2
...
POLYNOMIALS
41
Remark
...
e
...
Example
...
The first place in the first row is empty, then we list the
coefficients of f , including the coefficient 0 of x3
...
Then we compute the consecutive elements of the second line using (2
...
Exercise 2
...
Divide the polynomial f (x) by the monic linear polynomial g(x)
42
CHAPTER 2
...
2
...
6
...
ALGEBRAIC EXPRESSIONS
of the Euclidean division of f (x) by g(x):
a) f (x) = x6 + 4x5 − 11x4 − 31x3 − 4x2 + 11x + 30,
g(x) = x − 2
b) f (x) = x6 + 4x5 − 11x4 − 31x3 − 4x2 + 11x + 30,
g(x) = x + 2
c) f (x) = x6 + 4x5 − 11x4 − 31x3 − 4x2 + 11x + 30,
g(x) = x − 1
d) f (x) = x6 + 4x5 − 11x4 − 31x3 − 4x2 + 11x + 30,
g(x) = x + 1
e) f (x) = x6 + 4x5 − 11x4 − 31x3 − 4x2 + 11x + 30,
g(x) = x + 3
f) f (x) = x6 + 4x5 − 11x4 − 31x3 − 4x2 + 11x + 30,
g(x) = x − 3
g) f (x) = x6 + 4x5 − 11x4 − 31x3 − 4x2 + 11x + 30,
g(x) = x + 4
h) f (x) = x6 + 4x5 − 11x4 − 31x3 − 4x2 + 11x + 30,
g(x) = x + 5
i) f (x) = x7 + 3x6 − 7x5 − 28x4 − 21x3 + 7x2 + 27x + 18,
g(x) = x2 − 1
j) f (x) = x7 + 3x6 − 7x5 − 28x4 − 21x3 + 7x2 + 27x + 18,
g(x) = x2 − 4
k) f (x) = x7 + 3x6 − 7x5 − 28x4 − 21x3 + 7x2 + 27x + 18,
g(x) = x2 − 9
l) f (x) = x7 + 6x6 + 15x5 + 21x4 + 6x3 − 15x2 − 22x − 12,
g(x) = x − 1
m) f (x) = f (x) = x7 + 6x6 + 15x5 + 21x4 + 6x3 − 15x2 − 22x − 12,
g(x) = x + 1
n) f (x) = f (x) = x7 + 6x6 + 15x5 + 21x4 + 6x3 − 15x2 − 22x − 12,
g(x) = x2 − 1
o) f (x) = f (x) = x7 + 6x6 + 15x5 + 21x4 + 6x3 − 15x2 − 22x − 12,
g(x) = x − 2
p) f (x) = f (x) = x7 + 6x6 + 15x5 + 21x4 + 6x3 − 15x2 − 22x − 12,
g(x) = x + 2
q) f (x) = f (x) = x7 + 6x6 + 15x5 + 21x4 + 6x3 − 15x2 − 22x − 12,
g(x) = x2 − 4
r) f (x) = f (x) = x7 + 6x6 + 15x5 + 21x4 + 6x3 − 15x2 − 22x − 12,
g(x) = x − 3
s) f (x) = f (x) = x7 + 6x6 + 15x5 + 21x4 + 6x3 − 15x2 − 22x − 12,
g(x) = x + 3
t) f (x) = f (x) = x7 + 6x6 + 15x5 + 21x4 + 6x3 − 15x2 − 22x − 12,
g(x) = x2 − 9
u) f (x) = x8 + x7 − 11x6 − 12x5 + 11x4 + 23x3 + 59x2 + 36x + 36,
g(x) = x + 1
v) f (x) = x8 + x7 − 11x6 − 12x5 + 11x4 + 23x3 + 59x2 + 36x + 36,
g(x) = x − 1
w) f (x) = x8 + x7 − 11x6 − 12x5 + 11x4 + 23x3 + 59x2 + 36x + 36,
g(x) = x2 − 1
x) f (x) = x8 + x7 − 11x6 − 12x5 + 11x4 + 23x3 + 59x2 + 36x + 36,
g(x) = x2 − 4
y) f (x) = x8 + x7 − 11x6 − 12x5 + 11x4 + 23x3 + 59x2 + 36x + 36,
g(x) = x2 − 9
z) f (x) = x8 + x7 − 11x6 − 12x5 + 11x4 + 23x3 + 59x2 + 36x + 36,
g(x) = x2 + 5x + 6
2
...
FACTORIZATION OF POLYNOMIALS
45
Exercise 2
...
Compute the value of the following polynomial f (x) at the given
value of the indeterminate x:
a) f (x) = x8 + x7 − 11x6 − 12x5 + 11x4 + 23x3 + 59x2 + 36x + 36,
x=1
b) f (x) = x8 + x7 − 11x6 − 12x5 + 11x4 + 23x3 + 59x2 + 36x + 36,
x=2
c) f (x) = x8 + x7 − 11x6 − 12x5 + 11x4 + 23x3 + 59x2 + 36x + 36,
x=3
d) f (x) = x8 + x7 − 11x6 − 12x5 + 11x4 + 23x3 + 59x2 + 36x + 36,
x = −4
e) f (x) = x8 + x7 − 11x6 − 12x5 + 11x4 + 23x3 + 59x2 + 36x + 36,
x = −3
f) f (x) = x8 + x7 − 11x6 − 12x5 + 11x4 + 23x3 + 59x2 + 36x + 36,
x = −2
g) f (x) = x8 + x7 − 11x6 − 12x5 + 11x4 + 23x3 + 59x2 + 36x + 36,
x = −1
h) f (x) = x8 + x7 − 11x6 − 12x5 + 11x4 + 23x3 + 59x2 + 36x + 36,
x=0
i) f (x) = x6 − 13x4 − 9x3 + 40x2 + 81x − 36,
x=1
j) f (x) = x6 − 13x4 − 9x3 + 40x2 + 81x − 36,
x = −1
k) f (x) = x6 − 13x4 − 9x3 + 40x2 + 81x − 36,
x=2
l) f (x) = x6 − 13x4 − 9x3 + 40x2 + 81x − 36,
x = −2
m) f (x) = x6 − 13x4 − 9x3 + 40x2 + 81x − 36,
x=3
n) f (x) = x6 − 13x4 − 9x3 + 40x2 + 81x − 36,
x = −3
o) f (x) = x6 − 13x4 − 9x3 + 40x2 + 81x − 36,
x=4
2
...
Factorization of a polynomial is the process
of finding polynomials whose product is the original polynomial
...
In this sense factorization of polynomials is the inverse process,
i
...
changing sums into products
...
ALGEBRAIC EXPRESSIONS
Polynomials that cannot be factorized are called irreducible polynomials
...
So when we try to factorize a polynomial
we have to specify what kind of coefficients are allowed for the factors
...
g
...
Thus such a
2
decompositions is not considered a factorization
...
3
...
Example
...
3
...
Powers of sums and differences
2
...
FACTORIZATION OF POLYNOMIALS
47
(a + b)2 = a2 + 2ab + b2
(a − b)2 = a2 − 2ab + b2
(a + b)3 = a3 + 3a2 b + 3ab2 + b3
(a − b)3 = a3 − 3a2 b + 3ab2 − b3
n
∑ (n)
n
(a + b) =
an−i bi
i
i=0
Difference of perfect powers
a2 − b2 = (a + b)(a − b)
a3 − b3 = (a − b)(a2 + ab + b2 )
an − bn = (a − b)(an−1 + an−2 b + an−3 b2 +
...
− ab2k−1 + b2k )
Remark
...
The reason for this is that there are no such formulas over the real
numbers
...
Below we give some examples, where we use the above formulas to
factorize polynomials
...
ALGEBRAIC EXPRESSIONS
• x4 − 4y 2 = (x2 )2 − (2y)2 = (x2 + 2y)(x2 − 2y)
• x4 − 6x2 y + 8y 2 = (x2 )2 − 2x2 · 3y + (3y)2 − y 2 = (x2 − 3y)2 − y 2 =
(x2 − 3y + y)(x2 − 3y − y) = (x2 − 2y)(x2 − 4y)
2
...
3
Factorization by grouping terms
If the polynomial to be factored has at least four terms (or it has three, but
we can split one term in two), and neither factoring out a monomial, nor
using special factorization formulas leads to a factorization, we may try to
group the terms and factorize the groups so that, in the factorized form of
all these groups we will find a common polynomial which can be factored out,
this way leading to a factorized form of the original polynomial
...
Factorize the following polynomials over the integers:
1)
...
x3 + x2 y + xy 2 + y 3 =(x3 + x2 y) + (xy 2 + y 3 )
= x2 (x + y) + y 2 (x + y) = (x + y)(x2 + y 2 )
3)
...
3
...
3
...
1)
...
2)
...
3)
...
4)
...
5)
...
As the above strategy suggests, when factorizing polynomials in many cases, we
have to combine the methods presented in Sections 2
...
1, 2
...
2 and 2
...
3
...
Factorize the following polynomials using integer coefficients:
1)
...
Then we pull out the common
factor, finally we use the formula for difference of squares to factorize one of
the resulting factors
...
In the present example we first group the terms, we recognize that the second
group is a square of a binomial, then we use the formula for the difference of
50
CHAPTER 2
...
x4 −x2 + 6x − 9 = x4 − (x2 − 6x + 9) = (x2 )2 − (x − 3)2
(
)(
)
= x2 + (x − 3) x2 − (x − 3) = (x2 + x − 3)(x2 − x + 3)
Exercise 2
...
Factorize the following polynomials into factors with rational coefficients:
a) 5x2 y + 15y − 5
b) 8a − 8b + 16
c) 8x3 − 12x2 y 2 + 4x2 z
d) a3 b + a2 b3 + a2 b
e) a(x + y) − ab(x + y)
f)2m3 − 4m5 n + 2m2
g) x3 y 2 − 5x2 y + x4 y 2
h) 4a2 (b2 − 2) − 2ab(b2 − 2)
i) a(x − 1) + b(1 − x) − 7x + 7
j) x2 − 25
k) x2 − 9y 2
l) a4 − 16
m) 4a2 − 12ab + 9b2
n) a3 − 6a2 b + 12ab2 − 8b3
o) 125a3 + 8b3
p) 81a4 − 16b2
q) (5x − 3y)2 − 81y 2
r) 100(7x − 3y)2 − 9(4x + 5y)2
s) ac + ad + bc + bd
t) ac − ad + bc − bd
u) a3 + 2a2 + 2a + 4
v) a3 + a2 b − ab2 − b3
w) x3 − x2 z + 2xz 2 − 2z 3
x) a4 + a3 b − ab3 − b4
y) x2 − 4x + 3
z) x2 − x − 6
ω) x4 − 5x2 + 4
Exercise 2
...
Factorize the following polynomials into factors with rational coef-
2
...
FACTORIZATION OF POLYNOMIALS
51
ficients:
a) 3x + 18x3 y 3 + 27x5 y 6
b) x3 y 2 − 100x − x2 y 3 + 100y + x2 y 2 z − 100z
c) 9a4 + 41a2 − 20
d) (a2 b2 + 1)2 − (a2 + b2 )2
e) (x + 2y)3 + (3x − y)3
f) x8 + x4 + 1
g) (x + y)4 + x4 + y 4
h) x4 − 2(a2 + b2 )x2 + a4 + b4 − 2a2 b2
i) x2 + 3x − x4 − 3x
j) x5 − 5x4 + 4x3 − x2 + 5x − 4
k) x2 + 2xy + y 2 − xz − yz
l) (abc + abd + acd + bcd)2 − abcd(a + b + c + d)2
m) 3x4 y 4 − x8 − y 8
n) (ac + pbd)2 + p(ad − bc)2
o) ac2 − ab2 + b2 c − c3
p) (x2 + 4x + 8)2 + 3x(x2 + 4x + 8) + 2x2
q) a2 b4 c2 − a2 b2 c4 + a4 b2 c2 − a4 b4
r) a2 b2 + c2 d2 − a2 c2 − b2 d2 − 4abcd
s) x5 + 2x4 + 3x2 + 2x + 1
t) 9x6 + 18x5 + 26x4 + 16x3 + 6x2 − 2x − 1
u) (x + y)3 + 3(x + y)(x2 − y 2 ) + 3(x − y)(x2 − y 2 ) + (x − y)3 − 27y 3
v) (cx + by)(ax + cy)(bx + ay) − (bx + cy)(cx + ay)(ax + by)
w) (x2 + x + 1)(x2 + x + 2) − 12
x) abc(a + b + c) − ab − ac − bc − a2 b2 c2 + 1
y) (x2 + x + 1)(x3 + x2 + 1) − 1
z) (x − a)3 (b − c) + (x − b)3 (c − a) + (x − c)3 (a − b) + 3x(b − c)(c − a)(a − b)
2
...
5
Divisibility of polynomials
Definition 2
...
Let T be any of the sets Q, R
...
We say that Q divides P if there exists a polynomial R(x) ∈ T[x]
such that P (x) = Q(x)R(x)
...
Notation
...
52
CHAPTER 2
...
The fact that Q is a divisor of P in fact means that if we perform the
Euclidean division of P by Q then the remainder is 0
...
5
...
Then we have the following:
1)
...
if P | Q and Q | P then we have P = aQ with some a ∈ Q \ {0},
3)
...
if P | Q then P | QR,
5)
...
if P | Q and P | R then P | (S · P ± T · Q)
...
3
...
6
...
Then P (r) is just the remainder of the Euclidean division of
P (x) by x − r
...
7
...
Theorem 2
...
(The factor theorem) Let P (x) ∈ R[x] be a polynomial with real
coefficients and r ∈ R a real number
...
e
...
2
...
RATIONAL ALGEBRAIC EXPRESSIONS
2
...
9
...
Example
...
4
...
Example
...
It is very important that we can only simplify by a factor which is a
factor of the whole numerator and the whole denominator, but we cannot simplify
by an expression which is a factor of only a term of the numerator or (denominator),
but not of the whole numerator (or denominator)
...
(x + 1)(x2 + 3)
54
CHAPTER 2
...
10
...
Example
...
x−1
(x − 1)(x2 + x + 1)
x3 − 1
Exercise 2
...
Amplify the following expressions so that they have the same
denominator
1
3
and
2 b7
5a
a3 b
1
1
b) 3
and
2 (x + 1)2
a (x + 1)
a
x+3
x−1
c)
and
2x − 1
3x + 2
1
1
1
,
and
d)
3
x+1
a
3a
a)
2
...
RATIONAL ALGEBRAIC EXPRESSIONS
2
...
2
55
Multiplication and division of rational expressions
• If we have to multiply two rational algebraic expressions, we have
to multiply the numerator by the numerator and the denominator
by the denominator, however, if possible, we first simplify
...
Example
...
Here we present a simple example of division of two rational expressions:
(x − 1)(x2 + 1) (x − 1)(x + 5)
(x − 1)(x2 + 1)
x+4
(x2 + 1)(x + 4)
:
=
·
=
x−3
x+4
x−3
(x − 1)(x + 5)
(x − 3)(x + 5)
56
CHAPTER 2
...
4
...
If we have to add (subtract) two or more rational algebraic expressions, whose denominators are the same, then the result is a
fraction whose denominator is the same like the common denominator of
the summands, and the numerator is the sum (difference) of the original
numerators
...
If we have to add (subtract) two or more rational algebraic expressions, whose denominators are different then we first amplify the
fractions so that all denominators become the same expression, and we
use the rule described in 1
...
When choosing the common denominator, we have to try to find the
most simple such expression, i
...
The above rules can be summarized by the above formulas
...
d d
d
In the case when the denominators are different, then
ad bc
ad + bc
a c
+ =
+
=
,
b d
bd bd
bd
a
c
ad
bc
ad + bc
+
=
+
=
...
The product of the denominators is always a theoretically possible
choice for the common denominator, however we strongly discourage the student
to choose this way, since it may make the solution much more complicated
...
4
...
e
...
Example
...
(x − 1)(x + 2)
(x − 1)(x + 2)
In the next example (as always) it is also possible to take the product of the
denominators as the common denominator, however, this makes the computations
much more complicated
...
This shows that it is
always important to try to find the least common multiple of the denominators and
58
CHAPTER 2
...
x2
x
x+3
x
x+3
+
=
+
− 1 (x − 1)(x + 2)
(x + 1)(x − 1) (x − 1)(x + 2)
x(x + 2)
(x + 3)(x + 1)
=
+
(x + 1)(x − 1)(x + 2) (x + 1)(x − 1)(x + 2)
x2 + 2x
x2 + 3x + x + 3
=
+
(x + 1)(x − 1)(x + 2) (x − 1)(x + 2)(x2 − 1)
(2x2 + 6x + 3)
x2 + 2x + x2 + 3x + x + 3
=
=
(x + 1)(x − 1)(x + 2)
(x + 1)(x − 1)(x + 2)
In exercises generally our main goal is to transform complicated algebraic expressions to an equivalent, but much simpler algebraic expression
...
Simplify the following algebraic expression:
[
(
)] 3
1
1 1
1
2
a − b3
+ 2+
−
:
a ̸= 0, b ̸= 0, a ̸= −b
...
4
...
12
...
ALGEBRAIC EXPRESSIONS
[(
)2
]
2ab
16a4 + b4
) (
)
(
a
5x
10ax
x
2ax
5a
:
j)
+
+
+
+
a + x a − x a2 − x2
a + x a − x a2 − x2
(
) (
)
b2
1
a−b
a
k)
+
:
−
a3 − ab2 a + b
a2 + ab b2 + ab
(
)
) (
4xy
y
2xy
x
l) x −
+y :
−
−
x+y
x + y y − x x2 − y 2
) (
)
(
1
a+b
1 1
(a + b)2 + 2b2
−
+
·
−
m)
a 3 − b3
a − b a2 + ab + b2
b a
[( 2
) (
)]
x2 − 1
x − xy
2x2
y−1
n)
:
−
· 1−
y
xy
x2 y + y 3 y 3 − xy 2 + x2 y − x3
x − x2
) (
)
(
1
1
1
x3 − 1+ 1 + 1 + 1 : 1 −1 − 1 1 1
−x
2
x
x
x
x
x2
o)
3−1
x
( 2
)
x+y
x4 −y 4
x2 −y 2
− a3 +b3 : a2 −ab+b2 · (2a + 2b − ax − bx)
a+b
p)
x2 − 3x + 3
2 +ab+b2
2 +2ab+b2
2
2
a
− a a3 +b3 : a2a −b 2
a3 −b3
−ab+b
q)
(7a3 + ab + b5 )(2a − 3b) − b3 + a3
i)
2
...
Thus in many cases it is useful to rationalize the denominator
of such fractions (i
...
to get rid of the roots appearing in the denominator using
equivalent transformations of the expression)
...
Here we present the most frequently used methods for rationalizing the denominator of a fraction:
2
...
ALGEBRAIC EXPRESSIONS CONTAINING ROOTS
61
1)
...
Specially,
if the one-term denominator contains only one square-root, then we amplify
the fraction by that square-root
...
3
3
3· 3
2)
...
1
Indeed, if we wish to rationalize the denominator of √a−√b (a, b > 0) then
√
√
we amplify the fraction by a + b as follows:
√
√
√
√
1
a+ b
a+ b
√ = √
√ √
√ =
√
a−b
a− b
( a − b)( a + b)
Similarly, we have
√
√
√
√
1
a− b
a− b
√ = √
√ √
√ =
√
a−b
a+ b
( a + b)( a − b)
3)
...
62
CHAPTER 2
...
Rationalize the denominator of the following fractions
1)
...
1
√
253
3)
...
5
...
1
√
2− 3
5)
...
13
...
14
...
ALGEBRAIC EXPRESSIONS
Example
...
Solution 1
...
So we have
√
√
√
√
√
√
√
√
S = 4+2 3− 4−2 3= 3+2 3+1− 3−2 3+1=
√(
√
√
)2 √(√
)2
√
√
√ 2
√ 2
√
2−
2 =
3 +2· 3·1+1
3 +2· 3·1+1
3+1 −
3−1
=
(√
) (√
) √
√
√
√
3+1 −
3 − 1 = 3 + 1 − 3 + 1 = 2
...
The second solution is based on the idea to compute the square of S
...
(√
Further, since
√
√
√
√
√
4 + 2 3 > 4 − 2 3 we have S > 0
...
2
...
ALGEBRAIC EXPRESSIONS CONTAINING ROOTS
65
Exercise 2
...
Compute the exact value of the following expressions:
√
√
√
√
√
√
√
√
a) 6 + 2 5 − 6 − 2 5
b) 7 + 4 3 + 7 − 4 3
√
√
√
√
√
√
√
√
c) 16 + 8 3 − 16 − 8 3
d) 9 + 3 5 · 9 − 3 5
√
√
√
√
√
√
√
√
2+ 3
2− 3
√ +
√
e) 2 + 3 + 2 − 3
f)
2− 3
2+ 3
√
√
√
√
√
√
3
g) ( 3 + 3) 54 − 30 3
h) 17 − 4 9 + 4 5
√
√
√
√
√
√
√
√
i) 2 3 + 5 − 13 + 48
j) 13 + 30 2 + 9 + 4 2
√
√
√
√
6−2 5−
√
√
√
8−2 6+2 5
8+2
26 − 6 13 + 4
√
√
√
√
√
√
√
√
√
√
√
√
√
l) 2 + 3 2 + 2 + 3 2 + 2 + 2 + 3 2 − 2 + 2 + 3
k)
26 + 6
13 − 4
√
1
1
1
√ +√
√ + ··· + √
√
m) 1 +
1+ 2
2+ 3
2011 + 2012
√
√
√√
√√
√
√
4
8+
2−1− 48−
2−1
√
n)
√√
√
4
8−
2+1
Exercise 2
...
Simplify the following expressions:
√
√
3 a−2 2 a−4 a+1
a) √
− √
−
a−1
a+1
a−1
(
) √
√
√
√
2a + 3 a
3 a+2 4 a−1 2 a+3
√
√
b)
− √
+ √
4a + 12 a + 9 2 a + 3 2 a + 3 2 a − 3
√
√
)
√
√ ( √
√
x+ y−1
x− y
y
y
+
c)
√
√
√ +
√
x + xy
2 xy
x − xy x + xy
(
)
√
√
√ ) (
a + ab
1
2 ab
b
√ +
√ − √
√
√
√
d) √
: √
√
√
a− b a a−a b+b a−b b
a3 + a b + b a + b3 a + b
66
CHAPTER 2
...
1
Introduction to equations
Definition 3
...
An equation is a statement that two algebraic expressions are
equal
...
By the above definition the statement
√
√
1
√
√ = 3+ 2
3− 2
is an equation, however, in algebra we are more interested in equations which
contain variables (i
...
letters)
...
Here are three examples of equations which look completely similar:
1)
...
3x + 4 − 3(x + 3) + 5 = 1,
3)
...
67
68
CHAPTER 3
...
0 = 0,
2)
...
x = 0
...
the equation is true for all possible values of x,
2)
...
the equation is true for only a single value of x, namely 0, so the truth of
the equation depends on the value of x
...
Definition 3
...
An equation which is always false, independently of the value of
the variables involved, is called a contradiction
...
3
...
Definition 3
...
An equation whose truth depends on the value of the variables
involved is called a conditional equation
...
These values
are called solutions (or roots) of the equation
...
Only those
values are accepted as solutions of the original equation which give a true equality
when checked
...
1
...
5
...
Transformations that modify an equation to an equivalent equation
are called equivalent transformations of the equation
...
Equivalent Transformations of Equations
1)
...
2)
...
3)
...
4)
...
5)
...
6)
...
c
c
Remark
...
If we omit this restriction, then the resulting
70
CHAPTER 3
...
e
...
Such solutions are called extraneous solutions
...
Whenever writing a set of equations below each other we make the
convention that this means that they are in consequential relation, unless said
otherwise
...
In many countries the convention is different, i
...
the equations written below
each other are equivalent to each other, and then it is not necessary to check
the solutions
...
For the rest of this chapter we restrict our investigation to equations in one
variable
...
3
...
6
...
3
...
LINEAR EQUATIONS
71
Method of solving linear equations
1)
...
2)
...
3)
...
4)
...
If the coefficient c of x in the left hand side is non-zero, then
divide by it, to get the single solution of the equation, otherwise
decide if the equation is a contradiction or an identity
...
If there is a single solution check that solution, if there are infinitely
many solutions put the necessary conditions and mention that
we have used only equivalent transformations
...
• Solve the following linear equation over the real numbers
x−2 x
+ = 4
...
EQUATIONS I
Solution
...
• Solve the following linear equation over the real numbers
x−2 x
7x
+ =4+
...
We follow the general strategy described above:
x−2 x
7x
+ =4+
3
4
12
/ · 12
4(x − 2) + 3x = 48 + 7x
4x − 8 + 3x = 48 + 7x
7x − 8 = 48 + 7x
/ + 8 − 7x
0 = 56 contradiction
So there is no solution to this equation, i
...
the solutionset is the emptyset:
S = ∅
...
2
...
3
4
12
Solution
...
e
...
In our case every expression in the original equation has sense
for every real number
...
• Solve the following linear equation over the real numbers
7x − 8
x−2 1
+ =
...
First we have to put conditions:
3x ̸= 0 and 12x ̸= 0
which leads to x ∈ R \ {0}
...
EQUATIONS I
Now we follow the general strategy described above:
x−2 1
7x − 8
+ =
3x
4
12x
/ · 12x ̸= 0
4(x − 2) + 3x = 7x − 8
4x − 8 + 3x = 7x − 8
7x − 8 = 7x − 8
/ + 8 − 7x
0 = 0 identity
We have used equivalent transformations, and our equation reduces to an
identity, which means that every real number for which the original equation
is defined (i
...
the expressions in the equation have sense) is a solution to
this equation
...
• Solve the following equation over the real numbers
(x + 3) · (3x − 12) = 0
...
This equation is clearly not a linear equation
...
Indeed,
by Theorem 1
...
By
this, our equation can be replaced by the below two linear equations:
x+3=0
or
3x − 12 = 0
...
2
...
So the set of solution of our original equation is S = {−3, 4}
...
Solution
...
However,
when we simplify the expressions in our equation, we shall see that it reduces
to a linear equation:
2x2 − x + 2x − 1 − x2 − 4x − 4 + 1 = x2 − 1
x2 − 3x − 4 = x2 − 1
/x2 + 4
− 3x = 3
/ :3
x = −1
...
EQUATIONS I
so we get the set of solutions
S = {−1}
...
Solution
...
e
...
Since the only possible solution does not fulfill this
requirement the original equation has solution set S = ∅
...
2
...
1
...
3
CHAPTER 3
...
7
...
The quantity ∆ := b2 − 4ac is called the discriminant of the above equation
...
8
...
Consider the equation
ax2 + bx + c = 0
where a, b, c are constants, and a ̸= 0
...
1)
Then, depending on the sign of the discriminant ∆ := b2 − 4ac of the equation we
have the following possibilities:
1)
...
1) has no real solution;
2)
...
1) has two coinciding real solutions, namely
x1 = x2 =
−b
2a
(Sometimes we simply say that equation (3
...
if ∆ > 0 then equation (3
...
2a
3
...
QUADRATIC EQUATIONS
79
Proof
...
1) first we do the following equivalent transformations:
ax2 + bx + c = 0
(
)
b
c
2
a x + x+
=0
a
a
((
)
( )2 ) ( )2
b
c
b
b
a
x2 + 2 x +
−
+
=0
2a
2a
2a
a
(
)2
b
b2 − 4ac
a x+
−
=0
2a
4a
(
)2
b
b2 − 4ac
a x+
=
2a
4a
)2
(
b2 − 4ac
b
=
...
1)
...
This is clearly a contradiction, sine a non-negative number is never
equal to a negative one
...
If ∆ = 0 then the right hand side of the last equation is 0 and we get
(
)2
b
x+
=0
2a
which clearly gives
x1 = x2 = −
3)
...
2a
> 0 and by taking square root of both sides we get
b
x+
=
2a
√
b2 − 4ac
,
2|a|
80
CHAPTER 3
...
2a
General strategy for solving quadratic equations
1)
...
2)
...
3)
...
Example
...
x2 − 3x + 2 = 0
The coefficients are a = 1, b = −3, c = 2
...
Since ∆ is positive the equation has two solutions, which can be computed
by the ”almighty formula”:
3+1
2
x1,2
=2
3−1
2
=1
:
uu
uu
u
√
√
uu
−(−3) ± 1
3±1
−b ± ∆
uu
= II
=
=
=
II
2a
2·1
2
II
II
I$
3
...
QUADRATIC EQUATIONS
81
So the set of solutions of the above equation is S := {1, 2}
...
x2 + 4x + 4 = 0
The coefficients are a = 1, b = 4, c = 4
...
Since ∆ = 0 the equation has two coinciding solutions (or we may say that it
has only one solution), which can be computed by the ”almighty formula”:
√
√
−4) ± 0
−b ± ∆
=
= −2
...
3)
...
The discriminant is
∆ = b2 − 4ac = 02 − 4 · 1 · 2 = 0 − 8 = −8 < 0
...
e
...
82
CHAPTER 3
...
2
...
3
...
3
...
EQUATIONS I
Chapter 4
Inequalities
4
...
1
...
Example
...
e
...
The following statements are also examples of
inequalities:
x2 + y 2 ≤ 1,
3x > x + 2,
√
x2 + 1 ≤ 3,
|x| ≥ 3
...
Similarly to the case of equations, here are three examples of inequalities which look completely similar:
85
86
CHAPTER 4
...
3x + 4 − 3(x + 3) + 5 < 1,
2)
...
3x + 4 − 2(x + 3) + 2 < 0
...
0 < 1,
2)
...
x < 0
...
the inequality is true for all possible values of x,
2)
...
the inequality is true for a part of the possible values of x, so the truth of
the inequality depends on the value of x
...
The names given to the different types of
inequalities are the same like names of the corresponding cases of equations
...
2
...
Definition 4
...
An inequality which is always true, independently of the value of
the variables involved, is called an identity
...
1
...
4
...
To solve a conditional inequality
means to find the set of those values of the variables for which the equation is true
...
Definition 4
...
Two inequalities that have exactly the same set of solutions are
called equivalent inequalities
...
Equivalent Transformations of Inequalities
1)
...
2)
...
3)
...
Subtracting the same number or algebraic expression from both sides
a < b ⇐⇒ a − c < b − c
a > b ⇐⇒ a − c > b − c
a ≤ b ⇐⇒ a − c ≤ b − c
a ≥ b ⇐⇒ a − c ≥ b − c
5)
...
INEQUALITIES
6)
...
Dividing both sides by the same positive number or positive algebraic expression
a
b
<
with c < 0
c
c
a
b
a ≤ b ⇐⇒
≤
with c < 0
c
c
a < b ⇐⇒
a
b
>
with c < 0
c
c
a
b
a ≥ b ⇐⇒
≥
with c < 0
c
c
a > b ⇐⇒
8)
...
When solving inequalities, our goal is to give the set of all solutions,
i
...
that simplifying the inequality to a very simple equivalent inequality like
x < 2 is not a complete solution of the problem, since we are expected to solve
the resulting simple inequality and give the result in the form of an interval, like
x ∈] − ∞, 2[ in the case of the above example
...
2
Linear inequalities
Definition 4
...
An inequality is called a linear inequality if it is equivalent
to one of the following inequalities, where a, b ∈ R and a ̸= 0
ax − b < 0
ax − b > 0
ax − b ≤ 0
ax − b ≥ 0
4
...
LINEAR INEQUALITIES
89
Method of solving linear equations
1)
...
Add b to both sides of the resulting basic inequality
3)
...
Decide if the inequality is a contradiction or an identity, and if not,
then determine the set of solutions
...
Solve the following inequality in x ∈ R:
4x − 3 x − 4
−
< 2
...
90
CHAPTER 4
...
1
...
3
Table of signs
A special case of the inequalities is when the left hand side is a rational algebraic
expression in a single variable and the right hand side is 0
...
The result can be summarized in a so-called table
of signs
...
3
...
Remark
...
Later we shall see how
to simplify the solution of equations containing absolute values with the help of a
table of signs
...
92
CHAPTER 4
...
3
...
Theorem 4
...
The below simple table of signs describes the sign of the expression
ax + b where a, b ∈ R and a ̸= 0
...
Using a table of signs describe the sign of the expression 2x − 6
...
Clearly 2x − 6 takes zero if and only if
2x − 6 = 0
which means that x = 3
...
We fill by − (the sign opposite to
the sign of 3) the left half of the second row, and by + (the sign of 3) the right
half of the second row
...
Example
...
Solution
...
3
...
So we have to write −5 in the first row of our table,
and then below the −5 we put 0 in the second row
...
−∞
−5
∞
−2x − 10 + + + + + + + + + + + + 0 - - - - - - - - - - - - - - - - - - - Now it is easy to read the result from the above table
...
3
...
We have to split the discussion
in three cases depending on the sign of the discriminant
...
8
...
The below three simple tables of signs describe the sign of the expression ax2 +bx+c
where a, b, c ∈ R and a ̸= 0
...
If the discriminant is negative, i
...
∆ < 0, then the equation ax2 + bx + c = 0
has no real zeros, and we have
−∞
ax2 + bx + c
∞
sign of a
sign of a
sign of a
2)
...
e
...
INEQUALITIES
−∞
ax2 + bx + c
∞
x1
sign of a
sign of a
0
sign of a
sign of a
3)
...
e
...
Using a table of signs describe the sign of the expression x2 − 6x + 5
...
Clearly x2 − 6x + 5 takes zero if and only if
x2 − 6x + 5 = 0
which means that x1 = 1 or x2 = 5
...
We fill by − (the sign
opposite to the sign of the leading coefficient a = +1) the the second row between
the two zeros, and by + (the sign of a = +1) the two sides of the second row
...
4
...
TABLE OF SIGNS
95
Example
...
Solution
...
So we have to write 1 and 2 in the first row of
our table, and then below them we put 0 in the second row
...
−∞
1
∞
2
−2x2 + 6x − 4 - - - - - - - - - - - - - - - - 0 + + + + + + + 0 - - - - - - - - - - - - - Now it is easy to read the result from the above table
...
Using a table of signs describe the sign of the expression 2x2 + 8x + 8
...
Clearly 2x2 + 8x + 8 takes zero if and only if
2x2 + 8x + 8 = 0
which means that x1 = x2 = −2
...
Finally, we fill by +
(the sign of the leading coefficient a = 2) the the second row around the 0
...
96
CHAPTER 4
...
Using a table of signs describe the sign of the expression −x2 + 6x − 9
...
Clearly −x2 + 6x − 9 takes zero if and only if
−x2 + 6x − 9 = 0
which means that x1 = x2 = 3
...
Finally, we fill by − (the sign
of the leading coefficient a = −1) the the second row around the 0
...
Example
...
Solution
...
Thus we do not write anything in
the first row of our table (except for −∞ and ∞ at the two ends), and we fill by
+ (the sign of the leading coefficient a = 2) the the second row
...
Example
...
Solution
...
4
...
Thus we do not write
anything in the first row of our table (except for −∞ and ∞ at the two ends), and
we fill by − (the sign of the leading coefficient a = −1) the the second row
...
4
...
First we transform the inequality to the form ax2 + bx + c with
a, b, c ∈ R and a ̸= 0
...
We use the tables of signs described in Theorem 4
...
3)
...
Example
...
Solution
...
INEQUALITIES
which means that x1 = 2 or x2 = 3
...
Finally, we fill by
+ (the sign opposite to the sign of the leading coefficient a = −1) the second row
between the two zeros, and by − (the sign of a = −1) the two sides of the second
row
...
Example
...
Solution
...
4
...
1
Graphical approach of quadratic inequalities
We reconsider the previous example
...
Solve the following inequality in real values of x
−x2 + 5x − 6 < 0
...
Completing the square we have
1
−x2 + 5x − 6 = −(x − 2
...
4
4
...
QUADRATIC INEQUALITIES
99
Figure 4
...
5)2 + 4
...
1
...
5)2 on the Figure 4
...
Now one can easily obtain the graph of
f3 (x) = −(x − 2
...
3
...
4 shows the graph of the
original quadratic function
...
100
CHAPTER 4
...
2: Graph of the function (x − 2
...
4
...
3: Graph of the function −(x − 2
...
INEQUALITIES
Figure 4
...
5)2 +
1
4
4
...
SOLVING INEQUALITIES USING TABLE OF SIGNS
103
Exercise 4
...
Solve the following inequality in real values of x:
a) x2 − 4x + 3 > 0
c) − 2x2 − 10x − 6 > 0
d) x2 + x − 6 ≤ 0
e) x2 − 4 ≥ 0
f) − x2 + 9 > 0
g) x2 − 2x + 1 ≥ 0
h) − x2 + 6x − 9 ≥ 0
i) 2x2 + 8x + 8 > 0
j) − x2 − 8x − 10 < 0
k) x2 − x + 1 < 0
l) x2 − x + 1 ≥ 0
m) x2 − x + 1 > 0
n) − x2 + x − 3 ≤ 0
o) − x2 − 3 ≥ 0
4
...
Solve the following inequality in x ∈ R:
(x − 2)(3 − x)(x2 − x + 1)
≤0
(x2 + 4x + 3)(−x + 1)(x + 3)
Solution
...
To do that we first solve all factors of the
numerator and the denominator for zero:
x−2=0
3−x=0
x2 − x + 1 = 0
x2 + 4x + 3
−x+1
x1 = 2
x2 = 3
x ̸∈ R
x3,4 = −1; −3 x5 = 1
x+3
x6 = −3
We insert all these solutions in the first row of the table in increasing order
...
INEQUALITIES
composed using these factors
...
we put 0 if no factor of the denominator vanishes at that number
...
The
vertical line means that F has no sense, and the 0 means that F = 0 at the number
which is in the first row above the vertical line, and the 0, respectively
...
The intervals are closed if the endpoint of the interval is
above a zero of the last row
...
3
...
1
Equations containing absolute values
When we have to solve equations containing absolute values in most case we have
to ”get rid” of the absolute values involved, and solve the resulting equations
...
However, this means that we have to split the solution into subcases, depending
on the sign of the expressions appearing in absolute value
...
EQUATIONS II
for the first sight it seems that we have to distinguish four cases, according to the
sign of x − 1 and x − 3:
x − 1 ≤ 0
x − 1 ≤ 0
x − 3 ≤ 0
x − 1 ≥ 0
x − 3 ≤ 0
x − 3 ≥ 0
x − 1 ≥ 0
x − 3 ≥ 0
...
If the number of expressions
appearing in absolute value is larger, then the cases to consider ”a priori” is even
larger, and also the cases where the solution set of the system of inequalities is the
emptyset will increase
...
Let us solve step by step the above equation:
Example
...
Solution
...
We have to get rid of the
absolute values by the formulas
(x − 1)
if x − 1 ≥ 0
|x − 1| =
−(x − 1) if x − 1 ≤ 0
...
We build a table of signs containing all expressions appearing in absolute value:
x−1=0
x1 = 1
−∞
x−3=0
x2 = 3
1
3
∞
x−1 - - - - - - - - - - - - 0 + + + + + + + + + + + +
x−3 - - - - - - - - - - - - - - - - - - - - - - 0 + + + + + +
5
...
EQUATIONS CONTAINING ABSOLUTE VALUES
107
Now we split the solution into subcases
...
The easiest
way is to take closed ending at each finite endpoint of each interval, this way
considering these numbers twice
...
It is very important that
a solution of the resulting equation in a subcase is accepted as a solution of the
original equation only if the solution is an element of the defining interval of that
very case
...
Case 1
...
Case 2
...
EQUATIONS II
(x − 1) + (−(x − 3)) = 10
(x − 1) − (x − 3) = 10
x − 1 − x + 3 = 10
0 = 8 contradiction
So the solution set of the second case is
S2 = ∅
...
If x ∈ [3, ∞[ := I3 then
(x − 1) + (x − 3) = 10
(x − 1) + (x − 3) = 10
x − 1 + x − 3 = 10
2x = 14
x = 7 ∈ I3
So the solution set of the third case is
S3 = {7}
...
e
...
Example
...
5
...
EQUATIONS CONTAINING ABSOLUTE VALUES
109
Solution
...
We have to get rid of the
absolute values by the formulas
|x + 1| =
|x − 2| =
|3 − x| =
(x + 1)
if x + 1 ≥ 0
−(x + 1) if x + 1 ≤ 0
...
(3 − x)
if 3 − x ≥ 0
−(3 − x) if 3 − x ≤ 0
...
We have to consider all intervals appearing
in the first row of the table of signs (namely 4 intervals)
...
If x ∈ ]−∞, −1[ := I1 then
110
CHAPTER 5
...
Case 2
...
Case 3
...
5
...
EQUATIONS CONTAINING ABSOLUTE VALUES
111
Case 4
...
3
x=
To get the complete solution set of the equation we have to join the sets of
solutions of the above subcases, i
...
{
}
10
S = S1 ∪ S2 ∪ S3 ∪ S4 = 0,
...
1
...
Solve the following equation in x ∈ R
|x − 1| + |x + 1| = 2
...
We shall do equivalent transformations
...
|x + 1| =
(x + 1)
if x + 1 ≥ 0
−(x + 1) if x + 1 ≤ 0
...
EQUATIONS II
Figure 5
...
5
...
EQUATIONS CONTAINING ABSOLUTE VALUES
−∞
-1
1
113
∞
x−1 - - - - - - - - - - - - - - - - - - - - - - 0 + + + + + +
x+1 - - - - - - - - - - - - 0 + + + + + + + + + + + +
Now we split the solution into subcases
...
Case 1
...
Case 2
...
114
CHAPTER 5
...
If x ∈ [1, ∞[ := I3 then
(x − 1) + (x + 1) = 2
2x = 2
x = 1 ∈ I3
So the solution set of the third case is
S3 = {1}
...
e
...
For the graph of |x − 1| + |x + 1| − 2 see Figure 5
...
5
...
To solve polynomial equations of
higher degree is much harder
...
The present knowledge of the reader of this book
makes it possible to solve polynomial equations of higher degree only if the equation
is of some special shape, which allows the use of some clever method to reduce the
degree of the equation
...
5
...
POLYNOMIAL EQUATIONS OF HIGHER DEGREE
Figure 5
...
115
116
CHAPTER 5
...
2
...
The
following theorem makes possible to find all rational solutions, and if we are lucky
to have enough rational solutions, maybe we can completely solve our equation
...
1
...
If a rational number
u
v
∈ Q (u, v ∈ Z, v ̸= 0) in its lowest terms (i
...
with
gcd(u, v) = 1) is a solution to the equation
P (x) = 0
then the numerator u is a divisor of the free term a0 (u | a0 ) and the denominator
v is a divisor of the leading coefficient an (v | an )
...
1, we get the following corollaries:
Corollary 5
...
(The integer root theorem) Let P (x) ∈ Z[x] be a polynomial
with integer coefficients, having the form
P (x) = an xn + an−1 xn−1 + · · · + a1 x + a0
...
e
...
5
...
POLYNOMIAL EQUATIONS OF HIGHER DEGREE
117
Corollary 5
...
(The rational root theorem for monic polynomials) Let
P (x) ∈ Z[x] be a polynomial with integer coefficients, having the form
P (x) = xn + an−1 xn−1 + · · · + a1 x + a0
...
Example
...
(5
...
We try to find rational solutions of this equation using Corollary 5
...
If equation (5
...
e
...
=⇒
Now we try which of these is a solution of (5
...
We do this using Horner’s
scheme:
1
−1 −6
14
-12
8
−4
1
1
0
−6
−2
1
−3
0
2
1
1
−4
14 −40
6
0
So after trying 1 and −2 which are not solutions (the last value in the corresponding
Horner’s scheme is non-zero), we find that x1 = 2 is a solution of (5
...
EQUATIONS II
polynomial x4 − x3 − 6x2 + 14x − 12 can be factorized as (x − 2)(x3 + x2 − 4x + 6),
which means the (5
...
Since a product may be zero only if one of its factors is zero, thus we either have
x − 2 = 0, which gives the expected solution
x1 = 2,
or we have
x3 + x2 − 4x + 6 = 0
...
2)
This is a completely similar equation to the original equation (5
...
So we try the same procedure again
...
2) has a rational root x2 then it is also integer, and it divides 6,
i
...
we have
x2 | 6
=⇒
x2 ∈ {±1, ±2, ±3, ±4, ±6}
...
1), if any
...
On the other
hand, we should try x = 2 again, since it could be a double root of the original
polynomial
...
2
...
2)
...
2),
and thus also of (5
...
More precisely, (5
...
Thus the set of
solutions of the (5
...
Example
...
Solution
...
Thus we can
factorize the polynomial as
x6 − 6x + 5 = (x − 1)(x5 + x4 + x3 + x2 + x − 5)
...
We show that
x4 + 2x3 + 3x2 + 4x + 5 > 0
for every real x
...
120
CHAPTER 5
...
3: Graph of the polynomial x4 − x3 − 6x2 + 14x − 12
5
...
POLYNOMIAL EQUATIONS OF HIGHER DEGREE
Figure 5
...
EQUATIONS II
The strategy of solving polynomial equations having rational solutions
1)
...
1 (or one of its Corollaries) we determine
which are the possible rational solutions of the equation
...
Then using Horner’s scheme we try to decide if any of the above
determined rational numbers is a solution
...
If we find a solution we divide by the the corresponding linear factor,
and we reduce the original equation to a similar equation of smaller
degree
...
We repeat the procedure to this equation
...
We repeat the above steps until (hopefully) we get an equation which
we can solve directly
...
2
...
1
...
EQUATIONS II
5
...
2
Biquadratic equations
Definition 5
...
Equation of the form
ax4 + bx2 + c = 0,
where a ̸= 0 is called biquadratic
...
5
...
3
Reciprocal equations
Definition 5
...
A reciprocal equation of order n given by the polynomial equation
f (x) = 0,
where f (x) is a polynomial of degree n with
( )
1
f (x) = ±x f
...
Such equations may be reduced to equations of lower degree by the substitutions
y =x+
1
x
1
or x −
...
2
...
Solve the equation
2x4 − 5x3 + 6x2 − 5x + 2 = 0
...
x
x
1
Using the substitution x + x , we have the quadratic equation
2y 2 − 5y + 2
...
2
In the first case we get for x
x+
and so x = 1
...
Example
...
Solution
...
We
obtain
(
)4 (
)4
1
1
x+ +2 + x+
= 82
...
Rewriting the previous equation
we have
(y + 1)4 + (y − 1)4 = 82,
126
CHAPTER 5
...
5: Graph of the function 2x4 − 4x3 + 6x2 − 5x + 2
5
...
POLYNOMIAL EQUATIONS OF HIGHER DEGREE
127
and expanding it we have the following biquadratic equation
2(y 4 + 6y 2 − 40) = 0
...
It is clear that the first value is impossible
and the second value yields
x+
1
= −3
x
x+
1
= 1
...
So the real solutions of our original equations coincide
the solutions of the quadratic equation
x2 + 3x + 1 = 0,
i
...
x1,2 =
√
−3± 5
...
Solve the equation
1 + x4
3
=
4
(1 + x)
4
Solution
...
One can observe that this is a symmetric equation and x = 0 is not a solution
...
x
x
128
CHAPTER 5
...
Its solutions are y1,2 = 6 ±
√
56
...
x2 − (6 +
The solutions are
x1,2 = 3 +
√
√
14 ±
√
22 + 6 14,
see Figures 5
...
7!
Example
...
2
x
x
1
Solution
...
Then our equation leads to
y 2 + 4 = 4y
and we obtain y = 2 s x = 1, see Figure 5
...
3
Irrational equations
Definition 5
...
An equation with unknowns under the radical is called irrational
equation
...
2) Introducing new variable
...
3
...
6: The graph of
129
1+x4
(1+x)4
−
3
4
with x ∈ [0, 3]
130
CHAPTER 5
...
7: The graph of
1+x4
(1+x)4
−
3
4
with x ∈ [3, 15]
5
...
IRRATIONAL EQUATIONS
Figure 5
...
EQUATIONS II
Example
...
Solution
...
Square both sides we have
√
2x + 6 = 36 − 12 x − 1 + x − 1
...
Again square both sides we obtain
144(x − 1) = (29 − x)2 ,
that is
x2 − 202x + 895 = 0
...
After squaring, there could appear extraneous roots, so we need to check all roots
...
e
...
When x = 197 we have that
√
√
197 − 1 + 2 · 197 + 6 ̸= 6,
i
...
x = 197 is an extraneous root
...
5
...
IRRATIONAL EQUATIONS
Figure 5
...
EQUATIONS II
Example
...
2
Solution
...
However, note that this equation can be transformed into
quadratic
...
Now, we introduce a new variable, let y = 2x2 − 3x + 2, then equation can be
rewritten as
y 2 − 2y − 8 = 0
...
Thus, we obtained set of equations:
√
√
2x2 − 3x + 2 = 4, 2x2 − 3x + 2 = −2
...
2
Second equation does not have solutions, because square root cannot be negative
...
If x = 2 then we have
√ ( )
( )2
)
(
2
7
7
7
3 7
+3− 2·
−3· +2=
+4 ,
2
2
2
2 2
5
...
IRRATIONAL EQUATIONS
135
Figure 5
...
e
...
2
is a solution of the original equation
...
Example
...
Solution
...
Rewriting our equation we have
y 2 − 7 + y = 5,
136
CHAPTER 5
...
The negative value is impossible, the positive value
gives
x2 + x − 2 = 0
and we get
x1 = 1, x2 = −2
...
If x = 1, then we have
12 + 1 +
√
12 + 1 + 7 = 5,
this is a solution of our initial equation
...
5
...
7
...
The following theorem is our basic tool for solving exponential equation
...
8
...
Then
bx = by
implies
x=y
5
...
EXPONENTIAL EQUATIONS
Figure 5
...
EQUATIONS II
The strategy of solving exponential equations
1)
...
2)
...
3)
...
8 to get rid of the bases
...
We solve the resulting equations
...
We check the solutions by inserting them into the original equation
...
Example
...
Solution
...
In this
case both numbers can be written using the base 5, so we will rewrite the problem
using the same base, i
...
(52 )2x−1 = (53 )3x+4
and we have, using the properties of exponents,
54x−2 = 59x+12
...
5
5
...
EXPONENTIAL EQUATIONS
139
We are checking this potential solution, the left hand side is
25− 5 −1 = 5− 5 ,
28
66
and the right hand side is
125− 5 +4 = 125 5 = 5− 5 ,
42
22
66
so x = − 14 is the solution of the original equation
...
Solve the equation
e2x − 3ex + 2 = 0
...
We will introduce a new variable, let y = ex
...
Its solutions are
y1 = 1, y2 = 2,
and we get
ex1 = 1, ex2 = 2
...
One can check that x1 and x2 are solutions to the initial equation
...
Solve the equation
x2 10x − x10x = 6 · 10x
...
From the original equation we obtain
x2 10x − x10x − 6 · 10x ,
140
CHAPTER 5
...
The solutions of the equation
x2 − x − 6 = 0
are
x1 = −2, x2 = 3
...
For x = −2 the left hand side is
4 · 10−2 + 2 · 10−2 ,
the right hand side is 6 · 10−2
...
Example
...
Solution
...
The solutions of the equation
x2 − 7x + 12 = 0
are
x1 = 3, x2 = 4,
further, the solutions of the equation
x2 + x − 6 = 0
are
x3 = 2, x4 = −3
...
5
...
EXPONENTIAL EQUATIONS
141
Figure 5
...
Example
...
Solution
...
Rewriting our equation we
have
y 2y =
1
2
One can prove that the function f (y) = y 2y is strictly decreasing on the interval
0
1
e
and strictly increasing if y ≥ 1
...
See Figure 5
...
EQUATIONS II
Exercise 5
...
Solve the following equations in the set of real numbers:
√
a) 102x = 0
...
5x = 3
...
5 + 2x+2 = 4
g) 52x−1 + 4x = 52x − 4x+1
h) 42x+1 = 65 · 4x−1 − 1
i) 2x+1 + 1 = 3 · 22x
j) 3x + 3x+1 + 3x+2 + 3x+3 =
40
3
k) 10 · 2x − 4x = 16
1
l) 4x+ 2 + 31 · 2x−1 = 4
1
m) 7x+2 − 7x−1 + 2 · 7x = 48
7
n) 26 · 5
√
x+1
√
− 52
x+1+1
=5
o) 9x−1 − 3x+1 + 3x−3 = 1
p) 0
...
251−x + 0
...
2x−0
...
004x−1
5
( )4√x−4
√
5
x−1
r) 2
...
25
8
2
1 √ 3x−1
4
t)
4
= 8− 3
8
5
...
LOGARITHMIC EQUATIONS
5
...
5
...
We also may reverse the question: which exponent to use for a base,
to get a given result? This question motivates the definition of the logarithm
...
9
...
The number a for which ba = c is called the logarithm of c to the base
b, and it is denoted by logb c
...
e
...
Remark
...
Theorem 5
...
(Properties of logarithms) Let b be a positive real number
with b ̸= 1
...
The we
have the following properties of logarithms:
1) logb y = z ⇐⇒ bz = y
2) logb (xy) = logb (x) + logb (y)
( )
x
= logb x − logb y
3) logb
y
( )
1
4) logb
= − logb y
y
5) logb (xz ) = z logb x
6) logb 1 = 0
logb b = 1
...
EQUATIONS II
Remark
...
The logarithm of y to the base b is the exponent which has to be placed on
the base b to get the result y
...
)
2)
...
3)
...
4)
...
5)
...
6)
...
Remark
...
Theorem 5
...
(Changing the base of a logarithm) Let a, b, c be positive
real numbers with a ̸= 1, b ̸= 1 and c ̸= 1
...
Then we have
logc x
logc b
1
2) logb a =
loga b
1) logb x =
5
...
LOGARITHMIC EQUATIONS
145
Example
...
· logn (n + 1) = 10
...
logb a
Using this formula we have
log2 3 · log3 4 =
log3 4
= log2 4,
log3 2
log2 3 · log3 4 · log4 5 = log2 4 · log4 5 =
log4 5
= log2 5,
log4 2
and finally
log2 3 · log3 4 ·
...
logn 2
Now we have to solve the equation
log2 (n + 1) = 10
which gives n = 1023 as a solution
...
What is the exact value of the expression
2log6 18 · 3log6 3 ?
Solution
...
146
CHAPTER 5
...
Find (log2 x)2 if
log2 (log8 x) = log8 (log2 x)
...
Raise both as exponents with base 8,
8log2 (log8 x) = 8log8 (log2 x)
...
Applying the fact that k loga x = loga xk , we have
(log8 x)3 = log2 x
...
Finally, we have
(log2 x)2 = (log2 8)3 = 27
...
5
...
3
...
5
...
12
...
The following theorem is our basic tool for solving logarithmic equations
...
13
...
Then
logb x = logb y
implies
x = y,
and
logb x = a
implies
x = ba
...
EQUATIONS II
The strategy of solving logarithmic equations
1)
...
2)
...
3)
...
13 to get rid of the logarithms
...
We solve the resulting equations
...
We check the solutions by inserting them into the original equation
...
Example
...
Solution
...
Exponentiating both sides, we have
x(2x + 5) = 7
...
5
...
LOGARITHMIC EQUATIONS
149
Subtracting 7 from both sides, we obtain
2x2 + 5x − 7 = 0
...
And thus,
7
x1 = 1, x2 = −
...
If x = 1,
we get
ln 1 + ln(2 · 1 + 5) = 0 + ln 7 = ln 7,
and so x = 1 is a solution
...
However, neither expression is defined, since the domain of the logarithm function
does not contain negative numbers
...
Solve the equation
log2 (4x) = 2 − log2 x
...
Adding log2 x to both sides, we get
log2 (4x) − log2 x = 2
...
EQUATIONS II
we have
log2 (4x2 ) = 2
...
Using the fact that
aloga x = x,
we can simplify the left-hand side to get
4x2 = 4
...
Solving for x, we see that x = ±1
...
If
x = 1, we have
log2 (4 · 1) − log2 1 = log2 4 − 0 = 2,
so x = 1 is a solution
...
However, neither expression is defined, since the domain of the logarithm function
does not contain negative numbers
...
Example
...
5
...
Solution
...
Exponentiating both sides (with base 10) we have
(x + 2)(x − 1) = 10,
and
x2 + x − 12 = 0
...
We check these potential solutions in the original equation and find that x = −4 is
not a solution (because logarithms of negative numbers are undefined), but x = 3
is a solution
...
13
...
Example
...
Solution
...
152
CHAPTER 5
...
13: The graph of lg(x + 2) + lg(x − 1) − 1
5
...
LOGARITHMIC EQUATIONS
153
Figure 5
...
Now the solutions are the x-intercepts of the graph, see Figure 5
...
Example
...
Solution
...
We first use the change of base formula to write that
log16 (x + 1) =
log4 (x + 1)
log4 (x + 1)
1
=
= log4 (x + 1)
...
EQUATIONS II
We now rewrite the original equation as follows
...
2
We can use the product rule and we have
1
log4 (x + 1)(x + 1) 2 = log4 8
...
Finally, we get x = 3
...
Now the left hand
side is
log4 (3 + 1) + log16 (3 + 1) = 1 +
1
3
= ,
2
2
end the right hand side is
( 3) 3
log4 8 = log4 4 2 = ,
2
so the solution to the equation above is x = 3
...
Solve the logarithmic equation
log2 (x − 4) + log√2 (x3 − 2) + log0
...
Solution
...
log√2 (x3
log2 (x3 − 2)
√
− 2) =
= 2 log2 (x3 − 2)
...
5 (x − 4) = − log2 (x − 4)
...
5
...
Solve this equation for x, we get
x=
√
3
1026
...
Example
...
Solution
...
Divide by 64x3 and introduce the
new variable y = log2 x
...
Now the factorization of the left hand side is
22y
2 −y−6
− 2y
2 −4
− 2y
2 −y−2
+ 1 = (2y
2 −4
− 1)(2y
and we get
y1 = 1, y2 = −2, y3 = 2,
and
1
x1 = 2, x2 = 4, x3 =
...
15 and 5
...
156
CHAPTER 5
...
15: The graph of xlog2 (16x ) − 4xlog2 (4x)+1 − 16log2 (4x)+2 + 64x3 with 0
...
3
5
...
LOGARITHMIC EQUATIONS
157
Figure 5
...
3 ≤
2
x ≤ 4
...
158
CHAPTER 5
...
4
...
1
...
In solving a system of equations, we have to find
values for each of the unknowns that will satisfy every equation in the system
...
Example
...
Solution
...
1
...
One can check that this pair is solution of the system of equations above
...
Determine 3x4 + 2x5 if x1 , x2 , x3 , x4 , and x5 satisfy the system of equa159
160
CHAPTER 6
...
1: The graphs of − 2 x +
3
5
3
and −x + 1
...
2x + x + x + x + x = 6
1
2
3
4
5
x + 2x + x + x + x = 12
1
2
3
4
5
x + x2 + 2x3 + x4 + x5 = 24
1
x1 + x2 + x3 + 2x4 + x5 = 48
x1 + x2 + x3 + x4 + 2x5 = 96
Solution
...
On subtracting this from the fourth and fifth given equation we obtain x4 = 17
and x5 = 65, respectively
...
Example
...
Solution
...
Rewriting our original system of eqations we
obtain
a + b = 71
ab = 880
...
SYSTEMS OF EQUATIONS
which factors to
(a − 16)(a − 55) = 0
...
For the first case, it is
easy to see that (x, y) can be (5, 11) (or vice versa)
...
The solution is
52 + 112 = 146
...
Solve the following system of equations
x3 + 4y = y 3 + 16x,
2
1 + x = 5
...
From the first equation we have
x(x2 − 16) = y(y 2 − 4),
and the second equation gives
y 2 − 4 = 5x2
...
If x = 0, then y = ±2, otherwise we get
x2 − 16
,
y=
5x
which yields the equation
(
2
5x + 4 =
x2 − 16
5x
)2
...
2: The graph of 5x2 + 4 −
(
x2 −16
5x
)2
with 0
...
Introducing the new variable z = x2 we have
124z 2 + 132z − 256 = 0
...
This gives x = ±1 and y = ±3
...
See Figures 6
...
3!
164
CHAPTER 6
...
3: The graph of 5x2 + 4 −
(
x2 −16
5x
)2
with −3 ≤ x ≤ −0
...
165
Exercise 6
...
Solve the following systems of equations in the set of real numbers:
a) x + y = 7, xy = −18
b) 3x + 4y = −18, xy = 6
c) x2 + y 2 = 25, xy = 12
d) x2 + y 2 = 25, x + y = 5
e) x2 − y 2 = 20, x + y = 10
f) x2 − y 2 = 40, xy = 21
g) x2 + y 2 = 20, xy = 8
h) y − x2 = 3, y − x = 3
i) 4x2 + 4y 2 = 17xy, x + y = 10
1 1
+ = 1, x + y = 4
x y
4 6
3 4 17
k) + = 0, − −
x y
x y
6
j)
l) x − xy + y = 1, xy = 20
m) x + y + xy = 49, (x + y)xy = 468
n) log7 x + 2 log 1 x = log49 x − 3
7
o) 4y = x − 12, x + y = 2
p) y − 3x = 5, y + x = 3
166
CHAPTER 6
...
1
...
Let x = (a, b) and y = (c, d) be two complex numbers
...
Theorem 7
...
With these definitions of addition and multiplication the set of
complex number is a field with (0, 0) and (1, 0) in the role of 0 and 1 (cf
...
2)
Theorem 7
...
For any real numbers a and b we have
(a, 0) + (b, 0) = (a + b, 0),
167
168
CHAPTER 7
...
The previous theorem shows that the complex numbers of the form (a, 0) have
the same arithmetic structure as the corresponding real numbers a
...
This identification dives us the
field of real numbers as a subfield of the field of complex numbers
...
Definition 7
...
i = (0, 1)
...
Theorem 7
...
If a and b are real, then
(a, b) = a + bi
...
Definition 7
...
If a, b are real and z = a+bi, then the complex number z = a−bi is
called the conjugate of z
...
We will shortly write
a = Re(z), b = Im(z)
...
7
...
Using the last statement we can introduce the absolute value of a complex
number
...
Definition 7
...
If z is a complex number, its absolute value |z| is the nonnegative
square root of zz, that is
|z| =
√
zz
...
Indeed, the equation
x + iy = x − iy
implies y = 0
...
Theorem 7
...
Le z and w be complex numbers
...
170
CHAPTER 7
...
1
Identities of algebraic expressions
1)
...
Prove that
(1) (a + b + c)3 − (a3 + b3 + c3 ) = 3(b + c)(c + a)(a + b),
1
(2) a3 + b3 + c3 − 3abc = 2 (a + b + c) [(a − b)2 + (b − c)2 + (c − a)2 ]
...
Let a, b, c ∈ R be pairwise distinct non-zero real numbers
...
Let a, b, c, d ∈ R be pairwise distinct non-zero real numbers
...
EXERCISES FOR THE INTERESTED READER
(1)
1
(a−b)(a−c)
(2)
1
(a−b)(a−c)(a−d)
+
1
(b−a)(b−c)
+
+
1
(c−a)(c−b)
1
(b−a)(b−c)(b−d)
+
=0
1
(c−a)(c−b)(c−d)
+
1
(d−a)(d−b)(d−c)
=0
(3) Generalize the above statements
...
Let a, b, c, d ∈ R be real numbers, such that the below fractions have non-zero
denominators
...
5)
...
(1) If a + b = 1 then compute the value of the expression
a3 + b3 + 3(a3 b + ab3 ) + 6(a3 b2 + a2 b3 )
(2) If a, d ∈ Z then prove that the sum
a2 + 2(a + d)2 + 3(a + 2d)2 + 4(a + 3d)2
can be written as the sum of two perfect squares
...
(4) Prove that if the difference of two integers is 2 then the difference of
their cubes can be written as the sum of three perfect squares
...
2
...
Compute the value of the expression
(
) (
)
a−b b−c c−a
c
a
b
+
+
·
+
+
c
a
b
a−b b−c c−a
provided that
(1) a + b + c = 0,
(2) |c| = |a − b|
...
Let a, b, c ∈ R be real numbers
...
8)
...
Prove that if
1 1 1
1
+ + =
a b c
a+b+c
then for every n ∈ N we have
1
a2n+1
+
1
b2n+1
+
1
c2n+1
=
1
a2n+1
+
b2n+1
+ c2n+1
...
Let a, b, c ∈ R be real numbers
...
8
...
Let a, b, c ∈ R be positive real numbers
...
EXERCISES FOR THE INTERESTED READER
(3) 2(a3 + b3 + c3 ) ≥ (a + b)ab + (b + c)bc + (c + a)ca
...
Let a, b, c ∈ R be positive real numbers with a + b + c = 1
...
a b c
Under what conditions do we have equality above?
3)
...
Prove that
a4 + b4 ≥ 2
...
Let a, b ∈ R be positive real numbers, and m, n ∈ N be natural numbers of
the same type of parity
...
Let a, b, c ∈ R be positive real numbers
...
Let a, b, c, d ∈ R be positive real numbers
...
7)
...
Prove that
bc
ca
a+b+c
ab
+
+
=
...
Prove that
8
...
INEQUALITIES OF ALGEBRAIC EXPRESSIONS
175
(1) ab + ac + bc ≤ a2 + b2 + c2 for any a, b, c ∈ R
(2) ab + ac + ad + bc + bd + cd ≤ 3 (a2 + b2 + c2 + d2 ) for any a, b, c, d ∈ R,
2
n
n
n
∑ ∑
∑ 2
ai aj ≤ n−1
ai
...
Let
i=1
a1 a2
n
, ,
...
, n
...
Find the minimum value of
9x2 sin2 x + 4
x sin x
for 0 < x < π
...
Determine the minimum value of the function
x2
+ x cos x + cos 2x
...
The sum of the real numbers x and y is 1
...
13)
...
x x
x
x
x
x
x
x
14)
...
EXERCISES FOR THE INTERESTED READER
15)
...
+ 2n ) < n · 2n
in the set of positive integers
...
1
Chapter 1
Results of Exercise 1
...
3, −5, −4
...
99,
4, 4
...
a) − 7, −5
...
3, −5 ∈ ]−∞, −5]
√
c) − 4
...
99 ∈ ]−5, 4[
√
d) − 5, −4
...
99 ∈ [−5, 4[
√
e) − 4
...
99, 4 ∈ ]−5, 4]
√
f) − 5, −4
...
99, 4 ∈ [−5, 4]
√
g) − 5, −4
...
99, 4, 4
...
99, −π, −1, 0, 1, 3, 3
...
02, 5 ∈ ]−5, ∞[
[√ [
√
f) 3, 3
...
02 ∈
3, 5
177
178
CHAPTER 9
...
Results of Exercise 1
...
3
a) 213
b) 22
c) 26
d) 212
e) 24
f) 1
g) 32
h) 332
i) 310
j) 23 · 32 · 50 · 72
k) 24
l) 9a4 b8 c6
m) − 12a3 b5 c9
n) 9a4 b
o)
1
p) − a14 b4 c
3
1 −8 17 13
s) − x y z
8
q) 10a3 b2 d−3
r) a5 b2 c2
t) 6x11 y 12 z 7
u) a8 b2
Results of Exercise 1
...
2
...
5:
a) ab
b) a−1 b
c) a2 b− 2
d) a2 b−3
e) ab
f) a−2 b− 5
4
g) a 3 b 2
h) a− 3 b
j) 1
k) a 5 b 4
11
9
...
1:
a) 2x2 − 2xy 3 + 2xy + 3x + y 2 + y
b) 3x3 − 4x2 y + 4xy 2 + 5y 3
c) 2x5 − 7x4 y + 8x3 y 2 − 5x2 y 3 − 12xy 4 + 5y 5
d) − x2 + 4xy 3 − 13xy − 6x + y 2 − 2y
e) x4 − 2x3 y 3 + 2x3 y + 3x3 + 6x2 y 4 − 14x2 y 2 − 8x2 y − 2xy 5 + 5xy 3 + y 3
f) 2x3 y 3 − 8x3 y − 3x3 − 6x2 y 4 + 25x2 y 2 + 8x2 y + 2xy 5 − 11xy 3 + y 4 − y 3
Results of Exercise 2
...
3:
a) 4xy 3
d)
3 4
ab
5
b) 2y 4 z
e) 2a2 b4 c2
1 2 2
xy z
2
4
f) a2 c5 d2
5
c)
180
CHAPTER 9
...
4 Denoting the quotient by q(x) and the remainder by
9
...
CHAPTER 2
181
r(x) the results of the divisions are:
a) q(x) = x + 3,
r(x) := −2x − 1
b) q(x) = x3 − 2x2 + x + 5,
r(x) = −3
c) q(x) = x3 − 2x − 4,
r(x) = −12x + 10
d) q(x) = x3 − x − 1,
r(x) = −5x + 3
e) q(x) = x3 + 2x + 1,
r(x) = 0
f) q(x) = x4 + 2x3 − 8x − 16,
r(x) = 0
g) q(x) = x3 + x2 − x − 7,
r(x) = −20x + 30
h) q(x) = x3 + 7x2 + 28x + 126,
r(x) = 574x − 250
i q(x) = x3 + 2x2 − 4,
r(x) = −19x + 14
j) q(x) = x + 1,
r(x) = −3x3 + 5x2 − 2x − 1
k) q(x) = x5 + 4x4 + 2x3 − 3x2 − x − 4,
r(x) = 6
l) q(x) = x5 + 2x4 − 4x3 − x2 + 3x − 6,
r(x) = 16
m) q(x) = x5 + 5x4 + 8x3 + 11x2 + 24x + 45,
r(x) = 100
n) q(x) = x5 + x4 − 4x3 + 3x2 − 4x + 5,
r(x) = 0
o) q(x) = x5 − 2x3 + x2 − x,
r(x) = 10
p) q(x) = x4 + 3x3 − x2 − 2x + 1,
r(x) = −5x + 11
q) q(x) = x5 + x4 − 3x3 − 4x2 − 4x − 1,
r(x) = −3
r) q(x) = x5 − x4 − 3x3 + 2x2 − 2x + 5,
r(x) = −7
s) q(x) = x5 + 2x4 − x2 − 2x − 1,
r(x) = −4
t) q(x) = x5 − 2x4 − x2 + 2x − 1,
r(x) = 0
u) q(x) = x5 + 3x4 + 5x3 + 14x2 + 42x + 129, r(x) = 385
v) q(x) = x5 − 3x4 + 5x3 − 16x2 + 48x − 141, r(x) = 421
w) q(x) = x4 + x3 − 2x2 − 2x − 4,
r(x) = −3x − 6
x) q(x) = x4 − 3x2 − x − 3,
r(x) = 2x − 5
y) q(x) = x4 − x3 − 2x2 − 2,
r(x) = 5x − 4
z) q(x) = x4 − 2x3 + x2 − 5x + 11,
r(x) = −24x + 9
182
CHAPTER 9
...
5 Denoting the quotient by q(x) and the remainder by
9
...
CHAPTER 2
183
r(x) the results of the divisions are:
a) q(x) = x4 − 3x3 + 4x2 − 7x + 9,
r(x) = −14
b) q(x) = x4 − 3x3 − 3x2 − 8x − 14,
r(x) = −31
c) q(x) = x4 − 7x3 + 17x2 − 36x + 74,
r(x) = −151
d) q(x) = x6 − 4x5 − 2x4 − 6x3 − 7x2 − 7x − 4,
r(x) = −6
e) q(x) = x6 − 6x5 + 8x4 − 12x3 + 11x2 − 11x + 14,
r(x) = −16
f) q(x) = x6 − 3x5 − 4x4 − 12x3 − 25x2 − 50x − 97,
r(x) = −196
g) q(x) = x6 − 7x5 + 16x4 − 36x3 + 71x2 − 142x + 287,
r(x) = −576
r(x) = −1
h) q(x) = x5 + x4 + x3 + x2 + x + 1,
i) q(x) = x5 + 2x4 + 4x3 + 8x2 + 16x + 32
r(x) = 62
j) q(x) = x5 − 2x4 + 4x3 − 8x2 + 16x − 32,
r(x) = 62
k) q(x) = x5 + 4x4 + 2x3 − 3x2 − x − 4,
r(x) = 6
l) q(x) = x5 + 2x4 − 4x3 − x2 + 3x − 6,
r(x) = 16
m) q(x) = x5 + 5x4 + 8x3 + 11x2 + 24x + 45,
n) q(x) = x5 + x4 − 4x3 + 3x2 − 4x + 5,
o) q(x) = x5 − 2 ∗ x3 + x2 − x,
p) q(x) = x6 − x5 + x2 − x,
r(x) = 100
r(x) = 0
r(x) = 10
r(x) = 0
q) q(x) = x5 + x4 − 3x3 − 4x2 − 4x − 1,
r(x) = −3
r) q(x) = x5 − x4 − 3x3 + 2x2 − 2x + 5, r(x) = −7
s) q(x) = x5 + 2x4 − x2 − 2x − 1,
r(x) = −4
t) q(x) = x5 − 2x4 − x2 + 2x − 1,
r(x) = 0
u) q(x) = x5 + 3x4 + 5x3 + 14x2 + 42x + 129,
r(x) = 385
v) q(x) = x5 − 3x4 + 5x3 − 16x2 + 48x − 141,
r(x) = 421
w) q(x) = x6 + 2x5 − 3x3 + 4x2 − 9x + 6,
r(x) = −2
x) q(x) = x6 + 4x5 + 6x4 + 3x3 + 4x2 − x − 4,
y) q(x) = x6 + x5 − 3x3 + 7x2 − 19x + 35,
z) q(x) = x6 + 2x5 − 3x3 + 4x2 − 9x + 6,
r(x) = 0
r(x) = −66
r(x) = 0
184
CHAPTER 9
...
6 Denoting the quotient by q(x) and the remainder (when
9
...
CHAPTER 2
185
it is non-zero) by r(x) the results of the divisions are:
a) no
q(x) = x5 + 6x4 + x3 − 29x2 − 62x − 113,
r(x) = −196
b) yes
q(x) = x5 + 2x4 − 15x3 − x2 − 2x + 15
c) yes
q(x) = x5 + 5x4 − 6x3 − 37x2 − 41x − 30
d) no
q(x) = x5 + 3x4 − 14x3 − 17x2 + 13x − 2,
e) no
q(x) = x5 + x4 − 14x3 + 11x2 − 37x + 122,
f) yes
q(x) = x5 + 7x4 + 10x3 − x2 − 7x − 10
g) no
q(x) = x5 − 11x3 + 13x2 − 56x + 235,
h) yes
q(x) = x5 − x4 − 6x3 − x2 + x + 6
i) yes
q(x) = x5 + 3x4 − 6x3 − 25x2 − 27x − 18
j) no
q(x) = x5 + 3x4 − 3x3 − 16x2 − 33x − 57,
k) yes
q(x) = x5 + 3x4 + 2x3 − x2 − 3x − 2
l) yes
q(x) = x6 + 7x5 + 22x4 + 43x3 + 49x2 + 34x + 12
m) yes
q(x) = x6 + 5x5 + 10x4 + 11x3 − 5x2 − 10x − 12
n) yes
q(x) = x5 + 6x4 + 16x3 + 27x2 + 22x + 12
o) no
q(x) = x6 + 8x5 + 31x4 + 83x3 + 172x2 + 329x + 636,
p) no
q(x) = x6 + 4x5 + 7x4 + 7x3 − 8x2 + x − 24,
r(x) = 36
q) no
q(x) = x5 + 6x4 + 19x3 + 45x2 + 82x + 165,
r(x) = 306x + 648
r) no
q(x) = x6 + 9x5 + 42x4 + 147x3 + 447x2 + 1326x + 3956,
s) yes
q(x) = x6 + 3x5 + 6x4 + 3x3 − 3x2 − 6x − 4
t) no
q(x) = x5 + 6x4 + 24x3 + 75x2 + 222x + 660,
u) no
q(x) = x7 − 11x5 − x4 + 12x3 + 11x2 + 48x − 12,
v) no
q(x) = x7 + 2x6 − 9x5 − 21x4 − 10x3 + 13x2 + 72x + 108,
w) no
q(x) = x6 + x5 − 10x4 − 11x3 + x2 + 12x + 60,
x) yes
q(x) = x6 + x5 − 7x4 − 8x3 − 17x2 − 9x − 9
y) yes
q(x) = x6 + x5 − 2x4 − 3x3 − 7x2 − 4x − 4
z) yes
q(x) = x6 − 4x5 + 3x4 − 3x3 + 8x2 + x + 6
r(x) = 32
r(x) = −336
r(x) = −910
r(x) = −105x − 210
r(x) = 1260
r(x) = 11856
r(x) = 1976x + 5928
r(x) = 48
r(x) = 144
r(x) = 48x + 96
186
CHAPTER 9
...
7:
a) 144
b) 0
c) 0
d) 18564
e) 0
f) 0
g) 48
h) 36
i) 64
j) − 80
k) 70
l) − 110
m) 0
n) 0
o) 1120
Results of Exercise 2
...
2
...
9
a) 3x(1 + 3x2 y 3 )2
b) (x − y + z)(xy − 10)(xy + 10)
c) (3a + 2)(3a − 2)(a2 + 5)
d) (a − 1)(a + 1)(b − 1)(b + 1)(a2 + 1)(b2 + 1)
e) 7(4x + y)(x2 − xy + y 2 )
f) (x2 + x + 1)(x2 − x + 1)(x4 − x2 + 1)
g) 2(x2 + xy + y 2 )2
h) (x + a + b)(x + a − b)(x − a + b)(x − a − b)
i) x(1 + x)(1 − x)(x − 3)
j) (x − 1)2 (x − 4)(x2 + x + 1)
k) (x + y)(x + y − z)
l) (ab − cd)(bc − ad)(ac − bd)
m) (x2 y 2 + x4 − y 4 )(x2 y 2 − x4 + y 4 )
n) (a2 + pb2 )(c2 + pd2 )
o) (c + b)(c − b)(a − c)
p) (x + 2)(x + 4)(x2 + 5x + 8)
q) a2 c2 (b + c)(b − c)(a − c)
r) (ab − cd + ac + bd)(ab − cd − ac − bd)
s) x(x2 + x + 1)2
t) (3x − 1)(3x + 1)(x2 + x + 1)2
u) (2x − 3y)(4x2 − 6xy + 9y 2 )
v) (cx + by)(ax + cy)(bx + ay) − (bx + cy)(cx + ay)(ax + by)
v) xy(x − y)(a − b)(a − c)(c − b)
w) (x − 1)(x + 2)(x2 + x + 5)
x) (1 − ab)(1 − bc)(1 − ca)
y) x(x + 1)2 (x2 + 1)
z) (a − b)(a − c)(b − c)(a + b + c)
188
CHAPTER 9
...
10
5x
a)
y
−1
c)
2x2
1
2(x − 3)
3x2 + xy
g)
xy + 3y 2
5
i)
x+4
e)
y6z3
b)
2x2 w2
1
d)
−4x
5(x − 1)
f)
x+1
1
h)
x
7(x2 − y 2 )
j)
9y 2
Results of Exercise 2
...
2
...
12
−2
a−1
a)
b) x
a+3
a−3
c)
d) (a + b)(x + 1)
xy(x + y)
y−1
x+1
f)
x+y
1
g) 2
a
e)
h) a + b
i)
1
2ab
j) 5
b
b−a
k)
l) x − y
m)
1
ab
n) x + 1
o) 1
p) − x
q) 0
Results of Exercise 2
...
RESULTS OF THE EXERCISES
Results of Exercise 2
...
2
...
15
a) 2
b) 4
c) 4
√
e) 2
d) 6
g) 6
√
i) 3 − 1
f) 4
√
h) 5 − 2
√
j) 5 + 3 2
k) 6
√
m) 2012
l) 1
√
n) 2
Results of Exercise 2
...
3
CHAPTER 9
...
1
a) S = {4}
b) S = {4}
c) S = {3}
d) S = {4}
e) S = Q
f) S = ∅
g) S = {4}
h) S = ∅
{ }
17
iS=
9
j) S = {−2}
{ }
7
k) S =
8
l) S = {−2}
m) S = ∅
{ }
1
n) S =
9
o) S = {3}
{
}
271
p) S = −
10
q) S = {5}
r) S = {2}
{
}
2
s) S = −3,
5
t) S = {4}
}
{
1
u) S = −2, −3,
2
9
...
CHAPTER 3
193
Results of Exercise 3
...
RESULTS OF THE EXERCISES
Results of Exercise 3
...
4
...
4
195
Chapter 4
Results of Exercise 4
...
RESULTS OF THE EXERCISES
Results of Exercise 4
...
3:
a) x ∈ ]−1, 1[ ∪ ]2, 4[ ∩ ]5, 7[
]
]
1
c) x ∈ ]−∞, −3[ ∪ −1, − ∪ {1} ∪ [4, ∞[
2
e) x ∈ ]−8, −3] ∪ [−1, 2[ ∪ ]3, 4[ ∪ [5, 6]
b) x ∈ ]−5, −3] ∪ ]3, 7]
d) x ∈ ]−3, −2[ ∪ ]−2, −1[ ∩ ]1, 3[
f ) x ∈ ]−7, −5] ∪ ]−4, −3[ ∪ [1, 4[
9
...
CHAPTER 5
9
...
1:
a) x1 = −2, x2 = −1, x3 = 3
b) x1 = −2, x2 = 1, x3 = 2, x4 = 3
c) x1 = −4, x2 = −2, x3 = 1, x4 = 2, x5 = 3
d) x1 = 2, x2 = 2, x3 = −1, x4 = −5
e) x1 = −5, x2 = −3, x3 = −2, x4 = −2, x5 = 4
f ) x1 = −5, x2 = −2, x3 = 2, x4 = 3
g) x1 = −4, x2 = −2, x3 = 3, x4 = 5
h) x1 = −3, x2 = 2, x3 = 4, x4 = 5, x5 = −1 −
i) x1 = −4, x2 = −3, x3 = −1, x4 = 2, x5 = 5
−1 +
j) x1 = −5, x2 = 1, x3 = 2, x4 = 3, x5 =
2
√
√
k) x1 = −1, x2 = −2, x3 = −4, x4 = −8
−3 −
l) x1 = −1, x2 = 2, x3 = −4, x4 = 8, x5 =
2
3, x6 = −1 +
5
−1 −
, x6 =
2
√
5
√
3
√
5
−3 +
, x6 =
2
√
5
m) x1 = −4, x2 = −3, x3 = 1, x4 = 1, x5 = 2, x6 = 5
n) x1 = −4, x2 = −3, x3 = −2, x4 = 2, x5 = 3
o) x1 = −5, x2 = −2, x3 = −2, x4 = −1, x5 = 1, x6 = 5
p) x1 = −2, x2 = 2, x3 = 3, x4 = 3
√
√
q) x1 = −3, x2 = −3, x3 = −2, x4 = −1, x5 = 2, x6 = − 2, x7 = 2
√
√
r) x1 = −2, x2 = 1, x3 = 3, x4 = 9, x5 = − 3, x6 = 3
s) x1 = −25, x2 = −5, x3 = −1, x4 = 2
t) x1 = −3, x2 = −2, x3 = −2, x4 = 1, x5 = 1, x6 = 3
u) x1 = −4, x2 = −3, x3 = 3, x4 = 4
v) x1 = −5, x2 = −2, x3 = 2, x4 = 5
w) x1 = −5, x2 = 5
x) x1 = −6, x2 = 6
y) x1 = −6, x2 = 5, x3 = 7
198
CHAPTER 9
...
Farlow and H
...
M
...
´
´
´ ´ ´
´
[2] E
...
Gyapjas, V
...
Koranyi,
¨
´
´
L
...
Pogats, I
...
Scharnitzky, Osszefoglal´
o
feladatgy˝jtem´ny matematik´b´l , Nemzeti Tank¨nyvkiad´, Buu
e
a o
o
o
dapest, 1993
...
Gheba, Matematika gyakorlatok ´s feladatok l´
e
ıceumi felv´teli
e
versenyvizsg´k el˝k´sz´ es´re, Editura didacticˇ ¸i pedagogicˇ, Bua
o e ıt´ e
a s
a
cure¸ti, 1973
...
Gillet, College Algebra, Scott, Foresman and Company, Glenview, Illinois and London, 1989
...
Goodman and H
...
, Understanding Intermediate Algebra, West
Publishing Company, St
...
[6] C
...
Musat, Exercitii ¸i probleme de matematicˇ
¸
¸
¸ s
a
pentru clasele IX-X , Editura didacticˇ ¸i pedagogicˇ, Bucure¸ti,
a s
a
s
1978
...
Lial and C
...
ˇ ˇ
[8] C
...
Nita, M
...
Joita, Culegere
¸ˇ
¸
de probleme de algebrˇ pentru clasele IX-X , Editura didacticˇ ¸i
a
a s
pedagogicˇ, Bucure¸ti, 1981
...
Petrica, C
...
Alexe, Probleme de Matematicˇ pen¸
a
tru Gimnaziu, Editura didacticˇ ¸i pedagogicˇ, Bucure¸ti, 1985
...
Rudin, Principles of Mathematical Analysis, McGraw-Hill,
1976
...
Smith, R
...
Keedy, M
...
Orfan,
Algebra and Trigonometry , Addison Wesley, 1988
...
Stamate and I
...
as
a
s
Contents
1
Real Numbers
3
1
...
3
1
...
4
1
...
5
1
...
9
1
...
11
1
...
13
1
...
16
1
...
18
1
...
1
The graphical approach of absolute value function
19
Exponentiation
...
9
...
19
1
...
2
Radicals
...
9
...
27
1
...
1
Introduction to algebraic expressions
...
2
Polynomials
...
2
...
2
...
34
2
...
3
Euclidean division of polynomials in one variable
35
2
...
4
2
...
Horner’s scheme
...
45
2
...
1
Factorization by factoring out the greatest common monomial
...
3
...
3
...
48
2
...
4
General strategy of factorization of a polynomial
49
2
...
5
Divisibility of polynomials
...
3
...
4
46
Roots of polynomials
...
53
2
...
1
Simplification and amplification of rational algebraic expressions
...
4
...
4
...
2
...
60
67
3
...
67
3
...
70
3
...
78
Inequalities
85
CONTENTS
203
4
...
85
4
...
88
4
...
90
4
...
1
The sign of linear expressions
...
3
...
93
Quadratic inequalities
...
4
...
4
4
...
Solving inequalities using table of signs
...
1
Equations containing absolute values
...
2
Polynomial equations of higher degree
...
2
...
116
5
...
2
Biquadratic equations
...
2
...
124
5
...
128
5
...
136
5
...
143
5
...
1
Logarithms
...
5
...
147
6
Systems of equations
159
7
Complex numbers
167
8
Exercises for the interested reader
171
204
CONTENTS
8
...
2
9
Identities of algebraic expressions
...
173
Results of the Exercises
177
9
...
177
9
...
179
9
...
192
9
...
195
9
...
197
Title: College Algebra
Description: The note is for 1st year students. College Algebra lecture note by Attila Berczes.
Description: The note is for 1st year students. College Algebra lecture note by Attila Berczes.