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EDEN UNIVERSITY
MATHEMATICAL METHODS
NSC 1310
R. DOZVA (2020)
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TOPIC 1
INTRODUCTION TO SET THEORY
The concept of a set appears in all mathematics. This topic introduces the notation and terminology of
set theory which is basic.
SETS AND ELEMENTS, SUBSETS
A set may be viewed as any well-defined collection of objects, called the elements or members of the
set.
One usually uses capital letters, 𝐴, 𝐵, 𝑋, 𝑌, . . ., to denote sets, and lowercase letters, 𝑎, 𝑏, 𝑥, 𝑦, . . ., to
denote elements of sets.
Synonyms for “set” are “class,” “collection,” and “family.”
Membership in a set is denoted as follows:
𝑎 ∈ 𝑆 denotes that a belongs to a set S
𝑎, 𝑏 ∈ 𝑆 denotes that a and b belong to a set S
Here ∈ is the symbol meaning “is an element of.” We use ∉ to mean “is not an element of.”
Subsets
Suppose every element in a set A is also an element of a set B, that is, suppose 𝑎 ∈ 𝐴 implies 𝑎 ∈ 𝐵.
Then A is called a subset of B. We also say that A is contained in B or that B contains A. This relationship
is written
𝐴 ⊆ 𝐵 𝑜𝑟 𝐵 ⊇ 𝐴
Two sets are equal if they both have the same elements or, equivalently, if each is contained in the
other. That is:
𝐴 = 𝐵 if and only if 𝐴 ⊆ 𝐵 and 𝐵 ⊆ 𝐴
If 𝐴 is not a subset of 𝐵, that is, if at least one element of A does not belong to B, we write 𝐴 ⊈ 𝐵.
Theorem 1: Let 𝐴, 𝐵, 𝐶 be any sets. Then:
(i)
(ii)
(iii)
𝐴 ⊆ 𝐴
If 𝐴 ⊆ 𝐵 and 𝐵 ⊆ 𝐴, then 𝐴 = 𝐵
If A ⊆ B and B ⊆ C, then A ⊆ C
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Universal Set, Empty Set
All sets under investigation in any application of set theory are assumed to belong to some fixed large
set called the universal set which we denote by 𝑈 unless otherwise stated.
Given a universal set 𝑈 and a property 𝑃, there may not be any elements of 𝑈 which have property 𝑃.
For example, the following set has no elements:
𝑆 = {𝑥 |𝑥 𝑖𝑠 𝑎 𝑝𝑜𝑠𝑖𝑡𝑖𝑣𝑒 𝑖𝑛𝑡𝑒𝑔𝑒𝑟, 𝑥 2 = 3}
A set with no elements is called the empty set or null set and is denoted by ∅.
Theorem 2: For any set 𝐴, we have ∅ ⊆ 𝐴 ⊆ 𝑈.
Disjoint Sets
Two sets 𝐴 and 𝐵 are said to be disjoint if they have no elements in common. For example, suppose
𝐴 = {1, 2}, 𝐵 = {4, 5, 6}, and 𝐶 = {5, 6, 7, 8}
Then A and B are disjoint, and A and C are disjoint. But B and C are not disjoint since B and C have
elements in common.
VENN DIAGRAMS
A Venn diagram is a pictorial representation of sets in which sets are represented by enclosed areas in
the plane. The universal set 𝑈 is represented by the interior of a rectangle, and the other sets are
represented by disks lying within the rectangle. If 𝐴 ⊆ 𝐵, then the disk representing 𝐴 will be entirely
within the disk representing 𝐵 as in Fig. 1-1(a). If 𝐴 and 𝐵 are disjoint, then the disk representing 𝐴 will
be separated from the disk representing 𝐵 as in Fig. 1-1(b)
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If 𝐴 and 𝐵 are two arbitrary sets, it is possible that some objects are in 𝐴 but not in 𝐵, some are
in 𝐵 but not in 𝐴, some are in both 𝐴 and 𝐵, and some are in neither 𝐴 nor 𝐵; hence in general we
represent 𝐴 and 𝐵 as in Fig. 1-1(c)
SET OPERATIONS
We introduce a number of set operations, including the basic operations of union, intersection, and
complement.
Union and Intersection
The union of two sets 𝐴 and 𝐵, denoted by 𝐴 ∪ 𝐵, is the set of all elements which belong to 𝐴 or to 𝐵;
that is,
𝐴 ∪ 𝐵 = {𝑥 | 𝑥 ∈ 𝐴 𝑜𝑟 𝑥 ∈ 𝐵}
Here “or” is used in the sense of and/or. Figure 1-2(a) is a Venn diagram in which 𝐴 ∪ 𝐵 is shaded.
The intersection of two sets A and B, denoted by A ∩ B, is the set of elements which belong to both A
and B; that is,
𝐴 ∩ 𝐵 = {𝑥 | 𝑥 ∈ 𝐴 𝑎𝑛𝑑 𝑥 ∈ 𝐵}
Figure 1-2(b) is a Venn diagram in which 𝐴 ∩ 𝐵 is shaded.
Fig 1-2
Recall that sets A and B are said to be disjoint or nonintersecting if they have no elements in common
or, using the definition of intersection, if 𝐴 ∩ 𝐵 = ∅, the empty set. Suppose
𝑆 = 𝐴 ∪ 𝐵 and 𝐴 ∩ 𝐵 = ∅
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Then 𝑆 is called the disjoint union of 𝐴 and 𝐵.
Example
Let 𝐴 = {1, 2, 3, 4}, 𝐵 = {3, 4, 5, 6, 7}, 𝐶 = {2, 3, 8, 9}. Then
𝐴 ∪ 𝐵 = {1, 2, 3, 4, 5, 6, 7}, 𝐴 ∪ 𝐶 = {1, 2, 3, 4, 8, 9}, 𝐵 ∪ 𝐶 = {2, 3, 4, 5, 6, 7, 8, 9},
𝐴 ∩ 𝐵 = {3, 4},
𝐴 ∩ 𝐶 = {2, 3},
𝐵 ∩ 𝐶 = {3}.
Example
Let 𝑈 be the set of students at a university, and let 𝑀 denote the set of male students and let 𝐹 denote
the set of female students. The U is the disjoint union of 𝑀 of 𝐹; that is,
𝑈 = 𝑀 ∪ 𝐹 𝑎𝑛𝑑 𝑀 ∩ 𝐹 = ∅
This comes from the fact that every student in U is either in M or in F, and clearly no student belongs to
both M and F, that is, M and F are disjoint.
The following properties of union and intersection should be noted.
Property 1: Every element 𝑥 in 𝐴 ∩ 𝐵 belongs to both 𝐴 and 𝐵; hence 𝑥 belongs to A and 𝑥 belongs to B.
Thus 𝐴 ∩ 𝐵 is a subset of A and of B; namely 𝐴 ∩ 𝐵 ⊆ 𝐴 and 𝐴 ∩ 𝐵 ⊆ 𝐵
Property 2: An element 𝑥 belongs to the union 𝐴 ∪ 𝐵 if 𝑥 belongs to 𝐴 or 𝑥 belongs to 𝐵; hence every
element in 𝐴 belongs to 𝐴 ∪ 𝐵, and every element in B belongs to 𝐴 ∪ 𝐵. That is, 𝐴 ⊆ 𝐴 ∪ 𝐵
and 𝐵 ⊆ 𝐴 ∪ 𝐵
We state the above results formally:
Theorem 3: For any sets A and B, we have:
(i)
𝐴 ∩ 𝐵 ⊆ 𝐴 ⊆ 𝐴 ∪ 𝐵 and (ii) 𝐴 ∩ 𝐵 ⊆ 𝐵 ⊆ 𝐴 ∪ 𝐵
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Complements, Differences, Symmetric Differences
Recall that all sets under consideration at a particular time are subsets of a fixed universal set 𝑈. The
absolute complement or, simply, complement of a set 𝐴, denoted by 𝐴𝐶 , is the set of elements which
belong to 𝑈 but which do not belong to 𝐴. That is,
𝐴𝐶 = {𝑥 | 𝑥 ∈ 𝑈, 𝑥 ∉ 𝐴}
The relative complement of a set 𝐵 with respect to a set 𝐴 or, simply, the difference of 𝐴 and 𝐵,
denoted by 𝐴\𝐵, is the set of elements which belong to 𝐴 but which do not belong to 𝐵; that is
𝐴\𝐵 = {𝑥 | 𝑥 ∈ 𝐴, 𝑥 ∉ 𝐵}
The set 𝐴\𝐵 is read “𝐴 minus 𝐵.” Fig. 1-3(b) is a Venn diagram in which 𝐴\𝐵 is shaded.
The symmetric difference of sets 𝐴 and 𝐵, denoted by 𝐴 ⊕ 𝐵, consists of those elements which belong
to 𝐴 or B but not to both. That is,
𝐴 ⊕ 𝐵 = (𝐴 ∪ 𝐵)\(𝐴 ∩ 𝐵) or 𝐴 ⊕ 𝐵 = (𝐴\𝐵) ∪ (𝐵\𝐴)
Figure 1-3(c) is a Venn diagram in which 𝐴 ⊕ 𝐵 is shaded.
Fig.1-3
Example
Suppose 𝑈 = 𝑁 = {1, 2, 3, . . . } is the universal set. Let 𝐴 = {1, 2, 3, 4}, 𝐵 = {3, 4, 5, 6, 7},
𝐶 = {2, 3, 8, 9}, 𝐸 = {2, 4, 6, . . . } (Here 𝐸 is the set of even integers.) Then:
𝐴𝐶 = {5, 6, 7, . . . },
𝐵𝐶 = {1, 2, 8, 9, 10, . . . },
𝐸 𝐶 = {1, 3, 5, 7, . . . }
That is, 𝐸 𝐶 is the set of odd positive integers.
Also:
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𝐴\𝐵 = {1, 2},
𝐵\𝐴 = {5, 6, 7},
𝐴\𝐶 = {1, 4},
𝐶\𝐴 = {8, 9},
𝐵\𝐶 = {4, 5, 6, 7},
𝐶\𝐵 = {2, 8, 9},
𝐴\𝐸 = {1, 3},
𝐸\𝐴 = {6, 8, 10, 12, . . . }.
Furthermore:
𝐴 ⊕ 𝐵 = (𝐴\𝐵) ∪ (𝐵\𝐴) = {1, 2, 5, 6, 7},
𝐵 ⊕ 𝐶 = {2, 4, 5, 6, 7, 8, 9},
𝐴 ⊕ 𝐶 = (𝐴\𝐶) ∪ (𝐵\𝐶) = {1, 4, 8, 9},
𝐴 ⊕ 𝐸 = {1, 3, 6, 8, 10, . . . }.
ALGEBRA OF SETS
Sets under the operations of union, intersection, and complement satisfy various laws (identities) which
are listed in Table 1-1. These laws are formally stated as:
Theorem 5: Sets satisfy the laws in Table 1-1.
Consider, for example, the proof of De Morgan’s Law 10(a):
(𝐴 ∪ 𝐵)𝐶 = {𝑥 |𝑥 ∉ (𝐴 𝑜𝑟 𝐵)} = {𝑥 |𝑥 ∉ 𝐴 𝑎𝑛𝑑 𝑥 ∉ 𝐵} = 𝐴𝐶 ∩ 𝐵𝐶
TUTORIAL 1
1. Which of the following sets are equal?
𝐴 = {𝑥 ∶ 𝑥 2 − 4𝑥 + 3 = 0},
𝐵 = {𝑥: 𝑥 2 − 3𝑥 + 2 = 0 },
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𝐶 = {𝑥: 𝑥 ∈ ℕ, 𝑥 < 3} , 𝐷 = {𝑥: 𝑥 ∈ ℕ, 𝑥 𝑖𝑠 𝑜𝑑𝑑, 𝑥 < 5}
𝐸 = {1,2},
𝐹 = {1,2,1},
𝐺 = {3,1},
𝐻 = {1,1,3}.
2. List the elements of each set where ℕ = {1, 2, 3, … }.
(a) 𝐴 = {𝑥 ∈ ℕ ∶ 3 < 𝑥 < 9}
(b) 𝐵 = {𝑥 ∈ ℕ: 𝑥 𝑖𝑠 𝑒𝑣𝑒𝑛, 𝑥 < 11}
(c) C = {𝑥 ∈ ℕ ∶ 4 + 𝑥 = 3}.
3. Prove that 𝐵\𝐴 = 𝐵 ∩ 𝐴𝐶 .
4. Prove the Distributive Law: 𝐴 ∩ (𝐵 ∪ 𝐶) = (𝐴 ∩ 𝐵) ∪ (𝐴 ∩ 𝐶).
5. Illustrate De Morgan’s Laws on a Venn diagram.
6. Prove Theorem 4. The following are equivalent: 𝐴 ⊆ 𝐵, 𝐴 ∩ 𝐵 = 𝐴, 𝐴 ∪ 𝐵 = 𝐵.
7. Prove: (𝐴 ∪ 𝐵)\(𝐴 ∩ 𝐵) = (𝐴\𝐵) ∪ (𝐵\𝐴)
Counting Elements in Finite Sets
The notation 𝑛(𝑆) or |𝑆| will denote the number of elements in a set 𝑆. As an example, if
𝑛(𝐴) = 26, where 𝐴 is the letters in the English alphabet and 𝑛(𝐷) = 7, where 𝐷 is the days of the
week. Also, 𝑛(∅) = 0 since the empty set has no elements.
The following lemma applies.
Lemma 1: Suppose 𝐴 and 𝐵 are finite disjoint sets. Then 𝐴 ∪ 𝐵 is finite and
𝑛(𝐴 ∪ 𝐵) = 𝑛(𝐴) + 𝑛(𝐵).
(1)
This lemma may be restated as follows:
Lemma 2: Suppose 𝑆 is the disjoint union of finite sets 𝐴 and 𝐵. Then 𝑆 is finite and
𝑛(𝑆) = 𝑛(𝐴) + 𝑛(𝐵)
( 2)
PROOF. In counting the elements of 𝐴 ∪ 𝐵, first count those that are in 𝐴. There are 𝑛(𝐴) of these. The
only other elements of 𝐴 ∪ 𝐵 are those that are in 𝐵 but not in 𝐴. But since 𝐴 and 𝐵are disjoint, no
element of 𝐵 is in 𝐴, so there are 𝑛(𝐵) elements that are in 𝐵 but not in 𝐴. Therefore,
𝑛(𝐴 ∪ 𝐵) = 𝑛(𝐴) + 𝑛(𝐵).
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CLASSES OF SETS OF SETS, POWER SETS
Given a set 𝑆, we might wish to talk about some of its subsets. Thus we would be considering a set of
sets.
POWER SETS
For a given set 𝑆, we may speak of the class of all subsets of 𝑆. This class is called the power set of 𝑆, and
will be denoted by 𝑃(𝑆). If 𝑆 is finite, then so is 𝑃(𝑆). In fact, the number of elements in 𝑃(𝑆) is 2 raised
to the power 𝑛(𝑆). That is
𝑛(𝑃(𝑆)) = 2𝑛(𝑆)
(For this reason, the power set of 𝑆 is sometimes denoted by 2𝑆 . )
Note that the empty set ∅ belongs to 𝑃(𝑆) since ∅ is a subset of 𝑆. Similarly, 𝑆 belongs to 𝑃(𝑆).
EXAMPLE
Suppose 𝑆 = {1, 2, 3}. Then
𝑃(𝑆) = {∅, {1}, {2}, {3}, {1, 2}, {1, 3}, {2, 3}, 𝑆}
8. Determine the power set 𝑃(𝐴) of 𝐴 = {𝑎, 𝑏, 𝑐, 𝑑}.
9. Find the power set 𝑃(𝐴) of 𝐴 = {1, 2, 3, 4, 5}.
10. Given 𝐴 = {{𝑎, 𝑏}, {𝑐}, {𝑑, 𝑒, 𝑓}}
(a) List the elements of 𝐴.
(b) Find 𝑛(𝐴).
(c) Find the power set of 𝐴.
11. Suppose ℕ = {1, 2, 3, … } is the universal set, and
𝐴 = {𝑛: 𝑛 ≤ 6}, 𝐵 = {𝑛: 4 ≤ 𝑛 ≤ 9}, 𝐶 = {1, 3, 5, 7, 9}, 𝐷 = {2, 3, 5, 7, 8}.
Find
(a) 𝐴 ⨁ 𝐵
(b) 𝐵 ⨁ 𝐶 (c) 𝐴 ∩ (𝐵⨁𝐷)
(d) (𝐴 ∩ 𝐵) ⨁(𝐴 ∩ 𝐷).
12. Let ℕ = {1, 2, 3, … } and, for each 𝑛 ∈ ℕ, let
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𝐴𝑛 = {𝑛, 2𝑛, 3𝑛, … }.
Find :
(a) 𝐴3 ∩ 𝐴5
(b) 𝐴4 ∩ 𝐴5
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TOPIC 2: POLYNOMIALS
A polynomial function, 𝑃(𝑥), is an algebraic expression of the form
𝑃(𝑥) = 𝑎𝑛 𝑥 𝑛 + 𝑎𝑛−1 𝑥 𝑛−1 + 𝑎𝑛−2 𝑥 𝑛−2 + ⋯ + 𝑎1 𝑥 + 𝑎0 ,
where
𝑎𝑛 ≠ 0
and 𝑎𝑛 , 𝑎𝑛−1 , 𝑎𝑛−2 , … , 𝑎1 , 𝑎0
are constants.
The degree of a polynomial, Deg 𝑃(𝑥) is the highest power of 𝑥 in the expression.
Some Standard Polynomials
Degree
Name
0
1
2
3
4
constant
linear
quadratic
cubic
quartic
General Form
𝑎
𝑎𝑥 + 𝑏
2
𝑎𝑥 + 𝑏𝑥 + 𝑐
𝑎𝑥 3 + 𝑏𝑥 2 + 𝑐𝑥 + 𝑑
𝑎𝑥 4 + 𝑏𝑥 3 + 𝑐𝑥 2 + 𝑑𝑥 + 𝑒
QUADRATIC EQUATIONS
A quadratic polynomial is a polynomial of the form 𝑎𝑥 2 + 𝑏𝑥 + 𝑐.
A quadratic equation takes the form 𝑎𝑥 2 + 𝑏𝑥 + 𝑐 = 0 where 𝑎, 𝑏, 𝑐 are constants.
SOLUTION METHODS FOR QUADRATIC EQUATIONS
Quadratic equations are solved by using any one of the following three methods
1. Factorization
2. Completing the square
3. Quadratic formula
QUADRATIC FORMULA AND DISCRIMINANT.
The quadratic formula is given by
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𝑥=
−𝑏 ± √(𝑏 2 − 4𝑎𝑐)
2𝑎
KEY POINT SUMMARY
1. When the quadratic equation 𝑎𝑥 2 + 𝑏𝑥 + 𝑐 = 0 has roots 𝛼 and 𝛽:
𝑏
The sum of the roots, 𝛼 + 𝛽 = − 𝑎 and the product of roots, 𝛼𝛽 =
𝑐
𝑎
.
2. A quadratic equation can be expressed as
𝑥 2 − (𝑠𝑢𝑚 𝑜𝑓 𝑟𝑜𝑜𝑡𝑠)𝑥 + (𝑝𝑟𝑜𝑑𝑢𝑐𝑡 𝑜𝑓 𝑟𝑜𝑜𝑡𝑠) = 0.
3. Some useful results are
𝛼 2 + 𝛽 2 = (𝛼 + 𝛽)2 − 2𝛼𝛽
𝛼 3 + 𝛽 3 = (𝛼 + 𝛽)3 − 3𝛼𝛽(𝛼 + 𝛽)
TUTORIAL EXERCISE
1. Is 𝑃(𝑥) = 3𝑥 3 − 𝑥 2 + 2√𝑥 − 1 a polynomial? Give a reason for your answer.
2. Solve the equation
𝑥−7 =
4
𝑥
giving the answer in exact form.
3. Find the sum and product of roots for
(i)
𝑥(3 − 𝑥) = 5𝑥 − 2
(ii)
𝑎𝑥 2 − 𝑎2 𝑥 − 2𝑎3 = 0
4
𝑥−3
(iii)
=
𝑥+5
2
4. Given that 𝛼 + 𝛽 = 5 and 𝛼𝛽 =
2
3
, find the value of (𝛼 − 𝛽)2 .
5. The roots of the equation 𝑥 2 − 4𝑥 + 3 = 0 are 𝛼 and 𝛽. Without solving the equation, find the
value of
(i)
1
𝛼
(ii)
(iii)
𝛼 2 + 𝛽2
(𝛼 + 1)(𝛽 + 1).
+
1
𝛽
6. The roots of the equation 𝑥 2 + 4𝑥 + 1 = 0 are 𝛼 and 𝛽. Find
(a) 𝛼 + 𝛽
(b) 𝛼𝛽
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Hence find the value of
(𝛼 2 − 𝛽)(𝛽 2 − 𝛼)
(i)
(ii)
𝛼+2
𝛽+2
+
𝛽+2
.
𝛼+2
POLYNOMIAL DIVISION
SYNTHETIC DIVISION
Suppose that the polynomial
𝐴(𝑥) = 𝑎3 𝑥 3 + 𝑎2 𝑥 2 + 𝑎1 𝑥 + 𝑎0
is divided by (𝑥 − 𝑘) so that it results in a quotient
𝑄(𝑥) = 𝑏2 𝑥 2 + 𝑏1 𝑥 + 𝑏0
and a constant remainder 𝑅. Then we can write this as
𝐴(𝑥)
𝑅
= 𝑄(𝑥) +
𝑥−𝑘
𝑥−𝑘
or
𝐴(𝑥) = (𝑥 − 𝑘)𝑄(𝑥) + 𝑅.
Then substituting polynomial terms, we have
𝑎3 𝑥 3 + 𝑎2 𝑥 2 + 𝑎1 𝑥 + 𝑎0 = (𝑥 − 𝑘)(𝑏2 𝑥 2 + 𝑏1 𝑥 + 𝑏0 ) + 𝑅.
Expanding and equating coefficients, it can be seen that
𝑏2 = 𝑎3
𝑏1 = 𝑎2 + 𝑘𝑎3
𝑏0 = 𝑎1 + 𝑘𝑏1
𝑅 = 𝑎0 + 𝑘𝑏0
This can be set up in the table as follows:
𝑎3
𝑘
𝑎3
Example: Divide 𝑥 3 + 2𝑥 2 − 3𝑥 + 4 by (𝑥 − 2).
𝑎2
+ 𝑘𝑎3
𝑏1
𝑎1
+ 𝑘𝑏1
𝑏0
𝑎0
+ 𝑘𝑏0
𝑅
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Solution.
The coefficients of the polynomial are 1
2
−3
4 and the value of 𝑘 = 2.
Setting up the table we have
1
2
+2
4
2
1
𝑥 3 +2𝑥 2 −3𝑥+4
𝑥−2
Thus
= 𝑥 2 + 4𝑥 + 5 +
−3
+8
5
4
+10
14
14
.
𝑥−2
NB: Although it has been shown that the process works for a polynomial of degree three, the method
works for any polynomial.
The method relies on the relationship between the coefficients of 𝑥 in the product of the quotient and
divisor.
EXAMPLE 2: Divide 3𝑥 5 − 8𝑥 4 + 𝑥 2 − 𝑥 + 3 by (𝑥 − 2).
SOLUTION
The coefficients: 3
−8
3
2
3
Thus
0
1
−1
−8
+6
−2
3𝑥 5 −8𝑥 4 + 𝑥 2 −𝑥 +3
𝑥−2
3 and 𝑘 = 2.
0
−4
−4
=
1
−8
−7
3𝑥 4 − 2𝑥 3 − 4𝑥 2 − 7𝑥 − 15 −
−1
−14
−15
27
.
𝑥−2
3
−30
−27
15 | P a g e
POLYNOMIAL THEOREMS
THEOREM: REMAINDER THEOREM
For any polynomial 𝑃(𝑥), the remainder when divided by (𝑥 − 𝑎) is 𝑃(𝑎).
EXAMPLE
Find the remainder when 𝑃(𝑥) = 𝑥 3 + 𝑥 2 − 𝑥 + 1 is divided by (𝑥 + 1)
SOLUTION
𝑃(𝑥) = 𝑥 3 + 𝑥 2 − 𝑥 + 1
𝑃(−1) = (−1)3 + (−1)2 − (−1) + 1
= 2.
∴ the remainder when 𝑃(𝑥) is divided by (𝑥 + 1) is 2.
EXAMPLE
Find the remainder when 𝑃(𝑥) = 3𝑥 3 − 2𝑥 2 + 7𝑥 − 4 is divided by (3𝑥 − 2).
SOLUTION
2
2 3
2 2
2
2
𝑅 = 𝑃( ) = 3( ) − 2( ) + 7( ) − 4 =
3
3
3
3
3
THEOREM: FACTOR THEOREM
If (𝑥 − 𝑎) is a factor of 𝑃(𝑥), then 𝑃(𝑎) = 0.
EXAMPLE
Determine 𝑚 and 𝑛 so that 3𝑥 3 + 𝑚𝑥 2 − 5𝑥 + 𝑛 is divisible by both (𝑥 − 2) and (𝑥 + 1). Factorize
the resulting polynomial completely.
16 | P a g e
SOLUTION
By the factor theorem, (𝑥 − 𝛼) is a factor of 𝑃(𝑥) if 𝑃(𝛼) = 0.
∴ 𝑃(2) = 24 + 4𝑚 − 10 + 𝑛
As 𝑃(2) = 0, we have that 24 + 4𝑚 − 10 + 𝑛 = 0
𝑖. 𝑒.
4𝑚 + 𝑛 = −14
(1)
𝑃(−1) = −3 + 𝑚 + 5 + 𝑛 = 0
∴ 𝑚 + 𝑛 = −2
(2)
Solving (1) and (2) simultaneously we have :
(1) − (2):
Substituting into (2) :
3𝑚 = −12
− 4 + 𝑛 = −2
∴ 𝑚 = −4
∴ 𝑛 =2
Hence 𝑃(𝑥) = 3𝑥 3 − 4𝑥 2 − 5𝑥 + 2.
As we already know (𝑥 + 1) and (𝑥 − 2) are factors of 𝑃(𝑥), we also have that (𝑥 + 1)(𝑥 − 2)
is a factor of 𝑃(𝑥). All that remains is to find the third linear factor. We assume a linear factor of
the form (𝑎𝑥 + 𝑏) and determine the value of 𝑎 and 𝑏.
𝑃(𝑥) = (𝑥 + 1)(𝑥 − 2)(𝑎𝑥 + 𝑏).
Expanding and equating coefficients we have 𝑎 = 3, 𝑏 = −1
∴
3𝑥 3 − 4𝑥 2 − 5𝑥 + 2
= (𝑥 + 1)(𝑥 − 2)(3𝑥 − 1).
7. Determine 𝑚 and 𝑛 so that 3𝑥 3 + 𝑚𝑥 2 − 5𝑥 + 𝑛 is divisible by both (𝑥 − 2) and (𝑥 + 1).
Factorize the resulting polynomial completely.
𝑚 3
𝑛 2
8. Prove that if 𝑥 3 + 𝑚𝑥 + 𝑛 is divisible by (𝑥 − 𝑘)2 , then ( 3 ) + (2 ) = 0.
9. The expression 𝑥 3 − 𝛼𝑥 2 − 𝑥 + 𝛽 is exactly divisible by 𝑥 − 3 and on division by 𝑥 − 2
gives a remainder of 24. Calculate the values of 𝛼 and 𝛽.
10. Divide 𝑃(𝑥) = 𝑥 3 − 4𝑥 2 + 5𝑥 − 1 by (𝑥 − 2).
17 | P a g e
11. Divide 2𝑥 3 + 5𝑥 2 − 13 by 2𝑥 2 + 𝑥 − 2.
12. 𝑃(𝑥) = 𝑥 3 + 𝑚𝑥 2 + 𝑛𝑥 + 𝑘 is divisible by 𝑥 2 − 4 and leaves a remainder of 30 when divided
by (𝑥 − 3). Solve for 𝑚, 𝑛 and 𝑘 and hence fully factorize 𝑃(𝑥) into its three linear factors.
13. Use synthetic division to find the quotients and remainders below
(i)
(ii)
2𝑥 4 + 3𝑥 2 − 𝑥 ÷ (𝑥 + 2)
2𝑥 3 − 5𝑥 2 + 10𝑥 − 3 ÷ (2𝑥 − 1)
14. Prove that if 𝑥 3 + 𝑚𝑥 + 𝑛 and 3𝑥 2 + 𝑚 have a common factor (𝑥 − 𝑘) then
4𝑚3 + 27𝑛2 = 0.
15. The remainders when 𝑇(𝑥) = 𝑘𝑥 𝑛 − 3𝑥 2 + 6 is divided by (𝑥 − 1) and 𝑥 + 2) are 1 and 10
respectively. Find 𝑘 and 𝑛.
16. 𝑃(𝑥) = 𝑥 3 + 𝑚𝑥 2 + 𝑛𝑥 + 𝑘 is divisible by 𝑥 2 − 4 and leaves a remainder of 30 when divided
by (𝑥 − 3). Solve for 𝑚, 𝑛 and 𝑘 and hence fully factorize 𝑃(𝑥) into its three linear factors.
18 | P a g e
TOPIC 3
EQUATIONS
Polynomial Equations
The factor theorem has some very useful consequences, one of which allows us to solve equations of
the form 𝑃(𝑥) = 0.
The expression
𝑃(𝑥) = 0 where 𝑃(𝑥) = 𝑎𝑛 𝑥 𝑛 + 𝑎𝑛−1 𝑥 𝑛−1 + ⋯ + 𝑎1 𝑥 + 𝑎0 is called a polynomial equation.
The roots or solutions of this equation are the zeros of 𝑃(𝑥).
To solve a polynomial equation, the factor theorem is used alongside the three basic steps
1. Factorize the polynomial using the factor theorem (if necessary)
2. Use the null factor law
3. Solve for the unknown.
EXAMPLE
Solve 𝟐𝒙𝟑 + 𝟗𝒙𝟐 + 𝟕𝒙 − 𝟔 = 𝟎.
SOLUTION
Let 𝑷(𝒙) = 𝟐𝒙𝟑 + 𝟗𝒙𝟐 + 𝟕𝒙 − 𝟔.
Then, 𝑷(𝟏) = 𝟐 + 𝟗 + 𝟕 − 𝟔 ≠ 𝟎
𝑷(−𝟏) = −𝟐 + 𝟗 − 𝟕 − 𝟔 ≠ 𝟎
𝑷(−𝟐) = −𝟏𝟔 + 𝟑𝟔 − 𝟏𝟒 − 𝟔 = 𝟎. Therefore, 𝒙 + 𝟐 is a factor of 𝑷(𝒙).
Using synthetic division we have
𝟐
-2
𝟐
𝟗
−𝟒
𝟓
𝟕
−𝟏𝟎
−𝟑
−𝟔
𝟔
𝟎
19 | P a g e
Therefore, (𝒙) = (𝒙 + 𝟐)(𝟐𝒙𝟐 + 𝟓𝒙 − 𝟑) = (𝒙 + 𝟐)(𝟐𝒙 − 𝟏)(𝒙 + 𝟑).
Then 𝑷(𝒙) = 𝟎 ⟺ (𝒙 + 𝟐)(𝟐𝒙 − 𝟏)(𝒙 + 𝟑) = 𝟎
⟺ 𝒙 + 𝟐 = 𝟎 or 𝟐𝒙 − 𝟏 = 𝟎 or 𝒙 + 𝟑 = 𝟎
⟺ 𝒙 = −𝟐 or 𝒙 =
𝟏
𝟐
or 𝒙 = −𝟑.
TUTORIAL EXERCISE
Solve the following over the real number field
1.
2.
3.
4.
5.
𝑥 3 + 2𝑥 2 − 5𝑥 − 6 = 0
2𝑥 3 − 5𝑥 2 + 𝑥 + 2 = 0
2𝑥 3 + 9𝑥 2 + 8𝑥 + 2 = 0
Write down the equation of a polynomial with zeroes −2, 3 and −4.
Solve the equation
𝑥2 + 7 = 𝑥 −
1
8
+ 2.
𝑥 𝑥
6. Solve 2𝑥 3 + 𝑘𝑥 2 − 11𝑥 − 6 = 0 given that one solution is 𝑥 = −3.
7. Two solutions to the equation 2𝑥 4 + 𝑎𝑥 3 + 𝑥 2 + 6𝑥 + 𝑏 = 0 are 𝑥 = −3 and 𝑥 = 2. Find the
other two solutions, if they exist.
SIMULTANEOUS EQUATIONS
Solutions of Three Linear Equations
For three unknown quantities we need three connections (three equations). Then one unknown at a
time can be eliminated. One way to eliminate an unknown quantity is to add or subtract two of the
equations and then go on to eliminate the second unknown in a similar way.
EXAMPLE
Solve the equations 𝒙 + 𝒚 − 𝒛 = 𝟒
𝟐𝒙 + 𝒛 = 𝟕
𝟑𝒙 − 𝟐𝒚 = 𝟓
20 | P a g e
SOLUTION
𝒙+𝒚−𝒛 = 𝟒
𝟐𝒙 + 𝒛
𝟑𝒙 − 𝟐𝒚
=𝟕
=𝟓
(𝟏)
(𝟐)
(𝟑)
As 𝒛 appears only in equations (𝟏) and (𝟐) we can eliminate 𝒛 from these two equations in this
case by adding.
(𝟏) + (𝟐) gives
𝟑𝒙 + 𝒚 = 𝟏𝟏
(𝟒)
𝟑𝒙 − 𝟐𝒚 = 𝟓
(𝟑)
Now bring in (𝟑)
Subtracting (𝟑) from (𝟒) gives
𝟑𝒚 = 𝟔
∴ 𝒚 = 𝟐.
Using 𝒚 = 𝟐 in (𝟑),
𝟑𝒙 − 𝟒 = 𝟓
∴ 𝒙 = 𝟑 and using 𝒙 = 𝟑 in (𝟐)
𝟔+𝒛 = 𝟕
∴ 𝒛 = 𝟏.
Therefore the solution of the three simultaneous equations is 𝒙 = 𝟑, 𝒚 = 𝟐, 𝒛 = 𝟏.
It is not always easy to eliminate the first of the unknown quantities. If all three unknowns occur in all
three equations it is necessary to eliminate the same unknown from each of two different pairs of
equations.
EXAMPLE
Solve the equations
𝒙 − 𝒚 + 𝟐𝒛 = 𝟔
𝟐𝒙 + 𝒚 + 𝒛 = 𝟑
𝟑𝒙 − 𝒚 + 𝒛 = 𝟔
21 | P a g e
SOLUTION
𝒙 − 𝒚 + 𝟐𝒛 = 𝟔
(𝟏)
𝟐𝒙 + 𝒚 + 𝒛 = 𝟑
(𝟐)
𝟑𝒙 − 𝒚 + 𝒛
=𝟔
(𝟑)
The easiest letter to eliminate from two pairs of equations is 𝒚.
(𝟏) + (𝟐) gives
𝟑𝒙 + 𝟑𝒛 = 𝟗
Dividing by 𝟑 gives
Adding (𝟐) + (𝟑) gives
𝒙+𝒛=𝟑
(𝟒)
𝟓𝒙 + 𝟐𝒛 = 𝟗
(𝟓)
Now eliminate either 𝒙 or 𝒛 from (𝟒) and (𝟓) leads to
𝟑𝒛 = 𝟔
Using 𝒛 = 𝟐 in (𝟒) gives 𝒙 + 𝟐 = 𝟑
∴ 𝒛 = 𝟐.
⟹ 𝒙 = 𝟏.
Using 𝒙 = 𝟏 and 𝒛 = 𝟐 in (𝟐) gives 𝒚 = −𝟏.
Therefore the solution is 𝒙 = 𝟏, 𝒚 = −𝟏, 𝒛 = 𝟐.
TUTORIAL
Solve the following sets of equations
1. Solve
2. Solve the system
𝑥 + 2𝑦 = 4
𝑥 + 3𝑧 = 5
2𝑦 − 𝑧 = 1
22 | P a g e
𝑥+𝑦+𝑧 =6
2𝑥 − 𝑦 = 3
4𝑥 − 𝑧 = 2
2𝑥 − 3𝑦 + 𝑧 = 13
𝑥 + 𝑦 − 2𝑧 = −1
𝑥 + 2𝑦 − 3𝑧 = −4.
3. Solve the system
SIMULTANEOUS EQUATIONS INVOLVING LINEAR-QUADRATIC EQUATIONS
To solve a linear-quadratic system of equations it is often the case that the method of
substitution is most appropriate. The process is as follows:
STEP 1: Arrange the equations so that they are both in the form 𝑦 = ⋯ (𝑦 is expressed explicitly
in terms of 𝑥. )
STEP 2: Label the two equations
𝑦 = 𝑎𝑥 2 + 𝑏𝑥 + 𝑐
𝑦 = 𝑚𝑥 + 𝑘
STEP 3: Equate (1) and (2) :
(1)
(2)
𝑚𝑥 + 𝑘 = 𝑎𝑥 2 + 𝑏𝑥 + 𝑐
STEP 4: Transpose to obtain a new quadratic equation 𝑎𝑥 2 + (𝑏 − 𝑚)𝑥 + (𝑐 − 𝑘) = 0.
STEP 5: Solve for 𝑥 and then find 𝑦 by substituting into (1) and (2)
EXAMPLE
Solve the simultaneous system of equations
𝑦 = 𝑥 2 + 3𝑥 − 6
𝑦 = 2𝑥 − 4
SOLUTION
We label the equations as follows;
𝑦 = 𝑥 2 + 3𝑥 − 6
𝑦 = 2𝑥 − 4
Equating (1) and (2) gives
(1)
(2)
23 | P a g e
2𝑥 − 4 = 𝑥 2 + 3𝑥 − 6
0 = 𝑥2 + 𝑥 − 2
⟺ (𝑥 + 2)(𝑥 − 1) = 0
⟺ 𝑥 = −2, 1
Substituting 𝑥 = −2 into (2):
𝑦 = 2(−2) − 4 = −8
𝑥 = 1 into (2): 𝑦 = 2(1) − 4 = −2.
The solution can be expressed as (−2, −8), (1, −2).
To solve a quadratic-quadratic system of equations we use the same method of substitution that was
used for the linear-quadratic set-up.
TUTORIAL
1.
Solve the following pairs of simultaneous equations
(i)
𝑦 = 2𝑥 + 1
(ii) 𝑦 = 𝑥 + 1
2
𝑦 = 𝑥 + 2𝑥 − 3
𝑦 = 𝑥 2 + 2𝑥 − 1
(ii)
3𝑥 − 2𝑦 = 3
𝑦 = 𝑥 2 + 2𝑥 − 3
(iv)
𝑥 + 2𝑦 = 3
𝑦 = −𝑥 2 + 2𝑥 − 3
2. Solve the following simultaneous equations:
(i)
𝑦 = 𝑥 2 − 4𝑥 + 7
𝑦 = 2𝑥 2 + 2𝑥
(ii)
𝑦 = 𝑥 2 + 4𝑥 + 3
𝑦 = 4𝑥 − 𝑥 2
(iii)
𝑦 = 5𝑥 2 + 𝑥 + 4
𝑦 = 𝑥 2 + 5𝑥 + 3
(iv)
Solve the system of equations
where 𝑎 is a real number.
𝑦 = 4𝑥 2 + 3𝑎𝑥 − 2𝑎2
𝑦 = 2𝑥 2 + 2𝑎𝑥 − 𝑎2
24 | P a g e
3. Find the value of 𝑎 such that the line 𝑦 = 2𝑥 + 𝑎 has exactly one intersection point with the
parabola with equation
𝑦 = 𝑥 2 + 3𝑥 + 2.
4. If 𝑓(𝑥) = 4𝑥 2 − 20𝑥 − 4 intersects with 𝑔(𝑥) = 1 + 4𝑥 − 𝑥 2 at the point (𝑚, 𝑘), show that the
other point of intersection occurs where
1
𝑥= − .
𝑚
5. Show that the equations
𝑦 = 𝑘𝑥 − 1
𝑦 = 𝑥 2 + (𝑘 − 1)𝑥 + 𝑘 2
have no real solution for all values of 𝑘.
25 | P a g e
TOPIC 3
PARTIAL FRACTIONS
Let 𝑵(𝒙) ≠ 𝟎 and 𝑫(𝒙) ≠ 𝟎 be two polynomials. Then
𝑵(𝒙)
𝑫(𝒙)
is called a PROPER fraction if the degree of 𝑵(𝒙) is smaller than the degree of 𝑫(𝒙).
𝑵(𝒙)
Also 𝑫(𝒙) is called IMPROPER if the degree of 𝑵(𝒙) is greater than or equal to the degree of 𝑫(𝒙).
In such a case, we divide 𝑵(𝒙) by 𝑫(𝒙) obtaining a quotient 𝑸(𝒙) and a remainder 𝑹(𝒙) whose
degree is smaller than that of 𝑫(𝒙) i.e
𝑵(𝒙)
𝑫(𝒙)
= 𝑸(𝒙) +
𝑹(𝒙)
𝑫(𝒙)
.
Types of Proper Fractions into Partial Fractions
TYPE 𝟏
Linear and distinct factors in 𝑫(𝒙)
𝒂𝒙 + 𝒃
𝑨
𝑩
=
+
(𝒑𝒙 + 𝒒)(𝒓𝒙 + 𝒔)
𝒑𝒙 + 𝒒
𝒓𝒙 + 𝒔
TYPE 2
Repeated Linear Factors
𝒂𝒙𝟐 +𝒃𝒙+𝒄
(𝒑𝒙+𝒒)(𝒓𝒙+𝒔)𝟐
=
𝑨
𝒑𝒙+𝒒
+
𝑩
𝒓𝒙+𝒔
+
𝑪
(𝒓𝒙+𝒔)𝟐
TYPE 3
Quadratic Factor in the Denominator
𝒂𝒙𝟐 +𝒃𝒙+𝒄
(𝒑𝒙+𝒒)(𝒓𝒙𝟐 +𝒔)
=
𝑨
𝒑𝒙+𝒒
+
𝑩𝒙+𝑪
𝒓𝒙𝟐 +𝒔
26 | P a g e
TUTORIAL
1. What is a partial fraction?
2. Define a proper fraction and give an example.
3. Given that
𝒂
𝒙−𝟐
+
𝒃
𝒙+𝒄
=
𝟏𝟓𝒙
,
𝒙𝟐 +𝟐𝒙−𝟖
find the values of the constants 𝒂, 𝒃 and 𝒄.
4. Split
𝟏𝟐𝒙
(𝒙+𝟏)(𝟐𝒙+𝟑)(𝒙−𝟑)
into partial fractions.
5. Split
𝟖𝒙𝟐
(𝒙+𝟏)(𝒙−𝟏)(𝒙+𝟑)
into partial fractions.
6. Decompose
𝟒𝒂𝒙− 𝒂𝟐
𝒙𝟐 +𝒂𝒙−𝟐𝒂𝟐
into partial fractions.
7. Show that
𝒙𝟐 −𝟕𝒙−𝟔
𝒙𝟐 (𝒙−𝟑)
≡
𝟐
𝒙𝟐
+
𝟑
𝒙
−
8. Express
𝟗+𝟒𝒙𝟐
(𝟏−𝟐𝒙)𝟐 (𝟐+𝒙)
in partial fractions.
9. Resolve the fraction
𝒙
(𝒙−𝒂)(𝒙−𝒃)(𝒙−𝒄)
into partial fractions.
10. Show that
𝟏
𝒙𝟒 (𝒙+𝟏)
𝟏
𝟏
= − 𝒙 + 𝒙𝟐 −
𝟏
𝒙𝟑
+
𝟏
𝒙𝟒
+
𝟏
𝒙+𝟏
𝟐
𝒙−𝟑
27 | P a g e
IMPROPER FRACTIONS
Example:
𝟐𝒙𝟐 +𝟒𝒙−𝟑
(𝒙+𝟏)(𝟐𝒙−𝟑)
Split
into partial fractions
Solution
𝟐𝒙𝟐 +𝟒𝒙−𝟑
(𝒙+𝟏)(𝟐𝒙−𝟑)
Write
Multiplying by 𝒙 + 𝟏 gives
=𝑨+
𝟐𝒙𝟐 +𝟒𝒙−𝟑
𝟐𝒙−𝟑
𝑩
𝑪
+
𝒙+𝟏
𝟐𝒙−𝟑
≡ 𝑨(𝒙 + 𝟏) + 𝑩 +
𝑪(𝒙+𝟏)
.
𝟐𝒙−𝟑
Putting 𝒙 = −𝟏, gives
𝑩 = 𝟏.
Multiplying by 𝟐𝒙 − 𝟑 gives
𝟐𝒙𝟐 +𝟒𝒙−𝟑
𝒙+𝟏
Putting 𝒙 =
𝟑
𝟐
≡ 𝑨(𝟐𝒙 − 𝟑) +
𝑩(𝟐𝒙−𝟑)
+
𝒙+𝟏
𝑪.
𝟐𝒙𝟐 +𝟒𝒙−𝟑
gives 𝑪 = 𝟑. Therefore (𝒙+𝟏)(𝟐𝒙−𝟑) ≡ 𝑨 +
𝟏
𝟑
+ 𝟐𝒙−𝟑 .
𝒙+𝟏
At this point, any other value of 𝒙 can be used to find 𝑨. The simplest is 𝒙 = 𝟎. This gives
𝑨 = 𝟏. (𝒄𝒉𝒆𝒄𝒌)
Therefore
11. Split 𝒇(𝒙) =
𝟒𝒙𝟑 +𝟏𝟎𝒙𝟐 +𝟖𝒙−𝟏
(𝟐𝒙+𝟏)𝟐 (𝒙+𝟐)
𝟐𝒙𝟐 +𝟒𝒙−𝟑
(𝒙+𝟏)(𝟐𝒙−𝟑)
into partial fractions.
𝟔𝒙𝟑 +𝒙+ 𝟏𝟎
12. Split (𝒙−𝟐)(𝒙+𝟐)(𝟐𝒙−𝟏) into partial fractions.
13. Split
𝟖𝒙𝟑 −𝟏𝟐𝒙𝟐 −𝟏𝟖𝒙+𝟏𝟓
into
(𝟒𝒙𝟐 −𝟗)(𝟐𝒙−𝟑)
=𝟏+
partial fractions.
𝟏
𝒙+𝟏
+
𝟑
𝟐𝒙−𝟑
28 | P a g e
TOPIC 4 INEQUALITIES
Definition:
Quadratic inequalities in one variable are inequalities which can be written in one of the following
forms:
𝒂𝒙𝟐 + 𝒃𝒙 + 𝒄 < 𝟎
𝒂𝒙𝟐 + 𝒃𝒙 + 𝒄 > 𝟎
𝒂𝒙𝟐 + 𝒃𝒙 + 𝒄 ≤ 𝟎
𝒂𝒙𝟐 + 𝒃𝒙 + 𝒄 ≥ 𝟎
where a, b and c are real numbers.
Procedure
Solving Quadratic Inequalities
1.
2.
3.
4.
5.
Example 1
Solution
Move all terms to one side.
Simplify and factor the quadratic expression.
Find the roots of the corresponding quadratic equation.
Use the roots to divide the number line into regions.
Test each region using the inequality.
Solve the inequality, x2 > x +2.
x2 > x + 2
x2 − x −2 > 0
(x – 2)(x + 1) > 0
The corresponding equation is (x – 2)(x + 1) = 0 so…
x-2=0
x=2
or
x+1=0
x = -1
29 | P a g e
I
-7
-6
-5
-4
-3
II
-2
-1
0
III
1
2
3
4
5
6
7
Now we test one point in each region.
Region
I
II
III
-7
-6
Test Point
Inequality
Status
x = -2 (x - 2)(x + 1) = (-2 - 2)(-2 + 1) = 4 > 0
True
x=0
(x - 2)(x + 1) = (0 - 2)(0 + 1) = -2 > 0 False
x = 3 (x - 2)(x + 1) = (3 - 2)(3 + 1) = 4 > 0
True
-5
-4
-3
-2
-1
0
1
2
3
4
5
6
So the solution to this inequality is x < -1 or x > 2.
Example 2
Solution
Solve the inequality, (x+ 3)2 ≥ 2(x2 + 7).
(x+ 3)2 ≥ 2(x2 + 7)
x2 +6x + 9 ≥ 2x2 +14
−x2 + 6x −5 ≥ 0
−(x2 −6x+5) ≥ 0
x2 −6x +5 ≤ 0
(x - 1)(x - 5) ≤ 0
The corresponding equation is (x - 1)(x - 5) = 0 so…
x - 1 = 0 or x - 5 = 0
x = 1 or x =5
7
30 | P a g e
I
-7
-6
-5
-4
-3
-2
-1
II
0
1
2
3
III
4
5
6
7
Now we check one point in each region.
Region
I
II
III
Test Point
x=0
x=2
x=6
-7
Inequality
(x - 1)(x - 5) = (0 - 1)(0 - 5) = 5 < 0
(x - 1)(x - 5) = (2 - 1)(2 - 5) = -3 < 0
(x - 1)(x - 5) = (6 - 1)(6 - 5) = 5 < 0
-6
-5
-4
-3
-2
-1
0
1
2
3
4
So the solution to this inequality is 1 ≤ x ≤ 5.
Status
False
True
False
5
6
7