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SCHAUM’S OUTLINE OF
Mathematical Methods for Business and Economics
Schaum’s Outline of Mathematical Methods for Business
and Economics
Edward T. Dowling, Ph.D.
Chair and Professor
Department of Economics
Fordham University
Schaum’s Outline Series
Copyright © 1993 by The McGraw-Hill Companies, Inc. All rights reserved. Except as permitted
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Preface
Students of undergraduate business and economics and candidates for the M.B.A. and M.A. degrees
in economics today need a variety of mathematical skills to successfully complete their degree
requirements and compete effectively in their chosen careers. Unfortunately, the requisite
mathematical competence is not the subject of a single course in mathematics such as Calculus I or
Linear Algebra I, and many students, pressed with the demands from business and economics courses,
do not have space in their schedules for a series of math courses. Mathematical Methods for Business
and Economics is designed to cull the mathematical tools, topics, and techniques essential for success
in business and economics today. It is suitable for a one- or two-semester course in business
mathematics, depending on the previous background of the students. It can also be used profitably in
an introductory calculus or linear algebra course by professors and students interested in the business
connections and applications.
The theory-and-solved-problem format of each chapter provides concise explanations illustrated by
examples, plus numerous problems with fully worked-out solutions. No mathematical proficiency
beyond the high-school level is assumed. The learning-by-doing pedagogy will enable students to
progress at their own rate and adapt the book to their own needs.
Mathematical Methods for Business and Economics can be used by itself or as a supplement to
other texts for undergraduate and graduate students in business and economics. It is largely selfcontained. Starting with a basic review of high-school algebra in Chapter 1, the book consistently
explains all the concepts and techniques needed for the material in subsequent chapters.
This book contains 1,066 problems, all of them solved in considerable detail. To derive the most
from the book, students should strive as soon as possible to work independently of the solutions. This
can be done by solving problems on individual sheets of paper with the book closed. If difficulties
arise, the solution can then be checked in the book.
For best results, students should never be satisfied with passive knowledge—the capacity merely to
follow or comprehend the various steps presented in the book. Mastery of the subject and doing well
on exams require active knowledge—the ability to solve any problem, in any order, without the aid of
the book.
Experience has proved that students of very different backgrounds and abilities can be successful in
handling the subject matter of this text when the material is presented in the current format.
I wish to express my gratitude for help with this book to Dr. Dominick Salvatore and Dr. Timothy
Weithers of Fordham University; Maria Cristina Cacdac-Ampil, a doctoral candidate at Fordham;
Professor Henry Mark Smith, the reviewer; and John Carleo, John Aliano, Maureen Walker, Pat Koch,
and Patty Andrews of McGraw-Hill.
EDWARD T. DOWLTNG
Contents
Chapter 1 REVIEW
1.1 Exponents
1.2 Polynomials
1.3 Factoring
1.4 Fractions
1.5 Radicals
1.6 Order of Mathematical Operations
1.7 Use of a Pocket Calculator
Chapter 2 EQUATIONS AND GRAPHS
2.1 Equations
2.2 Cartesian Coordinate System
2.3 Linear Equations and Graphs
2.4 Slopes
2.5 Intercepts
2.6 The Slope-Intercept Form
2.7 Determining the Equation of a Straight-Line
2.8 Applications of Linear Equations in Business and Economics
Chapter 3 FUNCTIONS
3.1 Concepts and Definitions
3.2 Graphing Functions
3.3 The Algebra of Functions
3.4 Applications of Linear Functions for Business and Economics
3.5 Solving Quadratic Equations
3.6 Facilitating Nonlinear Graphing
3.7 Applications of Nonlinear Functions in Business and Economics
Chapter 4 SYSTEMS OF EQUATIONS
4.1 Introduction
4.2 Graphical Solutions
4.3 Supply-and-Demand Analysis
4.4 Break-Even Analysis
4.5 Elimination and Substitution Methods
4.6 Income Determination Models
4.7 IS-LM Analysis
4.8 Economic and Mathematical Modeling (Optional)
4.9 Implicit Functions and Inverse Functions (Optional)
Chapter 5 LINEAR (OR MATRIX) ALGEBRA
5.1 Introduction
5.2 Definitions and Terms
5.3 Addition and Subtraction of Matrices
5.4 Scalar Multiplication
5.5 Vector Multiplication
5.6 Multiplication of Matrices
5.7 Matrix Expression of a System of Linear Equations
5.8 Augmented Matrix
5.9 Row Operations
5.10 Gaussian Method of Solving Linear Equations
Chapter 6 SOLVING LINEAR EQUATIONS WITH MATRIX ALGEBRA
6.1 Determinants and Linear Independence
6.2 Third-Order Determinants
6.3 Cramer’s Rule for Solving Linear Equations
6.4 Inverse Matrices
6.5 Gaussian Method of Finding an Inverse Matrix
6.6 Solving Linear Equations with an Inverse Matrix
6.7 Business and Economic Applications
6.8 Special Determinants
Chapter 7 LINEAR PROGRAMMING: USING GRAPHS
7.1 Use of Graphs
7.2 Maximization Using Graphs
7.3 The Extreme-Point Theorem
7.4 Minimization Using Graphs
7.5 Slack and Surplus Variables
7.6 The Basis Theorem
Chapter 8 LINEAR PROGRAMMING: THE SIMPLEX ALGORITHM AND THE DUAL
8.1 The Simplex Algorithm
8.2 Maximization
8.3 Marginal Value or Shadow Pricing
8.4 Minimization
8.5 The Dual
8.6 Rules of Transformation to Obtain the Dual
8.7 The Dual Theorems
8.8 Shadow Prices in the Dual
8.9 Integer Programming
8.10 Zero-One Programming
Chapter 9 DIFFERENTIAL CALCULUS: THE DERIVATIVE AND THE RULES OF
DIFFERENTIATION
9.1 Limits
9.2 Continuity
9.3 The Slope of a Curvilinear Function
9.4 The Derivative
9.5 Differentiability and Continuity
9.6 Derivative Notation
9.7 Rules of Differentiation
9.8 Higher-Order Derivatives
9.9 Implicit Functions
Chapter 10 DIFFERENTIAL CALCULUS: USES OF THE DERIVATIVE
10.1 Increasing and Decreasing Functions
10.2 Concavity and Convexity
10.3 Relative Extrema
10.4 Inflection Points
10.5 Curve Sketching
10.6 Optimization of Functions
10.7 The Successive-Derivative Test
10.8 Marginal Concepts in Economics
10.9 Optimizing Economic Functions for Business
10.10 Relationship Among Total, Marginal, and Average Functions
Chapter 11 EXPONENTIAL AND LOGARITHMIC FUNCTIONS
11.1 Exponential Functions
11.2 Logarithmic Functions
11.3 Properties of Exponents and Logarithms
11.4 Natural Exponential and Logarithmic Functions
11.5 Solving Natural Exponential and Logarithmic Functions
11.6 Logarithmic Transformation of Nonlinear Functions
11.7 Derivatives of Natural Exponential and Logarithmic Functions
11.8 Interest Compounding
11.9 Estimating Growth Rates from Data Points
Chapter 12 INTEGRAL CALCULUS
12.1 Integration
12.2 Rules for Indefinite Integrals
12.3 Area under a Curve
12.4 The Definite Integral
12.5 The Fundamental Theorem of Calculus
12.6 Properties of Definite Integrals
12.7 Area between Curves
12.8 Integration by Substitution
12.9 Integration by Parts
12.10 Present Value of a Cash Flow
12.11 Consumers’ and Producers’ Surplus
Chapter 13 CALCULUS OF MULTIVARIABLE FUNCTIONS
13.1 Functions of Several Independent Variables
13.2 Partial Derivatives
13.3 Rules of Partial Differentiation
13.4 Second-Order Partial Derivatives
13.5 Optimization of Multivariable Functions
13.6 Constrained Optimization with Lagrange Multipliers
13.7 Income Determination Multipliers
13.8 Optimizing Multivariable Functions in Business and Economics
13.9 Constrained Optimization of Multivariable Economic Functions
13.10 Constrained Optimization of Cobb-Douglas Production Functions
13.11 Implicit and Inverse Function Rules (Optional)
INDEX
Chapter 1
Review
1.1 EXPONENTS
Given a positive integer n, xn signifies that x is multiplied by itself n number of times. Here x is
referred to as the base; n is called an exponent. By convention an exponent of one is not expressed:
x(1) = x, 8(1) = 8. By definition any nonzero number or variable raised to the zero power is equal to 1:
x0 = 1, 30 = 1, and 00 is undefined. Assuming a and b are positive integers and x and y are real
numbers for which the following exist, the rules of exponents are presented below, illustrated in
Examples 1 and 2, and treated in Problems 1.1, 1.24, 1.26, and 1.27.
EXAMPLE 1. In multiplication, exponents of the same variable are added; in division, exponents of
the same variable are subtracted; when raised to a power, the exponents of a variable are multiplied, as
indicated by the rules above and shown in the examples below followed by illustrations.
(a) x2 · x5 = x2+5 = x7 ≠ x10
(Rule 1)
x2 · x5 = (x · x)(x · x · x · x · x) = x7
(Rule 2)
(c) (x3)2 = x3·2 = x6 ≠ x9 or x5
(Rule 3)
(x3)2 = (x · x · x)(x · x · x) = x6
(d) (xy)3 = x3y3 ≠ xy3 or x3y
(Rule 4)
(xy)3 = (xy)(xy)(xy) = (x · x · x)(y · y · y) = x3y3
(Rule 5)
(Rules 2 and 6)
(Rule 7)
Since
and from Rule 1 exponents of a common base are added in
multiplication, the exponent of
, when added to itself, must equal 1. With
the exponent of
must equal . Thus,
,
.
Just as
so x1/4 · x1/4 · x1/4 · x1/4 = x1/4 + 1/4 + 1/4 + 1/4 = x1 = x.
See Problems 1.1, 1.24, 1.26, and 1.27.
EXAMPLE 2. From Rule 2, it can easily be seen why any variable or nonzero number raised to the
zero power equals one. For example, x3/x3 = x3−3 = x0 = 1; 85/85 = 85−5 = 80 = 1.
1.2 POLYNOMIALS
Given an expression such as 9x5, x is called a variable because it can assume any number of
different values, and 9 is referred to as the coefficient of x. Expressions consisting simply of a real
number or of a coefficient times one or more variables raised to the power of a positive integer are
called monomials. Monomials can be added or subtracted to form polynomials. Each monomial
constituting a polynomial is called a term. Terms that have the same variables and respective
exponents are called like terms. The degree of a monomial is the sum of the exponents of its variables.
The degree of a polynomial is the degree of its highest term. Rules for adding, subtracting,
multiplying, and dividing polynomials are explained below, illustrated in Examples 3 to 5, and treated
in Problems 1.3 and 1.4.
1.2.1 Addition and Subtraction of Polynomials
Like terms in polynomials can be added or subtracted by adding or subtracting their coefficients.
Unlike terms cannot be so added or subtracted.
EXAMPLE 3.
(a) 6x3 + 15x3 = 21x3
(b) 18xy − 7xy = 11xy
(c) (4x3 + 13x2 − 7x) + (11x3 − 8x2 − 9x) = 15x3 + 5x2 − 16x
(d) (22x − 19y) + (7x + 6z) = 29x − 19y) + 6z
See also Problem 1.3.
1.2.2 Multiplication and Division of Terms
Like and unlike terms can be multiplied or divided by multiplying or dividing both the coefficients
and variables.
EXAMPLE 4.
(a) 20x4 · 7y6 = 140x4y6
(b) 6x2y3 · 8x4y6 = 48x6y9
(c) 12x3y2 · 5y4z5 = 60x3y6z5
(d) 3x3y2z5 · 15x4y3z4 = 45x7y5z9
1.2.3 Multiplication of Polynomials
To multiply two polynomials, multiply each term in the first polynomial by each term in the second
polynomial and then add their products together.
EXAMPLE 5.
See also Problem 1.4.
1.3 FACTORING
Factoring reverses the process of polynomial multiplication in order to express a given polynomial
as a product of simpler polynomials called factors. A monomial such as the number 14 is easily
factored by expressing it as a product of its integer factors 1 · 14, 2 · 7, (−1) · (−14), or (−2) · (−7). A
binomial such as 5x4 − 45x3 is easily factored by dividing or factoring out the greatest common factor,
here 5x3, to obtain 5x3(x − 9). Factoring a trinomial such as mx2 + nx + p, however, generally requires
the following rules:
1. Given (mx2 + nx + p), the factors are (ax + c)(bx + d), where (1) ab = m; (2) cd = p; and (3) ad +
bc = n.
2. Given (mx2 + nxy + py2), the factors are (ax + cy)(bx + dy), where (1) ab = m; (2) cd = p; and (3)
ad + bc = n, exactly as above. For proof of these rules, see Problems 1.28 and 1.29.
EXAMPLE 6. To factor (x2 + 11x + 24), where in terms of Rule 1 (above) m = 1, n = 11, and p = 24,
we seek integer factors such that:
1) a · b = 1. Integer factors: 1 · 1, (−1) · (−1). For simplicity we shall consider only positive sets of
integer factors here and in step 2.
2) c · d = 24. Integer factors: 1 · 24, 2 · 12, 3 · 8, 4 · 6, 6 · 4, 8 · 3, 12 · 2, 24 · 1.
3) ad + bc = 11. With a = b = 1, c + d must equal 11.
Adding the different combinations of factors from step 2, we have 1 + 24 = 25, 2 + 12 = 14, 3 + 8 =
11, 4 + 6 = 10, 6 + 4 = 10, 8 + 3 = 11, 12 + 2 = 14, and 21 + 1 = 25. Since only 3 + 8 and 8 + 3 = 11 in
step 3, 3 and 8 are the only candidates for c and d from step 2 which, when used with a = b = 1 from
step 1, fulfill all the above requirements, and the order does not matter. Hence
(x2 + 11x + 24) = (x + 3)(x + 8) or (x + 8)(x + 3)
See Problems 1.5 to 1.13. For derivation of the rules, see Problems 1.28 and 1.29.
1.4 FRACTIONS
Fractions, or rational numbers, consist of polynomials in both numerator and denominator,
assuming always that the denominator does not equal zero. Reducing a fraction to lowest terms
involves the cancellation of all common factors from both the numerator and the denominator.
Raising a fraction to higher terms means multiplying the numerator and denominator by the same
nonzero polynomial. Assuming that A, B, C, and D are polynomials and C and D ≠ 0, fractions are
governed by the following rules:
The properties of fractions are illustrated in Example 7 and treated in Problems 1.14 to 1.21.
EXAMPLE 7.
(a) Multiplying or dividing both the numerator and the denominator of a fraction by the
same nonzero number or polynomial leaves the value of the fraction unchanged.
(Rule 1)
Rule 1 provides the basis for reducing a fraction to its lowest terms as well as for raising
a fraction to higher terms.
(b) To multiply fractions, simply multiply the numerators and the denominators separately.
The product of the numerators then forms the numerator of the product and the product of
the denominators forms the denominator of the product.
(Rule 2)
(c) To divide fractions, simply invert the divisor and multiply.
(Rule 3)
(d) Fractions can be added or subtracted only if they have exactly the same denominator,
called a common denominator. If a common denominator is present, simply add or
subtract the numerators and set the result over the common denominator. Remember
always to subtract all the terms within a given set of parentheses.
(Rule 4)
(e) To add or subtract fractions with different denominators, a common denominator must
first be found. Multiplication of one denominator by the other will always produce a
common denominator. Each fraction can then be restated in terms of the common
denominator using Rule 1 and the numerators added as in (d).
(Rule 5)
(f) Similarly,
(Rule 5)
The least common denominator (LCD) of two or more fractions is the polynomial of
lowest degree and smallest coefficient that is exactly divisible by the denominators of the
original fractions. Use of the LCD helps simplify the final sum or difference. See
Problems 1.19 to 1.21. Fractions are reviewed in Problems 1.14 to 1.21.
1.5 RADICALS
If bn = a, where b > 0, then by taking the nth root of both sides of the equation,
, where
is a radical (sign), a is the radicami, and n is the index. For square roots, the index 2 is not expressed.
Thus,
. From Rules 7 and 8 in Section 1.1, we should also be aware that
and
.
Assuming x and y are real nonnegative numbers and m and n are positive integers such that
exist, the rules of radicals are given below. For proof of Rule 1, see Problem 1.30.
and
EXAMPLE 8. The laws of radicals are used to simplify the following expressions. Note that for evennumbered roots, positive and negative answers are equally valid.
See also Problems 1.22, 1.23, and Problems 1.25 to 1.27.
1.6 ORDER OF MATHEMATICAL OPERATIONS
Given an expression involving multiple mathematical operations, computations within parentheses
are performed first. If there are parentheses within parentheses, computations on the innermost set
take precedence. Within parentheses, all constants and variables are first raised to the powers of their
respective exponents. Multiplication and division are then performed before addition and subtraction.
In carrying out operations of the same priority, the procedure is from left to right. In sum,
1. Start within parentheses, beginning with the innermost.
2. Raise all terms to their respective exponents.
3. Multiply and divide before adding and subtracting.
4. For similar priorities, move from left to right.
EXAMPLE 9. The following steps are performed to solve
1. 52 = 25
2. 25 · 6 = 150
4. 15 − 8 = 7
Thus
1.7 USE OF A POCKET CALCULATOR
Pocket calculators are helpful for checking one’s ordinary calculations and performing arduous or
otherwise time-consuming computations. Rules for the different mathematical operations are set forth
and illustrated below, including some rales which will not be used or needed until later in the text.
1.7.1 Addition of Two Numbers
To add two numbers, enter the first number, press the
press the
key, and enter the second number. Then
key to find the total.
EXAMPLE 10.
(a) To find 139 + 216, enter 139, press ihe
139 + 216 = 355.
key, enter 216, and press the
key to find
(b) To find 1025 + 38.75, enter 1025, press thethe
key, then enter 38.75, and hit the
key to find 1025 + 38.75 = 1063.75. Practice this and subsequent examples using simple
numbers to which you already know the answers to see if you are doing the procedure
correctly.
1.7.2 Addition of More Than Two Numbers
To add more than two numbers, simply follow each entry of a number by pressing the
all the numbers have been entered. Then press the
key to find the total. Pressing the
time after a number will give the subtotal at that point.
EXAMPLE 11. To find 139 + 216 + 187, enter 139, press the
key until
key at any
key, enter 216, press the
again, enter 187, and hit the
key to find 139 + 216 + 187 = 542. Hitting the
reveal the subtotal of 139 + 216 is 355, as in the example above.
key
key after 216 would
1.7.3 Subtraction
To find the difference A – B, enter A, press the
key, and enter B. Then press the
key to find
the remainder. Multiple subtractions can be done as multiple additions in 1.7.2 above, with the
key substituted for the
key.
EXAMPLE 12.
(a) To find 315 − 708, enter 315, press the
key, then enter 708 followed by the
key to
find 315 − 708 = −393.
(b) To find 528 − 79.62, enter 528, hit the
to find 528 − 79.62 = 448.38.
key, then enter 79.62 followed by the
key
1.7.4 Multiplication
To multiply two numbers, enter the first number, press the
press the
key, enter the second number, and
key to find the product. Serial multiplications can be done in the same way as multiple
additions in 1.7.2. with the
key substituted for the
key.
EXAMPLE 13.
(a) To find 486 · 27, enter 486, press the
486 · 27 = 13,122.
(b) To find 149 · −35, enter 149, press the
make it negative, and hit the
key, then enter 27, and hit the
key to learn
key, then enter 35 followed by the
key to
key to learn that 149 · −35 = −5215.
Note: Be aware of the distinction between the
key and the
key. The
key
initiates the process of subtraction; the
key simply changes the value of the previous
entry from positive to negative or negative to positive.
1.7.5 Division
Dividing A by B is accomplished by entering A, pressing the
the
key, then entering B and pressing
key.
EXAMPLE 14.
(a) To find 6715 ÷ 79, enter 6715, hit the
key, then enter 79 followed by the
display will show 85, indicating that 6715 ÷ 79 = 85.
(b) To find −297.36 ÷ 72.128, enter 297.36 followed by the
press the
key, enter 72.128, and hit the
−4.1226708.
key. The
key to make it negative, then
key to find −297.36 ÷ 72.128 =
1.7.6 Raising to a Power
To raise a number to a power, enter the number, hit the
the
key, then enter the exponent and press
key.
EXAMPLE 15.
(a) To find 85, enter 8, press the
key, then enter 5 followed by the
key to learn that 85
= 32,768. Continue to practice these and subsequent exercises by using simple numbers
for which you already know the answers.
(b) To find 360.25, enter 36, hit the
360.25 = 2.4494897.
key, then enter 0.25, and press the
key to see
(c) To find 2−3, enter 2, hit the
negative, and hit the
key, then press 3 followed by the
key to make it
key to discover 2−3 = 0.125.
See also Problem 1.24.
1.7.7 Finding a Square Root
To find the square root of a number, enter the number, then press the
key to find the square
root immediately without having to press the
key. Note that on many calculators the
the inverse (shift, or second function) of the
key, and to activate the
the
or
key followed by the
EXAMPLE 16. To find
square root of 529.
If the
enter 529, then press the
immediately that
key) followed by the
key, one must first press
key.
key to see immediately that ±23 is the
key is the inverse, shift, or second function of the
or
key is
key to activate the
= 23 without having to press the
key, enter 529, then press the
key, and you will see
key.
1.7.8 Finding the nth Root
To find the nth root of a number, enter the number, press the
root n and hit the
key to find the root. If the
is the inverse, shift, or second function of the
key, enter the number, press the
the value of the root n and hit the
key, then enter the value of the
, or
key followed by the
key, then enter
key to find the answer.
EXAMPLE 17.
(a) To find
hit the
(b) To find
hit the
, enter 17,576, press the
key to learn
(c) From Rule 8 in Section 1.1,
simply enter 32,768, press the
key, enter 3, then
.
, enter 32,768, hit the
key to learn
key followed by the
key followed by the
key, then enter 5 and
.
= 32,7681/5 = 32,7680.2. To use this latter form,
key, enter 0.2, and hit the
key to find 32,7680.2 = 8.
To make use of similar conversions, recall that
, and so forth. See Problems 1.25 to 1.27.
1.7.9 Logarithms
To find the value of the common logarithm log10x, enter the value of x and simply press the
key. The answer will appear without the need to press the
key.
EXAMPLE 18.
(a) To find the value of log 24, enter 24 and hit the
key. The screen will immediately
display 1.3802112, indicating that log 24 = 1.3802112.
(b) To find log 175, enter 175 and hit the
of log 175.
key. You will see 2.243038, which is the value
1.7.10 Natural Logarithms
To find the value of the natural logarithm lnx, enter the value of x and press the
answer will appear immediately without the need to press the
key. The
key.
EXAMPLE 19.
(a) To find In 20, enter 20 and hit the
key to see 2.9957323 = In 20.
(b) For In 0.75, enter 0.75 and hit the
key. You will find In 0.75 = −0.2876821.
1.7.11 Exponential Functions
To find the value of an exponential function y = ax, enter the value of a and press the
enter the value of x and hit the
key, then
key, similar to what was done in Section 1.7.6.
EXAMPLE 20.
(a) Given y = 1.53.2, enter 1.5, press the
1.53.2 = 3.6600922.
(b) For y = 256−1.25, enter 256, hit the
key, then enter 3.2, and hit the
key to get
key, then enter 1.25 followed immediately by the
key to make it negative, and then press the
key to leam 256−1.25 = 0.0009766.
1.7.12 Natural Exponential Functions
To find the value of a natural exponential function y = ex, enter the value of x press the
the answer will appear immediately without the need of hitting the
key, enter the value of x, hit the
inverse, shift, or second function of the
the
key. If the
key, and
key is the
key followed by
key, and the answer will also appear immediately.
EXAMPLE 21.
(a) Given y = e1.4, enter 1.4, press the
that e1.4 = 4.0552.
(b) For e−0.65, enter 0.65, hit the
followed by the
key followed by the
key, and you will see
key to make it negative, then press the
key to find e−0.65 = 0.5220458.
See also Problems 1.24 to 1.27.
Solved Problems
key
EXPONENTS
1.1. Simplify the following expressions using the rules of exponents from Section 1.1.
(a) x3 · x4
x3 · x4 = x3+4 = x7
(Rule 1)
(b) x5 · x−3
(Rule 1)
(c) x−2 · x−4
(Rule 1)
(d) x1/2 · x3
(Rules 7 and 1)
(Rule 2)
(Rules 2 and 6)
(Rule 2)
(Rules 2 and 7)
(i) (x4)−3
(Rules 3 and 6)
(Rules 8 and 3)
(Rules 6 and 4)
(Rules 3 and 5)
See also Problems 1.24, 1.26, and 1.27.
POLYNOMIALS
1.2. Perform the indicated arithmetic operations on the following polynomials:
(a) 35xy + 52xy
35xy + 52xy = 87xy
(b) 22yz2 − 46yz2
22yz2 − 46yz2 = −24yz2
(c) 79x2y3 − 46x2y3
79x2y3 − 46x2y3 = 33x2y3
(d) 16x1x2 + 62x1x2
16x1x2 + 62x1x2 = 78x1x2
(e) 57y1y2 − 70y1y2
57y1y2 − 70y1y2 = −13y1y2
(f) 0.5x2y3z5 + 0.9x2y3z5
0.5x2y3z5 + 0.9x2y3z5 = 1.4x2y3z5
1.3. Add or subtract the following polynomials as indicated. Note that in subtraction the sign of
every term within the parentheses must be changed before corresponding elements are added.
(a) (25x − 9y) + (32x + 16y)
(25x − 9y) + (32x + 16y) = 57x + 7y
(b) (84x − 31 y) − (76x + 43y)
Multiplying each term in the second set of parentheses by −1, which in effect changes the
sign of the said terms, and then simply adding, we have
(84x − 31y) − (76x + 43y) = 84x − 31y − 76x − 43y = 8x − 74y
(c) (9x2 + 7x) − (3x2 − 4x)
Changing the signs of each of the terms in the second set of parentheses and then adding,
we have
(9x2 + 7x) − (3x2 − 4x) = 9x2 + 7x − 3x2 + 4x = 6x2 + 11x
(d) (42x2 + 23x) − (5x2 + 11x − 82)
(42x2 + 23x) − (5x2 + 11x −82) = 42x2 + 23x − 5x2 − 11x + 82
= 37x2 + 12x + 82
1.4. Perform the indicated operations, recalling that each term in the first polynomial must be
multiplied by each term in the second and their products summed.
(a) (2x + 7)(4x − 5)
(2x + 7)(4x − 5) = 8x2 − 10x + 28x − 35 = 8x2 + 18x − 35
(b) (5x − 6y)(4x − 3y)
(5x − 6y)(4x − 3y) = 20x2 − 15xy − 18y2 = 20x2 − 39xy + 18y2
(c) (2x − 9)2
(2x − 9)2 = (2x − 9)(2x − 9) = 4x2 − 18x − 18x + 81 = 4x2 − 36x + 81
(d) (2x + 3y)(2x − 3y)
(2x + 3y)(2x − 3y) = 4x2 − 6xy + 6xy − 9y2 = 4x2 − 9y2
(e) (4x + 3y)(5x2 − 2xy + 6y2)
(4x + 3y)(5x2 − 2xy + 6y2) = 20x3 − 8x2y + 24xy2 + 15x2y − 6xy2 + 18y3
= 20x3 + 7x2y + 18xy2 + 18y3
(f) (3x2 − 5x2y2 − 2y3)(7x − 4y)
(3x3 − 5x2y2 − 2y3)(7x − 4y) = 21x4 − 12x3y − 35x3y2 + 20x2y3 − 14xy3 + 8y4
FACTORING
1.5. Simplify each of the following polynomials by factoring out the greatest common factor:
(a) 32x − 8
32x − 8 = 8(4x − 1)
(b) 18x2 + 27x
18x2 + 27x = 9x(2x + 3)
(c) 14x5 − 35x4
14x5 − 35x4 = 7x4(2x − 5)
(d) 45x2y5 − 75x4y3
45x2y5 − 75x4y3 = 15x2y3(3y2 − 5x2)
(e) 55x8y9 − 22x6y4 − 99x5y7
55x8y9 − 22x6y4 − 99x5y7 = 11x5y4(5x3y5 − 2x − 9y3)
1.6. Factor each of the following using integer coefficients:
(a) x2 + 10x + 21
Here, using the notation from Rule 1 in Section 1.3, m = 1, n = 10, and p = 21. For
simplicity we limit our search to positive integers such that:
(1) a · b = 1 [1, 1] Henceforth, when a = b = 1, this step will be omitted and only one order
will be considered.
(2) c · d = 21 [1,21; 3,7; 7,3; 21,1]
(3) ad + bc = 10. With a = 1 = b, (c + d) must equal 10. Since only 3 + 7 and 7 + 3 = 10,
x2 + 10x + 21 = (x + 3)(x + 7) or (x + 7)(x + 3)
Whenever a · b = 1, the order of factors is unimportant and only one ordering of a given pair
will be mentioned.
(b) x2 + 8x + 16
(1) c · d = 16 [1, 16; 2, 8; 4, 4]
(2) c + d = 8 [1 + 16 ≠ 8; 2 + 8 ≠ 8; 4 + 4 = 8]
x2 + 8x + 16 = (x + 4)(x + 4)