Lecture Notes: Intermediate Algebra - Faculty Site Listingfaculty.mccneb.edu/jdlee3/math_1310/oldnotes/lecture_notes_1310.pdf · Lecture Notes: Intermediate Algebra ... 4.6 Radicals

Lecture Notes: Intermediate Algebra

Joseph Lee

Metropolitan Community College

Contents

1 Linear Equations 51.1 Linear Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51.2 Linear Equations, Continued . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 91.3 Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 131.4 Systems of Linear Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 181.5 Systems of Linear Equations, Continued . . . . . . . . . . . . . . . . . . . . . . . . 241.6 Systems of Linear Equations: Applications . . . . . . . . . . . . . . . . . . . . . . . 30

2 Polynomials 322.1 Polynomials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 322.2 Polynomials, Continued . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 352.3 Factoring . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 392.4 Factoring, Continued . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 422.5 Factoring III . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 462.6 Factoring IV: An Application . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 50

3 Rational Expressions 533.1 Rational Expressions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 533.2 Rational Expressions, Continued . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 533.3 Rational Expressions III: Complex Rational Expressions . . . . . . . . . . . . . . . 533.4 Rational Expressions IV . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 53

4 Radicals 544.1 Radicals: An Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 544.2 Radicals: Rational Exponents . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 604.3 Radicals III . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 704.4 Radicals IV . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 804.5 Radicals V . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 854.6 Radicals VI . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 90

5 Quadratic Equations 945.1 The Complex Number System . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 945.2 Quadratic Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 995.3 Quadratic Equations: The Grand Finale . . . . . . . . . . . . . . . . . . . . . . . . 1055.4 Quadratic Equations III . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1095.5 Prelude To College Algebra: Graphing Quadratic Equations . . . . . . . . . . . . . 109

1

Preface

I created this set of notes during the winter quarter 2012 for my intermediate algebra students. Ihad taught this course for the two preceeding years, and honestly, intermediate algebra became myfavorite course to teach at Metropolitan Community College. Often times I find my appreciationfor this course is not shared by my students1, but I think its regrettable if a student misses thebeauty of the algebra presented.

The question I receive most often, regardless of the course, is, “When am I ever going to usethis?” I think the question misses the point entirely. While I don’t determine which classes studentsneed to get their degree, I do think it’s a good policy that students are required to take my course– for more reasons than just my continued employment, which I support as well.

If a student asked an English instructor why he or she had to read Willa Cather’s My Antonia,the instructor wouldn’t argue that understanding nineteenth century prairie life was essential tobecoming a competent tax specialist or licensed nurse. The instructor would not argue that readingMy Antonia would benefit the student directly through a future application. Instead, the benefitof reading this beautiful piece of American literature is entirely intrinsic. The mere enjoymentand appreciation is enough to justify its place in a post-secondary education. Moreover, the resultarrived to at the end of my course is as beautiful as any prose or poetry a student will encounterin his or her studies here at Metro or any other college.

Background

The spirit of this study was first undertaken by Greek mathematicians, most notably Diophantus.While these ideas would continue to be explored independently by Greek, Indian, and Chinesemathematicians, it was the Arab mathematician Al-Khwarizmi’s 9th century treatise Al-Kitab al-mukhtasar fi hisab al-jabr wa’l-muqabala2 that inspired the term “algebra.” Al-Khwarizmi’s solu-tion to a quadratic equation was not entirely complete, but it largely resembles the techniques wewill apply in Section 5.23.

About the Course

The course is divided into five units. The first unit covers linear equations. Graphing equations iscovered with special attention to finding the slope of a line. Point-slope form is used to write theequation of a line. Readers of other texts might prefer to only use only slope-intercept form, but Ichoose to stress point-slope form in all of my arguments. Functions are introduced in this course,and while I cover them thoroughly, the student should note these ideas will be strongly reinforced ifhe or she continues on to College Algebra. The second half of this unit deals with solving systems oflinear equations. These last sections are omitted from my notes and will be presented from the text.

The second unit covers polynomials with an emphasis on factoring. The first section on def-initions and the operations of addition and subtraction should be review. The second section onmultiplication is also review, but I emphasize what I consider are the more difficult ideas. Again,

1hopefully not due to my presentation2The Compendious Book on Calculation by Completion and Balancing3Al-Khwarizmi “completes the square” geometrically

2

the student might not enjoy my presentation at this point, but I feel the approach I present is theone intended by the course. The remainder of the unit covers factoring in all its glory.

The third unit covers rational expressions. As with systems of equations, I have omitted thesesections from the notes, and they will be covered from the text.

The fourth unit covers radicals. It starts with a basic introduction to the definitions. Rules forradicals are developed using rational exponents, which are defined in the second section. Sectionsthree through five focus on simplifying radicals following three basic rules. The last section of theunit covers solving equations with radicals.

The last unit brings the course to what I call “The Grand Finale.” The genius of the course isthat the student has learned all the tools to prove the famous quadratic formula. I feel that manyteachers, if not all, present the formula without proof. I think this causes me deep sadness in fact.Having learned some really great algebra – completing the square, adding rational expressions, rulesof radicals, factoring trinomials – it would be a complete waste not to use all these tools to prove thequadratic formula. I feel this last unit completely justifies learning all the algebra throughout thecourse. This fifth section introduces the complex number system. Quadratic equations are solvedby completing the square and the quadratic formula.

To The Student

I hope you are successful in this course. It will require a lot of work, and more for some than others.You should work on homework every night. Read your textbook, in addition to my notes. Whenyou get frustrated, take a break, but come back to it the next day. Come to my office hours or goto the math center for help. You should complete every homework problem.

I am glad you are taking my course, but you have to decide how important this class is to youand what your priorities are. If you decide you have more important things to do in life than doingevery single homework problem, I will not fault you for it. I have no children or other commitments,so it’s easy for me to spend all day doing math. I realize this may not be the case for you. However,you might find this course does require the kind of dedication I have outlined above.

Grading Policy

As an assessment tool, quizzes and tests are used to determine a student’s mastery of requiredcourse objectives. The six course objectives, listed in the syllabus, are as follows:

1. Graph linear functions and other basic functions: define a function and its notation.

2. Expand upon operations involving exponents, polynomials, and the methods of factoring.

3. Solve systems of equations and apply them to solving application problems.

4. Simplify rational expressions and solve rational equations.

5. Simplify radical expressions, solve radical equations, define rational exponents, (manipulateand convert from exponential to radical notation and visa/versa), and perform operationswith complex numbers.

3

6. Solve quadratic equations with real and complex solutions.

A test designed to assess the student’s mastery of objective 2, for example, should indicate thestudent’s proficiency at factoring. An 80% on a quiz or test should indicate that the student is ableto correctly factor 80% of the polynomials he or she is asked to factor – not that the student hasmastered 80% of the steps needed to factor a polynomial. Thus, in most cases, partial credit forincorrect answers will not be given.

Preface to the Spring 2013 Quarter

Since I first prepared these notes, my appreciation for this course has in no way been diminished.Admittedly, I am very evangelical about this course. Intermediate algebra is an amazing course, andall students should experience the ideas embedded in this course. In my opinion, it is the incorrectview of this course that it is designed to program the students with algorithms needed to answerquestions the instructor poses. Instead, this course is designed to teach the student fundamentalconcepts of algebra. These lectures are designed with that intent.

Critics would argue there are easier ways to teach a stucent how to answer specific exercisespresented throughout the course. I remain convinced, however, that the rationale for the courseis not simply to memorize a set of procedures for various mathematical questions, but instead tounderstand the concepts underlying these procedures. The examples presented in these lectures aretherefore intended as a means to discuss these ideas, and not an end in of themselves. Additionally,in many cases, a greater understanding of the underlying concept will improve a student’s proficiencyin completing a certain problem.

Preface to the Fall 2013 Quarter

I made some revisions to a few of my lectures. In particular, I removed some examples from Sections1.1 and 1.2 used to develop the ideas of slope-intercept and point-slope form. Instead, I simply givedefinitions and proceed straight into examples. I think students will find this more straight forwardapproach easier to follow in class.

4

1 Linear Equations

1.1 Linear Equations

.Definition: Linear Equation in Two Variables A linear equation in two variables is anyequation that can be written in the form

Ax + By = C,

where A, B, and C are real numbers.Examples

1. 3x + 2y = 6

2. y = −1

3x− 4

3. 4x = 20 + 5y

.Definition: Slope The slope, m, of a line is the ratio of the vertical change to the horizontalchange.

m =vertical change

horizontal change

.Definition: y-intercept The y-intercept of a line is the point where the line intersects the y-axis.

.Definition: Slope-Intercept Form The equation of a line is in slope-intercept form if it is writtenas

y = mx + b,

where m is the slope of the line and (0, b) is the y-intercept.

.Example 1.a For the equation y = −3

2x+ 3, determine the slope and the y-intercept. Then graph

the equation.

Solution: Since the equation is in slope-intercept form, we know the slope is m = −32

and they-intercept is (0, 3).

5

.Example 1.b For the equation y = 1

4x − 2, determine the slope and the y-intercept. Then graph

the equation.

Solution: Since the equation is in slope-intercept form, we know the slope is m = 14

and they-intercept is (0,−2).

.Example 1.c For the equation 4x− 3y = 9, determine the slope and the y-intercept. Then graphthe equation.

Solution:4x− 3y = 9

−3y = −4x + 9

y =4

3x− 3

.Example 1.d For the equation 5x + 2y = 6, determine the slope and the y-intercept. Then graphthe equation.

Solution:5x + 2y = 6

2y = −5x + 6

y = −5

2x + 3

6

.Example 2 Find the slope of the line passing through the points (x1, y1) and (x2, y2).

Solution:

m =vertical change

horizontal change

=y2 − y1x2 − x1

.Example 3.a Find the slope of the line connecting (1, 6) and (4, 2).

Solution:

m =y2 − y1x2 − x1

=2− 6

4− 1

=−4

3

= −4

3

.Example 3.b Find the slope of the line connecting (−2,−4) and (6, 12).

Solution:

m =y2 − y1x2 − x1

=12− (−4)

6− (−2)

=16

8

= 2

.Example 3.c Find the slope of the line connecting (−3, 2) and (4, 2).

7

Solution:

m =y2 − y1x2 − x1

=2− 2

4− (−3)

=0

7

= 0

.Example 3.d Find the slope of the line connecting (3,−2) and (3,−8).

Solution:

m =y2 − y1x2 − x1

=−8− (−2)

3− 3

=−6

0

However, −6 divided by 0 is undefined. (In other words, the ratio of the vertical change to thehorizontal change does not exist.)

8

1.2 Linear Equations, Continued

.Example 1.a Write the equation of the line, in slope-intercept form, that has a slope of −2

3and a

y-intercept of (0, 4).

Solution:

y = −2

3x + 4

.Example 1.b Write the equation of the line, in slope-intercept form, that has a slope of 3 and ay-intercept of

(0,−3

2

).

Solution:

y = 3x− 3

2

.Definition: Point-Slope Form An equation of the line with slope m passing through the point(x1, y1) is given by

y − y1 = m(x− x1).

The equation above is said to be in point-slope form.

.Example 2.a Write the equation of the line, in slope-intercept form, with a slope of −1

2that passes

though the point (−4, 3).

Solution:y − y1 = m(x− x1)

y − 3 = −1

2(x− (−4))

y − 3 = −1

2(x + 4)

y − 3 = −1

2x− 2

y = −1

2x + 1

.Example 2.b Write the equation of the line, in slope-intercept form, with a slope of −2 that passesthough the point (3,−8).

Solution:y − y1 = m(x− x1)

y − (−8) = −2(x− 3)

y + 8 = −2x + 6

y = −2x− 2

9

.Example 3.a Write the equation of the line, in slope-intercept form, that passes though the points(2,−3) and (−2, 1).

Solution: First we note the slope of the line is

m =1− (−3)

−2− 2=

4

−4= −1.

Then the equation of the line with slope −1 and passing through (2,−3) is

y − (−3) = −(x− 2)

y + 3 = −x + 2

y = −x− 1.

.Example 3.b Write the equation of the line, in slope-intercept form, that passes though the points(5,−2) and (−1, 1).


m =1− (−2)

−1− 5=

3

−6= −1

2.

Then the equation of the line with slope −12

and passing through (5,−2) is

y − (−2) = −1

2(x− 5)

y + 2 = −1

2x +

5

2

y = −1

2x +

1

2.

.Definition: Standard Form Recall our definition of a linear equation in two variables is anyequation that can be written in the form

Ax + By = C,

where A, B, and C are real numbers. The linear equation is said to be in standard form if it iswritten in this manner where A, B, and C are integers and A > 0.

Slope-Intercept Form Standard Form

y = 2x− 3 2x− y = 3

y = −2

3x + 1 2x + 3y = 3

10

.Example 4.a Write the equation of the line, in standard form, with slope 1

3that passes though

the point (−2, 2).

Solution:

y − 2 =1

3(x + 2)

y − 2 =1

3x +

2

3

y =1

3x +

8

3

−1

3x + y =

8

3

x− 3y = −8.

.Example 4.b Write the equation of the line, in standard form, that passes though the points (4, 0)and (−2, 3).


m =3− 0

−2− 4=

3

−6= −1

2.

Then the equation of the line with slope −12

and passing through (4, 0) is

y − 0 = −1

2(x− 4)

y = −1

2x + 2

1

2x + y = 2

x + 2y = 4.

.Definition: Parallel, Perpendicular Two lines are parallel if they have the same slope. Twolines are perpendicular if the slopes are opposite reciprocals (if their product is −1). Moreover:

• Any two horizontal lines are parallel.

• Any two vertical lines are parallel.

• A horizontal line and a vertical line are perpendicular.

.Example 5.a Write the equation of the line, in slope-intercept form, that is parallel to the line2x + y = 6 and passes though the point (3, 1).

11

Solution: Since our line is parallel to

2x + y = 6y = −2x + 6,

we know the slope of our line must be −2.Then the equation of the line with slope −2 and passing through (3, 1) is

y − 1 = −2(x− 3)

y − 1 = −2x + 6

y = −2x + 7.

.Example 5.b Write the equation of the line, in standard form, that is perpendicular to the line2x + 3y = 5 and passes though the point (0, 3).

Solution: Since our line is perpendicular to

2x + 3y = 5y = −2

3x + 5

3,

we know the slope of our line must be 32.

Then the equation of the line with slope 32

and y-intercept (0, 3) is

y =3

2x + 3

−3

2x + y = 3

3x− 2y = −6.

12

1.3 Functions

.Definition: Set A set is a collection. Members of the collection are called elements.

.Example 1.a A set is a collection. Members of the collection are called elements.

1. {black,white, pink} is the set of Joseph’s three favorite colors. This set has 3 elements.

2. {1, 2, 3, 4} is the set of positive integers less than 5.

3. {−2, 2} is the set of solutions to the equation x2 = 4.

4. ∅ is the set containing no elements. For example, the set of real solutions to the equationx2 = −4.

.Example 1.b We may also want to talk about sets with an infinite number of elements.

1. {1, 2, 3, 4, 5, ...} could be the set of positive integers.

2. {3, 5, 7, ...} could be the set of odd numbers greater than 1: {3, 5, 7, 9, 11, 13, ...}

3. On the other hand, {3, 5, 7, ...} could be the set of odd prime numbers: {3, 5, 7, 11, 13, 17, ...}

.Definition: Set-Builder Notation Set builder notation is a method to describe a set by itsproperties.

.Example 1.c The following sets are written using set-builder notation.

1. {x|x is one of Joseph’s three favorite colors} = {black,white, pink}

2. {x|x is an odd prime number} = {3, 5, 7, 11, 13, 17, ...}

3. {x|x is an integer & 0 < x < 5} = {1, 2, 3, 4}

.Definition: Relation A relation is a correspondence between two sets. Elements of the first setare called the domain. Elements of the second set are called the range. Relations are often expressedas sets of ordered pairs.

13

.Example 2.a A relation is a correspondence between two sets. Elements of the first set are calledthe domain. Elements of the second set are called the range.

1. {(Joseph, turkey), (Joseph, roast beef), (Michael, ham)} is a relation between math instructorsand sandwiches they enjoy. The domain of this relation is {Joseph, Michael}. The range ofthis relation is {turkey, roast beef, ham}.

2. {(1, 3), (2, 4), (−1, 1)} is a relation of x and y values that satisfy the equation y = x + 2. Thedomain of this relation is {−1, 1, 2}. The range of this relation is {1, 3, 4}.

3. {(3, 5), (4, 5), (5, 5)} is a relation of x and y values that satisfy the equation y = 5.

.Definition: Function A function is a specific type of a relation where each element in the thedomain corresponds to exactly one element in the range.

.Example 2.b Determine which of the following relations are functions.

1. {(Joseph, turkey), (Joseph, roast beef), (Michael, ham)}

2. {(1, 3), (2, 4), (−1, 1)}

3. {(3, 5), (4, 5), (5, 5)}.

Solution: The first relation is not a function, as Joseph is assigned to two elements in the range,turkey and roast beef. The last two relations are functions as each element in the domain is assignedto exactly one element in the range.

.Example 3.a The graph of the relation y = 3x− 2 is shown below.

1. Determine the domain of the relation. R (all real numbers)

2. Determine the range of the relation. R (all real numbers)

3. Determine if the relation is represents a function. Yes, it is a function.

14

.Example 3.b The graph of the relation x = y2 − 1 is shown below.

1. Determine the domain of the relation. {x|x ≥ −1}

2. Determine the range of the relation. R (all real numbers)

3. Determine if the relation is represents a function. No, it is not a function.

.Example 3.c The graph of the relation y = x2 − 2x− 3 is shown below.

1. Determine the domain of the relation. R (all real numbers)

2. Determine the range of the relation. {y| y ≥ −4}

3. Determine if the relation is represents a function. Yes, it is a function.

.Definition: Vertical Line Test If any vertical line intersects the graph of a relation more thanonce, the relation is not a function.

.Definition: Function Notation If a relation of x and y is a function, than you may solve for yand replace y with f(x). This notation is called function notation.

.Example 4.a Write the function 2x + 3y = 6 in function notation.

15

Solution:2x + 3y = 6

3y = −2x + 6

y = −2

3x + 2

f(x) = −2

3x + 2

.

Example 4.b Evaluate the function f(x) = −2

3x + 2 for the following values.

1. f(−3) = 4

2. f(0) = 2

3. f(1) =4

3

4. f(3) = 0

.Example 4.c Evaluate the function f(x) =

√x + 6 for the following values.

1. f(−2) = 2

2. f(19) = 5

3. f(−7) is undefined

.Example 5.a The graph of the function f is shown below.

Evaluate.

16

1. f(−4) = −3

2. f(−1) = 0

3. f(1) = 2

4. f(4) = 1

.Example 5.b The graph of the function f is shown below.

Evaluate.

1. f(−5) = 0

2. f(−1) = −2

3. f(2) = 0

17

1.4 Systems of Linear Equations

.Definition: System of Equations A system of equations is a group of two or more equations.

.Example 1.a Determine if (0,−4) is a solution to the following system of linear equations:

x− 2y = 84x + y = 5

Solution. First, we will check if (0,−4) is a solution of x− 2y = 8.

x− 2y = 8

(0)− 2(−4)?= 8

8 = 8

Thus, (0,−4) is a solution of x− 2y = 8, so now we will check if it is also a solution of 4x + y = 5.

4x + y = 5

4(0) + (−4)?= 5

−4 6= 5

Thus, (0,−4) is not a solution of this system of equations.

.Example 1.b Determine if (2,−3) is a solution to the following system of linear equations:

x− 2y = 84x + y = 5

Solution. First, we will check if (2,−3) is a solution of x− 2y = 8.

x− 2y = 8

(2)− 2(−3)?= 8

8 = 8

Thus, (2,−3) is a solution of x− 2y = 8, so now we will check if it is also a solution of 4x + y = 5.

4x + y = 5

4(2) + (−3)?= 5

5 = 5

Thus, (2,−3) is a solution of this system of equations.

.Example 1.c Graph both equations on the same coordinate plane.

x− 2y = 84x + y = 5

18

The line for x− 2y = 8 represents all the points that satisfy that equation. The line for 4x + y = 5represents all the points that satisfy that equation. Therefore, any point where the two linesintersect represents a point that will satisfy both equations – our solution.

.Example 2.a Determine if (3, 2) is a solution to the following system of linear equations:

2x + 3y = 12y = −2

3x + 4

Solution. First, we will check if (3, 2) is a solution of 2x + 3y = 12.

2x + 3y = 12

2(3) + 3(2)?= 12

12 = 12

Thus, (3, 2) is a solution of 2x+3y = 12, so now we will check if it is also a solution of y = −23x+4.

y = −23x + 4

(2)?= −2

3(3) + 4

2 = 2

Thus, (3, 2) is a solution of this system of equations.

.Example 2.b Determine if (9,−2) is a solution to the following system of linear equations:

2x + 3y = 12y = −2

3x + 4

Solution. First, we will check if (9,−2) is a solution of 2x + 3y = 12.

2x + 3y = 12

2(9) + 3(−2)?= 12

12 = 12

Thus, (9,−2) is a solution of 2x+3y = 12, so now we will check if it is also a solution of y = −23x+4.

y = −23x + 4

(−2)?= −2

3(9) + 4

−2 = −2

Thus, (9,−2) is a solution of this system of equations.

.Example 2.c Graph both equations on the same coordinate plane.

2x + 3y = 12y = −2

3x + 4

19

Notice the equations 2x + 3y = 12 and y = −23x + 4 represent the same line. The lines represent

all the points that satisfy the equation – therefore, any point on the lines satisfies both equations,i.e. the system. Thus, this system of linear equations has infinitely many solution. In particular,the solution is any point on the line 2x + 3y = 12.

.Example 3 Graph both equations on the same coordinate plane.

y = −3x + 43x + y = −1

Notice the lines represented by the equations y = −3x + 4 and 3x + y = −1 are parallel – meaningthe lines will never intersect. Since points of intersection are solutions to the system, this systemof linear equations has no solution.

.Definition: Dependent, Independent, Inconsistent Systems A system of equations is calledinconsistent if it has no solutions.A system of equations is called dependent if it has infinitely many solutions.

A system of equations is called independent if it has a single solution.

.Methods for Solving Systems of Linear Equations There are three methods we will discussfor solving systems of linear equations.

1. Graphing

2. Substitution

3. Elimination

.Example 4.a Solve the following system of equations using substitution.

x− 2y = 84x + y = 5

Solution. To solve by substitution, we will take one of our equations and solve for one of ourvariables. Let’s take our first equation x− 2y = 8 and solve for x.

x− 2y = 8x = 2y + 8

Now, we may substitute x = 2y + 8 into our other equation 4x + y = 5.

4x + y = 54(2y + 8) + y = 5

8y + 32 + y = 59y = −27y = −3

20

Since y = −3, we may substitute that value back into any equation to solve for x.

x = 2y + 8x = 2(−3) + 8x = 2

Thus, (2,−3) is the solution of this system of equations.

.Example 4.b Solve the following system of equations using substitution.

5x + 3y = 64x + 2y = 2

Solution. To solve by substitution, we will take one of our equations and solve for one of ourvariables. Let’s take our second equation 4x + 2y = 2 and solve for y.

4x + 2y = 22y = −4x + 2y = −2x + 1

Now, we may substitute y = −2x + 1 into our other equation 5x + 3y = 6.

5x + 3y = 65x + 3(−2x + 1) = 6

5x− 6x + 3 = 6−x = 3x = −3

Since x = −3, we may substitute that value back into any equation to solve for y.

y = −2x + 1y = −2(−3) + 1y = 7

Thus, (−3, 7) is the solution of this system of equations.

.Observation: Operations on Equations Observe that equations may be added together if wewanted...

x = 4y = 3

x + y = 7

orx + y = 7x− y = 1

(x + y) + (x− y) = 82x = 8

.Example 5.a Solve the following system of equations using elimination.

x− 2y = 84x + y = 5

21

Solution. To solve by elimination, we will eliminate one of our variables. Let’s eliminate y.

x− 2y = 84x + y = 5

x− 2y = 82(4x + y) = (5)2

x− 2y = 88x + 2y = 10

9x = 18x = 2

Since x = 2, we may substitute that value back into any equation to solve for y.

4x + y = 54(2) + y = 5

8 + y = 5y = −3

Thus, (2,−3) is the solution of this system of equations.

.Example 5.b Solve the following system of equations using elimination.

5x + 3y = 64x + 2y = 2

Solution. To solve by elimination, we will eliminate one of our variables. Let’s eliminate y.

5x + 3y = 64x + 2y = 2

2(5x + 3y) = (6)2−3(4x + 2y) = (2)(−3)

10x + 6y = 12−12x− 6y = −6

−2x = 6x = −3

Since x = −3, we may substitute that value back into any equation to solve for y.

4x + 2y = 24(−3) + 2y = 2−12 + 2y = 2

2y = 14y = 7

Thus, (−3, 7) is the solution of this system of equations.

22

.Example 6.a Solve the following system of equations using elimination.

4x + 6y = 24y = −2

3x + 4

Solution. First, let’s get rid of the fraction. Then we will move our variables over to the left side.Then we may choose a variable to eliminate.

4x + 6y = 243(y) =

(−2

3x + 4

)3

4x + 6y = 243y = −2x + 12

4x + 6y = 242x + 3y = 12

4x + 6y = 24−2(2x + 3y) = (12)(−2)

4x + 6y = 24−4x− 6y = −24

0 = 0

Arriving at this identity, we may conclude that this system is dependent. There are infinitely manysolutions, or more specifically, the solution is any point on the line 4x + 6y = 24.

.Example 6.b Solve the following system of equations using elimination.

y = −3x + 46x + 2y = −2

Solution. First, we will move our variables over to the left side. Then we may choose a variableto eliminate.

y = −3x + 46x + 2y = −2

3x + y = 46x + 2y = −2

−2(3x + y) = (4)(−2)6x + 2y = −2

−6x− 2y = −86x + 2y = −2

0 = −10

Arriving at this contradiction, we may conclude that this system is inconsistent. There is nosolution.

23

1.5 Systems of Linear Equations, Continued

.Example 1.a Solve the following system of equations.

x + y + z = 62x− 3y − z = 5

3x + 2y − 2z = −1

Solution. First, we will take two equations and eliminate a variable. Let’s take the first twoequations and eliminate z.

x + y + z = 62x− 3y − z = 5

3x− 2y = 11

Now, we want to take two different equations but eliminate the same variable: z. We will take thefirst and last equations.

x + y + z = 63x + 2y − 2z = −1

2(x + y + z) = (6)23x + 2y − 2z = −1

2x + 2y + 2z = 123x + 2y − 2z = −1

5x + 4y = 11

We now have two equations with two variables. We may proceed as usual from here.

3x− 2y = 115x + 4y = 11

2(3x− 2y) = (11)25x + 4y = 11

6x− 4y = 225x + 4y = 11

11x = 33x = 3

We substitute x = 3 into an equation containing only x and y.

3x− 2y = 113(3)− 2y = 11

9− 2y = 11−2y = 2

y = −1

24

Similarly, we substitute x = 3 and y = −1 into an equation containing x, y, and z.

x + y + z = 6(3) + (−1) + z = 6

2 + z = 6z = 4

Thus, (3,−1, 4) is the solution to this system of equations.

.Example 1.b Solve the following system of equations.

3x + 4y = −45y + 3z = 12x− 5z = 7

Solution. First, we will take two equations and eliminate a variable. Let’s take the last twoequations and eliminate z.

5y + 3z = 12x− 5z = 7

5(5y + 3z) = (1)53(2x− 5z) = (7)3

25y + 15z = 56x− 15z = 21

6x + 25y = 26

Notice we have two equations with two variables.

3x + 4y = −46x + 25y = 26

−2(3x + 4y) = (−4)(−2)6x + 25y = 26

−6x− 8y = 86x + 25y = 26

17y = 34y = 2

We may now substitute y = 2 into an equation only containing x and y to solve for x or into an

25

equation containing y and z to solve for z.

3x + 4y = −43x + 4(2) = −4

3x + 8 = −43x = −12x = −4

5y + 3z = 15(2) + 3z = 1

10 + 3z = 13z = −9z = −3

Thus, (−4, 2,−3) is the solution to the system of equations.

.Example 1.c Solve the following system of equations.

2x + 3y − z = 7−3x + 2y − 2z = 7

5x− 4y + 3z = −10

Solution. First, we will take two equations and eliminate a variable. Let’s take the first twoequations and eliminate z.

2x + 3y − z = 7−3x + 2y − 2z = 7

−2(2x + 3y − z) = (7)(−2)−3x + 2y − 2z = 7

−4x− 6y + 2z = −14−3x + 2y − 2z = 7

−7x− 4y = −7

Now, we want to take two different equations but eliminate the same variable: z. We will take thefirst and last equations.

3(2x + 3y − z) = (7)35x− 4y + 3z = −10

6x + 9y − 3z = 215x− 4y + 3z = −10

11x + 5y = 11

26


−7x− 4y = −711x + 5y = 11

5(−7x− 4y) = (−7)54(11x + 5y) = (11)4

−35x− 20y = −3544x + 20y = 44

9x = 9x = 1

We substitute x = 1 into an equation containing only x and y.

−7x− 4y = −7−7(1)− 4y = −7−7− 4y = −7−4y = 0

y = 0

Similarly, we substitute x = 1 and y = 0 into an equation containing x, y, and z.

2x + 3y − z = 72(1) + 3(0)− z = 7

2− z = 7−z = 5z = −5

Thus, (1, 0,−5) is the solution to this system of equations.

.Suggestion If you come across a system of equations not written with the variables on the left sideof the equal sign and the constants on the right side, you may perform basic algebraic manipulationto arrange the system in this more aesthetically pleasing manner.

.Example 2.a Solve the following system of equations.

x + 2y − z = 32x + 3y − 5z = 3

5x + 8y − 11z = 9

Solution. First, we will take two equations and eliminate a variable. Let’s take the first two

27

equations and eliminate x.x + 2y − z = 3

2x + 3y − 5z = 3

−2(x + 2y − z) = (3)(−2)2x + 3y − 5z = 3

−2x− 4y + 2z = −62x + 3y − 5z = 3

−y − 3z = −3

Now, we want to take two different equations but eliminate the same variable: x. We will take thefirst and last equations.

−5(x + 2y − z) = (3)(−5)5x + 8y − 11z = 9

−5x− 10y + 5z = −155x + 8y − 11z = 9

−2y − 6z = −6


y − 3z = −3−2y − 6z = −6

−2(−y − 3z) = (−3)(−2)−2y − 6z = −6

2y + 6z = 6−2y − 6z = 6

0 = 0

Arriving at this identity, we conclude that this is a dependent system of equations, and thus it hasinfinitely many solutions. Unlike in the previous section, we will not describe these solutions.4

.Example 2.b Solve the following system of equations.

3x− y + 2z = 4x− 5y + 4z = 3

6x− 2y + 4z = −8

Solution. First, we will take two equations and eliminate a variable. Let’s take the first and last

4If you are interested in these solutions, please do sign up for my college algebra course.

28

equations and eliminate x.3x− y + 2z = 4

6x− 2y + 4z = −8

−2(3x− y + 2z) = (4)(−2)6x− 2y + 4z = −8

−6x + 2y − 4z = −86x− 2y + 4z = −8

0 = −16

Arriving at this contradiction, we conclude that this is an inconsistent system of equations, andthus it has no solution.

29

1.6 Systems of Linear Equations: Applications

.Example 1 Joseph has a collection of quarters and dimes. A friend counts his change and deter-mines he has $3.70. Joseph knows he has 25 coins. How many quarters and dimes does Josephhave?

Solution. Our goal is to express this situation as a system of two equations. One equation shoulduse the fact that the coins total $3.70, and the other equations should use the fact that Joseph has25 coins. If x represents the number of quarters Joseph possesses and y represents the number ofdimes Joseph possesses, we may write:

.25x + .10y = 3.70x + y = 25

Solve the system using substitution or elimination.

.25x + .10y = 3.70x + y = 25

−10(.25x + .10y) = (3.70)(−10)x + y = 25

−2.5x− y = −37x + y = 25

−1.5x = −12x = 8

x + y = 258 + y = 25

y = 17

Joseph has 8 quarters and 17 dimes.

.Example 2 Ten liters of a 12% HCl solution is mixed with a 20% HCl solution to make a mixturethat is 15% HCl. How many liters of the 20% HCl solution were used to make the mixture?

Solution. Our goal is to express this situation as a system of two equations. Let x represents thenumber of liters of 20% solution used, and let y represents the number of liters in the 15% mixture.One equation should represent the total number of liters, and the other equation should addressthe HCl in the mixture.

x + 10 = y.20x + .12(10) = .15y

30


x + 10 = y.20x + .12(10) = .15y

.20x + .12(10) = .15(x + 10).20x + 1.2 = .15x + 1.5

.05x = 0.3x = 6

6 liters or 20% solution were used to make the mixture.

.Example 3 Against the wind, a plane can fly 2880 miles in 5 hours. Flying with the same wind atits tail, it only takes 4.5 hours. Determine the speed of the wind and the speed of the plane in stillair.

Solution. Our goal is to express this situation as a system of two equations. Let x representsthe speed of the plane in still air, and let y represents the speed of the wind. One equation shouldrepresent the plane flying against the wind, and the other equation should represent the plane flyingwith the wind. Keep in mind the distance formula:

rate · time = distance

(x− y)5 = 2880(x + y)(4.5) = 2880


(x− y)5 = 2880(x + y)(4.5) = 2880

5x− 5y = 28804.5x + 4.5y = 2880

9(5x− 5y) = (2880)910(4.5x + 4.5y) = (2880)10

45x− 45y = 2592045x + 45y = 28800

90x = 54720x = 608

5x− 5y = 28805(608)− 5y = 2880

3040− 5y = 2880−5y = −160

y = 32

The speed of the plane in still air is 608 miles per hour and the speed of the wind is 32 miles perhour.

31

2 Polynomials

2.1 Polynomials

.Definition: Polynomial A polynomial is a mathematical expression containing the operations ofaddition, subtraction, and multiplication.

Examples

1. 4x5 − 3x2 + 4

2. 3m4n2p3

3.1

2x− 3

4

4. x− 3y + 2z − 8

.Definition: Terms, Monomial, Binomial, Trinomial The parts of a polynomial that are addedor subtracted together are called terms.

A polynomial with exactly one term is called a monomial.

A polynomial with exactly two terms is called a binomial.

A polynomial with exactly three terms is called a trinomial.

.Definition: Coefficient, Degree of a Term, Degree of a Polynomial The coefficient is thenumerical part of the term without the variable factors.

The degree of a term is the number of variable factors in the term.

The degree of a polynomial is equal to the highest degree term it contains.

.Example 1.a Determine how many terms the polynomial contains and give its specific name, if itexists. For each term, determine its coefficient and its degree. Finally, determine the degree of thepolynomial.

7x4 + 1

Solution: The polynomial has 2 terms, so it is called a binomial.

Term Coefficient Degree7x4 7 41 1 0

32

Finally, 7x4 + 1 is a fourth degree polynomial.

.Example 1.b Determine how many terms the polynomial contains and give its specific name, if itexists. For each term, determine its coefficient and its degree. Finally, determine the degree of thepolynomial.

4x3 − 7x2 +1

2x− 2

Solution: The polynomial has 4 terms, so it does not have a special name.

Term Coefficient Degree4x3 4 3−7x2 −7 212x 1

21

−2 −2 0

Finally, 4x3 − 7x2 +1

2x− 2 is a third degree polynomial.

.Example 1.c Determine how many terms the polynomial contains and give its specific name, if itexists. For each term, determine its coefficient and its degree. Finally, determine the degree of thepolynomial.

−3.2x4y2z3 + 12.3x3yz2 − 7.9y2z4

Solution: The polynomial has 3 terms, so it is called a trinomial.

Term Coefficient Degree−3.2x4y2z3 −3.2 912.3x3yz2 12.3 6−7.9y2z4 −7.9 6

Finally, −3.2x4y2z3 + 12.3x3yz2 − 7.9y2z4 is a ninth degree polynomial.

.Definition: Descending Order A polynomial is written in descending order if the terms arewritten from highest degree to lowest degree. If two terms have the same degree, then they arewritten in alphabetical order.

Observe that each polynomial in Example 1 was written in descending order.

.Example 2.a Add the polynomials.

(4x2 − 7x + 12) + (3x2 − 8x− 9)

Solution:(4x2 − 7x + 12) + (3x2 − 8x− 9) = 7x2 − 15x + 3

.Example 2.b Subtract the polynomials.

(4x2 − 7x + 12)− (3x2 − 8x− 9)

33

Solution:(4x2 − 7x + 12)− (3x2 − 8x− 9) = 4x2 − 7x + 12− 3x2 + 8x + 9

= x2 + x + 21

.Example 2.c Add the polynomials.(

3

4x3 − 1

2x2 − 5x +

10

3

)+

(2

3x3 − 4x2 +

5

4x +

7

2

)Solution: (

3

4x3 − 1

2x2 − 5x +

10

3

)+

(2

3x3 − 4x2 +

5

4x +

7

2

)

=

(9

12x3 − 1

2x2 − 20

4x +

20

6

)+

(8

12x3 − 8

2x2 +

5

4x +

21

6

)

=17

12x3 − 9

2x2 − 15

4x +

41

6

.Example 2.d Subtract the polynomials.(

2x− 3

4

)−(

7

8x +

1

3

)Solution: (

2x− 3

4

)−(

7

8x +

1

3

)= 2x− 3

4− 7

8x− 1

3

=16

8x− 9

12− 7

8x− 4

12

=9

8x− 13

12

.Example 2.e Subtract the polynomials.

(−3.4x3y2 + 7.8x2y + 9.12x− 4.3y)− (2.78x3y2 − 2.9x2y + 4.75y − 5.3)

Solution:(−3.4x3y2 + 7.8x2y + 9.12x− 4.3y)−(2.78x3y2 − 2.9x2y + 4.75y − 5.3)

= −3.4x3y2 + 7.8x2y + 9.12x− 4.3y−2.78x3y2 + 2.9x2y − 4.75y + 5.3

= −6.18x3y2 + 10.7x2y + 9.12x− 9.05y + 5.3

34

2.2 Polynomials, Continued

.Example 1.a Multiply.

3x2(4x3 − 7x + 2)

Solution:3x2(4x3 − 7x + 2) = 3x2(4x3) + 3x2(−7x) + 3x2(2)

= 12x5 − 21x3 + 6x2

.Example 1.b Multiply.

(2x + 7)(3x− 2)

Solution:(2x + 7)(3x− 2) = 6x2 − 4x + 21x− 14

= 6x2 + 17x− 14

.Example 1.c Multiply.

(x3 + 1)(x3 − 8)

Solution:(x3 + 1)(x3 − 8) = x6 − 8x3 + x3 − 8

= x6 − 7x3 − 8

.Theorem. Square of a Binomial For any a and b,

(a + b)2 = a2 + 2ab + b2.

Proof.(a + b)2 = (a + b)(a + b)

= a2 + ab + ab + b2

= a2 + 2ab + b2


(x + 4)2

Solution:(x + 4)2 = (x)2 + 2(x)(4) + (4)2

= x2 + 8x + 16


(x− 4)2

35

Solution:(x− 4)2 =

[x + (−4)

]2= (x)2 + 2(x)(−4) + (−4)2

= x2 − 8x + 16


(3x2y + 7z)2

Solution:(3x2y + 7z)2 = (3x2y)2 + 2(3x2y)(7z) + (7z)2

= 9x4y2 + 42x2yz + 49z2


(a + b + c)2

Solution:(a + b + c)2 =

[(a + b) + c

]2= (a + b)2 + 2(a + b)(c) + (c)2

= (a + b)2 + 2(ac + bc) + c2

= (a + b)2 + 2ac + 2bc + c2

= a2 + 2ab + b2 + 2ac + 2bc + c2


(3x− 2y − 4z)2

Solution:(3x− 2y − 4z)2 =

[(3x− 2y) + (−4z)

]2= (3x− 2y)2 + 2(3x− 2y)(−4z) + (−4z)2

= (3x− 2y)2 + 2(−12xz + 8yz) + 16z2

= (3x− 2y)2 − 24xz + 16yz + 16z2

= 9x2 − 12xy + 4z2 − 24xz + 16yz + 16z2


(x− 2y + 4z + 7)2

36

Solution:(x− 2y + 4z + 7)2

=[(x− 2y) + (4z + 7)

]2= (x− 2y)2 + 2(x− 2y)(4z + 7) + (4z + 7)2

= (x− 2y)2 + 2(4xz + 7x− 8yz − 14y) + (4z + 7)2

= (x− 2y)2 + 8xz + 14x− 16yz − 28y + (4z + 7)2

= x2 − 4xy + 4y2 + 8xz + 14x− 16yz − 28y + 16z2 + 56z + 49

.Theorem. Product of Conjugates For any a and b,

(a + b)(a− b) = a2 − b2.

Proof.(a + b)(a− b) = (a + b)(a− b)

= a2 − ab + ab− b2

= a2 − b2


(x + 9)(x− 9)

Solution:(x + 9)(x− 9) = (x)2 − (9)2

= x2 − 81


(3m2 − 8n2)(3m2 + 8n2)

Solution:(3m2 − 8n2)(3m2 + 8n2) = (3m2)2 − (8n2)2

= 9m4 − 64n4


(6x2y − 5z3)(6x2y + 5z3)

Solution:(6x2y − 5z3)(6x2y + 5z3) = (6x2y)2 − (5z3)2

= 36x4y2 − 25z6


(x + y + z)(x + y − z)

37

Solution:(x + y + z)(x + y − z) =

((x + y) + z

)((x + y)− z

)= (x + y)2 − (z)2

= x2 + 2xy + y2 − z2


(x− y + z)(x + y − z)

Solution:(x− y + z)(x + y − z) =

(x− (y − z)

)(x + (y − z)

)= (x)2 − (y − z)2

= x2 − (y2 − 2yz + z2)

= x2 − y2 + 2yz − z2


(3x2 − y + z + 1)(3x2 − y − z − 1)

Solution:(3x2 − y + z + 1)(3x2 − y − z − 1)

=((3x2 − y) + (z + 1)

)((3x2 − y)− (z + 1)

)= (3x2 − y)2 − (z + 1)2

= (9x4 − 6x2y + y2)− (z2 + 2z + 1)

= 9x4 − 6x2y + y2 − z2 − 2z − 1

38

2.3 Factoring

.Definition: Factor, Greatest Common Factor Let a, b, c, m, and n be polynomials.If a · b = m, then a and b are called factors of m.

Example. Since 3x · 2y = 6xy, 3x and 2y are factors of 6xy.

If m = a · b and n = a · c, then a is called a common factor of m and n.

Example. Since 6xy = 3x · 2y and 12x2 = 3x · 4x, 3x is a common factor of 6xy and 12x2.

Observation: For any m and n, 1 is a common factor.

If m = a · b and n = a · c, and the only common factor between b and c is 1, then a is called the

greatest common factor of m and n.

.Example 1.a Factor.

15x3 − 25x2y

Solution: First, we note the greatest common factor of 15x3 and −25x2y is 5x2.

15x3 − 25x2y = 5x2(3x) + 5x2(−5y)

= 5x2(3x− 5y)

.Example 1.b Factor.

14x2 + 35x− 21

Solution: First, we note the greatest common factor of 14x2, 35x, and −21 is 7.

14x2 + 35x− 21 = 7(2x2) + 7(5x) + 7(−3)

= 7(2x2 + 5x− 3)

.Example 1.c Factor.

12p2q4r2 − 6p3q2r2 + 20pq4r3

Solution: First, we note the greatest common factor of 12p2q4r2, −6p3q2r2, and 20pq4r3 is 2pq2r2.

12p2q4r2 − 6p3q2r2 + 20pq4r3

= 2pq2r2(6pq2) + 2pq2r2(−3p2) + 2pq2r2(10q2r)

= 2pq2r2(6pq2 − 3p2 + 10q2r)


4x(3x− 1)− y(3x− 1)

39

Solution: First, we note the greatest common factor of 4x(3x− 1) and −y(3x− 1) is 3x− 1.

4x(3x− 1)− y(3x− 1) = (3x− 1)(4x) + (3x− 1)(−y)

= (3x− 1)(4x− y)


4x2(x2 + 2)− 3x(x2 + 2)

Solution: First, we note the greatest common factor of 4x2(x2 + 2) and −3x(x2 + 2) is x(x2 + 2).

4x2(x2 + 2)− 3x(x2 + 2) = [x(x2 + 2)](4x) + [x(x2 + 2)](−3)

= x(x2 + 2)(4x− 3)


(3x− 2)(x + 3) + y(x + 3)

Solution: First, we note the greatest common factor of (3x− 2)(x + 3) and y(x + 3) is x + 3.

(3x− 2)(x + 3) + y(x + 3) = (x + 3)(3x− 2) + (x + 3)(y)

= (x + 3)[(3x− 2) + y]

= (x + 3)(3x− 2 + y)


12x3 − 9x2 + 8x− 6

Solution:12x3 − 9x2 + 8x− 6 = (12x3 − 9x2) + (8x− 6)

= 3x2(4x− 3) + 2(4x− 3)

= (4x− 3)(3x2 + 2)


xy + 3y − 5x− 15

Solution:xy + 3y − 5x− 15 = (xy + 3y) + (−5x− 15)

= y(x + 3)− 5(x + 3)

= (x + 3)(y − 5)


7x2y − 14xy + 28xy − 56y

40

Solution:7x2y − 14xy + 28xy − 56y = 7y(x2 − 2x + 4x− 8)

= 7y[(x2 − 2x) + (4x− 8)]

= 7y[x(x− 2) + 4(x− 2)]

= 7y(x− 2)(x + 4)

.Example 3.d Factor.

8mn− 12m− 12n + 18

Solution:8mn− 12m− 12n + 18 = 2(4mn− 6m− 6n + 9)

= 2[(4mn− 6m) + (−6n + 9)]

= 2[2m(2n− 3)− 3(2n− 3)]

= 2(2n− 3)(2m− 3)

41

2.4 Factoring, Continued


(3x− 7)(2x + 5)

Solution:(3x− 7)(2x + 5) = 6x2 + 15x− 14x− 35

= 6x2 + x− 35


6x2 + x− 35

Solution:6x2 + x− 35 = (3x− 7)(2x + 5)


2x2 + 3x− 5

Solution:2x2 + 3x− 5 = (2x )(x )

We need factors of 5, and the only factors of 5 are 1 and 5. Observe, our two choices:

(2x 1)(x 5) = 2x2 10x x 5(2x 5)(x 1) = 2x2 2x 5x 5

The only way to come up with 2x2 + 3x− 5 would be

(2x 5)(x 1) = 2x2 − 2x + 5x 5.

Finally,(2x + 5)(x− 1) = 2x2 + 3x− 5.


3x2 + 11x− 20

Solution:3x2 + 11x− 20 = (3x )(x )

We need factors of 20:20 = 1 · 20

2 · 104 · 5

We have the following choices:

(3x 1)(x 20) = 3x2 60x x 20(3x 20)(x 1) = 3x2 3x 20x 20(3x 2)(x 10) = 3x2 30x 2x 20(3x 10)(x 2) = 3x2 6x 10x 20(3x− 4)(x + 5) = 3x2 + 15x− 4x− 20(3x 5)(x 4) = 3x2 12x 5x 20

42


5x2 − 13x + 6

Solution:5x2 − 13x + 6 = (5x )(x )


2 · 3We have the following choices:

(5x 1)(x 6) = 5x2 30x x 6(5x 6)(x 1) = 5x2 5x 6x 6(5x 2)(x 3) = 5x2 15x 2x 6(5x 3)(x 2) = 5x2 10x 3x 6

Our choice is:(5x− 3)(x− 2) = 5x2 − 10x− 3x + 6


4x2 + 12x− 7

Solution: Note we have two possibilities to begin with:

4x2 + 12x− 7 = (4x )(x )= (2x )(2x )

The only factors of 7, however, are 1 and 7.

Thus, we have the following choices:

(4x 1)(x 7) = 4x2 28x x 7(4x 7)(x 1) = 4x2 4x 7x 7

(2x 1)(2x 7) = 4x2 14x 2x 7

Our choice is:(2x− 1)(2x + 7) = 4x2 + 14x− 2x− 7

.Example 2.e Factor.

6x2 − 27x + 12

Solution: We should first observe that we may factor out a common factor.

6x2 − 27x + 12 = 3(2x2 − 9x + 4)= 3(2x )(x )


2 · 2

43


3(2x 1)(x 4) = 3(2x2 8x x 4)3(2x 4)(x 1) = 3(2x2 2x 4x 4)3(2x 2)(x 2) = 3(2x2 4x 2x 4)

Our choice is:3(2x− 1)(x− 4) = 3(2x2 − 8x− x + 4)

.Example 2.f Factor.

9m4n− 15m3n2 − 6m2n3

Solution: We should first observe that we may factor out a common factor.

9m4n− 15m3n2 − 6m2n3 = 3m2n(3m2 − 5mn− 2n2)= 3m2n(3m )(m )

The only factors of 2n2 that we need to consider are 2n and n.


3m2n(3m n)(m 2n) = 3m2n(3m2 6mn mn 2n2)3m2n(3m 2n)(m n) = 3m2n(3m2 3mn 2mn 2n2)

Our choice is:

3m2n(3m + n)(m− 2n) = 3m2n(3m2 − 6mn + mn− 2n2)


(x + 7)2 − 6(x + 7)− 16

Solution: Let u = x + 7.

(x + 7)2 − 6(x + 7)− 16 = u2 − 6u− 16

= (u− 8)(u + 2)

= [(x + 7)− 8][(x + 7) + 2]

= (x− 1)(x + 9)


4x6 − 4x3 − 15

Solution: Let u = x3.4x6 − 4x3 − 15 = 4u2 − 4u− 15

= (2u )(2u )

= (2u + 3)(2u− 5)

= (2x3 + 3)(2x3 − 5)

44


4(2n2 + 1)2 − 7(2n2 + 1) + 3

Solution: Let u = 2n2 + 1.

4(2n2 + 1)2 − 7(2n2 + 1) + 3 = 4u2 − 7u + 3

= (4u )(u )

= (4u− 3)(u− 1)

= [4(2n2 + 1)− 3][(2n2 + 1)− 1]

= (8n2 + 4− 3)(2n2)

= (8n2 + 1)(2n2)

45

2.5 Factoring III

.Factoring III Recall two theorems from multiplying polynomials:

(a + b)2 = a2 + 2ab + b2

(a + b)(a− b) = a2 − b2


4x2 + 28x + 49

Solution:4x2 + 28x + 49 = (2x)2 + 2(2x)(7) + (7)2

= (2x + 7)2


9x2 − 30x + 25

Solution:9x2 − 30x + 25 = (3x)2 + 2(3x)(−5) + (−5)2

= (3x− 5)2


100x4y2 + 60x2yz3 + 9z6

Solution:100x4y2 + 60x2yz3 + 9z6 = (10x2y)2 + 2(10x2y)(3z3) + (3z3)2

= (10x2y + 3z3)2


32x2 − 80xy + 50y2

Solution:32x2 − 80xy + 50y2 = 2(16x2 − 40xy + 25y2)

= 2[(4x)2 + 2(4x)(−5y) + (−5y)2]

= 2(4x− 5y)2


x2 − 16

Solution:x2 − 16 = (x)2 − (4)2

= (x + 4)(x− 4)

46


4m8 − 9n2

Solution:4m8 − 9n2 = (2m4)2 − (3n)2

= (2m4 + 3n)(2m4 − 3n)


12x3 − 3x

Solution:12x3 − 3x = 3x(4x2 − 1)

= 3x[(2x)2 − (1)2]

= 3x(2x + 1)(2x− 1)


x4 − 16

Solution:x4 − 16 = (x2)2 − (4)2

= (x2 + 4)(x4 − 4)

= (x2 + 4)[(x2)2 − (2)2]

= (x2 + 4)(x + 2)(x− 2)


(x + y)2 + 2(x + y) + 1

Solution: Let u = x + y.

(x + y)2 + 2(x + y) + 1 = u2 + 2u + 1

= (u + 1)2

= [(x + y) + 1]2

= (x + y + 1)2


25− (3z + 4)2

47

Solution: Let u = 3z + 4.

25− (3z + 4)2 = 25− u2

= (5 + u)(5− u)

= [5 + (3z + 4)][5− (3z + 4)]

= (5 + 3z + 4)(5− 3z − 4)

= (3z + 9)(−3z + 1)

= 3(z + 3)(−3z + 1)

.Theorem. Factoring the Sum of Cubes For any a and b,

a3 + b3 = (a + b)(a2 − ab + b2).

Proof.(a + b)(a2 − ab + b2) = (a + b)(a2) + (a + b)(−ab) + (a + b)(b2)

= a3 + a2b− a2b− ab2 + ab2 + b3

= a3 + b3

.Theorem. Factoring the Difference of Cubes For any a and b,

a3 − b3 = (a− b)(a2 + ab + b2).

Proof. Left for the student.


8x3 + 27

Solution:8x3 + 27 = (2x)3 + (3)3

= (2x + 3)[(2x)2 − (2x)(3) + (3)2]

= (2x + 3)(4x2 − 6x + 9)


1− 64x3

Solution:1− 64x3 = (1)3 − (4x)3

= (1− 4x)[(1)2 + (1)(4x) + (4x)2]

= (1− 4x)(1 + 4x + 16x2)

48


125x6 − 27y3z9

Solution:125x6 − 27y3z9 = (5x2)3 + (3yz3)3

= (5x2 + 3yz3)[(5x2)2 − (5x2)(3yz3) + (3yz3)2]

= (5x2 + 3yz3)(25x4 − 15x2yz3 + 9z6)


216− (a + b)3

Solution: Let u = a + b.

216− (a + b)3 = 216− u3

= (6)3 − u3

= (6− u)(36 + 6u + u2)

= [6− (a + b)][36 + 6(a + b) + (a + b)2]

= (6− a− b)(36 + 6a + 6b + a2 + 2ab + b2)


(3x− 2y)3 + (x + 1)3

Solution: Let u = a + b and let v = x + 1.

(3x− 2y)3 + (x + 1)3

= u3 + v3

= (u + v)(u2 − uv + v2)= [(3x− 2y) + (x + 1)]

[(3x− 2y)2 − (3x− 2y)(x + 1) + (x + 1)2]= (4x− 2y + 1)

[(9x2 − 12xy + 4y2)− (3x2 + 3x− 2xy − 2y) + (x2 + 2x + 1)]= (4x− 2y + 1)

(9x2 − 12xy + 4y2 − 3x2 − 3x + 2xy + 2y + x2 + 2x + 1)= (4x− 2y + 1)(7x2 − 10xy + 4y2 − x + 2y + 1)

49

2.6 Factoring IV: An Application

.Example 1.a Solve.

x2 − 5x + 6 = 0

Solution:x2 − 5x + 6 = 0

(x− 2)(x− 3) = 0

Thus, by the zero factor property, either

x− 2 = 0 or x− 3 = 0

x = 2 x = 3

The solution set is {2, 3}..Example 1.b Solve.

3x2 − 10x− 8 = 0

Solution:3x2 − 10x− 8 = 0

(3x + 2)(x− 4) = 0


3x + 2 = 0 or x− 4 = 0

3x = −2 x = 4

x = −2

3

The solution set is

{−2

3, 4

}.

.Example 1.c Solve.

3x3 − 3x2 − 6x = 0

Solution:3x3 − 3x2 − 6x = 0

3x(x2 − x− 2) = 0

3x(x + 1)(x− 2) = 0


3x = 0 or x + 1 = 0 or x− 2 = 0

x = 0 x = −1 x = 2

50

The solution set is {−1, 0, 2}..Example 2.a Solve.

2x2 = 5x + 3

Solution:2x2 = 5x + 3

2x2 − 5x− 3 = 0

(2x + 1)(x− 3) = 0


2x + 1 = 0 or x− 3 = 0

x = −1

2x = 3

The solution set is

{−1

2, 3

}.

.Example 2.b Solve.

(x− 3)(x− 5) = −1

Solution:(x− 3)(x− 5) = −1

x2 − 8x + 15 = −1

x2 − 8x + 16 = 0

(x− 4)2 = 0

Thus, by the zero factor property,x− 4 = 0

x = 4

The solution set is {4}..Example 2.c Solve.

3x(x− 2) = 4(x + 1) + 4

Solution:3x(x− 2) = 4(x + 1) + 4

3x2 − 6x = 4x + 4 + 4

3x2 − 6x = 4x + 8

3x2 − 10x− 8 = 0

(3x + 2)(x− 4) = 0

51

The solution set is

{−2

3, 4

}.

.Pythagorean Theorem If a right triangle has legs of lengths a and b and hypotenuse of length c,then

a2 + b2 = c2.

.Example 3.a

Solution:x2 + (x + 1)2 = (x + 2)2

x2 + x2 + 2x + 1 = x2 + 4x + 42x2 + 2x + 1 = x2 + 4x + 4x2 − 2x− 3 = 0

(x− 3)(x + 1) = 0

The solution set is {−1, 3}. Since x represents the length of a leg of the triangle, we know x = 3.Thus, the sides of the triangle are 3, 4, and 5.

.Example 3.b

Solution:(x− 8)2 + (x− 1)2 = x2

x2 − 16x + 64 + x2 − 2x + 1 = x2

2x2 − 18x + 65 = x2

x2 − 18x + 65 = 0(x− 5)(x− 13) = 0

The solution set is {5, 13}. Since x−8 represents the length of a leg of the triangle, we know x = 13(otherwise x− 8 = −3). Thus, the sides of the triangle are 5, 12, and 13.

52

3 Rational Expressions

3.1 Rational Expressions

The notes for this section will be presented from the text.

3.2 Rational Expressions, Continued


3.3 Rational Expressions III: Complex Rational Expressions


3.4 Rational Expressions IV


53

4 Radicals

4.1 Radicals: An Introduction

.Definition: nth Root, Principle nth Root The nth root of a is b if

bn = a.

The principle nth root of a is b if

bn = a and b ≥ 0 if possible.

The principle nth root of a is denoted by

n√a = b.

If n = 2, the principle square root of a is denoted by

√a = b.

.Example 1.a Evaluate. √

81

Solution: Since 92 = 81, √81 = 9.

.Example 1.b Evaluate. √

169

Solution: Since 132 = 169, √169 = 13.

.Example 1.c Evaluate. √

1

25

Solution: Since

(1

5

)2

=1

25, √

1

25=

1

5.

.Example 1.d Evaluate. √

−36

Solution: Notice (−6)2 = 36. In fact, for any real number x, x2 ≥ 0. Thus,√−36 is not a real

number.

.Example 1.e Evaluate.

3√

64

54

Solution: Since 43 = 64,3√

64 = 4.

.Example 1.f Evaluate.

3√−64

Solution: Since (−4)3 = 64,3√−64 = −4.

.Example 1.g Evaluate.

4√

16

Solution: Since 24 = 16,4√

16 = 2.

.Example 1.h Evaluate.

4√−16

Solution: Notice (−2)4 = 16. In fact, for any real number x, x4 ≥ 0. Thus, 4√−16 is not a real

number.

.Example 1.i Evaluate. √

15

Solution: Using a calculator, we can approximate

√15 ≈ 3.873.

.Example 1.j Evaluate.

3√

31

Solution: Using a calculator, we can approximate

3√

31 ≈ 3.141.

.Example 2.a Simplify. √

x2

Solution: √x2 = |x|

.Example 2.b Simplify. √

x10

Solution: √x10 = |x5|

55

.Example 2.c Simplify. √

16x6

Solution: √16x6 = 4|x3|

.Example 2.d Simplify. √

x4

Solution: √x4 = x2

.Example 2.e Simplify.

3√x3

Solution:3√x3 = x

.Example 2.f Simplify.

4√x4

Solution:4√x4 = |x|

.Example 2.g Simplify.

4√

81x8

Solution:4√

81x8 = 3x2

.Example 3.a Simplify. Assume all variables represent nonnegative values.

√x2

Solution: √x2 = x

.Example 3.b Simplify. Assume all variables represent nonnegative values.

√9x6

Solution: √9x6 = 3x3

.Example 3.c Simplify. Assume all variables represent nonnegative values.√

25x6y2

56

Solution: √25x6y2 = 5x3y

.Example 3.d Simplify. Assume all variables represent nonnegative values.√

4x2

y4

Solution: √4x2

y4=

2x

y2

.Example 3.e Simplify. Assume all variables represent nonnegative values.

3√

8x6

Solution:3√

8x6 = 2x2

.Example 3.f Simplify. Assume all variables represent nonnegative values.

4√

625x8y12

Solution:4√

625x8y12 = 5x2y3

.Example 3.g* Simplify. Assume all variables represent nonnegative values.

√x2 + 2x + 1

Solution: √x2 + 2x + 1 =

√(x + 1)2

= x + 1

.Example 4.a Let f(x) =

√x + 7. Find f(2).

Solution:

f(2) =√

(2) + 7

=√

9

= 3

.Example 4.b Let f(x) =

√x + 7. Find f(18).

Solution:

57

f(18) =√

(18) + 7

=√

25

= 5

.Example 4.c Let f(x) =

√x + 7. Find f(−7).

Solution:

f(−7) =√

(−7) + 7

=√

0

= 0

.Example 4.d Let f(x) =

√x + 7. Find f(−11).

Solution:

f(−11) =√

(−11) + 7

=√−4

However,√−4 is not a real number, so therefore f(−11) is undefined.

.Example 5.a State the domain of f(x) =

√x + 7.

Solution: For f(x) =√x + 7 to be defined,

x + 7 ≥ 0

x ≥ −7.

Thus, f(x) is defined for all real numbers x ≥ −7. The domain of f(x) is {x|x ≥ −7}..Example 5.b State the domain of g(x) =

√7x− 42.

Solution: For g(x) =√

7x− 42 to be defined,

7x− 42 ≥ 0

7x ≥ 42

x ≥ 6.

Thus, g(x) is defined for all real numbers x ≥ 6. The domain of g(x) is {x|x ≥ 6}..Example 5.c State the domain of h(x) =

√2x + 5.

Solution: For h(x) =√

2x + 5 to be defined,

58

2x + 5 ≥ 0

2x ≥ −5

x ≥ −5

2.

Thus, h(x) is defined for all real numbers x ≥ −52. The domain of h(x) is

{x|x ≥ −5

2

}.

59

4.2 Radicals: Rational Exponents

.Definition: x1/n We define

x1/n = n√x.

There are many reasons for this definition that have to do with our general intuition on howexponents should behave. We will develop these connections throughout the rest of the section andchapter. For now, we can just accept this definition as the one that works.

.Example 1.a Simplify.

91/2

Solution:91/2 =

√9

= 3

.Example 1.b Simplify.

641/3

Solution:641/3 = 3

√64

= 4

.Example 1.c Simplify.

−251/2

Solution:−251/2 = −

√25

= −5

.Example 1.d Simplify.

(−25)1/2

Solution:(−25)1/2 =

√−25

Thus, (−25)1/2 is not a real number.

.Example 1.e Simplify. Assume all variables represent positive values.

(81x2)1/2

Solution:(81x2)1/2 =

√81x2

= 9x

60

.Example 1.f Simplify. Assume all variables represent positive values.

4x1/2

Solution:4x1/2 = 4

√x

.Example 1.g Simplify. Assume all variables represent positive values.

(4x)1/2

Solution:(4x)1/2 =

√4x

= 2√x

.Example 1.h Simplify. Assume all variables represent positive values.(

144x8

169y6

)1/2

Solution: (144x8

169y6

)1/2

=

√144x8

169y6

=12x4

13y3

.Theorem: xm/n

xm/n =(

n√x)m

Proof.xm/n = x

1n·m

=(x

1n

)m= ( n√x)

m


813/4

Solution:813/4 =

(4√

81)3

= 33

= 27

61


642/3

Solution:642/3 =

(3√

64)2

= 42

= 16

.Example 2.c Simplify. Assume all variables represent nonnegative values.

323/5

Solution:323/5 =

(5√

32)3

= 23

= 8

.Example 2.d Simplify. Assume all variables represent nonnegative values.

(8x3)2/3

Solution:

(8x3)2/3 =(

3√

8x3)2

= (2x)2

= 4x2

.Theorem 2. xm/n

xm/n = n√xm

Proof.xm/n = xm· 1

n

= (xm)1/n

= n√xm


x2/3

62

Solution:x2/3 =

3√x2


(x2y3)211

Solution:(x2y3)

211 = 11

√(x2y3)2

= 11√x4y6


(x + 4)3/5

Solution:(x + 4)3/5 = 5

√(x + 4)3

.Definition: x−m

x−m =1

xm


16−1/2

Solution:

16−1/2 =1

161/2

=1√16

=1

4.Example 4.b Simplify.

−49−1/2

Solution:

−49−1/2 = − 1

491/2

= − 1√49

= −1

7.Example 4.c Simplify.

(−49)−1/2

63

Solution:

(−49)−1/2 =1

(−49)1/2

=1√−49

However,√−49 is not a real number, so neither is (−49)−1/2.


27−2/3

Solution:

27−2/3 =1

272/3

=1(

3√

27)2

=1

32

=1

9

.Example 4.e Simplify. (

1

125

)−2/3Solution: (

1

125

)−2/3= 1252/3

=(

3√

125)2

= 52

= 25


−64−2/3

64

Solution:

−64−2/3 = − 1

642/3

= − 1(3√

64)2

= − 1

42

= − 1

16.Basic Rules for Exponents

xm · xn = xm+n

xm

xn= xm−n

(xm)n = xmn

(ax

by

)m

=amxm

bmym

.Example 5.a Simplify. Write the answer with positive exponents. Assume all variables representpositive values.

x1/2 · x2/3

Solution:x1/2 · x2/3 = x

12+ 2

3

= x36+ 4

6

= x7/6

.Example 5.b Simplify. Write the answer with positive exponents. Assume all variables representpositive values.

x3/4

x2/3

Solution:x3/4

x2/3= x

34− 2

3

= x912− 8

12

= x1/12

65

.Example 5.c Simplify. Write the answer with positive exponents. Assume all variables representpositive values. (

x3/4)1/2

Solution: (x3/4

)1/2= x

34· 12

= x3/8

.Example 5.d Simplify. Write the answer with positive exponents. Assume all variables representpositive values.

(x2y3)2/3

Solution:(x2y3)2/3 = (x2)

2/3(y3)

2/3

= x4/3y2

.Example 5.e Simplify. Write the answer with positive exponents. Assume all variables representpositive values. (

x3

y4

)1/2

Solution: (x3

y4

)1/2

=(x3)

1/2

(y4)1/2

=x3/2

y2

.Example 5.f Simplify. Write the answer with positive exponents. Assume all variables representpositive values. (

9x6

y8

)1/2

Solution: (9x6

y8

)1/2

=(9x6)

1/2

(y8)1/2

=91/2x3

y4

=3x3

y4

.Example 6.a Simplify by writing as a root with a smaller index.

4√

25

66

Solution:4√

25 = (25)1/4

= (52)1/4

= 51/2

=√

5

.Example 6.b Simplify by writing as a root with a smaller index.

6√

8

Solution:6√

8 = (8)1/6

= (23)1/6

= 21/2

=√

2

.Example 6.c Simplify by writing as a root with a smaller index.

8√x6

Solution:8√x6 = (x6)1/8

= x3/4

=4√x3

.Example 6.d Simplify by writing as a root with a smaller index.

9√

x3y6

Solution:9√

x3y6 = (x3y6)1/9

= x1/3y2/3

= (xy2)1/3

= 3√

xy2

.Example 7.a Simplify. Write the answer as a radical. Assume all variables represent positivevalues. √

x 3√x

67

Solution: √x 3√x = x1/2 · x1/3

= x3/6 · x2/6

= x5/6

=6√x5

.Example 7.b Simplify. Write the answer as a radical. Assume all variables represent positivevalues.

4√x3 5√x2

Solution:4√x3 5√x2 = x3/4 · x2/5

= x15/20 · x8/20

= x23/20

=20√x23

.Example 7.c Simplify. Write the answer as a radical. Assume all variables represent positivevalues.

3√x2

√x

Solution:3√x2

√x

=x2/3

x1/2

= x23− 1

2

= x46− 3

6

= x1/6

= 6√x

.Example 7.d Simplify. Write the answer as a radical. Assume all variables represent positivevalues. √

5 · 3√

7

68

Solution: √5 · 3√

7 = 51/2 · 71/3

= 53/6 · 72/6

= (53 · 72) 1/6

= (6125)1/6

= 6√

6125

.Example 8.a Simplify. Write the answer as a single radical. Assume all variables represent positivevalues.

3

√√x

Solution:3√√

x = (√x)

1/3

=(x1/2

)1/3= x1/6

= 6√x

.Example 8.b Simplify. Write the answer as a single radical. Assume all variables represent positivevalues.

3

√4√x

Solution:3√

4√x = ( 4

√x)

1/3

=(x1/4

)1/3= x1/12

= 12√x

69

4.3 Radicals III

.Theorem: n

√x · n√y = n√xy For any x ≥ 0 and y ≥ 0,

n√x · n√y = n√xy.

Proof.n√x · n√y = x1/ny1/n

= (xy)1/n

= n√xy


3 ·√

27

Solution: √3 ·√

27 =√

81

= 9


5 ·√

125


125 =√

625

= 25


3√

2 · 3√

32

Solution:3√

2 · 3√

32 = 3√

64

= 4


4√

4 · 4√

4

Solution:4√

4 · 4√

4 = 4√

16

= 2


3√

5x · 3√

4y

70

Solution:3√

5x · 3√

4y = 3√

20xy


√x3 ·√x5

Solution: √x3 ·√x5 =

√x8

= x4

.Corollary:

√x ·√x = x For any x ≥ 0,

√x ·√x = x.

Proof. √x ·√x =√x2

= x


7 ·√

7


7 =√

72

= 7

.Example 2.b Simplify. Assume all variables represent nonnegative values.√

3x2y ·√

3x2y

Solution: √3x2y ·

√3x2y =

√(3x2y)2

= 3x2y

.Rules for Simplifying Radicals A radical expression is simplified if:

1. there are no factors inside any radical raised to a power greater than or equal to the index,

2. there are no fractions inside any radicals, and

3. there are no radicals in any denominator of any fraction.

71

.Theorem: n

√xy

=n√xn√y

For any x ≥ 0 and y ≥ 0,

n

√x

y=

n√x

n√y.

Proof.

n

√x

y=

(x

y

)1/n

=x1/n

y1/n

=n√x

n√y


5

9

Solution: √5

9=

√5√9

=

√5

3


14√2

Solution: √14√2

=√

7


3

√25

27

Solution:3

√25

27=

3√

253√

27

=3√

25

3


4√

484√

3

72

Solution:4√

484√

3= 4√

16

= 2


3

√9

x3

Solution:3

√9

x3=

3√

93√x3

=3√

9

x


8

Solution: √8 =√

23

=√

22 ·√

2

= 2√

2

Alternatively, √8 =√

4 ·√

2 = 2√

2


48

Solution: √48 =

√24 · 3

=√

22 · 22 · 3

= 2 · 2 ·√

3

= 4√

3

Alternatively, √48 =

√16 ·√

3 = 4√

3


72

73

Solution: √72 =

√23 · 32

=√

22 · 2 · 32

= 2 ·√

2 · 3

= 6√

2


√36 ·√

2 = 6√

2


56

Solution: √56 =

√23 · 7

=√

22 · 2 · 7

= 2 ·√

2 · 7

= 2√

14


√4 ·√

14 = 2√

14


4√

12

Solution:4√

12 = 4√

22 · 3

= 4 · 2 ·√

3

= 8√

3

Alternatively,4√

12 = 4 ·√

4 ·√

3 = 8√

3


−3√

54

Solution:−3√

54 = −3√

2 · 33

= −3√

2 · 32 · 3

= −3 · 3 ·√

2 · 3

= −9√

6

74

Alternatively,−3√

54 = −3 ·√

9 ·√

6 = −9√

6

.Example 4.g Simplify. √

252

Solution: √252 =

√22 · 32 · 7

= 2 · 3 ·√

7

= 6√

7


√36 ·√

7 = 6√

7

.Example 4.h Simplify. √

686

Solution: √686 =

√2 · 73

=√

2 · 72 · 7

=√

2 · 7 ·√

7

= 7√

14


√49 ·√

14 = 7√

14

.Example 4.i Simplify. √

720

Solution: √720 =

√24 · 32 · 5

=√

22 · 22 · 32 · 5

= 2 · 2 · 3 ·√

5

= 12√

5


√144 ·

√5 = 12

√5

.Example 4.j Simplify.

3√

32

75

Solution:3√

32 =√

25

=3√

23 · 22

= 2 · 3√

22

= 2 3√

4

Alternatively,3√

32 =3√

8 · 3√

4 = 23√

4

.Example 4.k Simplify.

3√

162

Solution:3√

162 =3√

2 · 34

=3√

2 · 33 · 3

= 3√

2 · 3 · 3√

3

= 3 3√

6

Alternatively,3√

162 =3√

27 · 3√

6 = 33√

6

.Example 4.l Simplify.

4√

208

Solution:4√

208 =4√

24 · 13

= 2 4√

13

Alternatively,4√

208 =4√

16 · 4√

13 = 24√

13


√x3

Solution: √x3 =

√x2 · x

= x√x


√x11

76

Solution: √x11 =

√x10 · x

= x5√x


3√x11

Solution:3√x11 =

3√x9 · x2

= x3 3√x2

.Example 5.d Simplify. Assume all variables represent nonnegative values.√

32x5y4

Solution: √32x5y4 =

√25 · x5 · y4

= 22x2y2√

2x

= 4x2y2√

2x


3√

54x8y7

Solution:3√

54x8y7 = 3√

2 · 33 · x8 · y7

= 3x2y2 3√

2x2y


4√

8x8y7

Solution:4√

8x8y7 = 4√

23 · x8 · y7

= x2y 4√

8y3


12 ·√

15

77


15 =√

22 · 3 ·√

3 · 5

=√

22 · 32 · 5

= 6√

5

.Example 6.b Simplify. Assume all variables represent nonnegative values.√

5x3y2 ·√

10xy3

Solution: √5x3y2 ·

√10xy3 =

√5x3y2 ·

√2 · 5 · xy3

=√

2 · 52 · x4 · y5

= 5x2y2√

2y

.Example 6.c Simplify. Assume all variables represent nonnegative values.(

4x2y√

2x2y)(

3xy4√

6xy)

Solution: (4x2y

√2x2y

) (3xy4√

6xy)

= 12x3y5√

2x2y · 6xy

= 12x3y5√

12x3y2

= 12x3y5√

22 · 3 · x3 · y2

= 12x3y5 · 2xy√

3x

= 24x4y6√

3x


540√12

Solution: √540√12

=

√540

12

=√

45

= 3√

5

.Example 6.e Simplify. Assume all variables represent positive values.√

72x5y7√9x2y

78

Solution: √72x5y7√9x2y

=

√72x5y7

9x2y

=√

8x3y6

= 2xy3√

2x

79

4.4 Radicals IV

.Example 1.a Add.

4√

3 + 7√

3

Solution:4√

3 + 7√

3 =√

3(4 + 7)

= 11√

3

.Example 1.b Subtract.

23√

7− 63√

7

Solution:2 3√

7− 6 3√

7 = 3√

7(2− 6)

= −4 3√

7

.Example 1.c Subtract. Assume all variables represent nonnegative values.

4√

5x− 3√

5x

Solution:4√

5x− 3√

5x =√

5x(4− 3)

=√

5x

.Example 1.d Subtract. Assume all variables represent nonnegative values.

2x2√x− 5x2

√x

Solution:2x2√x− 5x2

√x = x2

√x(2− 5)

= −3x2√x

.Example 2.a Add. √

18 +√

32

Solution: √18 +

√32 =

√9 · 2 +

√16 · 2

= 3√

2 + 4√

2

= 7√

2

.Example 2.b Subtract.

6√

20− 2√

80

80

Solution:6√

20− 2√

80 = 6√

4 · 5− 2√

16 · 5

= 6 · 2√

5− 2 · 4√

5

= 12√

5− 8√

5

= 4√

5

.Example 2.c Subtract.

3√

54− 53√

2

Solution:3√

54− 5 3√

2 = 3√

27 · 2− 5 3√

2

= 3 3√

2− 5 3√

2

= −2 3√

2

.Example 2.d Subtract. Assume all variables represent nonnegative values.

2√

48x5 − 7x√

12x3

Solution:2√

48x5 − 7x√

12x3 = 2√

16x4 · 3x− 7x√

4x2 · 3x

= 2 · 4x√

3x− 7x · 2x√

3x

= 8x2√

3x− 14x2√

3x

= −6x2√

3x


2x4√

16x− 4 4√x + 7

4√x5

Solution:2x 4√

16x− 4 4√x + 7

4√x5 = 2x

4√

24 · x− 4 4√x + 7

4√x4 · x

= 2x · 2 4√x− 4 4

√x + 7 · x 4

√x

= 4x 4√x− 4 4

√x + 7x 4

√x

= 11x 4√x− 4 4

√x

= (11x− 4) 4√x


6(√

5−√

3)

81

Solution: √6(√

5−√

3)

=√

6 ·√

5−√

6 ·√

3

=√

30−√

18

=√

30− 3√

2


6√

5(

2√

8− 3√

10)

Solution:6√

5(2√

8− 3√

10)

= 6√

5 · 2√

8− 6√

5 · 3√

10

= 12√

40− 18√

50

= 12 · 2√

10− 18 · 5√

2

= 24√

10− 90√

2

.Example 3.c Simplify. (

4 +√

3)(

6−√

7)

Solution: (4 +√

3) (

6−√

7)

= (4)(6) + (4)(−√

7)

+(√

3)

(6) +(√

3) (−√

7)

= 24− 4√

7 + 6√

3−√

21

.Example 3.d Simplify. (

2√

6− 3√

5)(

5√

6− 9√

5)

Solution: (2√

6− 3√

5) (

5√

6− 9√

5)

= 10√

36− 18√

30− 15√

30 + 27√

25

= 10 · 6− 18√

30− 15√

30 + 27 · 5

= 60− 18√

30− 15√

30 + 135

= 195− 33√

30

.Example 3.e Simplify. Assume all variables represent nonnegative values.(√

x + 3) (√

x + 7)

Solution:(√x + 3) (

√x + 7) =

√x2 + 7

√x + 3

√x + 21

= x + 10√x + 21

82

.Example 3.f Simplify. (

2 + 5√

3)2

Solution: (2 + 5

√3)2

=(2 + 5

√3) (

2 + 5√

3)

= 4 + 10√

3 + 10√

3 + 25√

9

= 4 + 20√

3 + 75

= 79 + 20√

3

.Example 3.g Simplify. (

3√

2− 6√

7)2

Solution: (3√

2− 6√

7)2

=(3√

2− 6√

7) (

3√

2− 6√

7)

= 9√

4− 18√

14− 18√

14 + 36√

49

= 18− 36√

14 + 252

= 270− 36√

14

.Example 3.h Simplify. Assume all variables represent nonnegative values.(√

x + 4)2

Solution:(√x + 4)

2= (√x + 4) (

√x + 4)

=√x2 + 4

√x + 4

√x + 16

= x + 8√x + 16

.Example 3.i Simplify. (

8 +√

3)(

8−√

3)

Solution: (8 +√

3) (

8−√

3)

= 64− 8√

3 + 8√

3−√

9

= 64− 3

= 61

.Example 3.j Simplify. (

3√

7− 4√

5)(

3√

7 + 4√

5)

83

Solution: (3√

7− 4√

5) (

3√

7 + 4√

5)

= 9√

49 + 12√

35− 12√

35− 16√

25

= 63− 80

= −17


8 ·√

6−√

5 ·√

15


6−√

5 ·√

15 =√

48−√

75

= 4√

3− 5√

3

= −√

3


3√

15003√

3+

3√

108

Solution:3√

15003√

3+ 3√

108 = 3√

500 + 3√

108

= 3√

125 · 4 + 3√

27 · 4

= 5 3√

4 + 3 3√

4

= 8 3√

4

84

4.5 Radicals V


1√2

Solution:1√2

=1√2·√

2√2

=

√2

2


4√3

Solution:4√3

=4√3·√

3√3

=4√

3

3


9

5

Solution: √9

5=

√9√5

=3√5

=3√5·√

5√5

=3√

5

5

.Example 1.d Simplify. Assume all variables represent positive values.√

2

x

85

Solution: √2

x=

√2√x

=

√2√x·√x√x

=

√2x

x


13√

2

Solution:13√

2=

13√

2·

3√

22

3√

22

=3√

4

2


13√

4

Solution:13√

4=

13√

22

=1

3√

22·

3√

23√

2

=3√

2

2

.Example 1.g Simplify. Assume all variables represent positive values.

3

√3x

4y2

86

Solution:

3

√3x

4y2=

3√

3x3√

4y2

=3√

3x3√

22y2

=3√

3x3√

22y2·

3√

2y3√

2y

=3√

6xy

2y

.Example 1.h Simplify.

14√

8

Solution:14√

8=

14√

23

=1

4√

23·

4√

24√

2

=4√

2

2

.Example 1.i Simplify.

3√

8√15

Solution:3√

8√15

=6√

2√15

=6√

2√15·√

15√15

=6√

30

15

=2√

30

5

.Example 2.a Multiply. (√

x +√y) (√

x−√y)

87

Solution: (√x +√y) (√

x−√y)

=√x2 −√xy +

√xy −

√y2

= x− y


3

4−√

2

Solution:3

4−√

2=

3

4−√

2· 4 +

√2

4 +√

2

=12 + 3

√2

16− 2

=12 + 3

√2

14


6√3 +√

18

Solution: √6√

3 +√

18=

√6√

3 +√

18·√

3−√

18√3−√

18

=

√18−

√108

3− 18

=3√

2− 6√

3

−15

= −3(√

2− 2√

3)

15= −√

2− 2√

3

5


2−√

3

2 +√

3

Solution:2−√

3

2 +√

3=

2−√

3

2 +√

3· 2−

√3

2−√

3

=4− 4

√3 + 3

4− 3

= 7− 4√

3

88


√x√

x +√

2y

Solution: √x√

x +√

2y=

√x√

x +√

2y·√x−√

2y√x−√

2y

=

√x2 −

√2xy

x− 2y

=x−√

2xy

x− 2y

.Example 3.a Rationalize the numerator. √

2

3

Solution: √2

3=

√2

3·√

2√2

=2

3√

2

.Example 3.b Rationalize the numerator.

3−√x√

2

Solution:3−√x√

2=

3−√x√

2· 3 +

√x

3 +√x

=9− x

3√

2 +√

2x

89

4.6 Radicals VI

.Example 1.a Solve. √

x = 4

Solution:(√x)

2= (4)2

x = 16

Check:√

16 = 4

.Example 1.b Solve. √

x = 13

Solution:(√x)

2= (13)2

x = 169

Check:√

169 = 13

.Example 1.c Solve. √

x = −5

Solution:(√x)

2= (−5)2

x = 25

Check:√

25 6= −5. Thus, x = 25 is an extraneous solution, so there is no solution.


x− 3 = 2

Solution: (√x− 3

)2= (2)2

x− 3 = 4

x = 7

Check:√

7− 3 = 2

.Example 2.b Solve. √

4x + 1 = 5

90

Solution: (√4x + 1

)2= (5)2

4x + 1 = 25

4x = 24

x = 6

Check:√

4(6) + 1 = 5

.Example 2.c Solve.

3√x + 8 = −3

Solution: (3√x + 8

)3= (−3)3

x + 8 = −27

x = −35

Check: 3√−35 + 8 = −3

.Example 2.d Solve.

4√

2x + 1 = −2

Solution: (4√

2x + 1)4

= (−2)4

2x + 1 = 16

2x = 15

x =15

2

Check: 4

√2(152

)+ 1 6= −2. Thus, x = 15

2is an extraneous solution, so there is no solution.


2x + 8 =√

3x + 2

Solution: (√2x + 8

)2=(√

3x + 2)2

2x + 8 = 3x + 2

6 = x

Check:√

2(6) + 8 =√

20 =√

3(6) + 2

91

.Example 3.b Solve.

3√

7x− 4 = 3√

1− 3x

Solution: (3√

7x− 4)3

=(

3√

1− 3x)3

7x− 4 = 1− 3x

10x = 5

x =1

2

Check: 3

√7(12

)− 4 = 3

√−1

2= 3

√1− 3

2

.Example 3.c Solve.

x + 1 =√x + 13

Solution:(x + 1)2 =

(√x + 13

)2x2 + 2x + 1 = x + 13

x2 + x− 12 = 0

(x + 4)(x− 3) = 0

We have solutions of x = −4 and x = 3. Checking those solutions however, we find that x = −4 isextraneous, as

−4 + 1 6=√−4 + 13.

Our solution is x = 3.

.Example 3.d Solve. √

x2 − x + 3− 1 = 2x

Solution: √x2 − x + 3 = 2x + 1(√

x2 − x + 3)2

= (2x + 1)2

x2 − x + 3 = 4x2 + 4x + 1

0 = 3x2 + 5x− 2

0 = (3x− 1)(x + 2)

We have solutions of x = 13

and x = −2. Checking those solutions however, we find that x = −2 isextraneous, as √

(−2)2 − (−2) + 3− 1 6= 2(−2).

92

Our solution is x = 13.

.Example 3.e Solve. √

x− 1 =√

2x + 2

Solution:(√x− 1)

2=(√

2x + 2)2

x− 2√x + 1 = 2x + 2

(−2√x)

2= (x + 1)2

4x = x2 + 2x + 1

0 = x2 − 2x + 1

0 = (x− 1)2

The solution, x = 1, is extraneous, as√

1− 1 = 0 6=√

2(1) + 2 = 2. Thus, there is no solution.

93

5 Quadratic Equations

5.1 The Complex Number System

.Definition: Complex Number We define i to be the number such that

i2 = −1.

We call i the imaginary unit.

The product of a real number b and the imaginary unit i,

bi,

is called an imaginary number.

The sum of any real number a and imaginary number bi,

a + bi,

is called a complex number.


−36

Solution: √−36 =

√−1 ·√

36

= 6i


−12


√−1 ·√

12

=√−1 ·√

4 · 3

= 2i√

3


−98


√−1 ·√

98

=√−1 ·√

49 · 2

= 7i√

2

94

.Example 2.a Add the complex numbers.

(−2 + 3i) + (6− 9i)

Solution:(−2 + 3i) + (6− 9i) = 4− 6i

.Example 2.b Subtract the complex numbers.

(5− 12i)− (8− 3i)

Solution:(5− 12i)− (8− 3i) = 5− 12i− 8 + 3i

= −3− 9i


(4i)(5i)

Solution:(4i)(5i) = 20i2

= −20


(−2i)(i)

Solution:(−2i)(i) = −2i2

= 2


4i(5− 2i)

Solution:4i(5− 2i) = 20i− 8i2

= 20i + 8

= 8 + 20i

.Example 3.d Multiply.

(3 + 4i)(5− i)

Solution:(3 + 4i)(5− i) = 15− 3i + 20i− 4i2

= 15 + 17i + 4

= 19 + 17i

95

.Example 3.e Multiply.

(1− 3i)(3− 2i)

Solution:(1− 3i)(3− 2i) = 3− 2i− 9i + 6i2

= 3− 11i− 6

= −3− 11i

.Example 3.f Multiply.

(4 + i)2

Solution:(4 + i)2 = (4 + i)(4 + i)

= 16 + 8i + i2

= 16 + 8i− 1

= 15 + 8i

.Example 3.g Multiply.

(2− 5i)2

Solution:(2− 5i) = 4− 20i + 25i2

= 4− 20i− 25

= −21− 20i

.Example 3.h Multiply.

(7− 2i)(7 + 2i)

Solution:(7− 2i)(7 + 2i) = 49 + 14i− 14i− 4i2

= 49 + 4

= 53

.Example 3.i Multiply.

(3 + 4i)(3− 4i)

Solution:(3 + 4i)(3− 4i) = 9− 12i + 12i− 16i2

= 9 + 16

= 25

96

.Example 4.a Divide.

3

4i

Solution:3

4i=

3

4i· ii

=3i

4i2

=3i

−4

= −3

4i

.Example 4.b Divide.

4− 5i

3i

Solution:4− 5i

3i=

4− 5i

3i· ii

=4i− 5i2

3i2

=4i + 5

−3

= −5

3− 4

3i

.Example 4.c Divide.

7 + 4i

2− 3i

Solution:7 + 4i

2− 3i=

7 + 4i

2− 3i· 2 + 3i

2 + 3i

=14 + 29i + 12i2

4− 9i2

=14 + 29i− 12

4 + 9

=2 + 29i

13

=2

13+

29

13i

97

.Example 4.d Divide.

5− i

4 + 5i

Solution:5− i

4 + 5i=

5− i

4 + 5i· 4− 5i

4− 5i

=20− 29i + 5i2

16− 25i2

=20− 29i− 5

16 + 25

=15− 29i

41

=15

41− 29

41i

98

5.2 Quadratic Equations

.Definition: Quadratic Equation A quadratic equation is an equation that can be written as

ax2 + bx + c = 0

where a, b, and c are real numbers and a 6= 0.

.Example 1 Solve.

x2 = 4

Solution:x2 = 4

x2 − 4 = 0

(x + 2)(x− 2) = 0

Then either:x + 2 = 0 or x− 2 = 0

x = −2 x = 2

.Theorem: Square Root Principal If x2 = a, then x =

√a or x = −

√a.

Proof.x2 = a

x2 − a = 0

(x +√a)(x−

√a) = 0

Then either:x +√a = 0 or x−

√a = 0

x = −√a x =

√a

.Example 2.a Solve.

x2 = 9

Solution: √x2 = ±

√9

x = ±3

.Example 2.b Solve.

x2 = 12


√12

x = ±2√

3

99

.Example 2.c Solve.

x2 = −25


√−25

x = ±5i

.Example 2.d Solve.

x2 − 4 = 32

Solution:x2 − 4 = 32

x2 = 36

√x2 = ±

√36

x = ±6

.Example 2.e Solve.

3x2 + 4 = 58

Solution:3x2 + 4 = 58

3x2 = 54

x2 = 18

√x2 = ±

√18

x = ±3√

2

.Example 2.f Solve.

(x− 3)2 = 4

Solution:(x− 3)2 = 4√(x− 3)2 = ±

√4

x− 3 = ±2

x = 3± 2

Thus, x = 5 or x = 1.

100

.Example 2.g Solve.

(2x− 1)2 = −5

Solution:(2x− 1)2 = −5√(2x− 1)2 = ±

√−5

2x− 1 = ±i√

5

2x = 1± i√

5

x =1± i

√5

2

Thus, x =1 + i

√5

2or x =

1− i√

5

2.

.Example 3.a Solve.

x2 + 6x + 9 = 49

Solution:x2 + 6x + 9 = 49

(x + 3)2 = 49√(x + 3)2 = ±

√49

x + 3 = ±7

x = −3± 7

Thus, x = 4 or x = −10.

.Example 3.b Solve.

x2 − 10x + 25 = 10

Solution:x2 − 10x + 25 = 10

(x− 5)2 = 10√(x− 5)2 = ±

√10

x− 5 = ±√

10

x = 5±√

10

Thus, x = 5 +√

10 or x = 5−√

10.

101

.Example 3.c Solve.

4x2 + 12x + 9 = −1

Solution:4x2 + 12x + 9 = −1

(2x + 3)2 = −1√(2x + 3)2 = ±

√−1

2x + 3 = ±i

2x = −3± i

x =−3± i

2

Thus, x = −3

2+

1

2i or x = −3

2− 1

2i.

.Example 4.a Solve.

x2 + 8x + 2 = −5

Solution:x2 + 8x + 2 = −5

x2 + 8x = −7

x2 + 8x + 16 = −7 + 16

(x + 4)2 = 9√(x + 4)2 = ±

√9

x + 4 = ±3

x = −4± 3

Thus, x = −1 or x = −7.

.Example 4.b Solve.

x2 − 6x + 1 = 0

102

Solution:x2 − 6x + 1 = 0

x2 − 6x = −1

x2 − 6x + 9 = −1 + 9

(x− 3)2 = 8√(x− 3)2 = ±

√8

x− 3 = ±2√

2

x = 3± 2√

2

.Example 4.c Solve.

x2 + x− 2 = 5

Solution:x2 + x = 7

x2 + x + 14

= 7 + 14(

x + 12

)2= 29

4√(x + 1

2

)2= ±

√294

x + 12

= ±√292

x = −1±√29

2

.Example 4.d Solve.

x2 − 5x +41

4= 0

Solution:x2 − 5x = −41

4

x2 − 5x + 254

= −414

+ 254(

x− 52

)2= −4√(

x− 52

)2= ±√−4

x− 52

= ±2i

x = 52± 2i

103

.Example 5.a Solve.

9x2 + 12x− 5 = 0

Solution:9x2 + 12x− 5 = 0

9x2 + 12x = 5

9x2 + 12x

9=

5

9

x2 +4

3x =

5

9

x2 +4

3x +

(2

3

)2

=5

9+

(2

3

)2

.Example 5.a (Continued)

x2 +4

3x +

4

9=

5

9+

4

9(x +

2

3

)2

= 1

√(x +

2

3

)2

= ±√

1

x +2

3= ±1

x = −2

3± 1

Thus, x = 13

or x = −53.

.Example 5.b Solve.

4x2 − 20x + 33 = 0

104

Solution:4x2 − 20x + 33 = 0

4x2 − 20x = −33

4x2 − 20x

4=−33

4

x2 − 5x =−33

4

x2 − 5x +

(5

2

)2

=−33

4+

(5

2

)2

.Example 5.b (Continued)

x2 − 5x +25

4=−33

4+

25

4(x− 5

2

)2

= −2

√(x− 5

2

)2

= ±√−2

x− 5

2= ±i

√2

x =5

2± i√

2

5.3 Quadratic Equations: The Grand Finale

.Definition: Quadratic Equation A quadratic equation is an equation that can be written as

ax2 + bx + c = 0

where a, b, and c are real numbers and a 6= 0.

.Theorem: Quadratic Formula For any equation ax2 + bx + c = 0, where a, b, and c are realnumbers, and a 6= 0,

x =−b±

√b2 − 4ac

2a.

105

Proof.ax2 + bx + c = 0

ax2 + bx = −c

ax2 + bx

a=−ca

x2 +b

ax = − c

a

x2 +b

ax +

(b

2a

)2

= − c

a+

(b

2a

)2

x2 +b

ax +

b2

4a2= − c

a+

b2

4a2(x +

b

2a

)2

= −4ac

4a2+

b2

4a2(x +

b

2a

)2

=b2

4a2− 4ac

4a2(x +

b

2a

)2

=b2 − 4ac

4a2√(x +

b

2a

)2

= ±√

b2 − 4ac

4a2

x +b

2a= ±√b2 − 4ac√

4a2

x +b

2a= ±√b2 − 4ac

2a

x = − b

2a±√b2 − 4ac

2a

x =−b±

√b2 − 4ac

2a

.Example 1.a Solve.

x2 + 2x− 8 = 0

106

Solution:

x =−b±

√b2 − 4ac

2a

=(−2)±

√22 − 4(1)(−8)

2(1)

=−2±

√4 + 32

2

=−2±

√36

2=−2± 6

2

Thus, x = 2 or x = −4.

.Example 1.b Solve.

3x2 − 5x− 2 = 0

Solution:

x =−b±

√b2 − 4ac

2a

=5±

√(−5)2 − 4(3)(−2)

2(3)

=5±√

25 + 24

6

=5±√

49

6=

5± 7

6

Thus, x = 2 or x = −13.

.Example 1.c Solve.

x2 − 3x− 7 = 0

Solution:

x =−b±

√b2 − 4ac

2a

=3±

√(−3)2 − 4(1)(−7)

2(1)

=3±√

9 + 28

2

=3±√

37

2

Thus, x =3 +√

37

2or x =

3−√

37

2.

107

.Example 1.d Solve.

2x2 + x + 1 = 0

Solution:

x =−b±

√b2 − 4ac

2a

=(−1)±

√12 − 4(2)(1)

2(2)

=−1±

√1− 8

4

=−1±

√−7

4=−1± i

√7

4

Thus, x =−1 + i

√7

4or x =

−1− i√

7

4.

.Example 1.e Solve.

4x2 = 7x− 3

Solution:4x2 = 7x− 3

4x2 − 7x + 3 = 0

x =7±

√(−7)2 − 4(4)(3)

2(4)

=7±√

49− 48

8

=7±√

1

8=

7± 1

8

Thus, x = 1 or x = 34.

.Example 1.f Solve.

3x− x2 = 1

Solution:3x− x2 = 1

−x2 + 3x− 1 = 0

108

x =(−3)±

√32 − 4(−1)(−1)

2(−1)

=−3±

√9− 4

−2

=−3±

√5

−2=

3∓√

5

2

Thus, x = 3−√5

2or x = 3+

√5

2.

5.4 Quadratic Equations III


5.5 Prelude To College Algebra: Graphing Quadratic Equations


109

Lecture Notes: Intermediate Algebra - Faculty Site Listingfaculty.mccneb.edu/jdlee3/math_1310/oldnotes/lecture_notes_1310.pdf · Lecture Notes: Intermediate Algebra ... 4.6 Radicals

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