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55Polynomials:Factoring
5.1 Introduction to Factoring
5.2 Factoring Trinomials of the Type
5.3 Factoring The FOIL Method
5.4 Factoring The ac-Method
5.5 Factoring Trinomial Squares andDifferences of Squares
5.6 Factoring Sums or Differences of Cubes
5.7 Factoring: A General Strategy
5.8 Solving Quadratic Equations byFactoring
5.9 Applications of QuadraticEquations
a � 1:ax2 � bx � c,
a � 1:ax2 � bx � c,
x2 � bx � c
Real-World ApplicationAn outdoor-education ropes course includes a 25-ftcable that slopes downward from a height of 37 ft toa height of 30 ft. How far is it between the trees thatthe cable connects?
We introduce factoring with a review of factoring natural numbers. Con-sider the product We say that 3 and 5 are factors of 15 and that
is a factorization of 15. Since we also know that 15 and 1 arefactors of 15 and that is a factorization of 15.
Finding the Greatest Common Factor
The numbers 20 and 30 have several factors in common, among them 2 and5. The greatest of the common factors is called the greatest common factor,GCF. One way to find the GCF is by making a list of factors of each number.
List all the factors of 20: 1, 2, 4, 5, 10, and 20.
List all the factors of 30: 1, 2, 3, 5, 6, 10, 15, and 30.
Now list the numbers common to both lists, the common factors:
1, 2, 5, and 10.
Then the greatest common factor, the GCF, is 10, the largest number in thecommon list.
The preceding procedure gives meaning to the notion of a GCF, but thefollowing method, using prime factorizations, is generally faster.
EXAMPLE 1 Find the GCF of 20 and 30.
We find the prime factorization of each number. Then we draw lines be-tween the common factors.
The GCF
EXAMPLE 2 Find the GCF of 180 and 420.
We find the prime factorization of each number. Then we draw lines be-tween the common factors.
The GCF Note how we can use the expo-nents to determine the GCF. There are 2 lines for the 2’s, 1 line for the 3, 1 line for the 5, and no line for the 7.
EXAMPLE 3 Find the GCF of 30 and 77.
We find the prime factorization of each number. Then we draw lines be-tween the common factors, if any exist.
Since there is no common prime factor, the GCF is 1.
77 � 7 � 11 � 71 � 111
30 � 2 � 3 � 5 � 21 � 31 � 51
� 2 � 2 � 3 � 5 � 22 � 31 � 51 � 60.
420 � 2 � 2 � 3 � 5 � 7 � 22 � 31 � 51 � 71
180 � 2 � 2 � 3 � 3 � 5 � 22 � 32 � 51
� 2 � 5 � 10.
30 � 2 � 3 � 5
20 � 2 � 2 � 5
15 � 115 � 15 � 1,3 � 5
15 � 3 � 5.
5.15.1 INTRODUCTION TO FACTORINGObjectivesFind the greatest commonfactor, the GCF, ofmonomials.
Factor polynomials when theterms have a common factor,factoring out the greatestcommon factor.
Factor certain expressionswith four terms usingfactoring by grouping.
First, we find a prime factorization of the coefficients, attaching a factorof �1 for the negative coefficients.
The greatest positive common factor of the coefficients is 3.Then we find the GCF of the powers of x. That GCF is because 2 is the
smallest exponent of x. Thus the GCF of the set of monomials is
What about the �1 factors in Example 5? Strictly speaking, both 1 and �1are factors of any number or expression. We see this as follows:
Because the coefficient �3 is less than the coefficient 3, we consider andnot the GCF.
EXAMPLE 6 Find the GCF of 2py, and
We have
The greatest positive common factor of the coefficients is 2, the GCF of thepowers of p is p, and the GCF of the powers of y is 1 since there is no y-factorin the last monomial. Thus the GCF is 2p.
TO FIND THE GCF OF TWO ORMORE MONOMIALS
1. Find the prime factorization of the coefficients, including �1 as afactor if any coefficient is negative.
2. Determine any common prime factors of the coefficients. For eachone that occurs, include it as a factor of the GCF. If none occurs,use 1 as a factor.
3. Examine each of the variables as factors. If any appear as a factorof all the monomials, include it as a factor, using the smallestexponent of the variable. If none occurs in all the monomials, use1 as a factor.
4. The GCF is the product of the results of steps (2) and (3).
The polynomials we consider most when factoring are those with more thanone term. To multiply a monomial and a polynomial with more than oneterm, we multiply each term of the polynomial by the monomial using thedistributive laws:
and
To factor, we do the reverse. We express a polynomial as a product usingthe distributive laws in reverse:
and
Compare.
Multiply Factor
� 3x3 � 6x2 � 12x� 3x � x2 � 3x � 2x � 3x � 4
3x�x2 � 2x � 4�
ab � ac � a�b � c�.ab � ac � a�b � c�
a�b � c� � ab � ac.a�b � c� � ab � ac
9. a) Multiply:
b) Factor:
10. a) Multiply:
b) Factor:
Answers on page A-20
2x3 � 10x2 � 8x.
2x�x2 � 5x � 4�.
3x � 6.
3�x � 2�.
311
5.1 Introduction to Factoring
� 3x�x2 � 2x � 4�� 3x � x2 � 3x � 2x � 3x � 4
3x3 � 6x2 � 12x
Caution!
Consider the following:
The terms of the polynomial, and have been factored butthe polynomial itself has not been factored. This is not what we mean by afactorization of the polynomial. The factorization is
A product
The expressions and are factors of
Do Exercises 9 and 10.
To factor, we first find the GCF of all terms. It may be 1.
EXAMPLE 7 Factor:
We have
Factoring each term
Factoring out the GCF, 7
Check: We multiply to check:
7�x2 � 2� � 7 � x2 � 7 � 2 � 7x2 � 14.
� 7�x2 � 2�. 7x2 � 14 � 7 � x2 � 7 � 2
7x2 � 14.
3x3 � 6x2 � 12x.x2 � 2x � 43x
3x�x2 � 2x � 4�.
�12x,6x2,3x3,
3x3 � 6x2 � 12x � 3 � x � x � x � 2 � 3 � x � x � 2 � 2 � 3 � x.
There are two important points to keep in mind as we study this chapter.
TIPS FOR FACTORING
• Before doing any other kind of factoring, first try to factor out the GCF.
• Always check the result of factoring by multiplying.
Factoring by Grouping: Four Terms
Certain polynomials with four terms can be factored using a method calledfactoring by grouping.
EXAMPLE 14 Factor:
The binomial is common to both terms:
The factorization is
Do Exercises 17 and 18 on the preceding page.
Consider the four-term polynomial
There is no factor other than 1 that is common to all the terms. We can, how-ever, factor and separately:
Factoring
Factoring
We have grouped certain terms and factored each polynomial separately:
as in Example 14. This method is called factoring by grouping. We began witha polynomial with four terms. After grouping and removing common factors,we obtained a polynomial with two parts, each having a common factor
Not all polynomials with four terms can be factored by this procedure,but it does give us a method to try.x � 1.
b) Explain why you need toconsider only positive factors,as in the following table.
c) Factor:
2. Factor:
Answers on page A-20
x 2 � 13x � 36.
x 2 � 7x � 12.
x 2 � 7x � 12.
317
5.2 Factoring Trinomials of the Type x2 � bx � c
Factoring
We now begin a study of the factoring of trinomials. We first factor trinomi-als like
by a refined trial-and-error process. In this section, we restrict our attention totrinomials of the type where The coefficient a is oftencalled the leading coefficient.
To understand the factoring that follows, compare the followingmultiplications:
F O I L
Note that for all four products:
• The product of the two binomials is a trinomial.
• The coefficient of x in the trinomial is the sum of the constant terms inthe binomials.
• The constant term in the trinomial is the product of the constant terms inthe binomials.
These observations lead to a method for factoring certain trinomials. The firsttype we consider has a positive constant term, just as in the first two multi-plications above.
CONSTANT TERM POSITIVETo factor we think of FOIL in reverse. We multiplied x times xto get the first term of the trinomial, so we know that the first term of each bi-nomial factor is x. Next, we look for numbers p and q such that
To get the middle term and the last term of the trinomial, we look for twonumbers p and q whose product is 10 and whose sum is 7. Those numbers are2 and 5. Thus the factorization is
Check:
� x 2 � 7x � 10.
�x � 2� �x � 5� � x 2 � 5x � 2x � 10
�x � 2� �x � 5�.
x 2 � 7x � 10 � �x � p� �x � q�.
x 2 � 7x � 10,
� x 2 � 4x � 21.
�x � 3� �x � 7� � x 2 � 7x � 3x � ��3�7
� x 2 � 4x � 21;
�x � 3� �x � 7� � x 2 � 7x � 3x � 3��7�
� x 2 � 7x � 10;
�x � 2� �x � 5� � x 2 � 5x � 2x � 2 � 5
� x 2 � 7x � 10;
�x � 2� �x � 5� � x 2 � 5x � 2x � 2 � 5
a � 1.ax 2 � bx � c,
x 2 � 5x � 6 and x 2 � 3x � 10
x2 � bx � c
5.25.2 FACTORING TRINOMIALS OF THETYPE x2 � bx � c
Think of FOIL in reverse. The first term of each factor is x: Next, we look for two numbers whose product is 6 and
whose sum is 5. All the pairs of factors of 6 are shown in the table on the leftbelow. Since both the product, 6, and the sum, 5, of the pair of numbers mustbe positive, we need consider only the positive factors, listed in the table onthe right.
The numbers we need are 2 and 3.
The factorization is We can check by multiplying to seewhether we get the original trinomial.
Check:
Do Exercises 1 and 2 on the preceding page.
Compare these multiplications:
TO FACTOR WHEN IS POSITIVE
When the constant term of a trinomial is positive, look for twonumbers with the same sign. The sign is that of the middle term:
EXAMPLE 2 Factor:
Since the constant term, 12, is positive and the coefficient of the middleterm, �8, is negative, we look for a factorization of 12 in which both factorsare negative. Their sum must be �8.
The factorization is The student should check by multiplying.
CONSTANT TERM NEGATIVEAs we saw in two of the multiplications earlier in this section, the product oftwo binomials can have a negative constant term:
and
Note that when the signs of the constants in the binomials are reversed, onlythe sign of the middle term in the product changes.
EXAMPLE 3 Factor:
The constant term, �20, must be expressed as the product of a negativenumber and a positive number. Since the sum of these two numbers must benegative (specifically, �8), the negative number must have the greater ab-solute value.
The numbers that we are looking for are 2 and �10. The factorization is
Check:
TO FACTOR WHEN IS NEGATIVE
When the constant term of a trinomial is negative, look for twonumbers whose product is negative. One must be positive and theother negative:
Consider pairs of numbers for which the number with the largerabsolute value has the same sign as b, the coefficient of the middle term.
Do Exercises 6 and 7. (Exercise 7 is on the following page.)
x 2 � 4x � 21 � �x � 3� �x � 7�.
x 2 � 4x � 21 � �x � 3� �x � 7�;
cx 2 � bx � c
� x 2 � 8x � 20.
�x � 2� �x � 10� � x 2 � 10x � 2x � 20
�x � 2� �x � 10�.
x 2 � 8x � 20.
�x � 3� �x � 7� � x 2 � 4x � 21.
�x � 3� �x � 7� � x 2 � 4x � 21
6. Consider
a) Explain why you would notconsider the pairs of factorslisted below in factoring
b) Explain why you wouldconsider the pairs of factorslisted below in factoring
c) Factor:
Answers on page A-20
x2 � 5x � 24.
x2 � 5x � 24.
x2 � 5x � 24.
x2 � 5x � 24.
319
5.2 Factoring Trinomials of the Type x2 � bx � c
1, �20 �19
2, �10 �8
4, �5 �1
5, �4 1
10, �2 8
20, �1 19
PAIRS OF FACTORS SUMS OF FACTORS
⎫⎪⎬⎪⎭
Because these sums areall positive, for thisproblem all of thecorresponding pairscan be disregarded.Note that in all threepairs, the positivenumber has the greaterabsolute value.
It helps to first write the trinomial in descending order:Since the constant term, �24, is negative, we look for a factorization of �24 inwhich one factor is positive and one factor is negative. Their sum must be 5,so the positive factor must have the larger absolute value. Thus we consideronly pairs of factors in which the positive term has the larger absolute value.
The factorization is The check is left to the student.
Do Exercises 8 and 9.
EXAMPLE 5 Factor:
Consider this trinomial as We look for numbers p andq such that
Since the constant term, �110, is negative, we look for a factorization of �110in which one factor is positive and one factor is negative. Their sum must be�1. The middle-term coefficient, �1, is small compared to �110. This tells usthat the desired factors are close to each other in absolute value. The numberswe want are 10 and �11. The factorization is
EXAMPLE 6 Factor:
We consider the trinomial in the equivalent form
This way we think of as the “constant” term and 4b as the “coefficient”of the middle term. Then we try to express as a product of two factorswhose sum is 4b. Those factors are and 7b. The factorization is
Check:
There are polynomials that are not factorable.
EXAMPLE 7 Factor:
Since 5 has very few factors, we can easily check all possibilities.
x 2 � x � 5.
� a2 � 4ab � 21b2.
�a � 3b� �a � 7b� � a2 � 7ab � 3ba � 21b2
�a � 3b� �a � 7b�.�3b
�21b2�21b2
a2 � 4ba � 21b2.
a2 � 4ab � 21b2.
�x 2 � 10� �x 2 � 11�.
x 4 � x 2 � 110 � �x 2 � p� �x 2 � q�.
�x 2�2 � x 2 � 110.
x 4 � x 2 � 110.
�t � 3� �t � 8�.
The numbers we needare �3 and 8.
t 2 � 5t � 24.
t 2 � 24 � 5t.7. Consider
a) Explain why you would notconsider the pairs of factorslisted below in factoring
b) Explain why you wouldconsider the pairs of factorslisted below in factoring
There are no factors whose sum is �1. Thus the polynomial is not factorableinto factors that are polynomials.
In this text, a polynomial like that cannot be factored furtheris said to be prime. In more advanced courses, polynomials like can be factored and are not considered prime.
Do Exercises 10–12.
Often factoring requires two or more steps. In general, when told to fac-tor, we should factor completely. This means that the final factorizationshould not contain any factors that can be factored further.
EXAMPLE 8 Factor:
Always look first for a common factor. This time there is one, 2x, which wefactor out first:
Now consider Since the constant term is positive and the co-efficient of the middle term is negative, we look for a factorization of 25 inwhich both factors are negative. Their sum must be �10.
The factorization of is or The finalfactorization is We check by multiplying:
Do Exercises 13–15.
Once any common factors have been factored out, the following sum-mary can be used to factor
TO FACTOR
1. First arrange in descending order.2. Use a trial-and-error process that looks for factors of c whose
sum is b.3. If c is positive, the signs of the factors are the same as the sign
of b.4. If c is negative, one factor is positive and the other is negative.
If the sum of two factors is the opposite of b, changing the sign of each factor will give the desired factors whose sum is b.
you with timemanagement. (See alsothe Study Tips on time
management inSections 2.2 and 5.7.)
� Are you a morning or an evening person? If you are an evening person, it might bebest to avoid scheduling early-morning classes. If you are a morning person, do theopposite, but go to bed earlier to compensate. Nothing can drain your study time andeffectiveness like fatigue.
� Keep on schedule. Your course syllabus provides a plan for the semester’s schedule.Use a write-on calendar, daily planner, Palm Pilot or other PDA, or laptop computer tooutline your time for the semester. Be sure to note deadlines involving term papersand exams so you can begin a task early, breaking it down into smaller segments thatcan be accomplished more easily.
� Balance your class schedule. You may be someone who prefers large blocks of timefor study on the off days. In that case, it might be advantageous for you to take coursesthat meet only three days a week. Keep in mind, however, that this might be a problemwhen tests in more than one course are scheduled for the same day.
“Time is our most important asset, yet we tend to waste it, kill it, and spend it rather thaninvest it.”
79. Arrests for Counterfeiting. In a recent year,29,200 people were arrested for counterfeiting. Thisnumber was down 1.2% from the preceding year. Howmany people were arrested the preceding year? [2.5a]
80. The first angle of a triangle is four times as large asthe second. The measure of the third angle is 30°greater than that of the second. Find the anglemeasures. [2.6a]
Solve.
Factor completely.
83. 84. 85. x2 �307 x �
257x2 �
14 x �
18x2 �
12 x �
316
86. 87. 88. a2m � 11am � 28b2n � 7bn � 1013 x3 �
13 x2 � 2x
Find a polynomial in factored form for the shaded area. (Leave answers in terms of �.)
In Section 5.2, we learned a trial-and-error method to factor trinomials ofthe type In this section, we factor trinomials in which the coeffi-cient of the leading term is not 1. The procedure we learn is a refined trial-and-error method.
The FOIL Method
We want to factor trinomials of the type Consider the followingmultiplication:
F O I L
� � � 20
� � 20
F O I L
To factor we reverse the above multiplication, using what wemight call an “unFOIL” process. We look for two binomials and whose product is The product of theFirst terms must be The product of the Outside terms plus the productof the Inside terms must be The product of the Last terms must be 20.We know from the preceding discussion that the answer is Generally, however, finding such an answer is a refined trial-and-errorprocess. It turns out that is also a correct answer, butwe generally choose an answer in which the first coefficients are positive.
We will use the following trial-and-error method.
THE FOIL METHOD
To factor using the FOIL method:
1. Factor out the largest common factor, if one exists.2. Find two First terms whose product is
FOIL
3. Find two Last terms whose product is c:
FOIL
4. Look for Outer and Inner products resulting from steps (2) and (3)for which the sum is bx:
I FOILO
5. Always check by multiplying.
� x � � � x � � � ax2 � bx � c.
� x � � � x � � � ax2 � bx � c.
� x � � � x � � � ax2 � bx � c.
ax2.
ax2 � bx � c, a � 1,
��2x � 5� ��3x � 4�
�2x � 5� �3x � 4�.23x.
6x2.�rx � p� �sx � q� � 6x2 � 23x � 20.
sx � qrx � p6x2 � 23x � 20,
5 � 45 � 32 � 42 � 3
�
23x6x2�
15x8x6x2�2x � 5� �3x � 4� �
ax2 � bx � c.
x2x2 � bx � c.
5.35.3 FACTORING THE FOIL METHOD
ax2 � bx � c, a � 1:
ObjectiveFactor trinomials of the type
, , usingthe FOIL method.
a � 1ax2 � bx � c
}
The ac-method in Section 5.4
To the student : In Section 5.4,we will consider an alternativemethod for the same kind offactoring. It involves factoringby grouping and is called theac-method.
To the instructor : We presenttwo ways to factor generaltrinomials in Sections 5.3 and5.4: the FOIL method inSection 5.3 and the ac-methodin Section 5.4. You can teachboth methods and let thestudent use the one that he orshe prefers or you can selectjust one.
1) First, we check for a common factor. Here there is none (other than 1 or�1).
2) Find two First terms whose product is
The only possibilities for the First terms are and x, so any factori-zation must be of the form
3) Find two Last terms whose product is �8.
Possible factorizations of �8 are
and
Since the First terms are not identical, we must also consider
and
4) Inspect the Outer and Inner products resulting from steps (2) and (3).Look for a combination in which the sum of the products is the middleterm,
Trial Product
Wrong middle term
Wrong middle term
Wrong middle term
Correct middle term!
Wrong middle term
Wrong middle term
Wrong middle term
Wrong middle term
The correct factorization is
5) Check:
Two observations can be made from Example 1. First, we listed all pos-sible trials even though we could have stopped after having found the correctfactorization. We did this to show that each trial differs only in the middleterm of the product. Second, note that as in Section 5.2, only the sign of themiddle term changes when the signs in the binomials are reversed:
1) First, we factor out the largest common factor, 4:
Now we factor the trinomial
2) Because can be factored as or we have these possibilitiesfor factorizations:
or
3) There are four pairs of factors of 10 and they each can be listed in twoways:
10, 1 �10, �1 5, 2 �5, �2
and
1, 10 �1, �10 2, 5 �2, �5.
4) The two possibilities from step (2) and the eight possibilities fromstep (3) give or 16 possibilities for factorizations. We look for Outerand Inner products resulting from steps (2) and (3) for which the sum isthe middle term, Since the sign of the middle term is negative, butthe sign of the last term, 10, is positive, the two factors of 10 must bothbe negative. This means only four pairings from step (3) need be consid-ered. We first try these factors with
If none gives the correct factorization, we will consider
Trial Product
Wrong middle term
Wrong middle term
Wrong middle term
Correct middle term!
Since we have a correct factorization, we need not consider
The factorization of is but do not forgetthe common factor ! We must include it in order to factor the original trinomial:
5) Check:
Caution!
When factoring any polynomial, always look for a common factor. Failureto do so is such a common error that this caution bears repeating.
In Example 2, look again at the possibility Without mul-tiplying, we can reject such a possibility. To see why, consider the following:
The expression has a common factor, 2. But we removed the largestcommon factor in the first step. If were one of the factors, then 2 wouldhave to be a common factor in addition to the original 4. Thus, can-not be part of the factorization of the original trinomial.
Given that the largest common factor is factored out at the outset, weneed not consider factorizations that have a common factor.
Do Exercises 3 and 4.
EXAMPLE 3 Factor:
1) There is no common factor (other than 1 or �1).
2) Because factors as or we have these possibilities forfactorizations:
or
3) There are two pairs of factors of 7 and they each can be listed in two ways:
1, 7 �1, �7 and 7, 1 �7, �1.
4) From steps (2) and (3), we see that there are 8 possibilities for factoriza-tions. Look for Outer and Inner products for which the sum is the middleterm. Because all coefficients in are positive, we need con-sider only positive factors of 7. The possibilities are
Correct middle term
The factorization is
5) Check:
Do Exercise 5.
TIPS FOR FACTORING
• Always factor out the largest common factor, if one exists.
• Once the common factor has been factored out of the originaltrinomial, no binomial factor can contain a common factor(other than 1 or �1).
• If c is positive, then the signs in both binomial factors mustmatch the sign of b. (This assumes that )
• Reversing the signs in the binomials reverses the sign of themiddle term of their product.
• Organize your work so that you can keep track of whichpossibilities have or have not been checked.
An important problem-solving strategy is to find a way to make newproblems look like problems we already know how to solve. (See Example 9 inSection 5.2.) The factoring tips above apply only to trinomials of the form
with This leads us to rewrite in descend-ing order:
Writing in descending order
Although looks similar to the trinomials we have factored,the tips above require a positive leading coefficient. This can be attained byfactoring out �1:
Factoring out �1 changes the signs of the coefficients.
Using the result from Example 1
The factorization of is Other correct an-swers are
Multiplying by �1
Multiplying by �1
Do Exercises 6 and 7.
EXAMPLE 5 Factor:
1) Factor out a common factor, if any.
There is none (other than 1 or �1).
2) Factor the first term,
Possibilities are 2p, 3p and 6p, p. We have these as possibilities forfactorizations:
or
3) Factor the last term, which has a negative coefficient.
There are six pairs of factors and each can be listed in two ways:
and
4) The coefficient of the middle term is negative, so we look for combinationsof factors from steps (2) and (3) such that the sum of their products has anegative coefficient. We try some possibilities:
Checking Factorizations A partial check of a factorization can beperformed using a table or a graph. To check the factorization
for example, we enterand on the
equation-editor screen (see page 175). Then we set up a table in AUTO
mode (see page 180). If the factorization is correct, the values of and will be the same regardless of the table settings used.
We can also graph andIf the graphs appear to coincide, the
factorization is probably correct.
Keep in mind that these procedures provide only a partial check sincewe cannot view all possible values of x in a table nor can we see the entiregraph.
Exercises: Use a table or a graph to determine whether thefactorization is correct.
Another method for factoring trinomials of the type , , in-volves the product, ac, of the leading coefficient a and the last term c. It iscalled the ac-method. Because it uses factoring by grouping, it is also referredto as the grouping method.
We know how to factor the trinomial . We look for factors ofthe constant term, 6, whose sum is the coefficient of the middle term, 5. Whathappens when the leading coefficient is not 1? To factor a trinomial like
, we can use a method similar to what we used for That method is outlined as follows.
THE ac-METHOD
To factor , , using the ac-method:
1. Factor out a common factor, if any.2. Multiply the leading coefficient a and the constant c.3. Try to factor the product ac so that the sum of the factors is b.
That is, find integers p and q such that and .4. Split the middle term. That is, write it as a sum using the factors
found in step (3).5. Factor by grouping.6. Check by multiplying.
EXAMPLE 1 Factor:
1) First, we factor out a common factor, if any. There is none (other than 1 or ).
2) We multiply the leading coefficient, 3, and the constant, :
3) Then we look for a factorization of in which the sum of the factors isthe coefficient of the middle term, .
4) Next, we split the middle term as a sum or a difference using the factorsfound in step (3): �10x � 2x � 12x.
105. The earth is a sphere (or ball) that is about 40,000 kmin circumference. Find the radius of the earth, inkilometers and in miles. Use 3.14 for . (Hint :
)
106. The second angle of a triangle is 10° less than twice thefirst. The third angle is 15° more than four times thefirst. Find the measure of the second angle.
1 km � 0.62 mi.�
107. 108. 24x2n � 22xn � 39x10 � 12x5 � 4
109. 110. �a � 4�2 � 2�a � 4� � 116x10 � 8x5 � 1
111.–120. Use the TABLE feature to check the factoring in Exercises 15–24.
EXAMPLE 3 Determine whether is a trinomial square.
It helps to first write the trinomial in descending order:
a) We know that and 49 are squares.
b) There is no minus sign before or 49.
c) If we multiply the square roots, 4x and 7, and double the product, we getthe opposite of the remaining term: 56x is the opposite of �56x.
Thus, is a trinomial square. In fact,
Do Exercises 1–8.
Factoring Trinomial Squares
We can use the factoring methods from Sections 5.2–5.4 to factor trinomialsquares, but there is a faster method using the following equations.
FACTORING TRINOMIAL SQUARES
We consider 3 to be a square root of 9 because Similarly, A is asquare root of We use square roots of the squared terms and the sign of theremaining term to factor a trinomial square.
a) The first expression is a square:The second expression is a square:
b) The terms have different signs, and
Thus we have a difference of squares,
EXAMPLE 11 Is a difference of squares?
a) The expression is not a square.
The expression is not a difference of squares.
EXAMPLE 12 Is a difference of squares?
a) The expressions and 16 are squares: and
b) The terms have different signs, and
Thus we have a difference of squares. We can also see this by rewriting in theequivalent form:
Do Exercises 18–24.
Factoring Differences of Squares
To factor a difference of squares, we use the following equation.
FACTORING A DIFFERENCE OF SQUARES
To factor a difference of squares we find A and B, which aresquare roots of the expressions and We then use A and B to form twofactors. One is the sum and the other is the difference
EXAMPLE 13 Factor:
EXAMPLE 14 Factor:
A2 � B2 � �A � B� �A � B�
9 � 16t 4 � 32 � �4t 2�2 � �3 � 4t 2� �3 � 4t 2�
9 � 16t 4.
A2 � B2 � �A � B� �A � B�
x2 � 4 � x2 � 22 � �x � 2� �x � 2�
x2 � 4.
A � B.A � B,B2.A2
A2 � B2,
A2 � B2 � �A � B� �A � B�
16 � 4x2.
�16.�4x2
16 � 42.4x2 � �2x�24x2
�4x2 � 16
t 3
25 � t 3
�3x�2 � 82.
�64.�9x2
64 � 82.9x2 � �3x�2.
9x2 � 64 Determine whether each is adifference of squares. Write “yes” or “no.”
18.
19.
20.
21.
22.
23.
24.
Answers on page A-22
�49 � 25t 2
9w 6 � 1
16x4 � 49
4x2 � 15
y2 � 36
t 2 � 24
x2 � 25
345
5.5 Factoring Trinomial Squares andDifferences of Squares
FACTORING COMPLETELYIf a factor with more than one term can still be factored, you should do so.When no factor can be factored further, you have factored completely. Al-ways factor completely whenever told to factor.
EXAMPLE 19 Factor:
Factoring a difference of squares
Factoring further; is adifference of squares.
The polynomial cannot be factored further into polynomials withreal coefficients.
Caution!
Apart from possibly removing a common factor, you cannot factor a sumof squares. In particular,
Consider Here a sum of squares has a common factor, 25.Factoring, we get where is prime. For example,
EXAMPLE 20 Factor:
factor is a differenceof squares.
The polynomial cannot be factored further into polynomials withreal coefficients.
EXAMPLE 21 Factor:
Factoring a differenceof squares
Factoring further. Thefactor is adifference of squares.
TIPS FOR FACTORING
• Always look first for a common factor! If there is one, factor it out.
• Be alert for trinomial squares and differences of squares. Oncerecognized, they can be factored without trial and error.
It is never too soon to beginreviewing for the finalexamination. The SkillMaintenance exercises foundin each exercise set review andreinforce skills taught inearlier sections. Include all ofthese exercises in your weeklypreparation. Answers to bothodd-numbered and even-numbered exercises appear atthe back of the book.
347
5.5 Factoring Trinomial Squares andDifferences of Squares
We can factor the sum or the difference of two expressions that are cubes.Consider the following products:
and
.
The above equations (reversed) show how we can factor a sum or a differenceof two cubes.
SUM OR DIFFERENCE OF CUBES
;
Note that what we are considering here is a sum or a difference of cubes. We are not cubing a binomial. For example, is not the same as . The table of cubes in the margin is helpful.
EXAMPLE 1 Factor: .
We have
.
In one set of parentheses, we write the cube root of the first term, x. Then we write the cube root of the second term, �3. This gives us the expres-sion :
.
To get the next factor, we think of and do the following:
Square the first term: .
Multiply the terms, , and thenchange the sign: 3x.
Square the second term: .
.
Note that we cannot factor . It is not a trinomial square nor can itbe factored by trial and error. Check this on your own.
� Treat every homework exercise as if it were a test question. If you had to work aproblem at your job with no backup answer provided, what would you do? You wouldprobably work it very deliberately, checking and rechecking every step. You might workit more than one time, or you might try to work it another way to check the result. Tryto use this approach when doing your homework. Treat every exercise as though itwere a test question with no answer at the back of the book.
� Be sure that you do questions without answers as part of every homework assignmentwhether or not the instructor has assigned them! One reason a test may seem such adifferent task is that questions on a test lack answers. That is the reason for taking atest: to see if you can do the questions without assistance. As part of your testpreparation, be sure you do some exercises for which you do not have the answers.Thus when you take a test, you are doing a more familiar task.
The purpose of doing your homework using these approaches is to give you more test-taking practice beforehand. Let’s use a sports analogy: At a basketball game, the playerstake lots of practice shots before the game. They play the first half, go to the locker room,and come out for the second half. What do they do before the second half, even thoughthey have just played 20 minutes of basketball? They shoot baskets again! We suggest thesame approach here. Create more and more situations in which you practice taking testquestions by treating each homework exercise like a test question and by doing exercisesfor which you have no answers. Good luck!
“He who does not venture has no luck.”
Mexican proverb
How often do you makethe following statement
after taking a test: “Iwas able to do the
homework, but I frozeduring the test”? This
can be an excuse forpoor study habits. Hereare two tips to help you
with this difficulty.Both are intended tomake test taking less
We now combine all of our factoring techniques and consider a gen-eral strategy for factoring polynomials. Here we will encounter poly-
nomials of all the types we have considered, in random order, so you will havethe opportunity to determine which method to use.
FACTORING STRATEGY
To factor a polynomial:
a) Always look first for a common factor. If there is one, factor outthe largest common factor.
b) Then look at the number of terms.
Two terms: Determine whether you have a difference of squares,Do not try to factor a sum of squares:
Three terms: Determine whether the trinomial is a square. If it is,you know how to factor. If not, try trial and error, using FOIL orthe ac-method.
Four terms: Try factoring by grouping.
c) Always factor completely. If a factor with more than one term canstill be factored, you should factor it. When no factor can befactored further, you have finished.
d) Check by multiplying.
EXAMPLE 1 Factor:
a) We look for a common factor:
b) The factor has only two terms. It is a difference of squares:. We factor and then include the common factor:
c) We see that one of the factors, is again a difference of squares. Wefactor it:
This is a sum of squares. It cannot be factored!
We have factored completely because no factor with more than one termcan be factored further.
d) Check:
� 5t 4 � 80. � 5�t 4 � 16�
5�t 2 � 4� �t � 2� �t � 2� � 5�t2 � 4� �t 2 � 4�
5�t 2 � 4� �t � 2� �t � 2�.
t2 � 4,
5�t 2 � 4� �t 2 � 4�.
t 4 � 16�t 2�2 � 42t 4 � 16
5t 4 � 80 � 5�t4 � 16�.
5t 4 � 80.
A2 � B2.A2 � B2.
5.75.7 FACTORING: A GENERAL STRATEGYObjectiveFactor polynomialscompletely using any of themethods considered in thischapter.
b) There are three terms in . We determine whether the trino-mial is a square. Since only is a square, we do not have a trinomialsquare. Can the trinomial be factored by trial and error? A key to theanswer is that x is only in the term . The polynomial might be in aform like , but there would be no x in the middle term.Thus, cannot be factored.
c) Have we factored completely? Yes, because no factor with more thanone term can be factored further.
d) The check is left to the student.
EXAMPLE 6 Factor:
a) We look for a common factor:
b) There are three terms in , but this trinomial cannot be factoredfurther.
c) Neither factor can be factored further, so we have factored completely.
d) The check is left to the student.
EXAMPLE 7 Factor:
a) We look first for a common factor. There isn’t one.
b) There are four terms. We try factoring by grouping:
c) Have we factored completely? Since neither factor can be factored further,we have factored completely.
d) Check:
EXAMPLE 8 Factor:
a) We look first for a common factor. There isn’t one.
b) There are three terms. We determine whether the trinomial is a square. Thefirst term and the last term are squares:
Since twice the product of 5x and 2y is the other term,
the trinomial is a perfect square.We factor by writing the square roots of the square terms and the sign
of the middle term:
c) Since cannot be factored further, we have factored completely.5x � 2y
� � p � q� �2x � y � 2�. � p � q� �x � 2� � � p � q� �x � y� � � p � q� ��x � 2� � �x � y��
�p � q� �x � 2� � �p � q� �x � y�.
2 � 7xy � y2�1 � y� �2 � y�
�7xy
y22 � 7xy � y2
6x2y4 � 21x3y5 � 3x2y6 � 3x2y4�2 � 7xy � y2�.
6x2y4 � 21x3y5 � 3x2y6.
Study Tips
TIME MANAGEMENT (PART 3)
Here are some additional tipsto help you with time manage-ment. (See also the Study Tipson time management inSections 2.2 and 5.2.)
� Avoid “time killers.” Welive in a media age, andthe Internet, e-mail,television, and movies areall time killers. Keep trackof the time you spend onsuch activities andcompare it to the timeyou spend studying.
� Prioritize your tasks. Becareful about taking ontoo many collegeactivities that fall outsideof academics. Theseactivities are importantbut keep them to aminimum to be sure thatyou have enough time foryour studies.
� Be aggressive about yourstudy tasks. Instead ofworrying over your mathhomework or test prepa-ration, do something toget yourself started. If thetask is large, break it downinto smaller parts, and doone at a time. You will besurprised at how quicklythe large task can then becompleted.
“Time is more valuable thanmoney. You can get moremoney, but you can’t get moretime.”
a) We look first for a common factor. There isn’t one.
b) There are three terms. We determine whether the trinomial is a square. Thefirst term is a square, but neither of the other terms is a square, so we donot have a trinomial square. We factor, thinking of the product pq as a sin-gle variable. We consider this possibility for factorization:
We factor the last term, 12. All the signs are positive, so we consider onlypositive factors. Possibilities are 1, 12 and 2, 6 and 3, 4. The pair 3, 4 givesa sum of 7 for the coefficient of the middle term. Thus,
c) No factor with more than one term can be factored further, so we have fac-tored completely.
d) Check:
EXAMPLE 10 Factor:
a) We look first for a common factor:
b) There are three terms in . We determine whether the tri-nomial is a square. Since none of the terms is a square, we do not have a trinomial square. We factor . Possibilities are , and , andothers. We also factor the last term, . Possibilities are , and
and others. We look for factors such that the sum of their productsis the middle term. The in the middle term, , should lead us to try
We try some possibilities:
c) No factor with more than one term can be factored further, so we have fac-tored completely. The factorization, including the common factor, is
In addition to the videotaped lecturesfor the books, there is a special VHStape, Math Problem Solving in the RealWorld. Check with your instructor tosee whether this tape is available onyour campus.
There is also a special Math StudySkills for Students Video on CD that isdesigned to help you make better useof your math study time and improveyour retention of concepts and pro-cedures taught in classes from basicmathematics through intermediatealgebra. (See the Preface for moreinformation.)
Second-degree equations like and are ex-amples of quadratic equations.
QUADRATIC EQUATION
A quadratic equation is an equation equivalent to an equation of the type
In order to solve quadratic equations, we need a new equation-solvingprinciple.
The Principle of Zero Products
The product of two numbers is 0 if one or both of the numbers is 0. Further-more, if any product is 0, then a factor must be 0. For example:
If then we know that
If then we know that or
If then we know that or
Caution!
In a product such as we cannot conclude with certainty that ais 24 or that b is 24, but if we can conclude that or
EXAMPLE 1 Solve:
We have a product of 0. This equation will be true when either factor is 0.Thus it is true when
or
Here we have two simple equations that we know how to solve:
or
Each of the numbers �3 and 2 is a solution of the original equation, as we cansee in the following checks.
Check: For �3:
TRUE 0 0��5�
��3 � 3� ��3 � 2� ? 0
�x � 3� �x � 2� � 0
x � 2.x � �3
x � 2 � 0.x � 3 � 0
�x � 3� �x � 2� � 0.
b � 0.a � 0ab � 0,ab � 24,
x � 2 � 0.x � 3 � 0�x � 3� �x � 2� � 0,
2x � 9 � 0.x � 0x�2x � 9� � 0,
x � 0.7x � 0,
a � 0.ax2 � bx � c � 0,
9 � x2 � 0x2 � x � 156 � 0
5.85.8 SOLVING QUADRATIC EQUATIONSBY FACTORING
ObjectivesSolve equations (alreadyfactored) using the principleof zero products.
Solve quadratic equations byfactoring and then using theprinciple of zero products.
Study Tips
WORKING WITH A CLASSMATE
If you are finding it difficult tomaster a particular topic orconcept, try talking about itwith a classmate. Verbalizingyour questions about thematerial might help clarify it.If your classmate is alsofinding the material difficult, itis possible that the majority ofthe people in your class areconfused and you can askyour instructor to explain theconcept again.
Using factoring and the principle of zero products, we can solve some newkinds of equations. Thus we have extended our equation-solving abilities.
EXAMPLE 4 Solve:
There are no like terms to collect, and we have a squared term. We firstfactor the polynomial. Then we use the principle of zero products.
Factoring
Using the principle of zero products
Check: For �2:
TRUE 0 �6 � 6
4 � 10 � 6 ��2�2 � 5��2� � 6 ? 0
x2 � 5x � 6 � 0
x � �2 or x � �3
x � 2 � 0 or x � 3 � 0
�x � 2� �x � 3� � 0
x2 � 5x � 6 � 0
x2 � 5x � 6 � 0.
5. Solve:
Solve.
6.
7.
Solve.
8.
9.
Answers on page A-23
9x2 � 16
x2 � 4x � 0
x2 � 6x � 9
x2 � 3x � 28
x2 � x � 6 � 0.
368
CHAPTER 5: Polynomials: Factoring
For �3:
TRUE 0 �6 � 6
9 � 15 � 6 ��3�2 � 5��3� � 6 ? 0
x2 � 5x � 6 � 0
The solutions are �2 and �3.
Caution!
Keep in mind that you must have 0 on one side of the equation before youcan use the principle of zero products. Get all nonzero terms on one sideand 0 on the other.
Do Exercise 5.
EXAMPLE 5 Solve:
We first add 16 to get a 0 on one side:
Adding 16
Factoring
Using the principle of zero products
Solving each equation
There is only one solution, 4. The check is left to the student.
Do Exercises 6 and 7.
EXAMPLE 6 Solve:
Factoring out a common factor
Using the principle of zero products
The solutions are 0 and �5. The check is left to the student.
The solutions are and The check is left to the student.
Do Exercises 8 and 9 on the preceding page.
EXAMPLE 8 Solve:
In this case, the leading coefficient of the trinomial is negative. Thus wefirst multiply by �1 and then proceed as we have in Examples 1–7.
Multiplying by �1
Simplifying
Factoring
Using the principle ofzero products
The solutions are and 1. The check is left to the student.
Do Exercises 10 and 11.
EXAMPLE 9 Solve:
Be careful with an equation like this one! It might be tempting to set eachfactor equal to 5. Remember: We must have a 0 on one side. We first carry outthe product on the left. Then we subtract 5 on both sides to get 0 on one side.Then we proceed with the principle of zero products.
Multiplying on the left
Subtracting 5
Simplifying
Factoring
Using the principle of zero products
The solutions are �3 and 3. The check is left to the student.
In Chapter 3, we graphed linear equations of the type andRecall that to find the x-intercept, we replaced y with 0 and
solved for x. This procedure can also be used to find the x-interceptswhen an equation of the form is to be graphed.Although the details of creating such graphs will be left to Chapter 11, we consider them briefly here from the standpoint of finding the x-intercepts. The graphs are shaped like the following curves. Note thateach x-intercept represents a solution of
EXAMPLE 10 Find the x-intercepts of thegraph of shown at right. (Thegrid is intentionally not included.)
To find the x-intercepts, we let andsolve for x :
Substituting 0for y
Factoring
Using the principleof zero products
The solutions of the equation are 5 and The x-interceptsof the graph of are and We can now label themon the graph.
Do Exercises 13 and 14.
x
y y � x 2 � 4x � 5
(�1, 0) (5, 0)
��1, 0�.�5, 0�y � x2 � 4x � 5�1.0 � x2 � 4x � 5
x � 5 or x � �1.
x � 5 � 0 or x � 1 � 0
0 � �x � 5� �x � 1�
0 � x2 � 4x � 5
y � x2 � 4x � 5
y � 0
y � x2 � 4x � 5
x
y
x
y
x
y
x
y
a �� 0 a �� 0
Nox-intercept
Twox-intercepts
Onex-intercept
ax2 � bx � c � 0.
a � 0,y � ax2 � bx � c,
Ax � By � C.y � mx � b
13. Find the x-intercepts of thegraph shown below.
Solving Quadratic Equations We can solve quadratic equationsgraphically. Consider the equation First, we must write theequation with 0 on one side. To do this, we subtract 8 on both sides of theequation; we get Next, we graph in awindow that shows the x-intercepts. The standard window works well inthis case.
The solutions of the equation are the values of x for whichThese are also the first coordinates of the x-intercepts
of the graph. We use the ZERO feature from the CALC menu to find thesenumbers. To find the solution corresponding to the leftmost x-intercept,we first press F m 2 to select the ZERO feature. The prompt “LeftBound?” appears. Next, we use the f or the g key to move the cursor tothe left of the intercept and press [. Now the prompt “Right Bound?”appears. Then we move the cursor to the right of the intercept and press[. The prompt “Guess?” appears. We move the cursor close to theintercept and press [ again. We now see the cursor positioned at theleftmost x-intercept and the coordinates of that point, aredisplayed. Thus, when This is one solution ofthe equation.
We can repeat this procedure to find the first coordinate of the otherx-intercept. We see that at that point. Thus the solutions of theequation are �4 and 2. Note that the x-intercepts ofthe graph of are and
Exercises:1. Solve each of the equations in Examples 4–8 graphically.
1. Dimensions of Picture. Arectangular picture is twice aslong as it is wide. If the area ofthe picture is , find itsdimensions.
Answer on page A-23
w
2w
288 in2
375
5.9 Applications of Quadratic Equations
Applied Problems, Quadratic Equations, and Factoring
We can now use our new method for solving quadratic equations and the fivesteps for solving problems.
EXAMPLE 1 Manufacturing Marble Slabs. Marble Supreme sells rectan-gular marble slabs used for tempering fudge in candy shops. The most popu-lar slab that Marble Supreme sells is twice as long as it is wide and has an areaof What are the dimensions of the slab?
1. Familiarize. We first make a drawing. Recall that the area of any rectan-gle is Length Width. We let the width of the slab, in inches. Thelength is then .
2. Translate. We reword and translate as follows:
Rewording : The area of the rectangle is .
Translating : 7200
3. Solve. We solve the equation as follows:
Subtracting 7200 to get 0 on one side
Removing a common factor of 2
Factoring a difference of squares
Dividing by 2
or Using the principle of zero products
or Solving each equation
4. Check. The solutions of the equation are 60 and . Since the widthmust be positive, cannot be a solution. To check 60 in., we note thatif the width is 60 in., then the length is and the area is
Thus the solution 60 checks.
5. State. The slab is 60 in. wide and 120 in. long.
Do Exercise 1.
60 in. � 120 in. � 7200 in2.2 � 60 in. � 120 in.,
�60�60
x � �60.x � 60
x � 60 � 0x � 60 � 0
�x � 60��x � 60� � 0
2�x � 60� �x � 60� � 0
2�x2 � 3600� � 0
2x2 � 7200 � 0
2x2 � 7200
2x � x � 7200
�2x � x
7200 cm2
x2x
2xx ��
7200 in2.
5.95.9 APPLICATIONS OF QUADRATIC EQUATIONS
ObjectivesSolve applied problemsinvolving quadraticequations that can be solved by factoring.
Solve applied problemsinvolving the Pythagoreantheorem and quadraticequations that can be solvedby factoring.
EXAMPLE 2 Racing Sailboat. The height of a triangular sail on a racingsailboat is 9 ft more than the base. The area of the triangle is . Find theheight and the base of the sail.Source: Whitney Gladstone, North Graphics, San Diego, CA
1. Familiarize. We first make a drawing. If you don’t remember the for-mula for the area of a triangle, look it up in the list of formulas at the backof this book or in a geometry book. The area is (base)(height).
We let the base of the triangle. Then the height.
2. Translate. It helps to reword this problem before translating:
times Base times Height is 110. Rewording
b 110 Translating
3. Solve. We solve the equation as follows:
Multiplying
Multiplying by 2
Simplifying
Subtracting 220 to get 0 on one side
Factoring
or Using the principle of zero products
or
4. Check. The base of a triangle cannot have a negative length, so cannot be a solution. Suppose the base is 11 ft. The height is 9 ft morethan the base, so the height is or 20 ft, and the area is
(11)(20), or . These numbers check in the original problem.
5. State. The height is 20 ft and the base is 11 ft.
Do Exercise 2.
110 ft212
11 ft � 9 ft,
�20
b � �20.b � 11
b � 20 � 0b � 11 � 0
�b � 11� �b � 20� � 0
b2 � 9b � 220 � 0
b2 � 9b � 220 � 220 � 220
b2 � 9b � 220
2 �12
�b2 � 9b� � 2 � 110
12
�b2 � 9b� � 110
12
� b � �b � 9� � 110
��b � 9���12
12
b + 9
b
b � 9 �b �
12
110 ft22. Dimensions of a Sail. The
mainsail of Stacey’s lightning-styled sailboat has an area of125 . The sail is 15 ft tallerthan it is wide. Find the heightand the width of the sail.
EXAMPLE 3 Games in a Sports League. In a sports league of x teams inwhich each team plays every other team twice, the total number N of gamesto be played is given by
Maggie’s basketball league plays a total of 240 games. How many teams are inthe league?
1., 2. Familiarize and Translate. We are given that x is the number of teams in a league and N is the number of games. To familiarizeyourself with this problem, reread Example 4 in Section 4.3 wherewe first considered it. To find the number of teams x in a league inwhich 240 games are played, we substitute 240 for N in theequation:
Substituting 240 for N
3. Solve. We solve the equation as follows:
Subtracting 240 to get 0 on one side
Factoring
or Using the principle of zero products
or
4. Check. The solutions of the equation are 16 and . Since thenumber of teams cannot be negative, cannot be a solution.But 16 checks, since .
5. State. There are 16 teams in the league.
Do Exercise 3.
162 � 16 � 256 � 16 � 240�15
�15
x � �15.x � 16
x � 15 � 0x � 16 � 0
�x � 16� �x � 15� � 0
x2 � x � 240 � 0
x2 � x � 240 � 240 � 240
x2 � x � 240
x2 � x � 240.
x2 � x � N.
3. Use for thefollowing.
a) Volleyball League. Amy’svolleyball league has 19 teams. What is the totalnumber of games to beplayed?
b) Softball League. Barry’sslow-pitch softball leagueplays a total of 72 games. How many teams are in the league?
Answers on page A-23
N � x2 � x
Study Tips FIVE STEPS FOR PROBLEM SOLVING
1. Familiarize yourself with the situation.
a) Carefully read and reread until you understandwhat you are being asked to find.
b) Draw a diagram or see if there is a formula thatapplies.
c) Assign a letter, or variable, to the unknown.
2. Translate the problem to an equation using theletter or variable.
3. Solve the equation.
4. Check the answer in the original wording of theproblem.
5. State the answer to the problem clearly withappropriate units.
“Most worthwhile achievements are the result of manylittle things done in a simple direction.”
EXAMPLE 4 Athletic Numbers. The product of the numbers of twoconsecutive entrants in a marathon race is 156. Find the numbers.
1. Familiarize. The numbers are consecutive integers. Recall that consecu-tive integers are next to each other, such as 49 and 50, or and . Let
the smaller integer; then the larger integer.
2. Translate. It helps to reword the problem before translating:
First integer times Second integer is 156. Rewording
x 156 Translating
3. Solve. We solve the equation as follows:
Multiplying
Subtracting 156 to get 0 on one side
Simplifying
Factoring
or Using the principle of zeroproducts
or
4. Check. The solutions of the equation are 12 and . When x is 12, thenis 13, and . The numbers 12 and 13 are consecutive in-
tegers that are solutions to the problem. When x is , then is and . The numbers and are consecutive inte-gers, but they are not solutions of the problem because negative numbersare not used as entry numbers.
5. State. The entry numbers are 12 and 13.
Do Exercise 4.
The Pythagorean Theorem
The following problems involve the Pythagorean theorem, which relates thelengths of the sides of a right triangle. A triangle is a right triangle if it has a90°, or right, angle. The side opposite the 90° angle is called the hypotenuse.The other sides are called legs.
�12�13��13� ��12� � 156�12,x � 1�13
12 � 13 � 156x � 1�13
x � �13.x � 12
x � 13 � 0x � 12 � 0
�x � 12� �x � 13� � 0
x2 � x � 156 � 0
x2 � x � 156 � 156 � 156
x2 � x � 156
x�x � 1� � 156
��x � 1��
x � 1 �x ��5�6
x � 1x
4. Page Numbers. The product ofthe page numbers on two facingpages of a book is 506. Find thepage numbers.
In any right triangle, if a and b are the lengths of the legs and c is thelength of the hypotenuse, then
EXAMPLE 5 Physical Education. An outdoor-education ropes course in-cludes a 25-ft cable that slopes downward from a height of 37 ft to a height of30 ft. How far is it between the trees that the cable connects?
1. Familiarize. We make a drawing as above, noting that when we subtract30 ft from 37 ft, we get the height of the right triangle that is formed. We let
the distance between the trees.
2. Translate. A right triangle is formed, so we can use the Pythagorean theorem:
Substituting 7 for the length of a leg and 25 for the length of the hypotenuse
3. Solve. We solve the equation as follows:
Squaring 7 and 25
Subtracting 625
Factoring
or Using the principle of zero products
or
4. Check. Since the distance between the trees cannot be negative, cannot be a solution. If the distance is 24 ft, we have
which is Thus, 24 checks and is a solution.
5. State. The distance between the trees is 24 ft.
Do Exercise 5.
252.49 � 576 � 625,72 � 242 �
�24
b � �24.b � 24
b � 24 � 0b � 24 � 0
�b � 24� �b � 24� � 0
b2 � 576 � 0
49 � b2 � 625
72 � b2 � 252
72 � b2 � 252.
a2 � b2 � c2
7 ft25 ft
b
b �
25 ft
?
37 ft
30 ft
ac
b
The symbol denotes a 90° angle.
a2 � b2 � c2.
5. Reach of a Ladder. Twila has a26-ft ladder leaning against herhouse. If the bottom of theladder is 10 ft from the base ofthe house, how high does theladder reach?
EXAMPLE 6 Ladder Settings. A ladder of length 13 ft is placedagainst a building in such a way that the distance from the top of theladder to the ground is 7 ft more than the distance from the bottom ofthe ladder to the building. Find both distances.
1. Familiarize. We first make a drawing. The ladder and the missingdistances form the hypotenuse and legs of a right triangle. We let
the length of the side (leg) across the bottom. Then thelength of the other side (leg). The hypotenuse has length 13 ft.
2. Translate. Since a right triangle is formed, we can use thePythagorean theorem:
Substituting
3. Solve. We solve the equation as follows:
Squaring the binomial and 13
Collecting like terms
Subtracting 169 to get 0 on one side
Simplifying
Factoring out a common factor
Dividing by 2
Factoring
or Using the principle of zero products
or
4. Check. The negative integer cannot be the length of a side.When , , and . So 5 and 12 check.
5. State. The distance from the top of the ladder to the ground is 12 ft. The distance from the bottom of the ladder to the building is 5 ft.
Do Exercise 6.
52 � 122 � 132x � 7 � 12x � 5�12
x � 5.x � �12
x � 5 � 0x � 12 � 0
�x � 12� �x � 5� � 0
x2 � 7x � 60 � 0
2�x2 � 7x � 60� � 0
2x2 � 14x � 120 � 0
2x2 � 14x � 49 � 169 � 169 � 169
2x2 � 14x � 49 � 169
x2 � �x2 � 14x � 49� � 169
x2 � �x � 7�2 � 132.
a2 � b2 � c2
x � 7
x
13 ft
x � 7 �x �
6. Right-Triangle Geometry. Thelength of one leg of a righttriangle is 1 m longer than theother. The length of thehypotenuse is 5 m. Find thelengths of the legs.
The goal of these matchingquestions is to practice step (2),Translate, of the five-step problem-solving process. Translate eachword problem to an equation andselect a correct translation fromequations A–O.
A.
B.
C.
D.
E.
F.
G.
H.
I.
J.
K.
L.
M.
N.
O.
Answers on page A-23
x � �x � 1� � �x � 2� � 180
x2 � 60 � 7021
x�x � 2� � 3599
2x2 � x � 288
x � 6% � x � 40,704
59% � x � 60
x �23
x � �x � 2� � 180
12
x�x � 1� � 1770
2�x � 2� � 2x � 240
6% � x � 40,704
x2 � �x � 70�2 � 1302
x2 � �x � 2�2 � 3599
59 � x � 60
x�x � 60� � 7021
2x � x � 288
Translatingfor Success
6. Cell-Phone Tower. A guy wireon a cell-phone tower is 130 ftlong and is attached to the topof the tower. The height of thetower is 70 ft longer than thedistance from the point on the ground where the wire isattached to the bottom of thetower. Find the height of thetower.
7. Sales Meeting Attendance.PTQ Corporation holds a salesmeeting in Tucson. Of the 60 employees, 59 of them attendthe meeting. What percentattend the meeting?
8. Dimensions of a Pool. Arectangular swimming pool istwice as long as it is wide. Thearea of the surface is Find the dimensions of the pool.
9. Dimensions of a Triangle. Theheight of a triangle is 1 cm lessthan the length of the base. Thearea of the triangle is Find the height and the lengthof the base.
10. Width of a Rectangle. Thelength of a rectangle is 60 ftlonger than the width. Find thewidth if the area of the rectangleis 7021 ft2.
1770 cm2.
288 ft2.
1. Angle Measures. The measuresof the angles of a triangle arethree consecutive integers. Findthe measures of the angles.
2. Rectangle Dimensions. Thearea of a rectangle is The length is 2 ft longer than thewidth. Find the dimensions ofthe rectangle.
3. Sales Tax. Claire paid $40,704for a new SUV. This included 6%for sales tax. How much did theSUV cost before tax?
4. Wire Cutting. A 180-m wire iscut into three pieces. The thirdpiece is 2 m longer than the first.The second is two-thirds as longas the first. How long is eachpiece?
5. Perimeter. The perimeter of arectangle is 240 ft. The length is2 ft greater than the width. Findthe length and the width.
3. Furnishings. A rectangular table in Arlo’s House ofTunes is six times as long as it is wide. The area of thetable is . Find the length and the width of the table.
4. Dimensions of Picture. A rectangular picture is threetimes as long as it is wide. The area of the picture is
. Find the dimensions of the picture.
3w
w
588 in2
w 6w
24 ft2
1. Design. The screen of the TI-84 Plus graphingcalculator is nearly rectangular. The length of therectangle is 2 cm more than the width. If the area of therectangle is find the length and the width.
2. Area of a Garden. The length of a rectangular garden is 4 m greater than the width. The area of the garden is . Find the length and the width.
ww + 4
96 m2
w + 2
w
24 cm2,
5. Dimensions of a Triangle. A triangle is 10 cm widerthan it is tall. The area is . Find the height and thebase.
6. Dimensions of a Triangle. The height of a triangle is3 cm less than the length of the base. The area of thetriangle is . Find the height and the length of thebase.
Games in a League. Use for Exercises 9–12.x2 � x � N
9. A chess league has 14 teams. What is the total numberof games to be played if each team plays every otherteam twice?
10. A women’s volleyball league has 23 teams. What is thetotal number of games to be played if each team playsevery other team twice?
11. A slow-pitch softball league plays a total of 132 games.How many teams are in the league if each team playsevery other team twice?
12. A basketball league plays a total of 90 games. How manyteams are in the league if each team plays every otherteam twice?
7. Road Design. A triangular traffic island has a base halfas long as its height. The island has an area of .Find the base and the height.
8. Dimensions of a Sail. The height of the jib sail on aLightning sailboat is 5 ft greater than the length of its“foot.” The area of the sail is . Find the length ofthe foot and the height of the sail.
x + 5
x
42 ft2
h
h
12
64 m2
Handshakes. A researcher wants to investigate the potential spread of germs by contact. She knows that the number ofpossible handshakes within a group of x people, assuming each person shakes every other person’s hand only once, is given by
Use this formula for Exercises 13–16.
N � 12 �x2 � x�.
13. There are 100 people at a party. How many handshakesare possible?
14. There are 40 people at a meeting. How manyhandshakes are possible?
17. Toasting. During a toast at a party, there were 190 “clicks” of glasses. How many people took part in the toast?
18. High-fives. After winning the championship, allDetroit Pistons teammates exchanged “high-fives.”Altogether there were 66 high-fives. How many playerswere there?
19. Consecutive Page Numbers. The product of the pagenumbers on two facing pages of a book is 210. Find thepage numbers.
20. Consecutive Page Numbers. The product of the pagenumbers on two facing pages of a book is 420. Find thepage numbers.
21. The product of two consecutive even integers is 168.Find the integers. (See Section 2.6.)
22. The product of two consecutive even integers is 224.Find the integers. (See Section 2.6.)
23. The product of two consecutive odd integers is 255.Find the integers.
24. The product of two consecutive odd integers is 143.Find the integers.
25. Right-Triangle Geometry. The length of one leg of aright triangle is 8 ft. The length of the hypotenuse is 2 ftlonger than the other leg. Find the length of thehypotenuse and the other leg.
26. Right-Triangle Geometry. The length of one leg of aright triangle is 24 ft. The length of the other leg is 16 ftshorter than the hypotenuse. Find the length of thehypotenuse and the other leg.
27. Roadway Design. Elliott Street is 24 ft wide when itends at Main Street in Brattleboro, Vermont. A 40-ft longdiagonal crosswalk allows pedestrians to cross MainStreet to or from either corner of Elliott Street (see thefigure). Determine the width of Main Street.
28. Sailing. The mainsail of a Lightning sailboat is a righttriangle in which the hypotenuse is called the leech. If a24-ft tall mainsail has a leech length of 26 ft and ifDacron® sailcloth costs $10 per square foot, find thecost of a new mainsail.
24 ft26 ft
24 ft
40 ft
Elliott St.
Main St.
15. Everyone at a meeting shook hands with each other.There were 300 handshakes in all. How many peoplewere at the meeting?
16. Everyone at a party shook hands with each other. Therewere 153 handshakes in all. How many people were atthe party?
29. Lookout Tower. The diagonal braces in a lookout towerare 15 ft long and span a distance of 12 ft. How highdoes each brace reach vertically?
30. Aviation. Engine failure forced Geraldine to pilot herCessna 150 to an emergency landing. To land, Geraldine’splane glided 17,000 ft over a 15,000-ft stretch of desertedhighway. From what altitude did the descent begin?
12 ft
15 ft h
31. Architecture. An architect has allocated a rectangularspace of for a square dining room and a 10-ftwide kitchen, as shown in the figure. Find thedimensions of each room.
32. Guy Wire. The guy wire on a TV antenna is 1 m longerthan the height of the antenna. If the guy wire isanchored 3 m from the foot of the antenna, how tall isthe antenna?
3 m
10 ft
264 ft2
Rocket Launch. A model rocket is launched with an initialvelocity of . Its height h, in feet, after t seconds isgiven by the formula
33. After how many seconds will the rocket first reach aheight of 464 ft?
34. After how many seconds from launching will the rocketagain be at that same height of 464 ft? (See Exercise 33.)
35. The sum of the squares of two consecutive odd positiveintegers is 74. Find the integers.
36. The sum of the squares of two consecutive odd positiveintegers is 130. Find the integers.
37. An archaeologist has measuring sticks of 3 ft, 4 ft,and 5 ft. Explain how she could draw a 7-ft by 9-ftrectangle on a piece of land being excavated.
38. Look closely at the problem-solving techniquesdeveloped in this chapter. What kinds of equations dowe use? In order to solve these equations, whatadditional new skill do we need? Compare the skillslearned in this chapter with those of Chapter 2.
DWDW
SKILL MAINTENANCE
39. To a polynomial is to express it as aproduct. [5.1b]
40. A(n) of a polynomial P is a polynomialthat can be used to express P as a product. [5.1b]
41. A factorization of a polynomial is an expression that namesthat polynomial as a(n) . [5.1b]
42. When factoring, always look first for the .[5.1b]
43. The expression is an example of a. [4.3i]
44. The asserts that when dividing withexponential notation, if the bases are the same, keep thebase and subtract the exponent of the denominator fromthe exponent of the numerator. [4.1e]
45. For the graph of the equation the pair is known as the . [3.3a]
46. For the graph of the equation the number is known as the . [3.4c]
434x � 3y � 12,
�0, �4�4x � 3y � 12,
�5x2 � 8x � 7
quotient rule
product rule
slope
common factor
common multiple
factor
x-intercept
y-intercept
binomial
trinomial
quotient
product
i VOCABULARY REINFORCEMENT
In each of Exercises 39–46, fill in the blank with the correct term from the given list. Some of the choices may not be used andsome may be used more than once.
47. Telephone Service. Use the information in the figurebelow to determine the height of the telephone pole.
48. Roofing. A square of shingles covers of surfacearea. How many squares will be needed to reshingle thehouse shown?
24 ft
16 ft
25 ft
32 ft
100 ft2
34 ft
5 ft
x
x � 112
49. Pool Sidewalk. A cement walk of constant width isbuilt around a 20-ft by 40-ft rectangular pool. The totalarea of the pool and the walk is . Find the widthof the walk.
50. Rain-Gutter Design. An open rectangular gutter ismade by turning up the sides of a piece of metal 20 in.wide. The area of the cross-section of the gutter is .Find the depth of the gutter.
20 in.
50 in2
50 in2
20 ft
40 ft
x
x
1500 ft2
51. Dimensions of an Open Box. A rectangular piece ofcardboard is twice as long as it is wide. A 4-cm square iscut out of each corner, and the sides are turned up tomake a box with an open top. The volume of the box is
. Find the original dimensions of the cardboard.
52. Solve for x.
63 cm
60 cm36 cm
x
44
V = 616 cm3
616 cm3
53. Dimensions of a Closed Box. The total surface area of aclosed box is . The box is 9 ft high and has asquare base and lid. Find the length of a side of the base.
54. The ones digit of a number less than 100 is 4 greaterthan the tens digit. The sum of the number and theproduct of the digits is 58. Find the number.
The review that follows is meant to prepare you for a chapter exam. It consists of three parts. The first part, ConceptReinforcement, is designed to increase understanding of the concepts through true/false exercises. The second part isa list of important properties and formulas. The third part is the Review Exercises. These provide practice exercises forthe exam, together with references to section objectives so you can go back and review. Before beginning, stop andlook back over the skills you have obtained. What skills in mathematics do you have now that you did not have beforestudying this chapter?
388
CHAPTER 5: Polynomials: Factoring
Summary and Review55
i CONCEPT REINFORCEMENT
Determine whether the statement is true or false. Answers are given at the back of the book.
1. The greatest common factor (GCF) of a set of natural numbers is at least 1 andalways less than or equal to the smallest number in the set.
2. To factor use a trial-and-error process that looks for factors of bwhose sum is c.
3.
4. A product is 0 if and only if all factors are 0.
5. In order that the principle of zero products can be used, one side of theequation must be 0.
40. Sharks’ Teeth. Sharks’ teeth are shaped like triangles.The height of a tooth of a great white shark is 1 cmlonger than the base. The area is Find the heightand the base.
b � 1
b
15 cm2.
41. The product of two consecutive even integers is 288.Find the integers.
42. The product of two consecutive odd integers is 323.Find the integers.
43. Tree Supports. A duckbill-anchor system is used tosupport a newly planted Bradford pear tree. The cable is2 ft longer than the distance from the base of the tree towhere the cable is attached to the tree. The cables areanchored 4 ft from the tree. How far from the groundare the cables attached to the tree?
44. If the sides of a square are lengthened by 3 km, the areaincreases to Find the length of a side of theoriginal square.
81 km2.
Find the x-intercepts for the graph of the equation. [5.8b]
45.
x
yy � x 2 � 9x � 20
46.
x
y y � 2x 2 � 7x � 15
47. Write a problem for a classmate to solve such thatonly one of the two solutions of a quadratic equationcan be used as an answer. [5.9a, b]
DW
48. On a quiz, Sheri writes the factorization of as Explain Sheri’s
mistake. [5.5d]�2x � 10� �2x � 10�.4x2 � 100
DW
SYNTHESIS
50. The cube of a number is the same as twice the square ofthe number. Find all such numbers.
51. The length of a rectangle is two times its width. Whenthe length is increased by 20 and the width decreased by1, the area is 160. Find the original length and width.
Solve. [5.8b]
52. x2 � 25 � 0
53. �x � 2� �x � 3� �2x � 5� � 0
54. �x � 3�4x2 � 3x�x � 3� � �x � 3�10 � 0
55. Find a polynomial for the shaded area in the figurebelow. [5.2a]
x
Solve. [5.9a]
49. The pages of a book measure 15 cm by 20 cm. Marginsof equal width surround the printing on each page andconstitute one-half of the area of the page. Find thewidth of the margins.
26. The length of a rectangle is 2 m more than the width.The area of the rectangle is Find the length andthe width.
27. The base of a triangle is 6 cm greater than twice theheight. The area is Find the height and the base.
2h � 6
h
28 cm2.48 m2.
28. Masonry Corner. A mason wants to be sure he has aright corner in a building’s foundation. He marks apoint 3 ft from the corner along one wall and anotherpoint 4 ft from the corner along the other wall. If thecorner is a right angle, what should the distance bebetween the two marked points?
3 ft
x
4 ft
Find the x-intercepts for the graph of the equation.
29. 30.
x
yy � 3x 2 � 5x � 2
x
y y � x 2 � 2x � 35
31. The length of a rectangle is five times its width. Whenthe length is decreased by 3 and the width is increasedby 2, the area of the new rectangle is 60. Find theoriginal length and width.