Coburn J.W. Algebra and Trigonometry
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30 CHAPTER R A Review of Basic Concepts and Skills R-30
them. Example 14 shows that when we multiply a binomial and its conjugate, the
“outers” and “inners” sum to zero and the result is a difference of two squares.
EXAMPLE 14
Multiplying Binomial Conjugates
Compute each product mentally:
a. b. c.
Solution
a. difference of squares
b. difference of squares:
c.
difference of squares:
Now try Exercises 117 through 124
The Product of a Binomial and Its Conjugate
Given any expression that can be written in the form , the conjugate of the
expression is and their product is a difference of two squares:
Binomial Squares
Expressions like are called binomial squares and are useful for solving many
equations and sketching a number of basic graphs. Note
using the F-O-I-L process. The expression is called a
perfect square trinomial because it is the result of expanding a binomial square. If
we write a binomial square in the more general form and
compute the product, we notice a pattern that helps us write the expanded form more
quickly.
repeated multiplication
F-O-I-L
simplify (perfect square trinomial)
The first and last terms of the trinomial are squares of the terms A and B. Also, the
middle term of the trinomial is twice the product of these two terms:
The F-O-I-L process shows us why. Since the outer and inner products are identical,
we always end up with two. A similar result holds for and the process can be
summarized for both cases using the symbol.
The Square of a Binomial
Given any expression that can be written in the form
1.
2. 1A B2
2
A
2
2AB B
2
1A B2
2
A
2
2AB B
2
1A B2
2
,
1A B2
2
AB AB 2AB.
A
2
2AB B
2
A
2
AB AB B
2
1A B2
2
1A B21A B2
1A B2
2
1A B21A B2
x
2
14x 49x
2
14x 49
1x 72
2
1x 721x 72
1x 72
2
1A B21A B2 A
2
B
2
A B
A B
x
2
a
2
5
b
2
ax
2
5
bax
2
5
b x
2
4
25
0
2
5
x
2
5
x
12x2
2
15y2
2
12x 5y212x 5y2 4x
2
25y
2
10xy (10xy) 0xy
1x2
2
172
2
1x 721x 72 x
2
49
7x 7x 0x
ax
2
5
bax
2
5
b12x 5y212x 5y21x 721x 72
LOOKING AHEAD
Although a binomial square
can always be found using
repeated factors and F-O-I-L,
learning to expand them
using the pattern is a valuable
skill. Binomial squares occur
often in a study of algebra
and it helps to find the
expanded form quickly.
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R-31 Section R.3 Exponents, Scientific Notation, and a Review of Polynomials 31
CAUTION
Note the square of a binomial always results in a trinomial (three terms). Specifically
.
EXAMPLE 15
Find each binomial square without using F-O-I-L:
a. b. c.
Solution
a.
simplify
b.
simplify
c.
simplify
Now try Exercises 125 through 136
With practice, you will be able to go directly from the binomial square to the result-
ing trinomial.
9 61x
x
1A B2
2
A
2
2AB B
2
13 1x2
2
9 213
#
1x2 x
9x
2
30x 25
1A B2
2
A
2
2AB B
2
13x 52
2
13x2
2
213x
#
52 5
2
a
2
18a 81
1A B2
2
A
2
2AB B
2
1a 92
2
a
2
21a
#
92 9
2
13 1x2
2
13x 52
2
1a 92
2
1A B2
2
Z A
2
B
2
F. You’ve just reviewed
how to compute special
products: binomial conjugates
and binomial squares
R.3 EXERCISES
CONCEPTS AND VOCABULARY
DEVELOPING YOUR SKILLS
Fill in each blank with the appropriate word or phrase.
Carefully reread the section, if necessary.
1. The equation is an example of the
property of exponents.
2. The equation is an example of the
property of exponents.
3. The sum of the “outers” and “inners” for
is , while the sum of outers and inners
for is .
12x 5212x 52
12x 52
2
12x
3
2
2
1
4x
6
13x
2
2
3
27x
6
4. The expression can be classified as
a of degree , with a leading
coefficient of .
5. Discuss/Explain why one of the following
expressions can be simplified further, while the
other cannot: (a) (b)
6. Discuss/Explain why the degree of is greater
than the degree of Include additional
examples for contrast and comparison.
2x
2
y
3
.
2x
2
y
3
7n
4
#
3n
2
.7n
4
3n
2
;
2x
2
3x 10
Determine each product using the product and/or power
properties.
7. 8.
9. 10.
11. 12. d
2
#
d
4
#
1c
5
2
2
#
1c
3
2
2
1a
2
2
4
#
1a
3
2
2
#
b
2
#
b
5
11.5vy
2
218v
4
y216p
2
q212p
3
q
3
2
24g
5
#
3
8
g
9
2
3
n
2
#
21n
5
Simplify each expression using the product to a power
property.
13. 14.
15. 16.
17. 18. a
b
3a
b
3
a
p
2q
b
2
12.5h
5
k2
2
13.2hk
2
2
3
13p
2
q2
2
16pq
2
2
3
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19. 20.
21. 22.
23. 24.
25. Volume of a cube: The formula
for the volume of a cube is
where S is the length of
one edge. If the length of each
edge is
a. Find a formula for volume in
terms of the variable x.
b. Find the volume of the cube if
26. Area of a circle: The formula
for the area of a circle is
where r is the length
of the radius. If the radius is
given as
a. Find a formula for area in
terms of the variable x.
b. Find the area of the circle if
Simplify using the quotient property or the property of
negative exponents. Write answers using positive
exponents only.
27. 28.
29. 30.
31. 32.
33. 34.
35. 36.
37. 38.
Simplify each expression using the quotient to a power
property.
39. 40.
41. 42.
43. 44.
45. 46. a
9p
3
q
2
r
3
12p
5
qr
2
b
3
a
5p
2
q
3
r
4
2pq
2
r
4
b
2
a
4p
3
3x
2
y
b
3
a
5m
2
n
3
2r
4
b
2
a
0.5a
3
0.4b
2
b
2
a
0.2x
2
0.3y
3
b
3
a
5v
4
7w
3
b
2
a
2p
4
q
3
b
2
A
2
3
B
2
A
1
2
B
3
142
2
122
3
3
m
2
2
h
3
A
5
6
B
1
A
2
3
B
3
5m
3
n
5
10mn
2
12a
3
b
5
4a
2
b
4
8z
7
16z
5
6w
5
2w
2
x 2.
5x
3
,
A r
2
,
5x
3
x 2.
3x
2
,
V S
3
,
3x
2
3x
2
3x
2
A
2
3
m
2
n
B
2
#
A
1
2
mn
2
BA
3
8
x
B
2
A
16xy
2
B
A
4
5
x
3
B
2
A
3
4
x
3
y
B
2
12.5a
3
2
2
13a
2
b
2
2
3
10.7c
4
2
2
110c
3
d
2
2
2
Use properties of exponents to simplify the following.
Write the answer using positive exponents only.
47. 48.
49. 50.
51. 52.
53. 54.
55. 56.
57. 58.
59. 60.
61. 62.
63. 64.
65. 66.
Convert the following numbers to scientific notation.
67. In mid-2007, the U.S. Census Bureau estimated the
world population at nearly 6,600,000,000 people.
68. The mass of a proton is generally given as 0.000
000 000 000 000 000 000 000 001 670 kg.
Convert the following numbers to decimal notation.
69. As of 2006, the smallest microprocessors in common
use measured across.
70. In 2007, the estimated net worth of Bill Gates, the
founder of Microsoft, was dollars.
Compute using scientific notation. Show all work.
71. The average distance between the Earth and the
planet Jupiter is 465,000,000 mi. How many hours
would it take a satellite to reach the planet if it
traveled an average speed of 17,500 mi per hour?
How many days? Round to the nearest whole.
72. In fiscal terms, a nation’s debt-per-capita is the
ratio of its total debt to its total population. In the
year 2007, the total U.S. debt was estimated at
$9,010,000,000,000, while the population was
estimated at 303,000,000. What was the U.S. debt-
per-capita ratio for 2007? Round to the nearest
whole dollar.
5.6 10
10
6.5 10
9
m
2n
0
12n2
0
5x
0
15x2
0
2
2
2
1
2
0
3
0
3
1
3
2
4
1
8
1
2
1
5
1
132
0
172
0
4
0
5
0
312x
3
y
4
z2
2
18x
2
yz
0
14a
3
bc
0
713a
2
b
2
c2
3
18n
3
813n
2
2
3
612x
3
2
2
10x
2
1p
4
q
8
2
2
p
5
q
2
a
a
3
#
b
c
2
b
4
15
3
2
4
5
9
1a
2
2
3
a
4
#
a
5
5k
3
20k
2
20h
2
12h
5
5m
5
n
2
10m
5
n
9p
6
q
4
12p
4
q
6
32 CHAPTER R A Review of Basic Concepts and Skills R-32
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R-33 Section R.3 Exponents, Scientific Notation, and a Review of Polynomials 33
Identify each expression as a polynomial or
nonpolynomial (if a nonpolynomial, state why); classify
each as a monomial, binomial, trinomial, or none of
these; and state the degree of the polynomial.
73.
74.
75. 76.
77. 78.
Write the polynomial in standard form and name the
leading coefficient.
79.
80.
81.
82.
83.
84.
Find the indicated sum or difference.
85.
86.
87.
88.
89.
90.
91. Subtract from
using a vertical format.
92. Find decreased by
using a vertical format.
Compute each product.
93.
94.
95. 13r 521r 22
2v
2
1v
2
2v 152
3x1x
2
x 62
4x
2
3x
x
4
3x
3
x
4
2x
3
x
2
2x
q
4
2q
3
q
6
2q
5
q
5
2q
4
q
2
2q
1
5
9
n
2
4n
1
2
2 1
2
3
n
2
2n
3
4
2
1
3
4
x
2
5x 22 1
1
2
x
2
3x 42
10.4n
2
5n 0.52 10.3n
2
2n 0.752
15.75b
2
2.6b 1.92 12.1b
2
3.2b2
15q
2
3q 42 13q
2
3q 42
13p
3
4p
2
2p 72 1p
2
2p 52
8 2n
2
7n
12
2
3
x
2
3v
3
14 2v
2
112v2
c
3
6 2c
2
3c
2k
2
12 k
7w 8.2 w
3
3w
2
q
3
2q
2
5qp
3
2
5
4
r
3
2.7r
2
r 15n
2
4n 117
2x
3
2
3
x
2
12x 1.2
35w
3
2w
2
112w2 14
96.
97.
98.
99.
100.
101. 102.
103. 104.
105. 106.
107. 108.
109. 110.
111. 112.
113. 114.
115. 116.
For each binomial, determine its conjugate and then
find the product of the binomial with its conjugate.
117. 118.
119. 120.
121. 122.
123. 124.
Find each binomial square.
125. 126.
127. 128.
129. 130.
131. 132.
Compute each product.
133.
134.
135.
136. 1a 621a 121a 52
1k 521k 621k 22
1a 321b 52
1x 321y 22
11x
72
2
14 1x2
2
15c 6d2
2
14p 3q2
2
15x 32
2
14g 32
2
1a 32
2
1x 42
2
p 12x 16
11 3r6 5k
c 37x 10
6n 54m 3
13y
2
2212y
2
1212x
2
521x
2
32
15x 3y212x 3y214c d213c 5d2
16a b21a 3b213x 2y212x 5y2
1n
2
5
21n
2
5
21m
3
4
21m
3
4
2
1z
1
3
21z
5
6
21x
1
2
21x
1
4
2
1q 4.921q 1.221p 2.521p 3.62
15 n215 n213 m213 m2
16w 1212w 5217v 4213v 52
12h
2
3h 821h 12
1b
2
3b 2821b 22
1z 521z
2
5z 252
1x 321x
2
3x 92
1s 3215s 42
WORKING WITH FORMULAS
137. Medication in the bloodstream:
If 400 mg of a pain medication are taken orally, the
number of milligrams in the bloodstream is
modeled by the formula shown, where M is the
number of milligrams and t is the time in hours,
Construct a table of values for
through 5, then answer the following.
t 10 t 6 5.
97t
2
348tM 0.5t
4
3t
3
a. How many milligrams are in the bloodstream
after 2 hr? After 3 hr?
b. Based on parts a and b, would you expect the
number of milligrams in the bloodstream
after 4 hr to be less or more than in part b?
Why?
c. Approximately how many hours until the
medication wears off (the number of
milligrams in the bloodstream is 0)?
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138. Amount of a mortgage payment:
The monthly mortgage payment required to pay off
(or amortize) a loan is given by the formula shown,
M
Aa
r
12
ba1
r
12
b
n
a1
r
12
b
n
1
where M is the monthly payment, A is the original
amount of the loan, r is the annual interest rate, and
n is the term of the loan in months. Find the
monthly payment (to the nearest cent) required to
purchase a $198,000 home, if the interest rate is
6.5% and the home is financed over 30 yr.
34 CHAPTER R A Review of Basic Concepts and Skills R-34
APPLICATIONS
139. Attraction between particles: In electrical theory,
the force of attraction between two particles P and
Q with opposite charges is modeled by ,
where d is the distance between them and k is a
constant that depends on certain conditions. This is
known as Coulomb’s law. Rewrite the formula
using a negative exponent.
140. Intensity of light: The intensity of illumination
from a light source depends on the distance from
the source according to , where I is the
intensity measured in footcandles, d is the distance
from the source in feet, and k is a constant that
depends on the conditions. Rewrite the formula
using a negative exponent.
141. Rewriting an expression: In advanced
mathematics, negative exponents are widely used
because they are easier to work with than rational
expressions. Rewrite the expression
using negative exponents.
142. Swimming pool hours: A swimming pool opens at
8
A.M. and closes at 6 P.M. In summertime, the
5
x
3
3
x
2
2
x
1
4
I
k
d
2
F
kPQ
d
2
number of people in the pool at any time can be
approximated by the formula ,
where S is the number of swimmers and t is the
number of hours the pool has been open (8
A.M.:
9
A.M.: 10 A.M.: etc.).
a. How many swimmers are in the pool at 6
P.M.?
Why?
b. Between what times would you expect the largest
number of swimmers?
c. Approximately how many swimmers are in the
pool at 3
P.M.?
d. Create a table of values for
and check your answer to part b.
143. Maximizing revenue: A sporting goods store finds
that if they price their video games at $20, they
make 200 sales per day. For each decrease of $1,
20 additional video games are sold. This means the
store’s revenue can be modeled by the formula
where x is the number
of $1 decreases. Multiply out the binomials and use
a table of values to determine what price will give
the most revenue.
144. Maximizing revenue: Due to past experience, a
jeweler knows that if they price jade rings at $60,
they will sell 120 each day. For each decrease of
$2, five additional sales will be made. This means
the jeweler’s revenue can be modeled by the
formula where x is the
number of $2 decreases. Multiply out the
binomials and use a table of values to determine
what price will give the most revenue.
R 160 2x21120 5x2,
R 120 1x21200 20x2,
4, . . .t 1, 2, 3,
t 2,t 1,t 0,
S1t2t
2
10t
EXTENDING THE CONCEPT
145. If
what is the
value of k?
12x
2
4x k2x
2
3x 2,
13x
2
kx 12 1kx
2
5x 72
146. If then the expression
is equal to what number?
4x
2
1
4x
2
a2x
1
2x
b
2
5,
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Section 0.0 Section Title 35
R.4 Factoring Polynomials
It is often said that knowing which tool to use is just as important as knowing how to
use the tool. In this section, we review the tools needed to factor an expression, an
important part of solving polynomial equations. This section will also help us decide
which factoring tool is appropriate when many different factorable expressions are
presented.
A. The Greatest Common Factor
To factor an expression means to rewrite the expression as an equivalent product. The
distributive property is an example of factoring in action. To factor we might
first rewrite each term using the common factor then
apply the distributive property to obtain We commonly say that we have
factored out 2x. The greatest common factor (or GCF) is the largest factor common
to all terms in the polynomial.
EXAMPLE 1
Factoring Polynomials
Factor each polynomial:
a. b.
Solution
a. 6 is common to all three terms:
mentally:
b. is common to both terms:
mentally:
Now try Exercises 7 and 8
B. Common Binomial Factors and Factoring by Grouping
If the terms of a polynomial have a common binomial factor, it can also be factored
out using the distributive property.
EXAMPLE 2
Factoring Out a Common Binomial Factor
Factor:
a. b.
Solution
a. b.
Now try Exercises 9 and 10
One application of removing a binomial factor involves factoring by grouping.
At first glance, the expression appears unfactorable. But by group-
ing the terms (applying the associative property), we can remove a monomial factor
from each subgroup, which then reveals a common binomial factor.
x
3
2x
2
3x 6
1x 221x
2
32 1x 321x
2
52
x
2
1x 22 31x 221x 32x
2
1x 325
x
2
1x 22 31x 221x 32x
2
1x 325
x
2
1x
3
12
x
2
#
x
3
x
2
#
1x
5
x
2
x
2
612x
2
3xy 5y2
6
#
2x
2
6
#
3xy 6
#
5y12x
2
18xy 30y
x
5
x
2
12x
2
18xy 30y
2x1x 32.
2x: 2x
2
6x 2x
#
x 2x
#
3,
2x
2
6x,
Learning Objectives
In Section R.4 you will review:
A. Factoring out the
greatest common factor
B. Common binomial
factors and factoring
by grouping
C. Factoring quadratic
polynomials
D. Factoring special forms
and quadratic forms
A. You’ve just reviewed
how to factor out the greatest
common factor
R-35 35
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36 CHAPTER R A Review of Basic Concepts and Skills R-36
EXAMPLE 3
Factoring by Grouping
Factor
Solution
Notice that all four terms have a common factor of 3. Begin by factoring it out.
original polynomial
factor out 3
group remaining terms
factor common monomial
Now try Exercises 11 and 12
When asked to factor an expression, first look for common factors. The resulting
expression will be easier to work with and help ensure the final answer is written in
completely factored form. If a four-term polynomial cannot be factored as written,
try rearranging the terms to find a combination that enables factoring by grouping.
C. Factoring Quadratic Polynomials
A quadratic polynomial is one that can be written in the form where a,
b, and One common form of factoring involves quadratic trinomials such
as and While we know
and using F-O-I-L, how can we
factor these trinomials without seeing the original problem in advance? First, it helps
to place the trinomials in two families—those with a leading coefficient of 1 and those
with a leading coefficient other than 1.
ax
2
bx c, where a 1
When the only factor pair for (other than is and the first term in
each binomial will be x: (x )(x ). The following observation helps guide us to the
complete factorization. Consider the product
F-O-I-L
distributive property
Note the last term is the product ab (the lasts), while the coefficient of the middle
term is (the sum of the outers and inners). Since the last term of
is 7 and the coefficient of the middle term is we are seeking two numbers with a
product of positive 7 and a sum of negative 8. The numbers are and so the fac-
tored form is It is also helpful to note that if the constant term is pos-
itive, the binomials will have like signs, since only the product of like signs is positive.
If the constant term is negative, the binomials will have unlike signs, since only the
product of unlike signs is negative. This means we can use the sign of the linear term
(the term with degree 1) to guide our choice of factors.
Factoring Trinomials with a Leading Coefficient of 1
If the constant term is positive, the binomials will have like signs:
or ,
to match the sign of the linear (middle) term.
If the constant term is negative, the binomials will have unlike signs:
with the larger factor placed in the binomial
whose sign matches the linear (middle) term.
1x 21x 2,
1x 21x 21x 21x 2
1x 721x 12.
1,7
8,
x
2
8x 7a b
x
2
1a b2x ab
1x b21x a2 x
2
ax bx ab
1x b21x a2:
x
#
x1
#
x
2
2x
2
a 1,
12x 321x 52 2x
2
13x 15x
2
7x 10
1x 521x 22 2x
2
13x 15.x
2
7x 10
a 0.c
ax
2
bx c,
31t 521t
2
22
33t
2
1t 52 21t 524
31t
3
5t
2
2t 102
31t
3
5t
2
2t 102
3t
3
15t
2
6t 30
3t
3
15t
2
6t 30.
factor common binomial
B. You’ve just reviewed
how to factor by grouping
WORTHY OF NOTE
Similarly, a cubic polynomial
is one of the form
It’s
helpful to note that a cubic
polynomial can be factored
by grouping only when
where a, b, c, and d
are the coefficients shown.
This is easily seen in
Example 3, where
gives
90 90✓.
1321302 1152162
ad bc,
ax
3
bx
2
cx d.
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R-37 Section R.4 Factoring Polynomials 37
EXAMPLE 4
Factoring Trinomials
Factor these expressions:
a. b.
Solution
a. First rewrite the trinomial in standard form as For
the constant term is positive so the binomials will have like
signs. Since the linear term is negative,
like signs, both negative
b. First rewrite the trinomial in standard form as The constant
term is negative so the binomials will have unlike signs. Since the linear term
is negative,
unlike signs, one positive and one negative
Now try Exercises 13 and 14
Sometimes we encounter prime polynomials, or polynomials that cannot be fac-
tored. For the factor pairs of 15 are and with neither pair
having a sum of We conclude that is prime.
ax
2
bx c, where a 1
If the leading coefficient is not one, the possible combinations of outers and inners are
more numerous. Furthermore, their sum will change depending on the position of the
possible factors. Note that and
result in a different middle term, even though identical numbers
were used.
To factor note the constant term is positive so the binomials must
have like signs. The negative linear term indicates these signs will be negative. We then
list possible factors for the first and last terms of each binomial, then sum the outer
and inner products.
2x
2
13x 15,
2x
2
15x 27
12x 921x 32 12x 321x 92 2x
2
21x 27
x
2
9x 159.
3
#
5,1
#
15x
2
9x 15,
1x 221x 52
x
2
3x 10 1x 21x 2
x
2
3x 10.
182132 24; 8 13211 11x 821x 32
11x
2
11x 242 11x 21x 2
x
2
11x 24,
11x
2
11x 242.
x
2
10 3xx
2
11x 24
is placed in the second binomial;
12215210; 2 1523
5 7 2, 5
Possible First and Last Terms Sum of
for and 15 Outers and Inners
1.
2.
3.
4. 6x 5x 11x12x 521x 32
d10x 3x 13x12x 321x 52
2x 15x 17x12x 1521x 12
30x 1x 31x12x 121x 152
2x
2
As you can see, only possibility 3 yields a linear term of and the correct
factorization is then With practice, this trial-and-error process can
be completed very quickly.
If the constant term is negative, the number of possibilities can be reduced by
finding a factor pair with a sum or difference equal to the absolute value of the linear
coefficient, as we can then arrange the sign of each binomial to obtain the needed result
(see Example 5).
12x 321x 52.
13x,
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38 CHAPTER R A Review of Basic Concepts and Skills R-38
EXAMPLE 5
Factoring a Trinomial Using Trial and Error
Factor .
Solution
Note the constant term is negative (binomials will have unlike signs),
and the factors of 35 are and Two possible first terms are: (6z )(z )
and (3z )(2z ), and we begin with 5 and 7 as factors of 35.
5
#
7.1
#
35
11
11,
6z
2
11z 35
(6z )(z ) Outers/Inners (3z )(2z ) Outers/Inners
Sum Diff Sum Diff
1. (6z 5)(z 7)47z 37z 3. (3z 5)(2z 7)31z 11z
2. (6z 7)(z 5)37z 23z 4. (3z 7)(2z 5)29z 1z
d
Since possibility 3 yields the linear term of 11z, we need not consider other
factors of 35 and write the factored form as
The signs can then be arranged to obtain a middle term of
✓.
Now try Exercises 15 and 16
D. Factoring Special Forms and Quadratic Forms
Next we consider methods to factor each of the special products we encountered in
Section R.3.
The Difference of Two Squares
Multiplying and factoring are inverse processes. Since we
know that In words, the difference of two squares will
factor into a binomial and its conjugate.To find the terms of the factored form, rewrite
each term in the original expression as a square:
Factoring the Difference of Two Perfect Squares
Given any expression that can be written in the form
Note that the sum of two perfect squares cannot be factored using real
numbers (the expression is prime). As a reminder, always check for a common factor
first and be sure to write all results in completely factored form. See Example 6(c).
EXAMPLE 6
Factoring the Difference of Two Perfect Squares
Factor each expression completely.
a. b. c. d. e.
Solution
a. write as a difference of squares
b. is prime.
c.
factor out
write as a difference of squares
A
2
B
2
1A B21A B2 31n 421n 42
3n
2
142
2
33n
2
48 31n
2
162
v
2
49
A
2
B
2
1A B21A B2 12w 9212w 92
4w
2
81 12w2
2
9
2
x
2
7z
4
1
81
3n
2
48v
2
494w
2
81
A
2
B
2
A
2
B
2
1A B21A B2
A
2
B
2
,
1 2
2
.
x
2
49 1x 721x 72.
1x 721x 72 x
2
49,
21z 10z 11z
11z: 13z 5212z 72,
6z
2
11z 35 13z
5212z
72.
C. You’ve just reviewed
how to factor quadratic
polynomials
WORTHY OF NOTE
In an attempt to factor a
sum of two perfect squares,
say let’s list all
possible binomial factors.
These are (1)
(2) and
(3) Note that
(1) and (2) are the binomial
squares and
with each product
resulting in a “middle” term,
whereas (3) is a binomial
times its conjugate, resulting
in a difference of squares:
With all possibilities
exhausted, we conclude that
the sum of two squares is
prime!
v
2
49.
1v 72
2
,
1v 72
2
1v 721v 72.
1v 721v 72,
1v 721v 72,
v
2
49,
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R-39 Section R.4 Factoring Polynomials 39
d. write as a difference of squares
write as a difference of squares
result
e. write as a difference of squares
Now try Exercises 17 and 18
Perfect Square Trinomials
Since we know that In
words, a perfect square trinomial will factor into a binomial square. To use this idea
effectively, we must learn to identify perfect square trinomials. Note that the first and
last terms of are the squares of x and 7, and the middle term is twice
the product of these two terms: These are the characteristics of a perfect
square trinomial.
Factoring Perfect Square Trinomials
Given any expression that can be written in the form
1.
2.
EXAMPLE 7
Factoring a Perfect Square Trinomial
Factor
Solution
check for common factors:
factor out 3m
For the remaining trinomial
1. Are the first and last terms perfect squares?
and ✓ Yes.
2. Is the linear term twice the product of and 1?
✓ Yes.
Factor as a binomial square:
This shows
Now try Exercises 19 and 20
CAUTION
As shown in Example 7, be sure to include the GCF in your final answer. It is a common
error to “leave the GCF behind.”
In actual practice, the tests for a perfect square trinomial are performed mentally,
with only the factored form being written down.
12m
3
12m
2
3m 3m12m 12
2
.
4m
2
4m 1 12m 12
2
2
#
2m
#
1 4m
2m
1 112
2
4m
2
12m2
2
4m
2
4m 1 . . .
3m14m
2
4m 12
GCF 3m12m
3
12m
2
3m
12m
3
12m
2
3m.
A
2
2AB B
2
1A B2
2
A
2
2AB B
2
1A B2
2
A
2
2AB B
2
,
217x2 14x.
x
2
14x 49
x
2
14x 49 1x 72
2
.1x 72
2
x
2
14x 49,
A
2
B
2
1A B21A B2 1x 1721x 172
x
2
7 1x2
2
1172
2
1z
1
3
21z
1
3
21z
2
1
9
2
z
2
1
1
3
2
2
1z
2
1
9
2
A
2
B
2
1A B21A B2
1z
2
1
9
21z
2
1
9
2
z
4
1
81
1z
2
2
2
1
1
9
2
2
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