Download Determine whether each expression is a polynomial. If it is a

a number and one or more variables with
nonnegative integer exponents. It has
only one term. is a division of two monomials, so
it is not a monomial.
8-1 Adding and Subtracting Polynomials
Determine whether each expression is a
polynomial. If it is a polynomial, find the degree
and determine whether it is a monomial,
binomial, or trinomial.
2
1. 7ab + 6b – 2a
SOLUTION: A polynomial is a monomial or the sum of
2 3
3
SOLUTION: A polynomial is a monomial or the sum of
2
3
monomials. 7ab + 6b – 2a is the sum of 3
monomials, so it is a polynomial. The degree of a polynomial is the greatest degree of
any term in the polynomial.The degree of each term
2
3
is 2, 2, and 3, so the degree of 7ab + 6b – 2a is 3.
The polynomial has three terms, so it is a trinomial.
2. 2y – 5 + 3y
2 3
5. 5m p + 6
2
monomials. 5m p + 6 is the sum of 2 monomials, so
it is a polynomial. The degree of a polynomial is the greatest degree of
any term in the polynomial. The degree of each term
2 3
is 5 and 0, so the degree of 5m p + 6 is 5. The
polynomial has two terms, so it is a binomial.
6. 5q
–4
+ 6q
SOLUTION: A monomial is a number, a variable, or the product of
a number and one or more variables with
nonnegative integer exponents. It has
SOLUTION: -4
2
2y – 5 + 3y is the sum of monomials, so it is a
polynomial. The degree of a polynomial is the greatest degree of
any term in the polynomial. The degree of each term
2
is 1, 0, and 2, so the degree of 2y – 5 + 3y is 2. The
polynomial has three terms, so it is a trinomial.
3. 3x
only one term. 5q is equal to , which is a
division of two monomials, so it is not a monomial.
Write each polynomial in standard form. Identify
the leading coefficient.
4
2
7. –4d + 1 – d
SOLUTION: Find the degree of each term.
4
2
SOLUTION: A polynomial is a monomial or the sum of monomials,
2
so 3x is a polynomial. The degree of a polynomial is the greatest degree of
2
any term in the polynomial. The degree of 3x is 2.
The polynomial has one term, so it is a monomial.
–4d → 4
1 → 0
2
–d → 2
4
The greatest degree is 4, from the term –4d , so the
4
2
leading coefficient of –4d + 1 – d is –4.
Rewrite the polynomial with each monomial in
descending order according to degree.
4
2
–4d – d + 1
4. SOLUTION: A monomial is a number, a variable, or the product of
a number and one or more variables with
nonnegative integer exponents. It has
only one term. is a division of two monomials, so
it is not a monomial.
2 3
5. 5m p + 6
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SOLUTION: A polynomial is a monomial or the sum of
5
8. 2x – 12 + 3x
SOLUTION: Find the degree of each term.
5
2x → 5
–12 → 0
3x → 1
Page 1
5
The greatest degree is 5, from the term 2x , so the
5
leading coefficient of 2x – 12 + 3x is 2.
Rewrite the polynomial with each monomial in
descending order according to degree.
4
2 Subtracting Polynomials
8-1 Adding
–4d – dand
+1
5
3
8. 2x – 12 + 3x
5
2x → 5
–12 → 0
3x → 1
5
The greatest degree is 5, from the term 2x , so the
5
leading coefficient of 2x – 12 + 3x is 2.
Rewrite the polynomial with each monomial in
descending order according to degree.
5
2x + 3x – 12
2
2
10. 2a + 4a – 5a – 1
SOLUTION: Find the degree of each term.
9. 4z – 2z – 5z
Rewrite the polynomial with each monomial in
descending order according to degree.
4
2
–5z – 2z + 4z
4
SOLUTION: Find the degree of each term.
2a → 1
3
4a → 3
2
– 5a → 2
– 1 → 0
3
The greatest degree is 3, from the term 4a , so the
3
2
leading coefficient of 2a + 4a – 5a – 1 is 4.
Rewrite the polynomial with each monomial in
descending order according to degree.
3
2
4a – 5a + 2a – 1
Find each sum or difference.
SOLUTION: Find the degree of each term.
4z → 1
2
– 2z → 2
3
3
11. (6x − 4) + (−2x + 9)
SOLUTION: 4
– 5z → 4
4
The greatest degree is 4, from the term – 5z , so the
2
4
leading coefficient of 4z – 2z – 5z is –5.
Rewrite the polynomial with each monomial in
descending order according to degree.
4
2
–5z – 2z + 4z
3
3
2
2
12. (g − 2g + 5g + 6) − (g + 2g)
SOLUTION: 2
10. 2a + 4a – 5a – 1
2
2
13. (4 + 2a − 2a) − (3a − 8a + 7)
SOLUTION: Find the degree of each term.
2a → 1
3
4a → 3
SOLUTION: 2
– 5a → 2
– 1 → 0
3
The greatest degree is 3, from the term 4a , so the
3
2
leading coefficient of 2a + 4a – 5a – 1 is 4.
Rewrite the polynomial with each monomial in
descending order according to degree.
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3
2
4a – 5a + 2a – 1
2
2
14. (8y − 4y ) + (3y − 9y )
SOLUTION: Page 2
8-1 Adding and Subtracting Polynomials
2
2
19. CCSS SENSE-MAKING The total number of
students T who traveled for spring break consists of
two groups: students who flew to their destinations F
and students who drove to their destination D. The
number (in thousands) of students who flew and the
total number of students who flew or drove can be
modeled by the following equations, where n is the
number of years since 1995.
14. (8y − 4y ) + (3y − 9y )
SOLUTION: 3
3
2
15. (−4z − 2z + 8) − (4z + 3z − 5)
SOLUTION: T = 14n + 21
F = 8n + 7
a. Write an equation that models the number of
students who drove to their destination for this time
period.
2
2
16. (−3d − 8 + 2d) + (4d − 12 + d )
SOLUTION: b. Predict the number of students who will drive to
their destination in 2012.
c. How many students will drive or fly to their
destination in 2015?
SOLUTION: a.
2
17. (y + 5) + (2y + 4y – 2)
D = 6n + 14
SOLUTION: b. n = 2012 – 1995 = 17
3
2
2
3
18. (3n − 5n + n ) − (−8n + 3n )
SOLUTION: The number of students who will drive to their
destination in 2012 is 116,000 students.
c. n = 2015 – 1995 = 20
19. CCSS SENSE-MAKING The total number of
students T who traveled for spring break consists of
two groups: students who flew to their destinations F
and students who drove to their destination D. The
number (in thousands) of students who flew and the
total number of students who flew or drove can be
eSolutions Manual - Powered by Cognero
modeled by the following equations, where n is the
number of years since 1995.
The number of students who will drive or fly to their
destination in 2015 is 301,000 students.
Determine whether each expression is a
polynomial. If it is a polynomial, find the degree
and determine whether it is a monomial,
Page 3
binomial, or trinomial.
monomials. A monomial is a number, a variable, or the product of
a number and one or more variables with
8-1 Adding
and of
Subtracting
Polynomials
The number
students who
will drive or fly to their
destination in 2015 is 301,000 students.
Determine whether each expression is a
polynomial. If it is a polynomial, find the degree
and determine whether it is a monomial,
binomial, or trinomial.
c
nonnegative integer exponents. 3d has a variable in
the exponent, so it is not a monomial.
24. a – a
2
SOLUTION: A polynomial is a monomial or the sum of
2
monomials. a – a is the sum of 2 monomials, so it is
a polynomial. 20. SOLUTION: A monomial is a number, a variable, or the product of
a number and one or more variables with
nonnegative integer exponents. It has
only one term. is a division of two monomials, so
it is not a monomial.
21. SOLUTION: A polynomial is a monomial or the sum of
monomials. 21 is a monomial, so it is also a polynomial. The degree of a polynomial is the greatest degree of
any term in the polynomial. The degree of each term
is 1 and 2, so the degree of a – a2 is 2. The
polynomial has two terms, so it is a binomial.
3
3
25. 5n + nq
SOLUTION: A polynomial is a monomial or the sum of
monomials. 5n3 + nq3 is the sum of 2 monomials, so
it is a polynomial. The degree of a polynomial is the greatest degree of
any term in the polynomial. The degree of each term
3
The degree of a polynomial is the greatest degree of
any term in the polynomial. The degree of 21 is 0.
The polynomial has only one term, so it is a
monomial.
4
2
22. c – 2c + 1
SOLUTION: A polynomial is a monomial or the sum of
4
2
monomials. c – 2c + 1 is the sum of 3 monomials,
so it is a polynomial. The degree of a polynomial is the greatest degree of
any term in the polynomial. The degree of each term
4
2
is 4, 2, and 0, so the degree of c – 2c + 1 is 4. The
polynomial has three terms, so it is a trinomial.
23. d + 3d
c
SOLUTION: A polynomial is a monomial or the sum of
monomials. A monomial is a number, a variable, or the product of
a number and one or more variables with
c
nonnegative integer exponents. 3d has a variable in
the exponent, so it is not a monomial.
24. a – a
2
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SOLUTION: A polynomial is a monomial or the sum of
3
is 3 and 4, so the degree of 5n + nq is 4. The
polynomial has two terms, so it is a binomial.
Write each polynomial in standard form. Identify
the leading coefficient.
2
26. 5x – 2 + 3x
SOLUTION: Find the degree of each term.
2
5x → 2
– 2 → 0
3x → 1
2
The greatest degree is 2, from the term 5x , so the
2
leading coefficient of 5x – 2 + 3x is 5.
Rewrite the polynomial with each monomial in
descending order according to degree.
2
5x + 3x – 2
27. 8y + 7y
3
SOLUTION: Find the degree of each term.
3
7y → 3
8y → 1
Page 4
Rewrite the polynomial with each monomial in
descending order according to degree.
2
8-1 Adding
5x + 3xand
– 2 Subtracting Polynomials
27. 8y + 7y
3
3
2
SOLUTION: Find the degree of each term.
3
7y → 3
8y → 1
3
The greatest degree is 3, from the term 7y , so the
3
leading coefficient of 8y + 7y is 7.
Rewrite the polynomial with each monomial in
descending order according to degree.
3
7y + 8y
3
–y → 3
3y → 1
2
3y → 2
2 → 0
3
The greatest degree is 3, from the term –y , so the
3
2
leading coefficient of –y + 3y – 3y + 2 is –1.
Rewrite the polynomial with each monomial in
descending order according to degree.
–y 3 – 3y 2 + 3y + 2
2
2
5
30. 11t + 2t – 3 + t
SOLUTION: Find the degree of each term.
4 → 0
3c → 1
2
– 5c → 2
SOLUTION: Find the degree of each term.
11t → 1
2
2t → 2
–3 → 0
t 5 → 5
2
The greatest degree is 2, from the term – 5c , so the
2
leading coefficient of 4 – 3c – 5c is –5.
Rewrite the polynomial with each monomial in
descending order according to degree.
2
–5c – 3c + 4
3
2
–5c – 3c + 4
29. –y + 3y – 3y + 2
SOLUTION: Find the degree of each term.
28. 4 – 3c – 5c
Rewrite the polynomial with each monomial in
descending order according to degree.
5
The greatest degree is 5, from the term t , so the
2
5
leading coefficient of 11t + 2t – 3 + t is 1.
Rewrite the polynomial with each monomial in
descending order according to degree.
5
2
2
t + 2t + 11t – 3
29. –y + 3y – 3y + 2
SOLUTION: Find the degree of each term.
31. 2 + r – r
3
SOLUTION: Find the degree of each term.
3
–y → 3
3y → 1
2
3y → 2
2 → 0
2 → 0
r → 1
3
The greatest degree is 3, from the term –y , so the
3
2
leading coefficient of –y + 3y – 3y + 2 is –1.
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Rewrite the polynomial with each monomial in
descending order according to degree.
– r3→ 3
Page 5
The greatest degree is 3, from the term – r3, so the
3
leading coefficient of 2 + r – r is –1.
Rewrite the polynomial with each monomial in
descending order according to degree.
8-1 Adding
and Subtracting Polynomials
5
2
t + 2t + 11t – 3
31. 2 + r – r
Rewrite the polynomial with each monomial in
descending order according to degree.
3
2
33. –9b + 10b – b
6
SOLUTION: Find the degree of each term.
SOLUTION: Find the degree of each term.
2 → 0
r → 1
–9b2 → 2
10b → 1
– r3→ 3
–b6→ 6
The greatest degree is 6, from the term – b6, so the
The greatest degree is 3, from the term – r3, so the
3
leading coefficient of 2 + r – r is –1.
2
6
leading coefficient of –9b + 10b – b is –1.
Rewrite the polynomial with each monomial in
descending order according to degree.
Rewrite the polynomial with each monomial in
descending order according to degree.
3
–r + r + 2
6
2
–b – 9b + 10b
Find each sum or difference.
2
34. (2c + 6c + 4) + (5c – 7)
32. SOLUTION: SOLUTION: Find the degree of each term.
→ 0
4
–3x → 4
7 → 0
2
2
35. (2x + 3x ) − (7 − 8x )
SOLUTION: The greatest degree is 4, from the term –3x 4, so the
leading coefficient of
is –3.
Rewrite the polynomial with each monomial in
descending order according to degree.
3
2
36. (3c − c + 11) − (c + 2c + 8)
SOLUTION: 2
33. –9b + 10b – b
6
SOLUTION: Find the degree of each term.
–9b2 → 2
10b → 1
2
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2
37. (z + z) + (z − 11)
–b6→ 6
SOLUTION: b6
The greatest degree is 6, from the term – , so the
2
6
leading coefficient of –9b + 10b – b is –1.
Page 6
8-1 Adding and Subtracting Polynomials
2
2
2
37. (z + z) + (z − 11)
2
42. (5n − 2p + 2np) − (4p + 4n)
SOLUTION: SOLUTION: 38. (2x − 2y + 1) − (3y + 4x)
2
SOLUTION: 2
2
2
43. (4rxt − 8r x + x ) − (6rx + 5rxt − 2x )
SOLUTION: 2
2
39. (4a − 5b + 3) + (6 − 2a + 3b )
SOLUTION: 2
2
44. PETS From 1999 through 2009, the number of dogs
D and the number of cats C (in hundreds) adopted
from animal shelters in the United States are
modeled by the equations D = 2n + 3 and C = n + 4,
where n is the number of years since 1999.
2
40. (x y − 3x + y) + (3y − 2x y)
SOLUTION: a. Write an equation that models the total number T
of dogs and cats adopted in hundreds for this time
period.
b. If this trend continues, how many dogs and cats
will be adopted in 2013?
2
SOLUTION: a.
2
41. (−8xy + 3x − 5y) + (4x − 2y + 6xy)
SOLUTION: 2
So, an equation that models the total number of dogs
and cats adopted is T = 3n + 7.
2
42. (5n − 2p + 2np) − (4p + 4n)
b. Evaluate the equation for the total number of dogs
and cats for n = 2013 – 1999 = 14.
SOLUTION: eSolutions Manual - Powered by Cognero
2
2
2
2
43. (4rxt − 8r x + x ) − (6rx + 5rxt − 2x )
Page 7
The number of cats and dogs adopted in 2013 will be
49 × 100 or 4900 cats and dogs.
5 → 0
The greatest degree is 2 and there are 3 terms, so 4x
2
– 3x + 5 is a quadratic trinomial.
8-1 Adding and Subtracting Polynomials
44. PETS From 1999 through 2009, the number of dogs
D and the number of cats C (in hundreds) adopted
from animal shelters in the United States are
modeled by the equations D = 2n + 3 and C = n + 4,
where n is the number of years since 1999.
a. Write an equation that models the total number T
of dogs and cats adopted in hundreds for this time
period.
b. If this trend continues, how many dogs and cats
will be adopted in 2013?
3
46. 11z
SOLUTION: Find the degree of each term.
11z 3 → 3
The greatest degree is 3 and there is one term, so
3
11z is a cubic monomial.
47. 9 + y
4
SOLUTION: SOLUTION: a.
4
Find the degree of each term of 9 + y .
9 → 1
y4→ 4
So, an equation that models the total number of dogs
and cats adopted is T = 3n + 7.
The greatest degree is 4 and there are 2 terms, so 9
+ y 4 is a quartic binomial.
3
b. Evaluate the equation for the total number of dogs
and cats for n = 2013 – 1999 = 14.
48. 3x – 7
SOLUTION: 3
Find the degree of each term of 3x – 7.
3
3x → 3
–7 → 0
The number of cats and dogs adopted in 2013 will be
49 × 100 or 4900 cats and dogs.
Classify each polynomial according to its degree
and number of terms.
2
45. 4x – 3x + 5
SOLUTION: The greatest degree is 3 and there are 2 terms, so
3
3x – 7 is a cubic binomial.
5
2
49. –2x – x + 5x – 8 SOLUTION: 2
2
Find the degree of each term of –2x 5 – x + 5x – 8.
Find the degree of each term of 4x – 3x + 5.
– 3x → 2
5 → 0
–2x → 5
–x2 → 2
5x → 1
– 8 → 0
The greatest degree is 2 and there are 3 terms, so 4x
2
– 3x + 5 is a quadratic trinomial.
The greatest degree is 5 and there are 4 terms, so –
5
4x → 1
2
2
5
2x – x + 5x – 8 is a quintic polynomial.
2
46. 11z
3
SOLUTION: Find the degree of each term.
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3
3
50. 10t – 4t + 6t
SOLUTION: 2
3
Find the degree of each term of 10t – 4t + 6t . Page 8
10t → 1
6t → 3
5x → 1
– 8 → 0
The greatest
degree is 5 and
there are 4 terms, so –
8-1 Adding
and Subtracting
Polynomials
2
5
2x – x + 5x – 8 is a quintic polynomial.
2
3
The greatest degree is 3 and there are 3 terms, so
2
3
10t – 4t + 6t is a cubic trinomial.
51. ENROLLMENT In a rapidly growing
50. 10t – 4t + 6t
SOLUTION: 2
3
Find the degree of each term of 10t – 4t + 6t .
10t → 1
2
4t → 2
3
6t → 3
school system, the numbers (in hundreds) of
total students N and K-5 students P
enrolled from 2000 to 2009 are modeled
by the equations N = 1.25t 2 – t + 7.5 and
P = 0.7t 2 – 0.95t + 3.8, where t is the
number of years since 2000.
a. Write an equation modeling the number of 6-12
students S enrolled for this time period.
The greatest degree is 3 and there are 3 terms, so
2
3
10t – 4t + 6t is a cubic trinomial.
51. ENROLLMENT In a rapidly growing
school system, the numbers (in hundreds) of
total students N and K-5 students P
enrolled from 2000 to 2009 are modeled
by the equations N = 1.25t 2 – t + 7.5 and
P = 0.7t 2 – 0.95t + 3.8, where t is the
number of years since 2000.
a. Write an equation modeling the number of 6-12
students S enrolled for this time period.
b. How many 6-12 students were enrolled in the
school system in 2007? SOLUTION: a. To write an equation that represents the number of
6-12 students enrolled, subtract the equations that
represent the total number of students and the
number of K-5 students.
b. Replace t with 7 in the equation for S to determine
the number of students enrolled in 6-12 in 2007.
In 2007 there were 30.3 hundreds or 3030 students
enrolled in 6-12.
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52. CCSS REASONING The perimeter of the figure
2
shown is represented by the expression 3x − 7x + 2.
b. How many 6-12 students were enrolled in the
school system in 2007? SOLUTION: a. To write an equation that represents the number of
6-12 students enrolled, subtract the equations that
represent the total number of students and the
number of K-5 students.
b. Replace t with 7 in the equation for S to determine
the number of students enrolled in 6-12 in 2007.
In 2007 there were 30.3 hundreds or 3030 students
enrolled in 6-12.
52. CCSS REASONING The perimeter of the figure
2
shown is represented by the expression 3x − 7x + 2.
Write a polynomial that represents the measure of
the third side.
SOLUTION: Page 9
8-1 Adding
Polynomials
In 2007 and
thereSubtracting
were 30.3 hundreds
or 3030 students
enrolled in 6-12.
52. CCSS REASONING The perimeter of the figure
2
shown is represented by the expression 3x − 7x + 2.
Write a polynomial that represents the measure of
the third side.
2
2
b. 2(4x + 2x – 1) + 2(2x – x + 3) is the sum of
twice the length and twice the width of a rectangle,
which is the formula for the perimeter of the
rectangle
Find each sum or difference.
54. (4x + 2y − 6z) + (5y − 2z + 7x) + (−9z − 2x − 3y)
SOLUTION: 2
2
2
55. (5a − 4) + (a − 2a + 12) + (4a − 6a + 8)
SOLUTION: SOLUTION: 2
2
56. (3c − 7) + (4c + 7) − (c + 5c − 8)
53. GEOMETRY Consider the rectangle.
2
SOLUTION: 2
a. What does (4x + 2x – 1)(2x – x + 3) represent?
b. What does 2(4x2 + 2x – 1) + 2(2x2 – x + 3)
represent?
3
2
3
2
57. (3n + 3n − 10) − (4n − 5n) + (4n − 3n − 9n + 4)
SOLUTION: SOLUTION: 2
2
a. (4x + 2x – 1)(2x – x + 3) is a multiplication of
the length and the width of the rectangle, which is the
formula for the area of a rectangle.
2
2
b. 2(4x + 2x – 1) + 2(2x – x + 3) is the sum of
twice the length and twice the width of a rectangle,
which is the formula for the perimeter of the
rectangle
Find each sum or difference.
54. (4x + 2y − 6z) + (5y − 2z + 7x) + (−9z − 2x − 3y)
SOLUTION: eSolutions Manual - Powered by Cognero
58. FOOTBALL The National Football League is
divided into two conferences, the American A and
the National N. From 2002 through 2009, the total
attendance T (in thousands) for both conferences and
for the American Conference games are modeled by
the following equations, where x is the number of
years since 2002.
3
2
3
2
T = –0.69x + 55.83x + 643.31x + 10,538
A = –3.78x + 58.96x + 265.96x + 5257
Determine how many people attended National
Conference football games in 2009.
Page 10
SOLUTION: 8-1 Adding and Subtracting Polynomials
58. FOOTBALL The National Football League is
divided into two conferences, the American A and
the National N. From 2002 through 2009, the total
attendance T (in thousands) for both conferences and
for the American Conference games are modeled by
the following equations, where x is the number of
years since 2002.
3
2
3
2
T = –0.69x + 55.83x + 643.31x + 10,538
A = –3.78x + 58.96x + 265.96x + 5257
Determine how many people attended National
Conference football games in 2009.
SOLUTION: attendance in standard form. 8829 ×1000 = 8,829,000.
So, about 8,829,000 people attended National
Conference football games in 2009.
59. CAR RENTAL The cost to rent a car for a day is
$15 plus $0.15 for each mile driven.
a. Write a polynomial that represents the cost of
renting a car for m miles.
b. If a car is driven 145 miles, how much would it
cost to rent?
c. If a car is driven 105 miles each day for four
days, how much would it cost to rent a car?
d. If a car is driven 220 miles each day for seven
days, how much would it cost to rent a car?
SOLUTION: a. The cost to rent a car is the daily rate and the
mileage cost or 15 + 0.15m.
b. Substitute 145 for m to find the daily cost to drive
145 miles.
Let x = 7 represent 2009, then find how many people attended a National Conference football game
in 2009.
The cost to rent the car would be $36.75.
c. The expression represents the cost per day. For 4
days, multiply the entire expression by 4 to find the
cost for a 4 day trip. Substitute 105 for m.
In 2009 the number of people who attended
National Conference football games was about
8829 thousand. Multiply by 1000 to find the
attendance in standard form. 8829 ×1000 = 8,829,000.
So, about 8,829,000 people attended National
Conference football games in 2009.
59. CAR RENTAL The cost to rent a car for a day is
$15 plus $0.15 for each mile driven.
The cost to rent the car would be $123.
d. The expression represents the cost per day. For 7
days, multiply the entire expression by 7 to find the
cost for a 7 day trip. Substitute 220 for m.
a. Write a polynomial that represents the cost of
renting a car for m miles.
The cost to rent the car would be $336.
b. If a car is driven 145 miles, how much would it
cost to rent?
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c. If a car is driven 105 miles each day for four
days, how much would it cost to rent a car?
60. MULTIPLE REPRESENTATIONS In this
problem, you will explore perimeter and area.
Page 11
a. Geometric Draw three rectangles that each have
a perimeter of 400 feet.
The cost to rent the car would be $336.
8-1 Adding and Subtracting Polynomials
60. MULTIPLE REPRESENTATIONS In this
problem, you will explore perimeter and area.
a. Geometric Draw three rectangles that each have
a perimeter of 400 feet.
and the area will be the y-values on the vertical. The
length cannot pass 200 since the sum of the length
and width is 200. Set the intervals for the x-axis to 25
feet. Extend the table of values to find more points to
plot on the graph. It appears that 10,000 is the
greatest area, so set the intervals for the y-axis to
2
1000 ft . After graphing, it appears that the highest
2
point on the graph is at an area of 10,000 ft . b. Tabular Record the width and length of each
rectangle in a table like the one shown below. Find
the area of each rectangle.
c. Graphical On a coordinate system, graph the area
of rectangle 4 in terms of the length, x. Use the
graph to determine the largest area possible.
d. Analytical Determine the length and width that
produce the largest area.
SOLUTION: a. For the perimeters to be 400, create lengths and
widths that sum to 200. Note that some lengths are
already provided in the table in part b.
d. The associated x-value with the maximum area is
x = 100, so the length must be 100 and the width
must be 200 – 100, or 100.
The length and width of the rectangle must be 100
feet each to have the largest area.
61. CCSS CRITIQUE Cheyenne and Sebastian are
2
2
finding (2x − x) − (3x + 3x − 2). Is either of them
correct? Explain your reasoning.
b. The area is length multiplied by width. Make sure
the units are squared for area.
The sum of the length and width must be 200, so if
the length is x, the width must be 200 – x.
SOLUTION: c. The length will be the x-values on the horizontal
and the area will be the y-values on the vertical. The
length cannot pass 200 since the sum of the length
and width is 200. Set the intervals for the x-axis to 25
feet. Extend the table of values to find more points to
plot on the graph. It appears that 10,000 is the
greatest area, so set the intervals for the y-axis to
2
1000Manual
ft . After
graphing,
it appears that the highest
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2
point on the graph is at an area of 10,000 ft . Neither is correct. Cheyenne, did not distribute the
negative to the 2nd and 3rd terms when she found
the additive inverse. Sebastian did not distribute the
negate to the 3rd terms when he found the additive
Page 12
inverse. To find the additive inverse, all terms should
be multiplied by −1.
x = 100, so the length must be 100 and the width
must be 200 – 100, or 100.
The length
width of the
rectangle must be 100
8-1 Adding
andand
Subtracting
Polynomials
feet each to have the largest area.
61. CCSS CRITIQUE Cheyenne and Sebastian are
2
2
finding (2x − x) − (3x + 3x − 2). Is either of them
correct? Explain your reasoning.
SOLUTION: Neither is correct. Cheyenne, did not distribute the
negative to the 2nd and 3rd terms when she found
the additive inverse. Sebastian did not distribute the
negate to the 3rd terms when he found the additive
inverse. To find the additive inverse, all terms should
be multiplied by −1.
62. REASONING Determine whether each of the
following statements is true or false . Explain your
reasoning.
a. A binomial can have a degree of zero.
b. The order in which polynomials are subtracted
does not matter.
SOLUTION: a. If a binomial has two terms that are each a degree
of 0, then those terms can be combined and the
binomial becomes a monomial.
For example, 18 + 7 = 25. If one of the terms of the binomial does not have a
degree of 0, then the binomial cannot have a degree
of 0, since the degree of a polynomial is the greatest
degree of any term in the polynomial.
b. Subtraction is not commutative. While 2 + 5 = 5 +
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2, 2 – 5 ≠ 5 – 2. This is also true for polynomials.
Sample answer: (2x – 3) – (4x – 3) = –2x, but (4x –
negative to the 2nd and 3rd terms when she found
the additive inverse. Sebastian did not distribute the
negate to the 3rd terms when he found the additive
inverse. To find the additive inverse, all terms should
be multiplied by −1.
62. REASONING Determine whether each of the
following statements is true or false . Explain your
reasoning.
a. A binomial can have a degree of zero.
b. The order in which polynomials are subtracted
does not matter.
SOLUTION: a. If a binomial has two terms that are each a degree
of 0, then those terms can be combined and the
binomial becomes a monomial.
For example, 18 + 7 = 25. If one of the terms of the binomial does not have a
degree of 0, then the binomial cannot have a degree
of 0, since the degree of a polynomial is the greatest
degree of any term in the polynomial.
b. Subtraction is not commutative. While 2 + 5 = 5 +
2, 2 – 5 ≠ 5 – 2. This is also true for polynomials.
Sample answer: (2x – 3) – (4x – 3) = –2x, but (4x –
3) – (2x – 3) = 2x
63. CHALLENGE Write a polynomial that represents
the sum of an odd integer 2n + 1 and the next two
consecutive odd integers.
SOLUTION: 64. WRITING IN MATH Why would you add or
subtract equations that represent real-world
situations? Explain.
SOLUTION: 65. WRITING IN MATH Describe how to add and
subtract polynomials using both the vertical and
horizontal formats. SOLUTION: To add polynomials in a horizontal format, you
combine like terms. For the vertical format, you write
the polynomials in standard form, align like terms in
columns, and combine like terms. Page 13
subtract equations that represent real-world
situations? Explain.
SOLUTION: 8-1 Adding
and Subtracting Polynomials
65. WRITING IN MATH Describe how to add and
subtract polynomials using both the vertical and
horizontal formats. SOLUTION: To add polynomials in a horizontal format, you
combine like terms. For the vertical format, you write
the polynomials in standard form, align like terms in
columns, and combine like terms. 66. Three consecutive integers can be represented by x,
x + 1, and x + 2. What is the sum of these three
integers?
A x(x + 1)(x + 2)
3
B x + 3
C 3x + 3
D x + 3
SOLUTION: The correct choice is C.
67. SHORT RESPONSE What is the perimeter of a
square with sides that measure 2x + 3 units?
SOLUTION: To subtract polynomials in a horizontal format you
find the additive inverse of the polynomial you are
subtracting, and then combine like terms. For the
vertical format, you write the polynomials in standard
form, align like terms in columns, and subtract by
adding the additive inverse.
The perimeter of the square is 8x + 12 units.
68. Jim cuts a board in the shape of a regular hexagon
and pounds in a nail at each vertex, as shown. How
many rubber bands will he need to stretch a rubber
band across every possible pair of nails?
F 15
66. Three consecutive integers can be represented by x,
x + 1, and x + 2. What is the sum of these three
integers?
G 14
H 12
A x(x + 1)(x + 2)
J 9
3
B x + 3
C 3x + 3
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D x + 3
SOLUTION: SOLUTION: The first nail would connect to 5 others, the second
to 4 others, the third to 3 others, etc.
5 + 4 + 3 + 2 + 1 = 15
The correct choice is F.
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69. Which ordered pair is in the solution set of the
system of inequalities shown in the graph?
8-1 Adding
and Subtracting Polynomials
The perimeter of the square is 8x + 12 units.
68. Jim cuts a board in the shape of a regular hexagon
and pounds in a nail at each vertex, as shown. How
many rubber bands will he need to stretch a rubber
band across every possible pair of nails?
SOLUTION: The first nail would connect to 5 others, the second
to 4 others, the third to 3 others, etc.
5 + 4 + 3 + 2 + 1 = 15
The correct choice is F.
69. Which ordered pair is in the solution set of the
system of inequalities shown in the graph?
F 15
A (−3, 0)
G 14
B (0, −3)
H 12
C (5, 0)
J 9
D (0, 5)
SOLUTION: The first nail would connect to 5 others, the second
to 4 others, the third to 3 others, etc.
5 + 4 + 3 + 2 + 1 = 15
The correct choice is F.
SOLUTION: Choice A is outside the shaded area for both
inequalities. Choices B and D are inside the shaded
area for only one inequality. Choice C is the only
point in the solution for both inequalities. So, the correct choice is C.
69. Which ordered pair is in the solution set of the
system of inequalities shown in the graph?
A (−3, 0)
B (0, −3)
C (5, 0)
D (0, 5)
SOLUTION: Choice A is outside the shaded area for both
inequalities. Choices B and D are inside the shaded
area for only one inequality. Choice C is the only
point in the solution for both inequalities. eSolutions Manual - Powered by Cognero
So, the correct choice is C.
70. COMPUTERS A computer technician charges by
70. COMPUTERS A computer technician charges by
the hour to fix and repair computer equipment. The
total cost of the technician for one hour is $75, for
two hours is $125, for three hours is $175, for four
hours is $225, and so on. Write a recursive formula
for the sequence.
SOLUTION: Write out the terms.
$75, $125, $175, $225, ...
The first term is 75, and 50 is added to form each
following term.
Therefore, we have a 1 = 75, a n = a n – 1 + 50, n ≥ 2.
Determine whether each sequence is
arithmetic, geometric, or neither. Explain.
71. 8, –32, 128, –512, ...
SOLUTION: Check for a common difference.
–32 – 8 = –40
128 – (–32) = 160
There is no common difference. Check for a
Page 15
The first term is 75, and 50 is added to form each
following term.
8-1 Adding
and Subtracting Polynomials
Therefore, we have a 1 = 75, a n = a n – 1 + 50, n ≥ 2.
Determine whether each sequence is
arithmetic, geometric, or neither. Explain.
71. 8, –32, 128, –512, ...
SOLUTION: Check for a common difference.
–32 – 8 = –40
128 – (–32) = 160
There is no common difference. Check for a
common ratio.
–32 ÷ 8 = –4
128 ÷ (–32) = –4
Geometric; the common ratio is –4.
There is no common ratio, so the sequence is not
arithmetic or geometric.
74. 43, 52, 61, 70, ...
SOLUTION: Check for a common difference.
52 – 43 = 9
61 – 52 = 9
Arithmetic; the common difference is 9.
75. –27, –16, –5, 6, ...
SOLUTION: Check for a common difference.
–16 – (–27) = 11
–5 – (–16) = 11
72. 25, 8, –9, –26, ...
SOLUTION: Check for a common difference.
8 – 25 = –17
–9 – 8 = –17
Arithmetic; the common difference is –17.
Arithmetic; the common difference is 11.
76. 200, 100, 50, 25, …
SOLUTION: Check for a common difference.
100 – 200 = –100
50 – 100 = –50
There is no common difference. Check for a
common ratio.
73. SOLUTION: Check for a common difference.
There is no common difference. Check for a
common ratio.
100 ÷ 200 = 0.5
50 ÷ 100 = 0.5
Geometric; the common ratio is 0.5 or
.
77. JOBS Kimi received an offer for a new job. She
wants to compare the offer with her current job.
What is total amount of sales that Kimi must get
each month to make the same income at either job?
There is no common ratio, so the sequence is not
arithmetic or geometric.
74. 43, 52, 61, 70, ...
SOLUTION: Check for a common difference.
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eSolutions
52 – 43 = 9
61 – 52 = 9
SOLUTION: Let s be Kimi’s monthly sales.
Page 16
100 ÷ 200 = 0.5
50 ÷ 100 = 0.5
13 –
26 – 13 = 13
There is not a common different. The sequence is
not an arithmetic sequence 8-1 Adding and Subtracting Polynomials
Geometric; the common ratio is 0.5 or .
77. JOBS Kimi received an offer for a new job. She
wants to compare the offer with her current job.
What is total amount of sales that Kimi must get
each month to make the same income at either job?
=
80. 7, 6, 5, 4, …
SOLUTION: The sequence is an arithmetic sequence because the
items differ by a constant. The common difference is
–1, because 6 – 7 = –1; 5 – 6 = –1; 4 – 5 = –1;etc.
81. 10, 12, 15, 18, …
SOLUTION: Let s be Kimi’s monthly sales.
SOLUTION: Find the difference between the terms. 12 – 10 = 2
15 – 12 = 3
18 – 15 = 3
There is no common difference. The sequence is not
an arithmetic sequence.
82. −15, −11, −7, −3, …
SOLUTION: The sequence is an arithmetic sequence because the
items differ by a constant. The common difference is
4, because –11 – (–15) = 4; –7 – (–11) = 4; –3 – (–
7) = 4;etc.
Kimi must sell $80,000 each month to make the same
income at either job.
Determine whether each sequence is an
arithmetic sequence. If it is, state the common
difference.
78. 24, 16, 8, 0, …
SOLUTION: The sequence is an arithmetic sequence because the
items differ by a constant. The common difference is
–8, because 16 – 24 = –8; 8 –16 = –8; etc.
83. −0.3, 0.2, 0.7, 1.2, …
SOLUTION: The sequence is an arithmetic sequence because the
items differ by a constant. The common difference is
0.5, because 0.2 – (–0.3) = 0.5; 0.7 – 0.2 = 0.5; 1.2 –
0.7 = 0.5; etc.
Simplify.
5
7
84. t(t )(t )
SOLUTION: 79. , 13, 26, …
SOLUTION: Find the difference between the terms. =
–
3
2
3
85. n (n )(−2n )
SOLUTION: 13 –
=
26 – 13 = 13
There is not a common different. The sequence is
not an arithmetic sequence 80. 7, 6, 5, 4, …
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SOLUTION: The sequence is an arithmetic sequence because the
items differ by a constant. The common difference is
5 2
3 4
86. (5t v )(10t v )
SOLUTION: Page 17
85. n (n )(−2n )
SOLUTION: 8-1 Adding and Subtracting Polynomials
5 2
3 4
86. (5t v )(10t v )
SOLUTION: 4 5
4
87. (−8u z )(5uz )
SOLUTION: 2 3
88. [(3) ]
SOLUTION: 3 2
89. [(2) ]
SOLUTION: 4 3 2
2 3
90. (2m k ) (−3mk )
SOLUTION: 2 2
2 2 2 3
91. (6xy ) (2x y z )
SOLUTION: eSolutions Manual - Powered by Cognero
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