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

Transcription

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
3
SOLUTION: 2
3
A polynomial is a monomial or the sum of 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 is 2, 2, and
2
3
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
SOLUTION: 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 is 1, 0, and
2
2, so the degree of 2y – 5 + 3y is 2. The polynomial has three terms, so it is a trinomial.
3. 3x
2
SOLUTION: 2
A polynomial is a monomial or the sum of monomials, so 3x is a polynomial. 2
The degree of a polynomial is the greatest degree of any term in the polynomial. The degree of 3x is 2. The
polynomial has one term, so it is a monomial.
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
SOLUTION: 2 3
A polynomial is a monomial or the sum of 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 is 5 and 0,
2 3
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.
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only one term. 5q is equal to , which is a division of two monomials, so it is not a monomial.
2 3
A polynomial is a monomial or the sum of monomials. 5m p + 6 is the sum of 2 monomials, so it is a polynomial. The degree
a polynomial
is the greatest degree of any term in the polynomial. The degree of each term is 5 and 0,
8-1 Adding
andofSubtracting
Polynomials
2 3
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
-4
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
–4d → 4
1 → 0
2
–d → 2
4
4
2
The greatest degree is 4, from the term –4d , so the 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
5
8. 2x – 12 + 3x
5
2x → 5
–12 → 0
3x → 1
5
5
The greatest degree is 5, from the term 2x , so the leading coefficient of 2x – 12 + 3x is 2.
5
2x + 3x – 12
2
9. 4z – 2z – 5z
4
4z → 1
2
– 2z → 2
4
– 5z → 4
4
2
4
The greatest degree is 4, from the term – 5z , so the leading coefficient of 4z – 2z – 5z is –5.
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8-1 Adding
and Subtracting Polynomials
5
2x + 3x – 12
2
9. 4z – 2z – 5z
4
4z → 1
2
– 2z → 2
4
– 5z → 4
4
2
4
The greatest degree is 4, from the term – 5z , so the leading coefficient of 4z – 2z – 5z is –5.
4
2
–5z – 2z + 4z
3
2
10. 2a + 4a – 5a – 1
2a → 1
3
4a → 3
2
– 5a → 2
– 1 → 0
3
3
2
The greatest degree is 3, from the term 4a , so the leading coefficient of 2a + 4a – 5a – 1 is 4.
3
2
4a – 5a + 2a – 1
Find each sum or difference.
3
3
11. (6x − 4) + (−2x + 9)
SOLUTION: 3
2
2
12. (g − 2g + 5g + 6) − (g + 2g)
SOLUTION: eSolutions Manual - Powered by Cognero
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3
2
2
12. (g − 2g + 5g + 6) − (g + 2g)
SOLUTION: 2
2
13. (4 + 2a − 2a) − (3a − 8a + 7)
SOLUTION: 2
2
14. (8y − 4y ) + (3y − 9y )
SOLUTION: 3
3
2
15. (−4z − 2z + 8) − (4z + 3z − 5)
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3
3
2
15. (−4z − 2z + 8) − (4z + 3z − 5)
SOLUTION: 2
2
16. (−3d − 8 + 2d) + (4d − 12 + d )
SOLUTION: 2
17. (y + 5) + (2y + 4y – 2)
SOLUTION: 3
2
2
3
18. (3n − 5n + n ) − (−8n + 3n )
SOLUTION: SENSE-MAKING 19. CCSS
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total number of students T who traveled for spring break consists of two groups:Page 5
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.
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.
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.
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.
D = 6n + 14
b. n = 2012 – 1995 = 17
The number of students who will drive to their destination in 2012 is 116,000 students.
c. n = 2015 – 1995 = 20
The number of students who will drive or fly to their destination in 2015 is 301,000 students.
exponents. It has
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only one term. 21. is a division of two monomials, so it is not a monomial.
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The number of students who will drive or fly to their destination in 2015 is 301,000 students.
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 21 is 0. The
polynomial has only one term, so it is a monomial.
4
2
22. c – 2c + 1
SOLUTION: 4
2
A polynomial is a monomial or the sum of 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 is 4, 2, and
4
2
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 nonnegative integer
c
exponents. 3d has a variable in the exponent, so it is not a monomial.
24. a – a
2
SOLUTION: 2
A polynomial is a monomial or the sum of monomials. a – a 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 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 is 3 and 4,
3
3
so the degree of 5n + nq is 4. The polynomial has two terms, so it is a binomial.
2
26. 5x – 2 + 3x
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2
A polynomial is a monomial or the sum of monomials. a – a is the sum of 2 monomials, so it is a polynomial. The degree
a polynomial
is the greatest degree of any term in the polynomial. The degree of each term is 1 and 2,
8-1 Adding
andofSubtracting
Polynomials
2
so the degree of a – a 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 is 3 and 4,
3
3
so the degree of 5n + nq is 4. The polynomial has two terms, so it is a binomial.
2
26. 5x – 2 + 3x
2
5x → 2
– 2 → 0
3x → 1
2
2
2
5x + 3x – 2
27. 8y + 7y
3
3
7y → 3
8y → 1
3
3
The greatest degree is 3, from the term 7y , so the leading coefficient of 8y + 7y is 7.
3
7y + 8y
28. 4 – 3c – 5c
2
4 → 0
3c → 1
2
– 5c → 2
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2
2
The greatest degree is 3, from the term 7y , so the leading coefficient of 8y + 7y is 7.
8-1 Adding
3
7y + 8y
28. 4 – 3c – 5c
2
4 → 0
3c → 1
2
– 5c → 2
2
2
The greatest degree is 2, from the term – 5c , so the leading coefficient of 4 – 3c – 5c is –5.
2
–5c – 3c + 4
3
2
29. –y + 3y – 3y + 2
3
–y → 3
3y → 1
2
3y → 2
2 → 0
3
3
2
The greatest degree is 3, from the term –y , so the leading coefficient of –y + 3y – 3y + 2 is –1.
–y 3 – 3y 2 + 3y + 2
2
5
30. 11t + 2t – 3 + t
11t → 1
2
2t → 2
–3 → 0
t 5 → 5
5
2
5
The greatest degree is 5, from the term t , so the leading coefficient of 11t + 2t – 3 + t is 1.
5
2
t + Manual
2t + 11t
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31. 2 + r – r
3
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8-1 Adding
–y 3 – 3y 2 + 3y + 2
2
5
30. 11t + 2t – 3 + t
11t → 1
2
2t → 2
–3 → 0
t 5 → 5
5
2
5
The greatest degree is 5, from the term t , so the leading coefficient of 11t + 2t – 3 + t is 1.
5
2
t + 2t + 11t – 3
31. 2 + r – r
3
2 → 0
r → 1
– r3→ 3
3
3
The greatest degree is 3, from the term – r , so the leading coefficient of 2 + r – r is –1.
3
–r + r + 2
32. SOLUTION: Find the degree of each term.
→ 0
4
–3x → 4
7 → 0
The greatest degree is 4, from the term –3x 4, so the leading coefficient of
is –3.
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Rewrite
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with each
monomial in descending order according to degree.
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The greatest degree is 3, from the term – r , so the leading coefficient of 2 + r – r is –1.
8-1 Adding
3
–r + r + 2
32. SOLUTION: Find the degree of each term.
→ 0
4
–3x → 4
7 → 0
The greatest degree is 4, from the term –3x 4, so the leading coefficient of
is –3.
2
33. –9b + 10b – b
6
–9b2 → 2
10b → 1
–b6→ 6
The greatest degree is 6, from the term – b6, so the leading coefficient of –9b2 + 10b – b6 is –1.
6
2
–b – 9b + 10b
2
34. (2c + 6c + 4) + (5c – 7)
SOLUTION: 2
2
35. (2x + 3x ) − (7 − 8x )
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2
2
35. (2x + 3x ) − (7 − 8x )
SOLUTION: 3
2
36. (3c − c + 11) − (c + 2c + 8)
SOLUTION: 2
2
37. (z + z) + (z − 11)
SOLUTION: 38. (2x − 2y + 1) − (3y + 4x)
SOLUTION: 2
2
39. (4a − 5b + 3) + (6 − 2a + 3b )
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2
2
39. (4a − 5b + 3) + (6 − 2a + 3b )
SOLUTION: 2
2
2
40. (x y − 3x + y) + (3y − 2x y)
SOLUTION: 2
2
41. (−8xy + 3x − 5y) + (4x − 2y + 6xy)
SOLUTION: 2
2
42. (5n − 2p + 2np) − (4p + 4n)
SOLUTION: 2
2
2
2
43. (4rxt − 8r x + x ) − (6rx + 5rxt − 2x )
SOLUTION: eSolutions
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2
2
2
2
43. (4rxt − 8r x + x ) − (6rx + 5rxt − 2x )
SOLUTION: 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?
SOLUTION: a.
So, an equation that models the total number of dogs and cats adopted is T = 3n + 7.
b. Evaluate the equation for the total number of dogs and cats for n = 2013 – 1999 = 14.
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
2
Find the degree of each term of 4x – 3x + 5.
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8-1 Adding
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: 2
Find the degree of each term of 4x – 3x + 5.
4x → 1
2
– 3x → 2
5 → 0
2
The greatest degree is 2 and there are 3 terms, so 4x – 3x + 5 is a quadratic trinomial.
3
46. 11z
11z 3 → 3
The greatest degree is 3 and there is one term, so 11z 3 is a cubic monomial.
47. 9 + y
4
SOLUTION: Find the degree of each term of 9 + y 4.
9 → 1
y4→ 4
4
The greatest degree is 4 and there are 2 terms, so 9 + y is a quartic binomial.
3
48. 3x – 7
SOLUTION: 3
Find the degree of each term of 3x – 7.
3
3x → 3
–7 → 0
3
The greatest degree is 3 and there are 2 terms, so 3x – 7 is a cubic binomial.
5
2
49. –2x – x + 5x – 8 SOLUTION: 2
5
Find the degree of each term of –2x – x + 5x – 8.
–2x 5 → 5
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–x → 2
5x → 1
– 8 → 0
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3
3x → 3
–7 → 0
8-1 Adding
3
The greatest degree is 3 and there are 2 terms, so 3x – 7 is a cubic binomial.
5
2
49. –2x – x + 5x – 8 SOLUTION: 2
5
Find the degree of each term of –2x – x + 5x – 8.
–2x 5 → 5
–x2 → 2
5x → 1
– 8 → 0
2
The greatest degree is 5 and there are 4 terms, so –2x 5 – x + 5x – 8 is a quintic polynomial.
2
3
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
2
3
The greatest degree is 3 and there are 3 terms, so 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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3
6t → 3
8-1 Adding
2
3
The greatest degree is 3 and there are 3 terms, so 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.
2
52. CCSS REASONING The perimeter of the figure shown is represented by the expression 3x − 7x + 2. Write a
polynomial that represents the measure of the third side.
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8-1 Adding
2
52. CCSS REASONING The perimeter of the figure shown is represented by the expression 3x − 7x + 2. Write a
polynomial that represents the measure of the third side.
SOLUTION: 53. GEOMETRY Consider the rectangle.
a. What does (4x2 + 2x – 1)(2x2 – x + 3) represent?
2
2
b. What does 2(4x + 2x – 1) + 2(2x – x + 3) represent?
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
54. (4x + 2y − 6z) + (5y − 2z + 7x) + (−9z − 2x − 3y)
SOLUTION: Page 18
the area of a rectangle.
2
2
8-1 Adding
b. 2(4x and
+ 2xSubtracting
– 1) + 2(2x Polynomials
– 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
54. (4x + 2y − 6z) + (5y − 2z + 7x) + (−9z − 2x − 3y)
SOLUTION: 2
2
2
55. (5a − 4) + (a − 2a + 12) + (4a − 6a + 8)
SOLUTION: 2
2
56. (3c − 7) + (4c + 7) − (c + 5c − 8)
SOLUTION: 3
2
3
2
57. (3n + 3n − 10) − (4n − 5n) + (4n − 3n − 9n + 4)
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3
2
3
2
57. (3n + 3n − 10) − (4n − 5n) + (4n − 3n − 9n + 4)
SOLUTION: 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
T = –0.69x + 55.83x + 643.31x + 10,538
3
2
A = –3.78x + 58.96x + 265.96x + 5257
Determine how many people attended National Conference football games in 2009.
SOLUTION: Let x = 7 represent 2009, then find how many people attended a National Conference football game in 2009.
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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
T = –0.69x + 55.83x + 643.31x + 10,538
3
2
A = –3.78x + 58.96x + 265.96x + 5257
Determine how many people attended National Conference football games in 2009.
SOLUTION: Let x = 7 represent 2009, then find how many people attended a National Conference football game in 2009.
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.
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?
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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?
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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.
8-1 Adding
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.
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.
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.
The cost to rent the car would be $336.
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.
b. Tabular Record the width and length of each rectangle in a table like the one shown below. Find the area of each
rectangle.
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a. Geometric Draw three rectangles that each have a perimeter of 400 feet.
b. Tabular Record the width and length of each rectangle in a table like the one shown below. Find the area of each
rectangle.
8-1 Adding
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.
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.
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
2
2
for the y-axis to 1000 ft . After graphing, it appears that the highest point on the graph is at an area of 10,000 ft . eSolutions Manual - Powered by Cognero
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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
8-1 Adding and Subtracting
Polynomials
2
2
for the y-axis to 1000 ft . After graphing, it appears that the highest point on the graph is at an area of 10,000 ft . 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.
2
2
61. CCSS CRITIQUE Cheyenne and Sebastian are 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 by
terms
should be multiplied by −1.
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62. REASONING Determine whether each of the following statements is true or false . Explain your reasoning.
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.
8-1 Adding
The length and width of the rectangle must be 100 feet each to have the largest area.
2
2
61. CCSS CRITIQUE Cheyenne and Sebastian are 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 + 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.
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b. Subtraction is not commutative. While 2 + 5 = 5 + 2, 2 – 5 ≠ 5 – 2. This is also true for polynomials.
8-1 Adding
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. 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.
66. Three consecutive integers can be represented by x, x + 1, and x + 2. What is the sum of these three integers?
Page 26
64. WRITING IN MATH Why would you add or subtract equations that represent real-world situations? Explain.
SOLUTION: 8-1 Adding
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. 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.
A x(x + 1)(x + 2)
3
B x + 3
C 3x + 3
D x + 3
SOLUTION: Page 27
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: 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
G 14
H 12
J 9
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?
Page 28
SOLUTION: The first nail would connect to 5 others, the second to 4 others, the third to 3 others, etc.
5 + 4 + 3and
+ 2Subtracting
+ 1 = 15
8-1 Adding
Polynomials
The correct choice is F.
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. So, the correct choice is C.
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 common ratio.
–32 ÷ 8 = –4
128 ÷ (–32) = –4
Geometric; the common ratio is –4.
72. 25, 8, –9, –26, ...
Page 29
$75, $125, $175, $225, ...
The first term is 75, and 50 is added to form each following term.
8-1 Adding
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, ...
–32 – 8 = –40
128 – (–32) = 160
–32 ÷ 8 = –4
128 ÷ (–32) = –4
Geometric; the common ratio is –4.
72. 25, 8, –9, –26, ...
8 – 25 = –17
–9 – 8 = –17
Arithmetic; the common difference is –17.
73. SOLUTION: Check for a common difference.
There is no common ratio, so the sequence is not arithmetic or geometric.
74. 43, 52, 61, 70, ...
52 – 43 = 9
61 – 52 = 9
eSolutions
Manual - Powered
by Cognero
Arithmetic;
the common
difference is 9.
75. –27, –16, –5, 6, ...
Page 30
8-1 Adding
There is no common ratio, so the sequence is not arithmetic or geometric.
74. 43, 52, 61, 70, ...
52 – 43 = 9
61 – 52 = 9
Arithmetic; the common difference is 9.
75. –27, –16, –5, 6, ...
–16 – (–27) = 11
–5 – (–16) = 11
Arithmetic; the common difference is 11.
76. 200, 100, 50, 25, …
100 – 200 = –100
50 – 100 = –50
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?
SOLUTION: Let s be Kimi’s monthly sales.
Page 31
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?
SOLUTION: Let s be Kimi’s monthly sales.
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.
79. , 13, 26, …
SOLUTION: Find the difference between the terms. =
–
13 –
=
26 – 13 = 13
There is not a common different. The sequence is not an arithmetic sequence 80. 7, 6, 5, 4, …
because 6 – 7 = –1; 5 – 6 = –1; 4 – 5 = –1;etc.
81. 10, 12, 15, 18, …
SOLUTION: Find the difference between the terms. Page 32
becauseand
6 – Subtracting
7 = –1; 5 – 6 Polynomials
= –1; 4 – 5 = –1;etc.
8-1 Adding
81. 10, 12, 15, 18, …
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.
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: 3
2
3
85. n (n )(−2n )
SOLUTION: 5 2
3 4
86. (5t v )(10t v )
SOLUTION: 4 5
4
87. (−8u z )(5uz )
SOLUTION: 3
eSolutions2Manual
- Powered by Cognero
88. [(3) ]
SOLUTION: Page 33
87. (−8u z )(5uz )
SOLUTION: 8-1 Adding and Subtracting Polynomials
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 )
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Determine whether each expression is a polynomial. If it is a

Transcription

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