Chapter 7 Maintaining Mathematical Proficiency (p. 335) 1. 3x 7 x = 3x x 7 = (3 )x 7 = 5x 7. 4r 6 9r 1 = 4r 9r 6 1 = (4 9)r 6 1 = 5r 5 3. 5t 3 t 4 8t = 5t t 8t 3 4 = ( 5 1 8)t 3 4 = ()t ( 1) = t 1 4. 3(s 1) 5 = 3(s) 3(1) 5 = 3s 3 5 = 3s 5. m 7(3 m) = m 7(3) 7( m) = m 1 7m = 9m 1 6. 4(h 6) (h ) = 4(h) 4(6) h ( ) = 4h 4 h = 4h h 4 = (4 1)h 4 = 3h 6 7. 0 = 5 36 = 3 3 The GCF of 0 and 36 is = 4. 8. 4 = 3 7 63 = 3 3 7 The GCF of 4 and 63 is 3 7 = 1. 9. 54 = 3 3 3 81 = 3 3 3 3 The GCF of 54 and 81 is 3 3 3 = 7. 10. 7 = 3 3 84 = 3 7 The GCF of 7 and 84 is 3 = 1. 11. 8 = 7 64 = The GCF of 8 and 64 is = 4 1. 30 = 3 5 77 = 7 11 The numbers have no common prime factors. So, the GCF of 30 and 77 is 1. 13. It is not possible for two integers to have no common factors because 1 is a factor of every integer. Chapter 7 Mathematical Practices (p. 356) 1. The algebraic expression is 3x x 1.. The algebraic expression is x x. 3. The algebraic expression is x x. 4. The algebraic expression is x x 1. 5. The algebraic expression is x. 6. The algebraic expression is x 6. 7. The algebraic expression is x x. 8. The expression is 9. 9. The algebraic expression is x. 7.1 Explorations (p. 357) 1. Step 1 (3x ) (x 5) Step 4x 5 Step 3 4x ( ) 3 Step 4 4x 3. Step 1 ( x x ) (x 1) Step ( x x ) ( x 1) Step 3 x x x 3 Step 4 x x (x x) 3 Step 5 x x 3 3. Sample answer: If a polynomial is being subtracted, find the opposite of each of its terms. Use the Commutative and Associative Properties to rearrange the terms so that like terms are grouped together. Then add or subtract like terms. 4. a. ( x x 1 ) ( x x 1 ) = x x x x 1 1 = ( x x ) (x x) ( 1 1) = 3x 0 0 = 3x b. (4x 3) (x ) = 4x x 3 = (4x x) (3 ) = 5x 1 Copyright Big Ideas Learning, LLC Algebra 1 379
c. ( x ) ( 3x x 5 ) = ( x ) ( 3x x 5 ) = x 3x x 5 = ( 1x 3x ) x ( 5) = x x 3 d. (x 3x) ( x x 4 ) = ( x) ( x x 4 ) = x x x 4 = x x x 4 = x ( x x) 4 = x x 4 7.1 Monitoring Progress (pp. 358 361) 1. In the monomial 3x 4, the exponent of x is 4. So, the degree of the monomial is 4.. In the monomial 7c 3 d, the exponent of c is 3, and the exponent of d is. So, the degree of the monomial is 3, or 5. 3. In the monomial 5 y, the exponent of y is 1. So, the degree of 3 the monomial is 1. 4. You can rewrite 0.5 as 0.5x 0. So, the degree of the monomial is 0. 5. You can write the polynomial 4 9z in standard form as 9z 4. The greatest degree is 1, so the degree of the polynomial is 1. The leading coefficient is 9. The polynomial has terms, so it is a binomial. 6. You can write the polynomial t t 3 10 t in standard form as t 3 t 10 t. The greatest degree is 3, so the degree of the polynomial is 3. The leading coefficient is 1. The polynomial has 3 terms, so it is a trinomial. 7. You can write the polynomial.8x x 3 in standard form as x 3.8x. The greatest degree is 3, so the degree of the polynomial is 3. The leading coefficient is 1. The polynomial has terms, so it is a binomial. 8. (b 10) (4b 3) = b 4b 10 3 = (b 4b) ( 10 3) = 5b ( 13) = 5b 13 The sum is 5b 13. 9. x x 7x x 8x x The sum is 8x x. 10. ( p p 3 ) ( 4p p 3 ) = p p 3 4p p 3 = p 4p p p 3 3 = ( p 4p ) ( p p) (3 3) = 5p p 0 = 5p p The difference is 5p p. 11. k 5 k 5 ( 3k 6 ) 3k 6 3k k 11 The difference is 3k k 11. 1. a. Penny: 16t 5t 00 16t 5t 00 Paintbrush: ( 16t 100 ) 16t 100 5t 100 The polynomial 5t 100 represents the distance between the objects after t seconds. b. When t = 0, the distance between the objects is 5(0) 100 = 100 feet. So, the constant term 100 indicates the distance between the penny and the paintbrush is 100 feet when they begin to fall. As the value of t increases by 1, the value of 5t 100 decreases by 5. This means that the objects become 5 feet closer to each other after each second. So, the coefficient 5 of the linear term represents how much the distance between the objects changes each second. 7.1 Exercises (pp. 36 364) Vocabulary and Core Concept Check 1. A polynomial in one variable is in standard form when the exponents of the terms decrease from left to right.. The polynomial must have three terms, and the highest degree must be 5. Sample answer: One possible polynomial is x 5 x 7. 3. To determine whether a set of numbers is closed under an operation, determine if the operation performed on any two numbers in the set results in a number that is also in the set. 4. The expression that does not belong with the other three is x 8 x. This expression has a variable in one of its exponents, so it is not a polynomial. The other three expressions are polynomials because the terms are constants or other monomials whose variables have only whole number exponents. 380 Algebra 1 Copyright Big Ideas Learning, LLC
Monitoring Progress and Modeling with Mathematics 5. In the monomial 4g, the exponent of g is 1. So, the degree of the monomial is 1. 6. In the monomial 3x 4, the exponent of x is 4. So, the degree of the monomial is 4. 7. In the monomial 1.75k, the exponent of k is. So, the degree of the monomial is. 8. You can rewrite 4 9 as 4 9 x 0. So, the degree of the monomial is 0. 9. In the monomial 7s 8 t, the exponent of s is 8, and the exponent of t is 1. So, the degree of the monomial is 8 1, or 9. 10. In the monomial 8m n 4, the exponent of m is, and the exponent of n is 4. So, the degree of the monomial is 4, or 6. 11. In the monomial 9x y 3 z 7, the exponent of x is 1, the exponent of y is 3, and the exponent of z is 7. So, the degree of the monomial is 1 3 7, or 11. 1. In the monomial 3q 4 rs 6, the exponent of q is 4, the exponent of r is 1, and the exponent of s is 6. So, the degree of the monomial is 4 1 6, or 11. 13. You can write the polynomial 6c c 4 c in standard form as c 4 6c c. The greatest degree is 4, so the degree of the polynomial is 4. The leading coefficient is. The polynomial has 3 terms, so it is a trinomial. 14. You can write the polynomial 4w 11 w 1 in standard form as w 1 4w 11. The greatest degree is 1, so the degree of the polynomial is 1. The leading coefficient is 1. The polynomial has terms, so it is a binomial. 15. You can write the polynomial 7 3p in standard form as 3p 7. The greatest degree is, so the degree of the polynomial is. The leading coefficient is 3. The polynomial has terms, so it is a binomial. 16. You can write the polynomial 8d 4d 3 in standard form as 4d 3 8d. The greatest degree is 3, so the degree of the polynomial 3. The leading coefficient is 4. The polynomial has 3 terms, so it is a trinomial. 18. You can write the polynomial 5z z 3 3z 4 in standard form as 3z 4 z 3 5z. The greatest degree is 4, so the degree of the polynomial is 4. The leading coefficient is 3. The polynomial has 3 terms, so it is a trinomial. 19. You can write the polynomial πr 5 7 r 8 r 5 in standard form as 5 7 r 8 r 5 πr. The greatest degree is 8, so the degree of the polynomial is 8. The leading coefficient is 5 7. The polynomial has 3 terms, so it is a trinomial. 0. The polynomial 7 n 4 is in standard form. The only term has a degree of 4, so the degree of the polynomial is 4. The leading coefficient is 7. The polynomial has 1 term, so it is a monomial. 1. The expression 4 3 πr 3 is a monomial because it is the product of a number, 4 π, and a variable with a whole number 3 exponent, r 3. The only variable has an exponent of 3, so the degree of the monomial is 3.. The expression 400x 8 600x 6 has two terms, so it is a binomial. The greatest degree is 8, so the degree of the binomial is 8. 3. (5y 4) ( y 6) = 5y y 4 6 = (5y y) (4 6) = 3y 10 4. ( 8x 1) (9x 4) = 8x 9x 1 4 Alternate solution: 8x 1 9x 4 x 8 = ( 8x 9x) ( 1 4) = x ( 8) = x 8 5. ( n 5n 6 ) ( n 3n 11 ) = n n 5n 3n 6 11 = ( n n ) ( 5n 3n) ( 6 11) = n 8n 5 17. The polynomial 3t 8 is in standard form. The only term has a degree of 8, so the degree of the polynomial is 8. The leading coefficient is 3. The polynomial has 1 term, so it is a monomial. Copyright Big Ideas Learning, LLC Algebra 1 381
6. ( 3p 3 5p p ) ( p 3 8p 15p ) = 3p 3 p 3 5p 8p p 15p = ( 3p 3 p 3 ) ( 5p 8p ) ( p 15p) = 4p 3 ( 3p ) ( 17p) = 4p 3 3p 17p Alternate solution: 3p 3 5p p p 3 8p 15p 4p 3 3p 17p 7. ( 3g g ) ( 3g 8g 4 ) = 3g 3g g 8g 4 = ( 3g 3g ) ( g 8g) 4 = 6g ( 9g) 4 = 6g 9g 4 8. ( 9r 4r 7 ) ( 3r 3r ) = 9r 3r 4r 3r 7 = ( 9r 3r ) (4r 3r) 7 = 1r r 7 Alternate solution: 9r 4r 7 3r 3r 1r r 7 9. ( 4a a 3 3 ) ( a 3 5a 8 ) = a 3 a 3 5a 4a 3 8 = ( a 3 a 3 ) 5a 4a ( 3 8) = a 3 5a 4a 5 30. ( s 3 s 9 ) ( s 6s 3 s ) = s 3 6s 3 s s s 9 = ( s 3 6s 3 ) s ( s s) 9 = 5s 3 s s 9 Alternate solution: s 3 s 9 6s 3 s s 5s 3 s s 9 31. (d 9) (3d 1) = d 9 3d 1 = (d 3d) ( 9 1) = d 8 3. (6x 9) (7x 1) = 6x 9 7x 1 = (6x 7x) (9 1) = x 8 Alternate solution: 6x 9 6x 9 (7x 1) 7x 1 x 8 33. ( y 4y 9 ) ( 3y 6y 9 ) = y 4y 9 3y 6y 9 = ( y 3y ) ( 4y 6y) (9 9) = y y 18 34. ( 4m m ) ( 3m 10m 4 ) = 4m m 3m 10m 4 = ( 4m 3m ) ( m 10m) ( 4) = 7m 11m Alternate solution: 4m m 4m m ( 3m 10m 4 ) 3m 10m 4 7m 11m 35. ( k 3 7k ) ( k 1 ) = k 3 7k k 1 36. ( r 10) ( 4r 3 r 7r ) = r 10 4r 3 r 7r = 4r 3 r ( r 7r) 10 = 4r 3 r 8r 10 Alternate solution: = k 3 k 7k ( 1) = k 3 k 7k 14 r 10 r 10 ( 4r 3 r 7r ) 4r 3 r 7r 4r 3 r 8r 10 37. ( t 4 t t ) ( 1 9t 7t ) = t 4 t t 1 9t 7t = t 4 ( t 9t ) (t 7t) 1 = t 4 8t 8t 1 Alternate solution: t 4 t t t 4 t t ( 9t 7t 1 ) 9t 7t 1 t 4 8t 8t 1 38. ( 4d 6d 3 3d ) (10d 3 7d ) = 4d 6d 3 3d 10d 3 7d = ( 6d 3 10d 3 ) 3d (4d 7d) = 16d 3 3d 3d 39. When writing the subtraction as addition, the last term of the polynomial was not multiplied by 1. ( x x ) ( x 3x ) = x x x 3x = ( x x ) (x 3x) = x 4x 38 Algebra 1 Copyright Big Ideas Learning, LLC
40. The terms 4x and 8x are not like terms, so they cannot be added. x 3 4x 3 3x 3 8x x 3 4x 8x 1 41. (8b 6) (4 5b) = 8b 6 4 5b = (8b 5b) (6 4) = 3b A polynomial that represents how much more it costs to make b necklaces than b bracelets is 3b. 4. (14 1m) (5 6m) = 14 5 1m 6m = (14 5) (1m 6m) = 194 18m A polynomial that represents the total number of memberships at the fitness center is 194 18m. 43. s 5st t s 5st t ( s 7st t ) s 7st t s 1st 44. ( a 3ab b ) ( 4a 5ab b ) = a 4a 3ab 5ab b b = ( a 4a ) ( 3ab 5ab) ( b b ) = 3a ab b 45. c 6d c cd d c cd 4d 46. ( x 9xy ) ( x 6xy 8y ) = x 9xy x 6xy 8y = ( x x ) (9xy 6xy) 8y = x 3xy 8y 47. The terms of a polynomial are always monomials. A polynomial is a monomial or a sum of monomials, and each monomial is a term of the polynomial. 48. The difference of two trinomials is sometimes a trinomial. If like terms have the same coefficient, they will cancel when subtracted, so the difference will have fewer than 3 terms. Or, if the terms in the trinomial are not all of the same degree, then the difference could have more than 3 terms. 49. A binomial is sometimes a polynomial of degree. The two terms in the binomial can be of any degree. 50. The sum of two polynomials is always a polynomial. Polynomials are closed under addition. 51. 16t v 0 t s 0 = 16t ( 45)t 00 = 16t 45t 00 Let t = 1. 16t 45t 00 = 16(1) 45(1) 00 = 16 45 00 = 139 A polynomial that represents the height of the water balloon is 16t 45t 00, and the water balloon is 139 feet from the ground after 1 second. 5. 16t v 0 t s 0 = 16t 16t 3 Let t = 1. 16t 16t 3 = 16(1) 16(1) 3 = 16(1) 16 3 = 16 16 3 = 3 The polynomial 16t 16t 3 represents the height of the tennis ball after t seconds, and the tennis ball is 3 feet high after 1 second. So, it is back to the initial height where you hit it with the racket. 53. a. ( 16t 98 ) ( 16t 46t 6 ) = 16t 98 16t 46t 6 = ( 16t 16t ) 46t (98 6) = 46t 9 The polynomial 46t 9 represents the distance between the objects after t seconds. b. When t = 0, the distance between the objects is 46(0) 9 = 9 feet. So, the constant term 9 indicates that the distance between the two balls is 9 feet when they are thrown. As the value of t increases by 1, the value of 46t 9 decreases by 46. This means that the two balls become 46 feet closer to each other after each second. So, the coefficient, 46, of the linear term represents how much the distance between objects changes each second. 54. a. 0.08t 3 0.06t 0.1t 17 0.38t 1.5t 4 0.08t 3 0.3t 1.6t 59 The polynomial 0.08t 3 0.3t 1.6t 59 represents the total amount spent each year on buying new and used vehicles in the 7-year period. b. Let t = 5. 0.08t 3 0.3t 1.6t 59 = 0.08(5) 3 0.3(5) 1.6(5) 59 = 0.08(15) 0.3(5) 8 59 = 3.5 8 8 59 = 55.5 The total amount spent on buying new and used vehicles in the fifth year was $55.5 million. Copyright Big Ideas Learning, LLC Algebra 1 383
55. P = x (3x ) (x 1) (5x ) = (x 3x x 5x) ( 1 ) = 1x 3 The polynomial 1x 3 represents the perimeter of the quadrilateral. 56. a. x 1 (x ) or x ( x 1) = x 1 x = x x 1 = ( x x) (1 ) = (x x) ( 1) = 3x 3 or = 3x 3 The vertical distance between points on the lines with the same x-value can be represented by the absolute value of 3x 3 or 3x 3. b. 3x 3 = 0 or 3x 3 = 0 3 3 3 3 3x = 3 3x = 3 3x 3 = 3 3x 3 3 = 3 3 x = 1 x = 1 Each expression equals 0 when x = 1, which is the x-coordinate of the point of intersection of the two lines. 57. Your friend is correct. Addition is commutative and associative, so you can add in any order. 58. Sample answer: The polynomials 1 x 1 and 1 x 1 have a sum of x and a difference of 1. 1 x 1 1 x 1 1 x 1 1 x 1 ( 1 x 1 ) = 1 x 1 x 1 59. a. The set of negative integers is not closed under multiplication because the product of two negative integers is always a positive integer. b. The set of whole is closed under addition because the sum of two whole numbers is always a whole number. 60. a. Level 1: A = w = x ( 10 (x 1) ) = x ( x (10 1) ) = x(x ) = x(x) x() = x x Level : A = w = x(x 1) = x(x) x(1) = x 1x Total area: ( x x ) ( x 1x ) = x x x 1x = ( x x ) ( x 1x) b. x 14x = (0) 14(0) = (400) 80 = 800 80 = 50 When x = 0 feet, the total area of the deck is 50 square feet. c. Divide the total area of the deck by the area that can be covered by 1 gallon of sealant to get the amount of sealant needed. 50 = 1.3 400 So, you need 1.3 gallons of sealant to cover the deck once. 61. a. Area of patio = x 14x 384 Algebra 1 Copyright Big Ideas Learning, LLC = Area of rectangular area Area of pool Area of hot tub A = w w s A = (8x 10)(x x) (6x 14)(x) x = (8x 10)(4x) ( 6x(x) 14(x) ) x = 8x(4x) 10(4x) (1x 8x) x = 3x 40x 1x 8x x = ( 3x 1x x ) ( 40x 8x) = 19x 1x b. 19x 1x = 19(9) 1(9) = 19(81) 108 = 1539 108 = 1431 So, when x = 9, the area of the patio is 1431 square feet. The patio costs 1431(10) = $14,310. Maintaining Mathematical Proficiency 6. (x 1) 3(x ) = (x) (1) 3(x) 3() = x 3x 6 = (x 3x) ( 6) = 5x 4 63. 8(4y 3) ( y 5) = 8(4y) 8(3) ( y) (5) = 3y 4 y 10 = (3y y) ( 4 10) = 34y 34 64. 5(r 1) 3( 4r ) = 5(r) 5(1) 3( 4r) 3() = 10r 5 1r 6 = (10r 1r) (5 6) = r 1
7. Explorations (p. 365) 1. a. 1; The product of 1 and 1 is 1. b. 1; The product of 1 and 1 is 1. c. 1; The product of 1 and 1 is 1. d. x; The product of any number and 1 is that number. e. x; The product of any number and 1 is the opposite of that number. f. x; The product of any number and 1 is that number. g. x; The product of the opposite of a number and 1 is the number. h. x ; The product of any number multiplied by itself is the number squared. i. x ; The product of any number and its opposite is the opposite of the number squared. j. x ; The product of any number multiplied by itself is the number squared.. a. (x 3)(x ) = x ( x 3x) 6 = x x 6 b. (x 1)(x 1) = 4x (x x) 1 = 4x 1 c. (x )(x 1) = x ( x 4x) = x 3x d. ( x )(x 3) = x (3x x) 6 = x x 6 3. Multiply each term in one polynomial by each term in the other polynomial, then combine like terms. 4. Sample answer: (x 1)(x ) = x ( x x) = x x 7. Monitoring Progress (pp. 366 368) 1. ( y 4)( y 1) = y( y 1) 4( y 1) = y( y) y(1) 4( y) 4(1) = y y 4y 4 = y 5y 4 The product is y 5y 4.. z z 6 6z 1 z z z 4z 1 The product is z 4z 1. 3. ( p 3)( p 8) = ( p 3) [ p ( 8) ] p 3 p p 3p 8 8p 4 The product is p 3p 8p 4, or p 5p 4. 4. (r 5)(r 1) = [ r ( 5) ] [ r ( 1) ] r 5 r r 10r 1 r 5 The product is r 10r r 5, or r 11r 5. 5. First Outer Inner Last (m 3)(m 7) = m(m) m( 7) ( 3)(m) ( 3)( 7) = m ( 7m) ( 3m) 1 = m 10m 1 The product is m 10m 1. 6. First Outer Inner Last (x 4)(x ) = x(x) x() ( 4)(x) ( 4)() = x x ( 4x) ( 8) = x x 8 The product is x x 8. 7. First Outer Inner Last ( u 1 ) ( u 3 ) = u(u) u ( 3 ) 1 (u) 1 ( 3 The product is u 5 u 3 4. = u ( 3u) 1 u 3 4 = u 5 u 3 4 ) Copyright Big Ideas Learning, LLC Algebra 1 385
8. First Outer Inner Last (n ) ( n 3 ) = n ( n ) n(3) ( n ) (3) = n 3 3n n 6 = n 3 n 3n 6 The product is n 3 n 3n 6. 9. x 5x 8 x 1 x 5x 8 x 3 5x 8x x 3 6x 13x 8 The product is x 3 6x 13x 8. 10. n n 4 n 3 3n 6n 1 n 3 n 4n n 3 5n 10n 1 The product is n 3 5n 10n 1. 11. 1 h ( b 1 b ) = 1 (x 7) [ x(x 11) ] = 1 (x 7)(x 11) F O I L = 1 [ x 11x ( 14x) ( 77) ] = 1 ( x 3x 77 ) = x 3 x 77 The polynomial that represents the area of the trapezoidal region becomes x 3 x 77. The longer base becomes x 11. Substituting this value in the formula for the area of a trapezoid along with the other unchanged values changes the linear and constant terms in the polynomial. 7. Exercises (pp. 369 370) Vocabulary and Core Concept Check 1. Sample answer: Distribute one of the binomials over each term in the other binomial, and simplify by combining like terms. Or, write each binomial as a sum of terms and make a table of products. Then, write the sum of the products and simplify by combining like terms.. The letters stand for the sets of terms to multiply: first, outer, inner, and last. Monitoring Progress and Modeling with Mathematics 3. (x 1)(x 3) = x(x 3) 1(x 3) = x(x) x(3) 1(x) 1(3) = x 3x x 3 = x 4x 3 4. ( y 6)( y 4) = y( y 4) 6( y 4) = y( y) y(4) 6( y) 6(4) = y 4y 6y 4 = y 10y 4 5. (z 5)(z 3) = z(z 3) 5(z 3) = z(z) z(3) 5(z) 5(3) = z 3z 5z 15 = z z 15 6. (a 8)(a 3) = a(a 3) 8(a 3) = a(a) a(3) 8(a) 8(3) = a 3a 8a 4 = a 5a 4 7. (g 7)(g ) = g(g ) 7(g ) = g(g) g() 7(g) 7( ) = g g 7g 14 = g 9g 14 8. (n 6)(n 4) = n(n 4) 6(n 4) = n(n) n(4) 6(n) 6( 4) = n 4n 6n 4 = n 10n 4 9. (3m 1)(m 9) = 3m(m 9) 1(m 9) = 3m(m) 3m(9) 1(m) 1(9) = 3m 7m m 9 = 3m 8m 9 10. (5s 6)(s ) = 5s(s ) 6(s ) = 5s(s) 5s() 6(s) 6() = 5s 10s 6s 1 = 5s 4s 1 11. (x 3)(x ) x 3 x x 3x x 6 x 3x x 6 = x 5x 6 1. ( y 10)( y 5) = ( y 10) [ y ( 5) ] y 10 y y 10y 5 5y 50 y 10y 5y 50 = y 5y 50 386 Algebra 1 Copyright Big Ideas Learning, LLC
13. (h 8)(h 9) = [ h ( 8) ] [ h ( 9) ] h 8 h h 8h 9 9h 7 h 8h 9h 7 = h 17h 7 14. (c 6)(c 5) = [ c ( 6) ] [ c ( 5) ] c 6 c c 6c 5 5c 30 c 6c 5c 30 = c 11c 30 15. (3k 1)(4k 9) = [ 3k ( 1) ] (4k 9) 3k 1 4k 1k 4k 9 7k 9 1k 4k 7k 9 = 1k 3k 9 16. (5g 3)(g 8) 5g 3 g 5g 3g 8 40g 4 5g 3g 40g 4 = 5g 43g 4 17. ( 3 j )(4j 7) = [ j ( 3) ] [ 4j ( 7) ] j 3 4j 8j 1j 7 14j 1 8j 1j 14j 1 = 8j 6j 1 18. (5d 1)( 7 3d) = [ 5d ( 1) ] [ 3d ( 7) ] 5d 1 3d 15d 36d 7 35d 84 15d 36d 35d 84 = 15d 71d 84 19. The first term t also should be multiplied by t 5. (t )(t 5) = t (t 5) (t 5) = t(t) t(5) (t) (5) = t 5t t 10 = t 3t 10 0. The two terms that represent x 5 on the left side of the table should be x and 5, not 5. (x 5)(3x 1) = [ x ( 5) ] (3x 1) 3x 1 x 3x x 5 15x 5 3x x 15x 5 = 3x 14x 5 1. First Outer Inner Last (b 3)(b 7) = b(b) b(7) 3(b) 3(7) = b 7b 3b 1 = b 10b 1. First Outer Inner Last (w 9)(w 6) = w(w) w(6) 9(w) 9(6) = w 6w 9w 54 = w 15w 54 3. First Outer Inner Last (k 5)(k 1) = k(k) k( 1) 5(k) 5( 1) = k k 5k 5 = k 4k 5 4. First Outer Inner Last (x 4)(x 8) = x(x) x(8) ( 4)(x) ( 4)(8) = x 8x 4x 3 = x 4x 3 5. First Outer Inner Last ( q 3 4 ) ( q 1 4 ) = q(q) q ( 1 4 ) ( 3 4 ) (q) ( 3 4 ) ( 1 4 ) = q 1 4 q 3 4 q 3 16 = q 1 q 3 16 6. First Outer Inner Last ( z 5 3 ) ( z 3 ) = z(z) z ( 3 ) ( 5 3 ) (z) ( 5 3 ) ( 3 ) = z 3 z 5 3 z 10 9 = z 7 3 z 10 9 7. First Outer Inner Last (9 r)( 3r) = 9() 9( 3r) ( r)() ( r)( 3r) = 18 7r r 3r = 18 9r 3r = 3r 9r 18 Copyright Big Ideas Learning, LLC Algebra 1 387
8. First Outer Inner Last (8 4x)(x 6) = 8(x) 8(6) ( 4x)(x) ( 4x)(6) = 16x 48 8x 4x = 8x 16x 4x 48 = 8x 8x 48 9. First Outer Inner Last (w 5) ( w 3w ) = w ( w ) w(3w) 5 ( w ) 5(3w) = w 3 3w 5w 15w = w 3 8w 15w 30. First Outer Inner Last (v 3) ( v 8v ) = v ( v ) v(8v) ( 3)( v ) ( 3)(8v) = v 3 8v 3v 3 4v = v 3 5v 4v 31. A = w = (x 9)(x 5) F O I L = x(x) x(5) ( 9)(x) ( 9)(5) = x 10x 9x 45 = x x 45 The polynomial x x 45 represents the area of the rectangular region. 3. A = 1 bh = 1 (p 1)(p 6) F O I L = 1 [ p(p) p( 6) 1(p) 1( 6) ] = 1 ( p 6p p 6 ) = 1 ( p 4p 6 ) = 1 ( p ) 1 (4p) 1 (6) = p p 3 The polynomial p p 3 represents the area of the triangular region. 33. 34. Area of shaded region = Area of rectangular region Area of triangular region A = w 1 bh = (x 6)(x 5) 1 (x 6)(x 5) F O I L F O I L = [ x(x) x(5) 6(x) 6(5) ] 1 [ x(x) x(5) 6(x) 6(5) ] = ( x 5x 6x 30 ) 1 ( x 5x 6x 30 ) = (x 11x 30) 1 ( x 11x 30 ) = x 11x 30 1 ( x ) 1 (11x) 1 (30) = x 11x 30 1 x 11 x 15 = ( x 1 x ) (30 15) x ) ( 11x 11 = 1 x 11 x 15 The polynomial 1 x 11 x 15 represents the area of the shaded region. Area of shaded region = Area of square region Area of rectangular region A = s w = (x 1) (x 7)(5) = (x 1)(x 1) (x 7)(5) F O I L = x(x) x(1) 1(x) 1(1) [ x(5) 7(5) ] = x x x 1 (5x 35) = x x 1 5x 35 = x x 5x 1 35 = x 3x 36 The polynomial x 3x 36 represents the area of the shaded region. 35. x 3x x 4 4x 1x 8 x 3 3x x x 3 7x 14x 8 36. f 4f 8 f 1 f 4f 8 f 3 4f 8f f 3 5f 1f 8 37. y 8y y 3 3y 4y 6 y 3 8y y y 3 11y y 6 388 Algebra 1 Copyright Big Ideas Learning, LLC
38. t 5t 1 t t 10t t 3 5t t t 3 7t 11t 39. 5b 5b 4 b 4 0b 0b 16 5b 3 5b 4b 5b 3 15b 4b 16 40. d d 7 d 6 1d 6d 4 d 3 d 7d d 3 11d d 4 41. 3e 5e 7 6e 1 3e 5e 7 18e 3 30e 4e 18e 3 7e 37e 7 4. 6v v 9 5v 4 4v 8v 36 30v 3 10v 45v 30v 3 14v 53v 36 43. a. A = w = (10x 10)(4x 0) F O I L =10x(4x) 10x(0) 10(4x) 10(0) = 40x 00x 40x 00 = 40x 40x 00 A polynomial that represents the area of the football field is ( 40x 40x 00 ) square feet. b. 4x 0 = 160 0 0 4x = 140 4x 4 = 140 4 x = 35 40x 40x 00 = 40(35) 40(35) 00 = 40(15) 8400 00 = 49,000 8400 00 = 57,600 When the width of a football field is 160 feet, the area is 57,600 square feet. 44. a. A = w = (x x)(x 0 x) = (x )(x 0) F O I L = x(x) x(0) (x) (0) = 4x 40x 44x 440 = 4x 84x 440 A polynomial that represents the combined area of the photo and the frame is ( 4x 84x 440 ) square inches. b. Let x = 4. 4x 84x 440 = 4(4) 84(4) 440 = 4(16) 336 440 = 64 336 440 = 840 When the width of the frame is 4 inches, the combined area of the photo and the frame is 840 square inches. 45. The degree of the product is the sum of the degrees of each binomial. 46. Sample answer: (x 6) ( x 3x 4 ) x 3x 4 x 6 6x 18x 4 x 3 6x 8x x 3 10x 4 The product is x 3 10x 4, which is a trinomial of degree 3. 47. The FOIL method can only be used for multiplying two binomials, because each of the four letters represent one of the products when two binomials are multiplied. When two trinomials are multiplied, however, there are 6 products. The FOIL method would leave out the products that include the middle terms of the two trinomials. 48. a. The two binomials being multiplied are ( 4x 3) and ( 8x 9). The binomial ( 4x 3) is comprised of the two terms on top of the table, and the binomial ( 8x 9) is comprised of the two terms on the left side of the table. b. When x > 0, a is positive because it is the product of two negative terms, b is negative because it is the product of one negative term and one positive term, c is positive because it is the product of two negative terms, and d is negative because it is the product of one negative term and one positive term. 49. Your answers should be equivalent. You are both multiplying the same binomials, and neither the order in which you multiply nor the method used will make a difference. Copyright Big Ideas Learning, LLC Algebra 1 389
50. V = wh = (4x 3)(x 1)(x ) = (4x 3) [ x(x ) 1(x ) ] = (4x 3) [ x(x) x() 1(x) 1() ] = (4x 3) ( x x x ) = (4x 3) ( x 3x ) = 4x ( x 3x ) 3 ( x 3x ) = 4x ( x ) 4x(3x) 4x() 3 ( x ) 3(3x) 3() = 4x 3 1x 8x 3x 9x 6 = 4x 3 ( 1x 3x ) (8x 9x) 6 = 4x 3 9x x 6 A polynomial that represents the volume of the container is ( 4x 3 9x x 6 ) cubic feet. 51. a. The product of m and n is c. So, when c > 0, m and n have the same signs because the product of two numbers with the same sign is positive. b. The product of m and n is c. So, when c < 0, m and n have opposite signs because the product of one positive number and one negative number is negative. Maintaining Mathematical Proficiency 5. y = x 4 A piecewise function for y = x 4 is g(x) = { x 4, if x < 0 x 4, if x 0. 53. y = 6 x 3 g(x) = { 6[ (x 3) ], if x 3 < 0 6(x 3), if x 3 0 g(x) = 6 [ (x 3) ], if x 3 < 0 g(x) = 6( x 3), 3 3 g(x) = 6( x) 6(3), g(x) = 6x 18, if x < 3 g(x) = 6(x 3), if x 3 0 g(x) = 6(x) 6(3) 3 3 g(x) = 6x 18, if x 3 So, a piecewise function for y = 6 x 3 is g(x) = { 6x 18, if x < 3 6x 18, if x 3. 54. y = 4 x g(x) = { 4[ (x ) ], if x < 0 4(x ), if x 0 g(x) = 4 [ (x ) ], if x < 0 g(x) = 4( x ), g(x) = 4( x) 4( ), g(x) = 4x 8, if x < g(x) = 4(x ), if x 0 g(x) = 4(x) 4(), g(x) = 4x 8, if x So, a piecewise function for y = 4 x is g(x) = { 4x 8, if x < 4x 8, if x. 55. 10 10 9 = 10 9 = 10 11 56. x 5 x = x 5 1 x 8 x 8 = x 6 x 8 = x 6 8 = x = 1 x 57. (3z 6 ) 3 1 = (3z 6 ) 3 = 1 3 3 (z 6 ) 3 = 1 7z 6 3 = 1 7z 18 58. ( y ) 4 = (y 4 3 ) = (y) = y 1 = y 3 y = 1 4y 7.3 Explorations (p. 371) 1. a. x x x 4 = x 4 b. 4x x x 1 = 4x 1. a. x x x 4 = x 4x 4 b. 4x x x 1 = 4x 4x 1 390 Algebra 1 Copyright Big Ideas Learning, LLC
3. F O I L (a b)(a b) = a(a) a( b) b(a) b( b) = a ab ab b = a b So, (a b)(a b) = a b. (a b) = (a b)(a b) F O I L = a(a) a(b) b(a) b(b) = a ab ab b = a ab b So, (a b) = a ab b. (a b) = (a b)(a b) F O I L = a(a) a( b) ( b)(a) ( b)( b) = a ab ab b = a ab b So, (a b) = a ab b. 4. a. ( a b)( a b) = a b (x 3)(x 3) = x 3 = x 9 Check x 3x 3x 9 = x 9 The product is x 9. b. (a b)(a b) = a b (x 4)(x 4) = x 4 = x 16 Check x 4x 4x 16 = x 16 The product is x 16. c. (a b)(a b) = a b (3x 1)(3x 1) = (3x) 1 = 9x 1 Check 9x 3x 3x 1 = 9x 1 The product is 9x 1. d. (a b) = a ab b (x 3) = x (x)(3) 3 = x 6x 9 Check x 3x 3x 9 = x 6x 9 The product is x 6x 9. e. (a b) = a ab b (x ) = x (x)() = x 4x 4 Check x x x 4 = x 4x 4 The product is x 4x 4. Copyright Big Ideas Learning, LLC Algebra 1 391
f. (a b) = a ab b (3x 1) = (3x) (3x)(1) 1 Check = 9x 6x 1 9x 3x 3x 1 = 9x 6x 1 The product is 9x 6x 1. 7.3 Monitoring Progress (pp. 37 374) 1. (x 7) = x (x)(7) 7 = x 14x 49 The product is x 14x 49.. (7x 3) = (7x) (7x)(3) 3 = 49x 4x 9 The product is 49x 4x 9. 3. (4x y) = (4x) (4x)(y) y = 16x 8xy y The product is 16x 8xy y. 4. (3m n) = (3m) (3m)(n) n = 9m 6mn n The product is 9m 6mn n. 5. (x 10)(x 10) = x 10 = x 100 The product is x 100. 6. (x 1)(x 1) = (x) 1 = 4x 1 The product is 4x 1. 7. (x 3y)(x 3y) = x (3y) = x 9y The product is x 9y. 8. Rewrite 1 as (0 1). Then use the square of a binomial pattern. (0 1) = 0 (0)(1) 1 = 400 40 1 = 441 9. a. The Punnett square shows four possible gene combinations of the offspring. Of these combinations, one results in black. So, 1 = 5% of the possible gene 4 combinations result in black. b. Model the gene from each parent with the expression 0.5B 0.5W. There is an equal chance that the offspring inherits a black or a white gene from each parent. You can model the possible gene combinations of the offspring with the expression (0.5B 0.5W ). Notice that this product also represents the area of the Punnett square. Expand the product to find the possible colors of the offspring. (0.5B 0.5W ) = (0.5B) (0.5B)(0.5W ) (0.5W) = 0.5B 0.5BW 0.5W Consider the coefficients in the polynomial. So, 5% of the offspring are BB, or black; 50% of the offspring are BW, or gray; and 5% of the offspring are WW, or white. 7.3 Exercises (pp. 375 376) Vocabulary and Core Concept Check 1. Substitute the first term of the binomial for a and the second term of the binomial for b in the square of a binomial pattern a ab b. Then simplify.. The expression that does not belong is (x )(x 3). It is the only one that cannot be simplified using the sum and difference pattern. The pattern for this one is (a b)(a c), but the pattern for the others is (a b)(a b). Monitoring Progress and Modeling with Mathematics 3. (x 8) = x (x)(8) 8 = x 16x 64 4. (a 6) = a (a)(6) 6 = a 1a 36 5. (f 1) = (f ) (f )(1) 1 = 4f 4f 1 6. (5p ) = (5p) (5p)() = 5p 0p 4 7. ( 7t 4) = ( 7t) ( 7t)(4) 4 = 49t 56t 16 8. ( 1 n) = ( 1) ( 1)(n) n = 144 4n n = n 4n 144 9. (a b) = (a) (a)(b) b = 4a 4ab b 10. (6x 3y) = (6x) (6x)(3y) (3y) = 36x 36xy 9y 39 Algebra 1 Copyright Big Ideas Learning, LLC
11. (x 4) = x (x)(4) 4 = x 8x 16 1. (x 7 x) = (x 7) = (x) (x)(7) 7 = 4x 8x 49 13. (7n 5) = (7n) (7n)(5) 5 = 49n 70n 5 14. (4c 4d ) = (4c) (4c)(4d ) (4d ) = 16c 3cd 16d 15. (t 7)(t 7) = t 7 = t 49 16. (m 6)(m 6) = m 6 = m 36 17. (4x 1)(4x 1) = (4x) 1 = 16x 1 18. (k 4)(k 4) = (k) 4 = 4k 16 19. (8 3a)(8 3a) = 8 (3a) = 64 9a 0. ( 1 c ) ( 1 c ) = ( 1 ) c = 1 4 c 1. ( p 10q)( p 10q) = p (10q) = p 100q. (7m 8n)(7m 8n) = (7m) (8n) = 49m 64n 3. ( y 4)( y 4) = ( y) 4 = y 16 4. ( 5g h)( 5g h) = ( 5g) (h) 5. 16 4 = (0 4)(0 4) = 0 4 = 400 16 = 384 = 5g 4h 7. 4 = (40 ) = 40 (40)() = 1600 160 4 = 1764 8. 9 = (30 1) = 30 (30)(1) 1 = 900 60 1 = 841 9. 30.5 = (30 0.5) = 30 (30)(0.5) 0.5 = 900 30 0.5 = 930.5 30. 10 1 3 9 3 = ( 10 1 3 ) ( 10 1 = 10 ( 1 3 ) = 100 1 9 = 99 8 9 31. The middle term in the square of a binomial pattern was not included. (k 4) = k (k)(4) 4 = k 8k 16 3. There is no middle term in the sum and difference pattern. (s 5)(s 5) = s 5 = s 5 33. a. (x 50) = x (x)(50) 50 3 ) = x 100x 500 A polynomial that represents the area of the house after the renovation is ( x 100x 500 ) square feet. b. x 100x 500 = (15) 100(15) 500 = 5 1500 500 = 45 The area of the house after the renovation is 45 square feet. The original area of the house was 50 = 500 square feet. So, the area of the renovation is 45 500 = 175 square feet. 6. 33 7 = (30 3)(30 3) = 30 3 = 900 9 = 891 Copyright Big Ideas Learning, LLC Algebra 1 393
34. a. (100 x)(100 x) = 100 x = 10,000 x The product that represents the area of the new parking lot is ( 10,000 x ) square feet. b. The area of the parking lot decreases. The original area is 10,000 square feet. So, the new area is the original area decreased by x. c. 10,000 x = 10,000 1 = 10,000 441 = 9559 When x = 1, the area of the new parking lot is 9559 square feet. 35. a. The Punnett square shows four possible gene combinations of the offspring. Of these combinations, one results in albino coloring. So, 1 = 5% of the possible 4 gene combinations result in albino coloring. b. (0.5N 0.5a) = (0.5N) (0.5N )(0.5a) (0.5a) = 0.5N 0.5Na 0.5a The coefficients show that 5% 50% = 75% of the possible gene combinations result in normal coloring, while 5% of the possible gene combinations result in albino coloring. 36. a. A = π r = π (6 x) = π ( 6 (6)(x) x ) = π ( 36 1x x ) = π (36) π (1x) π ( x ) = 36π 1π x π x = π x 1π x 36π A polynomial that represents the area of your pupil is ( π x 1π x 36π ) square millimeters. b. Let x = 4. Find the area of your pupil before entering the room. π x 1π x 36π = π (4) 1π (4) 36π = π (16) 48π 36π = 16π 48π 36π = 4π The area of your pupil before entering the room is 4π millimeters. Let x =. Find the area of your pupil after entering the room. π x 1π x 36π = π () 1π () 36π = π (4) 4π 36π = 4π 4π 36π = 16π The area of your pupil after entering the room is 16π millimeters. So, the area of your pupil is 16π 4π = 4 times greater after entering the room than before. 37. x 11 fits the product side of the sum and difference pattern, so working backward, a and b are the square roots of a and b. x = x and 11 = 11 So, x 11 = (x 11)(x 11). 38. The Punnett square shows four possible gene combinations of the offspring. Of these combinations, three result in green pods. So, 3 = 75% of the possible gene combinations 4 result in green pods. Alternately, the coefficients of the polynomial model show that GG accounts for 5% of the possible gene combinations, and Gy accounts for 50% of the possible gene combinations. So, 5% 50% = 75% of the possible gene combinations result in green pods. 39. ( x 1 ) ( x 1 ) = ( x ) = 1 = x 4 1 40. ( y 3 4 ) = ( y 3 ) ( y 3 ) (4) 4 = y 6 8y 3 16 41. ( m 5n ) = ( m ) ( m ) ( 5n ) ( 5n ) = m 0m n 5 n = 4m 4 0m n 5n 4 4. ( r 3 6t 4 ) ( r 3 6t 4 ) = ( r 3 ) ( 6t 4 ) = r 3 6 t 4 = r 6 36t 8 43. Your friend is incorrect. The expression ( 4 1 3 ) can be written as ( 4 1 3 ). However, using the square of a binomial pattern results in 4 (4) ( 1 3 ) ( 1 3 ) = 16 8 3 1, which is 9 18 7 9, not 16 1 9. 44. Sample answer: One way to modify the dimensions of the lake is by increasing one dimension by 4 and decreasing the other dimension by 4, which can be modeled by (x 4)(x 4). This follows a sum and difference pattern. A second way to modify the dimensions of the lake is by increasing both dimensions by, which can be modeled by (x ). This follows a square of a binomial pattern. A third way to modify the dimensions of the lake is by decreasing both dimensions by, which can be modeled by (x ). This also follows a square of a binomial pattern, but this one has subtraction. 394 Algebra 1 Copyright Big Ideas Learning, LLC
45. Let 9x 48x k = a ab b. So, 9x = a 9x = a 3x = a. Then, 48x = ab 48x = (3x)b 48x = 6xb 48x 6x = 6xb 6x 8 = b. Finally, k = b k = 8 k = 64. So, when k = 64, 9x 48x k = 9x 48x 64 is the square of the binomial 3x 8. 46. (x 1) 3 = (x 1)(x 1)(x 1) = [x(x 1) 1(x 1)](x 1) = [x(x) x(1) 1(x) 1(1)](x 1) = ( x x x 1 ) (x 1) = ( x x 1 ) (x 1) = x (x 1) x(x 1) 1(x 1) = x (x) x (1) x(x) x(1) 1(x) 1(1) = x 3 ( x x ) (x x) 1 = x 3 3x 3x 1 (x ) 3 = (x )(x )(x ) = [x(x ) (x )](x ) = [x(x) x() (x) ()](x ) = ( x x x 4 )(x ) = ( x 4x 4 )(x ) = x (x ) 4x(x ) 4(x ) = x (x) x () 4x(x) 4x() 4(x) 4() = x 3 x 4x 8x 4x 8 = x 3 6x 1x 8 So, (a b) 3 = a 3 3a b 3ab b 3. 47. Sample answer: The statement (a b)(a b) < (a b) < (a b) is true when a = 3 and b = 4. Check (a b)(a b) < (a b) < (a b) (3 4)(3 4) <? (3 4) <? (3 4) 50. 49s 35t = 7(7s 5t) 51. 15x 10y = 5(3x y) 7.4 Explorations (p. 377) 1. a. C; 5; (x 1)(x 3) = 0 Factored Form F O I L x(x) x( 3) ( 1)(x) ( 1)( 3) = 0 x 3x x 3 = 0 x 4x 3 = 0 Standard Form, C 3 3 x 4x = 3 Nonstandard Form, 5 So, the equation (x 1)(x 3) = 0 is equivalent to x 4x 3 = 0 (C) and x 4x = 3 (5). b. D; 1; (x )(x 3) = 0 Factored Form F O I L x(x) x( 3) ( )(x) ( )( 3) = 0 x 3x x 6 = 0 x 5x 6 = 0 Standard Form, D 6 6 x 5x = 6 Nonstandard Form, 1 So, the equation (x )(x 3) = 0 is equivalent to x 5x 6 = 0 (D) and x 5x = 6 (1). c. A; 3; (x 1)(x ) = 0 Factored Form F O I L x(x) x( ) 1(x) 1( ) = 0 x x x = 0 x x = 0 Standard Form, A x x = Nonstandard Form, 3 So, the equation (x 1)(x ) = 0 is equivalent to x x = 0 (A) and x x = (3). (7)( 1) <? ( 1) <? (7) 7 < 1 < 49 Maintaining Mathematical Proficiency 48. 1y 18 = 6(y 3) 49. 9r 7 = 9(r 3) Copyright Big Ideas Learning, LLC Algebra 1 395
. d. B; 4; (x 1)(x ) = 0 Factored Form F O I L x(x) x() ( 1)(x) ( 1)() = 0 x x x = 0 x x = 0 Standard Form, B x(x 1) = Nonstandard Form, 4 x(x) x(1) = x x = x x = x x = 0 Standard Form, B So, the equation (x 1)(x ) = 0 is equivalent to x x = 0 (B) and x(x 1) = (4). e. E; ; (x 1)(x 3) = 0 Factored Form F O I L x(x) x( 3) 1(x) 1( 3) = 0 x 3x x 3 = 0 x x 3 = 0 Standard Form, E (x 1) = 4 Nonstandard Form, x (x)(1) 1 = 4 x x 1 = 4 4 4 x x 3 = 0 Standard Form, E So, the equation (x 1)(x 3) = 0 is equivalent to x x = 3 (E) and (x 1) = 4 (). a. b. c. x = 1 (x )(x 3) = 0 ( 1)( ) 0 False (0)( 1) = 0 True 3 4 5 6 (1)(0) = 0 True ()(1) 0 False (3)() 0 False x = 1 (x 3)(x 4) = 0 ( )( 3) 0 False (4)(3) 0 False ( 1)( ) 0 False 3 4 5 6 (0)( 1) = 0 True x = 1 (x 1)(x ) = 0 (1)(0) = 0 True (0)( 1) = 0 True ()(1) 0 False (1)(0) = 0 True 3 4 5 6 ()(1) 0 False (3)() 0 False (4)(3) 0 False (5)(4) 0 False (3)() 0 False d. e. f. x = 1 (x 4)(x 5) = 0 ( 3)( 4) 0 False ( )( 3) 0 False 3 4 5 6 ( 1)( ) 0 False (0)( 1) = 0 True (1)(0) = 0 True x = 1 (x 5)(x 6) = 0 ( 4)( 5) 0 False ()(1) 0 False ( 3)( 4) 0 False 3 4 5 6 ( )( 3) 0 False ( 1)( ) 0 False (0)( 1) = 0 True x = 1 (x 6)(x 1) = 0 ( 5)(0) = 0 True (1)(0) = 0 True ( 4)(1) 0 False 3 4 5 6 ( 3)() 0 False ( )(3) 0 False ( 1)(4) 0 False (0)(5) = 0 True If the product of two values is 0, then at least one of the values must be 0. If (x a) is a factor of an equation, then x = a is a solution. 3. a. When you add 0 to a number n, you get n. Adding 0 does not change the value, but adding 1 increases the value. b. If the product of two numbers is 0, then at least one of the numbers is 0. The product of 0 and any number is always 0, never 1. c. The square of 0 is equal to itself, and the square of 1 is equal to itself. Both are true: 0 = 0 and 1 = 1. d. When you multiply a number n by 1, you get n. The product of any number and 1 is the number. On the other hand, the product of any number and 0 is 0. e. When you multiply a number n by 0, you get 0. The product of any number and 0 is 0. On the other hand, the product of any number and 1 is the number. f. The opposite of 0 is equal to itself. Zero is neither positive nor negative. So, it is its own opposite. The opposite of 1, however, is 1. 4. When an equation is in factored form, the product of the factors equals 0. So, set each polynomial factor equal to 0, and solve. 5. b; If the product of two numbers is 0, then at least one of the numbers is 0. This property is called the Zero-Product Property because it describes what happens when you have a product that is equal to zero. The Zero-Product Property is used in algebra to solve polynomial equations by setting each polynomial factor equal to 0. It is important because it provides an easy way to solve polynomial equations. 396 Algebra 1 Copyright Big Ideas Learning, LLC
7.4 Monitoring Progress (pp. 378 380) 1. x(x 1) = 0 x = 0 or x 1 = 0 1 1 x = 1 Check x(x 1) = 0 x(x 1) = 0 0(0 1) =? 0 1(1 1) =? 0 0( 1) =? 0 1(0) =? 0 0 = 0 0 = 0 The roots are x = 0 and x = 1.. 3t(t ) = 0 3t = 0 or t = 0 3t 3 = 0 3 t = 0 t = Check 3t(t ) = 0 3t(t ) = 0 3(0)(0 ) =? 0 3( )( ) =? 0 0() =? 0 6(0) =? 0 0 = 0 0 = 0 The roots are t = 0 and t =. 3. (z 4)(z 6) = 0 z 4 = 0 or z 6 = 0 4 4 6 6 z = 4 z = 6 Check (z 4)(z 6) = 0 (z 4)(z 6) = 0 (4 4)(4 6) =? 0 (6 4)(6 6) =? 0 0( ) =? 0 (0) =? 0 0 = 0 0 = 0 The roots are z = 4 and z = 6. 4. (3s 5)(5s 8) = 0 3s 5 = 0 or 5s 8 = 0 5 5 8 8 3s = 5 5s = 8 3s 3 = 5 5s 3 5 = 8 5 s = 5 s = 8 3 5 Check (3s 5)(5s 8) = 0 [ 3 ( 5 3 ) 5 ] [ 5 ( 5 3) 8 ] =? 0 ( 5 5) ( 5 3 8 ) =? 0 0 ( 1 3) =? 0 0 = 0 (3s 5)(5s 8) = 0 [ 3 ( 8 5 5) ] [ 5 ( 8 5 ) 8 ] =? 0 ( 4 5 ) 5 ( 8 8) =? 0 ( 1 5) (0) =? 0 0 = 0 The roots are s = 5 3 and s = 8 5. 5. (b 7) = 0 (b 7)(b 7) = 0 b 7 = 0 or b 7 = 0 7 7 7 7 b = 7 b = 7 Check (b 7) = 0 ( 7 7) =? 0 (0) =? 0 0 = 0 The equation has repeated roots of b = 7. Copyright Big Ideas Learning, LLC Algebra 1 397
6. (d )(d 6)(d 8) = 0 d = 0 or d 6 = 0 or d 8 = 0 6 6 8 8 d = d = 6 d = 8 Check (d )(d 6)(d 8) = 0 ( )( 6)( 8) =? 0 0(8)(10) =? 0 0 = 0 (d )(d 6)(d 8) = 0 ( 6 )( 6 6)( 6 8) =? 0 ( 8)(0)() =? 0 0 = 0 (d )(d 6)(d 8) = 0 ( 8 )( 8 6)( 8 8) =? 0 ( 10)( )(0) =? 0 0 = 0 The roots are d =, d = 6, and d = 8. 7. The GCF of 8 and 4 is 8. The GCF of y and y is y. So, the greatest common monomial factor of the terms is 8y. So, 8y 4y = 8y( y 3). 8. a 5a = 0 a(a 5) = 0 a = 0 or a 5 = 0 5 5 a = 5 Check a 5a = 0 a 5a = 0 (0) 5(0) =? 0 ( 5) 5( 5) =? 0 0 0 =? 0 5 5 =? 0 0 = 0 0 = 0 The roots are a = 0 and a = 5. 9. 3s 9s = 0 3s(s 3) = 0 3s = 0 or s 3 = 0 3s 3 = 0 3 3 3 s = 0 s = 3 Check 3s 9s = 0 3s 9s = 0 3(0) 9(0) =? 0 3(3) 9(3) =? 0 0 0 =? 0 3(9) 7 =? 0 0 = 0 7 7 =? 0 0 = 0 The roots are s = 0 and s = 3. 10. 4x = x 4x x = x x 4x x = 0 x(x 1) = 0 x = 0 or x 1 = 0 x = 0 1 1 x = 0 x = 1 Check 4x = x 4(0) =? (0) 4(0) =? 0 0 = 0 x = 1 x = 1 The roots are x = 0 and x = 1. 11. Let y = 0. y = 1 (x 4)(x 4) 0 = 1 (x 4)(x 4) (0) = ( 1 )(x 4)(x 4) 0 = (x 4)(x 4) x 4 = 0 or x 4 = 0 4 4 4 4 x = 4 x = 4 So, the width of the entrance at ground level is 4 4 = 8 feet. 7.4 Exercises (pp. 381 38) 4x = x ) =? ( ) 1 4 ( 1 4 ( 1 4) =? 1 1 = 1 Vocabulary and Core Concept Check 1. The product of 3x and x 6 equals 0. According to the Zero-Product Property, at least one of the factors equals 0. So, let 3x = 0 and x 6 = 0. Then solve each equation to get the solutions x = 0 and x = 6. 398 Algebra 1 Copyright Big Ideas Learning, LLC
. The one that is different is Find the value of k for which (k 4) (k 3) = 0. (k 4) (k 3) = 0 (k k) (4 3) = 0 3k 1 = 0 1 1 3k = 1 3k 3 = 1 3 k = 1 3 The solution is k = 1 3. (k 4)(k 3) = 0 k 4 = 0 or k 3 = 0 4 4 3 3 k = 4 k = 3 k = 4 k = The roots are k = and k = 3. Monitoring Progress and Modeling with Mathematics 3. x(x 7) = 0 x = 0 or x 7 = 0 7 7 x = 7 The roots are x = 0 and x = 7. 4. r(r 10) = 0 r = 0 or r 10 = 0 10 10 r = 10 The roots are r = 0 and r = 10. 5. 1t(t 5) = 0 1t = 0 or t 5 = 0 1t 1 = 0 5 5 1 t = 0 t = 5 The roots are t = 0 and t = 5. 6. v(v 1) = 0 v = 0 or v 1 = 0 v = 0 1 1 v = 0 v = 1 The roots are v = 0 and v = 1. 7. (s 9) (s 1) = 0 s 9 = 0 or s 1 = 0 9 9 1 1 s = 9 s = 1 The roots are s = 9 and s = 1. 8. ( y ) ( y 6) = 0 y = 0 or y 6 = 0 6 6 y = y = 6 The roots are y = and y = 6. 9. (a 6) (3a 15) = 0 a 6 = 0 or 3a 15 = 0 6 6 15 15 a = 6 3a = 15 a = 6 a = 3 The roots are a = 3 and a = 5. 3a 3 = 15 3 a = 5 10. (4q 3) (q ) = 0 4q 3 = 0 or q = 0 3 3 4q = 3 q = 4q 4 = 3 4 q = 3 4 The roots are q = 3 and q =. 4 11. (5m 4) = 0 (5m 4) (5m 4) = 0 5m 4 = 0 or 5m 4 = 0 4 4 4 4 5m = 4 5m = 4 5m 5 = 4 5 5m 5 = 4 5 m = 4 5 m = 4 5 The equation has repeated roots of m = 4 5. Copyright Big Ideas Learning, LLC Algebra 1 399
1. (h 8) = 0 (h 8) (h 8) = 0 h 8 = 0 or h 8 = 0 8 8 8 8 h = 8 h = 8 The equation has repeated roots of h = 8. 13. (3 g) (7 g) = 0 3 g = 0 or 7 g = 0 g g g g 3 = g 7 = g 3 = g 3 = g The roots are g = 3 and g = 7. 14. ( 4d ) ( 4d ) = 0 4d = 0 or 4d = 0 4d = 4d = 4d 4 = 4 d = 1 The roots are d = 1 and d = 1. 4d 4 = 4 d = 1 15. z (z )(z 1) = 0 z = 0 or z = 0 or z 1 = 0 1 1 z = z = 1 The roots are z = 0, z =, and z = 1. 16. 5p(p 3)(p 7) = 0 5p = 0 or p 3 = 0 or p 7 = 0 5p 5 = 0 5 p = 0 3 3 7 7 p = 3 p = 7 p = 3 p = 3 The roots are p = 0, p = 3, and p = 7. 17. (r 4) (r 8) = 0 (r 4)(r 4)(r 8) = 0 r 4 = 0 or r 4 = 0 or r 8 = 0 4 4 4 4 8 8 r = 4 r = 4 r = 8 One root of the equation is r = 8. The equation also has repeated roots of r = 4. 18. w(w 6) = 0 w(w 6)(w 6) = 0 w = 0 or w 6 = 0 or w 6 = 0 6 6 6 6 w = 6 w = 6 One root of the equation is w = 0. The equation also has repeated roots of w = 6. 19. (15 5c)(5c 5)( c 6) = 0 15 5c = 0 or 5c 5 = 0 or c 6 = 0 15 15 5 5 6 6 5c = 15 5c = 5 c = 6 5c 5 = 15 5 5c 5 = 5 5 c 1 = 6 1 c = 3 c = 1 c = 6 The roots are c = 3, c = 1, and c = 6. 0. ( n) ( 6 3 ) n (n )= 0 n = 0 or 6 3 n = 0 or n = 0 6 6 n = n 1 = 1 n = 3 n = 6 n = 3 3 n = 3 ( 6 ) n = 9 One root of the equation is n = 9. The equation also has repeated roots of n =. 1. y = (x 8)(x 8) 0 = (x 8)(x 8) x 8 = 0 or x 8 = 0 8 8 8 8 x = 8 x = 8 The x-coordinates of the points where the graph crosses the x-axis are the roots x = 8 and x = 8. 400 Algebra 1 Copyright Big Ideas Learning, LLC
. y = ( x 1 )( x 7 ) 0 = ( x 1 )( x 7 ) x 1 = 0 or x 7 = 0 1 1 7 7 x = 1 x = 7 The x-coordinates of the points where the graph crosses the x-axis are the roots x = 1 and x = 7. 3. y = ( x 14 ) ( x 5 ) 0 = ( x 14 ) ( x 5 ) 1 ( 0 ) = 1 [ ( x 14 ) ( x 5 ) ] 0 = ( x 14 )( x 5 ) x 14 = 0 or x 5 = 0 14 14 5 5 x = 14 x = 5 The x-coordinates of the points where the graph crosses the x-axis are the roots x = 14 and x = 5. 4. y = 0. ( x )( x 15 ) 0 = 0. ( x )( x 15 ) 0 0. ( x )( x 15 ) = 0. 0. 0 = ( x )( x 15 ) x = 0 or x 15 = 0 15 15 x = x = 15 The x-coordinates of the points where the graph crosses the x-axis are the roots x = and x = 15. 5. 5z 45z = 5z ( z 9 ) 6. 6d 1d = 3d ( d 7 ) 7. 3y 3 9y = 3y ( y 3 ) 8. 0x 3 30x = 10x ( x 3 ) 9. 5n 6 n 5 = n 5 ( 5n ) 30. 1a 4 8a = 4a ( 3a 3 ) 31. 4p p = 0 p ( 4p 1 ) = 0 p = 0 or 4p 1 = 0 1 1 4p = 1 4p 4 = 1 4 3. 6m 1m = 0 6m ( m ) = 0 6m = 0 or m = 0 6m 6 = 0 6 m = 0 m = The roots are m = 0 and m =. 33. 5c 10c = 0 5c ( 5 c ) = 0 5c = 0 or 5 c = 0 5c 5 = 0 5 5 5 c = 0 c = 5 c = 5 c = 5 The roots are c = 0 and c = 5. 34. 18q q = 0 q ( 9 q ) = 0 q = 0 or 9 q = 0 q = 0 q q q = 0 9 = q The roots are q = 0 and q = 9. 35. 3n = 9n 3n 9n = 9n 9n 3n 9n = 0 3n ( n 3 ) = 0 3n = 0 or n 3 = 0 3n 3 = 0 3 3 3 n = 0 n = 3 The roots are n = 0 and n = 3. 36. 8r = 4r 8r 8r = 4r 8r 0 = 4r 8r 0 = 4r ( r 7 ) 4r = 0 or r 7 = 0 4r 4 = 0 7 7 4 r = 0 r = 7 The roots are r = 0 and r = 7. p = 1 4 The roots are p = 0 and p = 1 4. Copyright Big Ideas Learning, LLC Algebra 1 401
37. The other factor, 6x, should also be set equal to 0. Then each equation should be solved to get two roots of the original equation. 6x ( x 5 ) = 0 x = 0 or x 5 = 0 6x 6 = 0 6 5 5 x = 0 x = 5 The roots are x = 0 and x = 5. 38. Each side of the equation cannot be divided by y because y could be 0, and division by 0 is undefined. Instead, 1y should be subtracted from each side, so that the equation is 3y 1y = 0, and the greatest monomial factor should be factored out on the left side of the equal sign. Then each factor should be set equal to zero, and each equation should be solved to find both roots of the original equation. 3y = 1y 3y 1y = 1y 1y 3y 1y = 0 3y ( y 7 ) = 0 3y = 0 or y 7 = 0 3y 3 = 0 3 7 7 y = 0 y = 7 The roots are y = 0 and y = 7. 39. Let y = 0. y = 11 50 ( x 4 )( x 4 ) 0 = 11 50 ( x 4 )( x 4 ) 50 11 ( 0 ) = 50 11( 11 ( x 4 )( x 4 ) 50) 0 = ( x 4 )( x 4 ) x 4 = 0 or x 4 = 0 4 4 4 4 x = 4 x = 4 So, the width of the tunnel at ground level is 4 4 = 0 feet. 40. a. Let y = 0. y = 315 ( x 315 )( x 315 ) 0 = 315 ( x 315 )( x 315 ) 315 ( 0 ) = 315 ( 315) ( x 315 )( x 315 ) 0 = ( x 315 )( x 315 ) x 315 = 0 or x 315 = 0 315 315 315 315 x = 315 x = 315 So, the width of the arch at ground level is 315 315 = 630 feet. b. Let x = 0. y = 315 ( x 315 )( x 315 ) y = 315 ( 0 315 )( 0 315 ) y = 315 ( 315 )( 315 ) = ( )( 315 ) = 630 So, the height of the arch is 630 feet. 41. y = 16x 4.8x 0 = 16x 4.8x 0 = x ( 16x 4.8 ) x = 0 or 16x 4.8 = 0 4.8 4.8 16x = 4.8 16x 16 = 4.8 16 x = 0.3 Because y = 0, the roots are the x-values that represent the times when the penguin is at water level. So, when x = 0, the penguin emerges from the water. Then it is above the water until 0.3 second later, the moment it breaks the surface of the water again after the leap. So, the penguin is airborne for 0.3 second with each leap. 4. The point of intersection on the positive x-axis is farther from the origin than the point of intersection on the negative x-axis. So, the positive solution to the equation is 5, because it is the one with the greater absolute value. Therefore, the equation is y = ( x 5 ) ( x 3 ). 43. The graph has two x-intercepts because x-intercepts occur when y = 0, and this equation has two roots (one of which is a repeated root) when y = 0. 44. no; The equation y = ( x a ) ( x b ) will not have two x-intercepts if a = b. In this case, there is a repeated root, and the graph will only have one x-intercept. 45. no; The equation ( x 3 ) ( x 4 1 ) = 0 does not have any real roots. Roots will occur if x 3 3 = 0 or x 4 1 = 0. However, solving these equations results in x = 3 or x 4 = 1, and even powers of any number cannot be negative. x 3 3 = 0 x 4 1= 0 3 3 1 1 x 3 = 3 x 4 = 1 46. Sample answer: A polynomial equation of degree 4 that only has three roots must have a repeated root. So, one possible equation is ( x 1 ) ( x ) ( x 3 ) = 0. 40 Algebra 1 Copyright Big Ideas Learning, LLC
47. a. ( x y ) ( x y ) = 0 x y = 0 or x y = 0 x y y = 0 y x = y x y y = 0 y x = y x = y x = 1 y So, the roots of the equation occur when x = y and x = 1 y. b. ( x y ) ( 4x 16y ) = 0 x y = 0 or 4x 16y = 0 x y y = 0 y x = y ± x = ± y x = ± y 4x 4x 16y = 0 4x 16y = 4x 16y 4 = 4x 4 4y = x So, the roots of the equation occur when x = ± y and x = 4y. 48. ( 4 x 5 16 ) ( 3 x 81 ) = 0 4 x 5 16 = 0 or 3 x 81 = 0 16 16 81 81 4 x 5 = 16 3 x = 81 4 x 5 = 4 3 x = 3 4 x 5 = x = 4 5 5 x = 7 The solutions are x = 7 and x = 4. Maintaining Mathematical Proficiency 49. 10 50. 18 51. 30 1 ( 10 ) = 10 ( 5 ) = 10 So, the factor pairs of 10 are 1 and 10, and and 5. 1 ( 18 ) = 18 ( 9 ) = 18 3 ( 6 ) = 18 So, the factor pairs of 18 are 1 and 18, and 9, and 3 and 6. 1 ( 30 ) = 30 ( 15 ) = 30 3 ( 10 ) = 30 5 ( 6 ) = 30 So, the factor pairs of 30 are 1 and 30, and 15, 3 and 10, and 5 and 6. 5. 48 1 ( 48 ) = 48 ( 4 ) = 48 3 ( 16 ) = 48 4 ( 1 ) = 48 6 ( 8 ) = 48 So, the factor pairs of 48 are 1 and 48, and 4, 3 and 16, 4 and 1, and 6 and 8. 7.1 7.4 What Did You Learn? (p. 383) 1. Sample answer: Use x 4 to represent each side of the square. So, an expression that represents the area is ( x 4 ). Then use the square of a binomial pattern to simplify this expression into a polynomial. As another method, you could calculate the area of each shaded region and add those areas together.. Sample answer: The solutions are the constant terms with the opposite sign. This method does not work when the coefficient is not 1. 7.1 7.4 Quiz (p. 384) 1. The polynomial 8q 3 is in standard form. The only term has a degree of 3. So, the degree of the polynomial is 3. The leading coefficient is 8. The polynomial has 1 term. So, it is a monomial.. You can write the polynomial 9 d 3d in standard form as d 3d 9. The greatest degree is. So, the degree of the polynomial is. The leading coefficient is 1. The polynomial has 3 terms. So, it is a trinomial. 3. You can write the polynomial 3 m4 5 6 m6 in standard form as 5 6 m6 3 m4. The greatest degree is 6. So, the degree of the polynomial is 6. The leading coefficient is 5 6. The polynomial has terms. So, it is a binomial. 4. You can write the polynomial 1.3z 3z 4 7.4z in standard form as 3z 4 7.4z 1.3z. The greatest degree is 4. So, the degree of the polynomial is 4. The leading coefficient is 3. The polynomial has 3 terms. So, it is a trinomial. 5. ( x 5 ) ( x 4 ) = x x 5 4 = ( x x ) ( 5 4 ) = x 9 The sum is x 9. Copyright Big Ideas Learning, LLC Algebra 1 403
6. ( 3n n ) ( c 7 ) = 3n n n 7 = ( 3n n ) n 7 = 5n n 7 The difference is 5n n 7. 7. ( p 4p ) ( p 3p 15 ) = p 4p p 3p 15 = ( p p ) ( 4p 3p ) 15 = p 7p 15 The difference is p 7p 15. 8. ( a 3ab b ) ( a ab b ) = a a 3ab ab b b = ( a a ) ( 3ab ab ) ( b b ) = 0 ab b = ab b The sum is ab b. 9. ( w 6 )( w 7 ) = w ( w 7 ) 6 ( w 7 ) = w ( w ) w ( 7 ) 6 ( w ) 6 ( 7 ) = w 7w 6w 4 = w 13w 4 The product is w 13w 4. 10. ( 3 4d )( d 5 ) F O I L = 3 ( d ) 3 ( 5 ) ( 4d )( d ) ( 4d )( 5 ) = 6d 15 8d 0d = 8d ( 6d 0d ) 15 = 8d 6d 15 The product is 8d 6d 15. 11. y y 3 y 9 9y 18y 7 y 3 y 3y y 3 11y 15y 7 The product is y 3 11y 15y 7. 1. (3z 5)(3z 5) = (3z) 5 = 9z 5 The product is 9z 5. 13. (t 5) = t (t)(5) 5 =t 10t 5 The product is t 10t 5. 14. (q 6) = (q) (q)(6) 6 = 4q 4q 36 The product is 4q 4q 36. 15. 5x 15x = 0 5x(x 3) = 0 5x = 0 or x 3 = 0 5x 5 = 0 5 3 3 x = 0 3 = 0 The roots are x = 0 and x = 3. 16. (8 g)(8 g) = 0 8 g = 0 or 8 g = 0 g g g g 8 = g 8 = g The equation has repeated roots of g = 8. 17. (3p 7)(3p 7)(p 8) = 0 3p 7 = 0 or 3p 7 = 0 or p 8 = 0 7 7 7 7 8 8 3p = 7 3p = 7 p = 8 3p 3 = 7 3 p = 7 3 The roots are p = 7 3, p = 7 3 18. 3y(y 8)(y 1) = 0 3p 3 = 7 3 404 Algebra 1 Copyright Big Ideas Learning, LLC p = 7 3, and p = 8. 3y = 0 or y 8 = 0 or y 1 = 0 3y 3 = 0 8 8 1 1 3 y = 0 y = 8 y = 1 y = 1 y = 1 The roots are y = 0, y = 8, and y = 1. 19. a. P = w = (x 7 x) (x 48 x) = (x 7) (x 48) = (x) (7) (x) (48) = 4x 144 4x 96 = 8x 40 A polynomial that represents the perimeter of the blanket including the fringe is (8x 40) inches. b. A = w = (x 7 x)(x 48 x) = (x 7)(x 48) F O I L = x(x) x(48) 7(x) 7(48) = 4x 96x 144x 3456 = 4x 40x 3456 A polynomial that represents the area of the blanket including the fringe is (4x 40x 3456) square inches.