Algebra 1 Common Core Student Edition, Grade 8-9

(Marvins-Underground-K-12) #1
The solutions are the coordinates of the points on the
line with equation 5x + 4y = 20.


  1. 6 x 2 - 11 x - 10 63. 24m2 - 34m + 7

  2. 5x 2 + 53x + 72 65. decrease of 25% 66. increase
    of 25% 67. increase of 25% 68. decrease of 12.5%

  3. 6x(2x 3 + 5x 2 + 7) 70. 9(8x 3 + 6 x 2 + 3)

  4. 7x(5x 2 + x + 9)


Lesson 8-5 pp. 51 2-5 17
Go t It? 1.(r+8)(r+3) 2a. (y - 4)(y - 2)
b. No. There are no factors of 2 with sum - 1.
3a. (n + 12)(n - 3) b. (c - 7)(c + 3) 4. x + 8
and x - 9 5. (m + 9n)(m - 3ri)
Lesso n Ch eck 1. (x + 4)(x + 3) 2. (r - 7 )(r- 6 )


  1. (p + 8 )(p - 5) 4. (a + 4b)(a + 8 b) 5. n - 7 and
    n + 4 6. positive 7. positive 8. negative 9. when the
    constant term is positive and the coefficient of the
    second term is negative
    Ex e r c i se s 1 1. 2 1 3. 2 1 5. (f + 2)(f + 8 )

  2. (n - 7 ){n - 8 ) 19. (q - 6 )(q - 2) 21. 6 23. 1

  3. (w+ 1)(w - 8 ) 2 7. (x+ 6 )(x -1)

  4. (n + 2)(n - 5) 31. r-4 and r+ 1 33. A

  5. (r+9s)(/-+10s) 37. (m - 7n)(m + An)

  6. (w— 1 0 z )(w - 4z) 41a. p and q must have the
    same sign. b. p and q must have opposite signs.

  7. x- 12 45. 4x 2 + 12x + 5; (2 x + 5 ) (2 x + 1 )
    47a. They are opposites, b. Since the coefficient of the
    middle term is negative, the number with the greater
    absolute value must be negative. So, p must be a negative
    integer. 49. (x + 25)(x + 2) 51. (k - 21)(k + 3)

  8. (s + 5t)(s - 15f) 55. (x 6 + 7)(x 6 + 5)

  9. (r3 - 16)(r3 - 5) 59. (x 6 - 24)(x6 + 5) 61. C

  10. A 65. c 2 + 8 c + 16 66. 4v2 - 36v + 81

  11. 9w2 - 49 68. y 69. §r 70. mn-c 71. 7x

  12. 6 73. 3


Lesson 8-6 pp. 5 18-5 22

Go t It? 1a. (3x + 5)(2x + 1) b. The factors are both
negative. 2. (2x + 7)(5x - 2) 3. 2x + 3 and 4x + 5



  1. 4(2x + 1 )(x - 5)
    Lesso n Ch eck 1. (3x + 1 )(x + 5) 2. (5g + 2)(2g + 1)
    3. (2w - 1 )(2w + 3) 4. 3x + 8 and 2x - 9 5. There are
    no factors of 20 with sum 7. 6. 24 7. Answers may vary.
    Sample: If a = 1, you look for factors of c whose sum is
    b. If a + 1, you look for factors of ac whose sum is b.
    Ex e r c i se s 9. (3d + 2)(d + 7 ) 11. (4p + 3)(p + 1)
    13. (2g-3)(4g- 1) 15. {2k + 3){k - 8 )
    17. (3x — 4)(x + 9) 19. (2c/+5)(2d-7) 21. 5x+2
    and 3x - 4 23. 2{4v - 3){v + 5) 25. 5 (w - 2)(4w - 1)

  2. 3(3r- 5 ){r+ 2) 29-33. Answers may vary. Samples
    are given. 29. -31, (5v+ 3){3v- 8 ); 31,


(5v — 3)(3v + 8 ) 31. 20, (3g + 2)(3g + 2); 15,
(3g + 1 )(3g + 4) 33. 41, ( 8 r - 7){r + 6 ); - 5 ,
(8r-21)(r+ 2) 35. 6 x + 4 37a. (2x + 2)(x + 2);
(x + 1 )(2x + 4) b. yes c. Answers may vary. Sample:
Neither factoring is complete. Each one has a common
factor, 2. 39. 3(11k + 4)(2k + 1) 41. 28(b - 1)(/? + 2)


  1. (11 n- 6)(5n - 2) 45. (9g - 5)(7g - 6 ) 47. 2;
    explanations may vary. Sample: ax 2 + b x + c factors to
    (ax + 1 )(x + c) or (ax + c)(x + 1 ) so b = ac + 1 or
    b = a + c. 49.(7p - 3q)(7p + 12q) 51a. -2, -3
    b. (x + 2)(x + 3) c. Answers may vary. Sample: if you
    set each factor equal to 0 and solve the resulting
    equations, you get the x-intercepts.


Lesson 8-7 pp. 523-528
Go t It? 1a. (x + 3 )2 b. (x - 7 )2 2. 4m - 9
3a. {v - 1 0)(v +10) b. (s - 4)(s + 4)
4a. (5d + 8)(5d - 8) b. No; 2 5d2 + 64 is not a
difference of two squares. 5a. 12(f + 2)(f - 2)
b. 3(2x + 1)2
Lesso n Ch eck 1. (y - 8)2 2. (3q + 2)2


  1. (p + 6 )(p - 6 ) 4. 6w + 5 5. perfect-square trinomial

  2. perfect-square trinomial 7. difference of two squares

  3. In a difference of two squares, both terms are perfect
    squares separated by a subtraction symbol.
    Ex e r c i se s 9. (h + 4 )2 11.(d-10)2 13. (g+1)2

  4. (8x+7)2 17. (3n-7)2 19. (5z + 4)2
    21.10r— 11 23. 5c+3 25. (a + 7)(a - 7)

  5. (f + 5)(f - 5) 29. (m + 15)(m - 15)

  6. (9r + 1 )(9r - 1) 33. (8q + 9)(8q - 9)

  7. (3n + 20)(3n - 20) 37. 3(3w+ 2){3w-2)

  8. (x2)2 - (y2)2; (x - y)(x + y)(x^2 + y2)^41. Answers
    may vary. Sample: Rewrite the absolute value of both
    terms as squares. The factorization is the product of two
    binomials. The first is the sum of square roots of the
    squares. The second is the difference of the square roots
    of the squares. Example 1: x 2 - 4 = (x + 2)(x - 2);
    Example 2: 4y 2 - 25 = (2y + 5 )(2 y - 5)

  9. [1] S u b tract by com bining like term s.
    (49x 2 - 56x + 16) - (1 6x2 + 24x + 9) =
    (49x 2 - 16x2) + (—56x - 24x) + (16 - 9) =
    33x 2 - 80x + 7
    [2] Factor each expression, then use the rule for factoring
    the difference of two squares. (49x 2 - 56x + 1 6 ) -
    (1 6x2 + 24x + 9) = (7x - 4)2 - (4x + 3)2 =
    [(7x - 4) - (4x + 3)] - [(7x - 4) + (4x + 3)] =
    (3x— 7)(11x— 1) = 33x 2 — 80x + 7

  10. 11,9 47. 14, 6 49a. Answers may vary. Sample:
    x 2 + 6 x + 9 b. because the first term x 2 is a square, the
    last term 3 2 is a square, and the middle term is
    2(x)(3) 51. ( 8 r3 - 9)2 53. ( 6 m 2 + 7)2 55. (x 10 - 2y5)2


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