Polynomials: Factoring - XYZ Custom Plus

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776Chapter 12 Polynomials: Factoring43. 49a 7 − 9a 5 44. 8a 3 + 1 45. 49x 2 + 9y 246. 12x 4 − 62x 3 + 70x 2 47. 25a 3 + 20a 2 + 3a 48. 36a 4 − 100a 249. xa − xb + ay − by 50. xy − bx + ay − ab 51. 48a 4 b − 3a 2 b52. 18a 4 b 2 − 12a 3 b 3 + 8a 2 b 4 53. 20x 4 − 45x 2 54. 16x 3 + 16x 2 + 3x55. 3x 2 + 35xy − 82y 2 56. 3x 2 + 37xy − 86y 2 57. 16x 5 − 44x 4 + 30x 358. 16x 2 + 16x − 1 59. 2x 2 + 2ax + 3x + 3a 60. 2x 2 + 2ax + 5x + 5a61. y 4 − 1 62. a 7 + 8a 4 b 3 63. 12x 4 y 2 + 36x 3 y 3 + 27x 2 y 464. 16x 3 y 2 − 4xy 2 65. 16t 2 − 64t + 48 66. 16t 2 − 32t + 1267. 54x 2 + 111x + 56 68. 63x 2 + 110x + 48Getting Ready for the Next SectionSolve each equation.69. 3x − 6 = 9 70. 5x − 1 = 14 71. 2x + 3 = 072. 4x − 5 = 0 73. 4x + 3 = 0 74. 3x − 1 = 0Maintaining Your SkillsSolve each equation.75. −2(x + 4) = −10 76. 3 _4 ( −4x − 8) = 21 77. 3 _5 x + 4 = 22 78. −10 = 4 − 7 _4 x79. 6x − 4(9 − x) = −96 80. −2(x − 5) + 5x = 4 − 9 81. 2x − 3(4x − 7) = −3x 82. _3 4 (8 + x) = 1 _( 5 x − 15)583. 1 _2 x − 5 _12 = 1 _12 x + 5 _12 84. 3 _10 x + 5 _2 = 3 _5 x − 1 _2
Solving Equations by Factoring 12.7In this section we will use the methods of factoring developed in previous sections,along with a special property of 0, to solve quadratic equations.ObjectivesA Solve an equation by writing it instandard form and then factoring.DefinitionAny equation that can be put in the form ax 2 + bx + c = 0, where a, b, andc are real numbers (a ≠ 0), is called a quadratic equation. The equationax 2 + bx + c = 0 is called standard form for a quadratic equation:Examples now playing atMathTV.com/booksan x 2 term an x term and a constant terma(variable) 2 + b(variable) + c(absence of the variable) = 0The number 0 has a special property. If we multiply two numbers and the productis 0, then one or both of the original two numbers must be 0. In symbols, thisproperty looks like this.Zero-Factor PropertyLet a and b represent real numbers. If a ⋅ b = 0, then a = 0 or b = 0.Suppose we want to solve the quadratic equation x 2 + 5x + 6 = 0. We can factorthe left side into (x + 2)(x + 3). Then we have:x 2 + 5x + 6 = 0(x + 2)(x + 3) = 0Now, (x + 2) and (x + 3) both represent real numbers. Their product is 0; therefore,either (x + 3) is 0 or (x + 2) is 0. Either way we have a solution to our equation. Weuse the property of 0 stated to finish the problem:x 2 + 5x + 6 = 0(x + 2)(x + 3) = 0x + 2 = 0 or x + 3 = 0x = −2 or x = −3Our solution set is {−2, −3}. Our equation has two solutions. To check our solutionswe have to check each one separately to see that they both produce a truestatement when used in place of the variable:Whenx = −3the equation x 2 + 5x + 6 = 0becomes (−3) 2 + 5(−3) + 6 ≟ 09 + (−15) + 6 = 00 = 0Whenx = −2the equation x 2 + 5x + 6 = 0becomes (−2) 2 + 5(−2) + 6 ≟ 04 + (−10) + 6 = 00 = 0We have solved a quadratic equation by replacing it with two linear equations inone variable.12.7 Solving Equations by Factoring777
Page 1 and 2: Polynomials: Factoring12Introductio
Page 3 and 4: The Greatest Common Factorand Facto
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Page 7 and 8: 12.1 Problem Set745Problem Set 12.1
Page 9 and 10: 12.1 Problem Set747B Factor by grou
Page 11 and 12: 12.2Factoring TrinomialsIn this sec
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Page 17 and 18: 12.3More Trinomials to FactorWe now
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Page 37: 12.6 Problem Set775Problem Set 12.6
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Page 45 and 46: 12.7 Problem Set78358. (x + 7) 2 =
Page 47 and 48: Applications 12.8In this section we
Page 49 and 50: 12.8 Applications787Since the area
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Page 55 and 56: 12.8 Problem Set793Business Problem
Page 57: 12.8 Problem Set795Skyscrapers The
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