APPENDICES Appendices Appendi A.1: Factoring Polynomials Techniques for Factoring Trinomials Techniques for Factoring Trinomials Factorability Test for Trinomials: Eample: Solution: 696
APPENDIX A.1 Factoring Polynomials Factoring Trinomials with Leading Coefficient 1: MATH 10 Precalculus 697
APPENDICES 698
APPENDIX A.1 Factoring Polynomials Eample: Solution: MATH 10 Precalculus 699
APPENDICES Factoring Trinomials with Leading Coefficient Different from 1: 700
APPENDIX A.1 Factoring Polynomials MATH 10 Precalculus 701
APPENDICES Eample: Solution: Additional Eample 1: (a) (b) 8 4 5 70
APPENDIX A.1 Factoring Polynomials Solution: Additional Eample : Solution: MATH 10 Precalculus 70
APPENDICES Additional Eample : Solution: 704
APPENDIX A.1 Factoring Polynomials Additional Eample 4: Solution: MATH 10 Precalculus 705
APPENDICES Additional Eample 5: Solution: 706
Eercise Set A.1: Factoring Polynomials At times, it can be difficult to tell whether or not a quadratic of the form a b c can be factored into the form d e f g, where a, b, c, d, e, f, and g are integers. If b 4ac is a perfect square, then the quadratic can be factored in the above manner. For each of the following problems, (a) Compute b 4ac. (b) Use the information from part (a) to determine whether or not the quadratic can be written as factors with integer coefficients. (Do not factor; simply answer Yes or No.) 1... 4. 5. 6. 7. 8. 9. 5 7 10 6 16 6 4 9 7 7 4 6 1 5 1... 4. 5. 6. 7. 8. 9. 0. 1... 4. 5. 6. 16 64 6 9 15 56 6 7 11 60 19 48 17 4 1 64 49 6 8 9 5 16 81 5 16 15 10. 5 4 1 7. 8 Factor the following polynomials. If the polynomial can not be rewritten as factors with integer coefficients, then write the original polynomial as your answer. 11. 1. 1. 14. 4 5 9 14 5 6 6 8. 9. 40. 41. 4. 4. 44. 4 16 15 9 9 4 5 17 6 4 10 9 1 10 1 17 6 8 6 7 15. 16. 17. 18. 19. 0. 7 1 8 15 1 0 7 18 5 4 9 6 Factor the following. Remember to first factor out the Greatest Common Factor (GCF) of the terms of the polynomial, and to factor out a negative if the leading coefficient is negative. MATH 10 Precalculus 707 45. 46. 47. 9 16 5 0
Eercise Set A.1: Factoring Polynomials 48. 49. 50. 51. 5. 5. 54. 55. 56. 57. 58. 59. 60. 61. 6. 6. 64. 65. 66. 4 8 18 8 8 4 5 0 75 10 8 1 6 10 10 40 4 40 100 9 7 6 4 4 5 4 10 1 4 6 6 80 5 9 100 1 10 49 64 50 55 15 0 4 7 Factor the following polynomials. (Hint: Factor first by grouping, and then continue to factor if possible.) 67. 68. 69. 70. 71. 7. 5 50 4 1 5 4 0 9 18 5 50 4 6 9 9 7 4 1 708
APPENDIX A. Dividing Polynomials Appendi A.: Dividing Polynomials Polynomial Long Division and Synthetic Division Polynomial Long Division and Synthetic Division Long Division of Polynomials: Eample: Solution: MATH 10 Precalculus 709
APPENDICES 710
APPENDIX A. Dividing Polynomials MATH 10 Precalculus 711
APPENDICES 71
APPENDIX A. Dividing Polynomials MATH 10 Precalculus 71
APPENDICES Synthetic Division of Polynomials: Eample: Solution: 714
APPENDIX A. Dividing Polynomials A Comparison of Long Division and Synthetic Division Let us now analyze the previous two eamples, both of which solved the same problem using long division and then synthetic division. Long Division 4 5 0 8 5 Constant: 5 4 5 0 8 5 Notice the coefficients of the dividend:, 0, 8,, 5 4 10 4 1 4 5 0 8 5 10 10 8 10 50 4 4 10 1 5 1 1065 1060 Synthetic Division 5 0 8 5 Change the sign of the constant term when performing synthetic division. 5 0 8 5 Write the coefficients of the dividend (without changing any signs). Do not forget the placeholder for 0. 5 0 8 5 10 50 10 1065 10 4 1 1060 Notice that the coefficients in each column of the subtraction problems under the division sign (at the left) are similar to the numbers in each column of the synthetic division problem (above). Remember that at the left, the signs are changed when the epressions are subtracted. MATH 10 Precalculus 715
APPENDICES 10 4 1 4 5 0 8 5 4 10 10 8 10 50 In the long division problem, there is one column for each power of, and the arithmetic in each column is done with the coefficients. 4 4 10 1 5 1 1065 1060 5 0 8 5 10 50 10 1065 10 4 1 100 6 Notice that the numbers in the answer line of the synthetic division problem are the same as the coefficients of the quotient plus the final remainder in the long division problem. Synthetic division is a shortcut for doing the arithmetic with the coefficients without having to write down all the variables. Remember that this synthetic division procedure ONLY works when the divisor is of the form D c. The Remainder Theorem: 716
APPENDIX A. Dividing Polynomials Additional Eample 1: Solution: Additional Eample : Solution: MATH 10 Precalculus 717
APPENDICES Additional Eample : Solution: 718
APPENDIX A. Dividing Polynomials Additional Eample 4: Solution: MATH 10 Precalculus 719
APPENDICES 70
APPENDIX A. Dividing Polynomials Additional Eample 5: Solution: MATH 10 Precalculus 71
Eercise Set A.: Dividing Polynomials Use long division to find the quotient and the remainder. Use synthetic division to find the quotient and the remainder. 1. 611 15. 84 10. 51 16. 4 6. 7 1 17. 1 6 8 5 4. 65 4 18. 1 4 5. 19 1 19. 4 4 1 6. 5 0. 5 4 7 8 1 7. 6 5 6 1 1 1. 4 11 7 5 18 10 8. 1 1 14 4. 4 18 5 1 9. 1 8 1 1. 8 10. 4 7 4 4 1 7 4. 4 81 11. 5 4 44 14 4 6 5. 4 7 5 1 1. 8 6 4 10 0 8 4 4 6 6. 4 6 10 9 1 1. 14. 4 15 5 5 4 7 Evaluate P(c) using the following two methods: (a) Substitute c into the function. (b) Use synthetic division along with the Remainder Theorem. 7. P ( ) 4 5 ; c 8. P ( ) 5 7 8 ; c 1 7
Eercise Set A.: Dividing Polynomials 9. P ( ) 7 8 4 1; c 1 4 0. P ( ) 7 6 14; c Evaluate P(c) using synthetic division along with the Remainder Theorem. (Notice that substitution without a calculator would be quite tedious in these eamples, so synthetic division is particularly useful.) 41. 4. 75 1 5 41 7 6 1. P ( ) 8 8 70 1 ; c 5 5 6 5. P ( ) 10 5 11; c 4. 4. 4 P ( ) 4 5 1; c 6 5 4 P ( ) 6 19 59 1; c 4 7 When the remainder is zero, the dividend can be written as a product of two factors (the divisor and the quotient), as shown below. 0 6, so 0 5 6 5. 6, so 6 In the following eamples, use either long division or synthetic division to find the quotient, and then write the dividend as a product of two factors. 5. 6. 7. 8. 9. 40. 114 8 40 5 718 101 4 51 7 4 6 MATH 10 Precalculus 7
APPENDICES Appendi A.: Geometric Formulas Geometric Formulas Geometric Formulas The following two pages contain geometric formulas which may be helpful to you in this course. 74
Appendi A.: Geometric Formulas Rectangle Perimeter: P w Area: A w w Trapezoid Perimeter: Area: Add the side lengths A b b h 1 h b 1 b Square Perimeter: P 4s Area: A s Parallelogram Perimeter: Area: A bh Add the side lengths b Triangle Perimeter: Area: Add the side lengths bh A Equilateral Triangle Perimeter: P s Area: s A 4 Circle Circumference: C r d Area: A r d h h b s r s Right Circular Cylinder Lateral Area: L rh Ch Total Surface Area: S L B, where B represents the area of the base, so S rh r Volume: V Bh r h Right Circular Cone Lateral Area: L r C Total Surface Area: S L B, where B represents the area of the base, so S r r Bh r h Volume: V Rectangular Prism Lateral Area: L h wh Ph, where P represents the perimeter of the base. Total Surface Area: S L B, where B represents the area of the base, so S h wh w Volume: V Bh wh Sphere Surface Area: Volume: S 4 r 4 r V r h w r r h h MATH 10 Precalculus 75
Appendi A.: Geometric Formulas Pythagorean Theorem a b c a b Distance Formula Distance between the points y 1 1 d y y y 1, 1 d y, 1, 1 c y and, y : 0 o -60 o -90 o Triangle In a 0 o -60 o -90 o triangle, the length of the hypotenuse is twice the length of the shorter leg, and the length of the longer leg is times the length of the shorter leg. 45 o -45 o -90 o Triangle In a 45 o -45 o -90 o triangle, the legs are congruent, and the length of the 45 o hypotenuse is times the length of either leg. 45 o 60 o 0 o Midpoint Formula Midpoint of the segment joining the points 1, y 1 and, y : 1 1,, M y y y y y, y 1, 1 M, y 76