Appendices. Appendix A.1: Factoring Polynomials. Techniques for Factoring Trinomials Factorability Test for Trinomials:
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1 APPENDICES Appendices Appendi A.1: Factoring Polynomials Techniques for Factoring Trinomials Techniques for Factoring Trinomials Factorability Test for Trinomials: Eample: Solution: 696
2 APPENDIX A.1 Factoring Polynomials Factoring Trinomials with Leading Coefficient 1: MATH 10 Precalculus 697
3 APPENDICES 698
4 APPENDIX A.1 Factoring Polynomials Eample: Solution: MATH 10 Precalculus 699
5 APPENDICES Factoring Trinomials with Leading Coefficient Different from 1: 700
6 APPENDIX A.1 Factoring Polynomials MATH 10 Precalculus 701
7 APPENDICES Eample: Solution: Additional Eample 1: (a) (b)
8 APPENDIX A.1 Factoring Polynomials Solution: Additional Eample : Solution: MATH 10 Precalculus 70
9 APPENDICES Additional Eample : Solution: 704
10 APPENDIX A.1 Factoring Polynomials Additional Eample 4: Solution: MATH 10 Precalculus 705
11 APPENDICES Additional Eample 5: Solution: 706
12 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.) Factor the following polynomials. If the polynomial can not be rewritten as factors with integer coefficients, then write the original polynomial as your answer 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
13 Eercise Set A.1: Factoring Polynomials Factor the following polynomials. (Hint: Factor first by grouping, and then continue to factor if possible.)
14 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
15 APPENDICES 710
16 APPENDIX A. Dividing Polynomials MATH 10 Precalculus 711
17 APPENDICES 71
18 APPENDIX A. Dividing Polynomials MATH 10 Precalculus 71
19 APPENDICES Synthetic Division of Polynomials: Eample: Solution: 714
20 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 Constant: Notice the coefficients of the dividend:, 0, 8,, Synthetic Division Change the sign of the constant term when performing synthetic division Write the coefficients of the dividend (without changing any signs). Do not forget the placeholder for 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
21 APPENDICES In the long division problem, there is one column for each power of, and the arithmetic in each column is done with the coefficients 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
22 APPENDIX A. Dividing Polynomials Additional Eample 1: Solution: Additional Eample : Solution: MATH 10 Precalculus 717
23 APPENDICES Additional Eample : Solution: 718
24 APPENDIX A. Dividing Polynomials Additional Eample 4: Solution: MATH 10 Precalculus 719
25 APPENDICES 70
26 APPENDIX A. Dividing Polynomials Additional Eample 5: Solution: MATH 10 Precalculus 71
27 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 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 ( ) ; c 1 7
28 Eercise Set A.: Dividing Polynomials 9. P ( ) ; c P ( ) ; 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.) P ( ) ; c P ( ) ; c P ( ) 4 5 1; c P ( ) ; 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 , 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 MATH 10 Precalculus 7
29 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
30 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
31 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
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