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# Notes solving polynomials using synthetic division

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• ### Notes solving polynomials using synthetic division

1. 1. Solving Polynomials Using Synthetic Division
2. 2. Simplify Polynomials using Synthetic Substitution• Recall Synthetic Division can be used to divide a polynomial by a binomial.
3. 3. Simplify Polynomials using Synthetic Substitution• Recall Synthetic Division can be used to divide a polynomial by a binomial.• Synthetic Division can also be used to evaluate a polynomial at a speciﬁc value.
4. 4. Simplify Polynomials using Synthetic Substitution• Recall Synthetic Division can be used to divide a polynomial by a binomial.• Synthetic Division can also be used to evaluate a polynomial at a speciﬁc value.• Such as given f ( x ) = x − x + 5 , ﬁnd f ( −2 ) 2
5. 5. Simplify Polynomials using Synthetic Substitution• Recall Synthetic Division can be used to divide a polynomial by a binomial.• Synthetic Division can also be used to evaluate a polynomial at a speciﬁc value.• Such as given f ( x ) = x − x + 5 , ﬁnd f ( −2 ) 2• The x value goes in the box and the coefﬁcients of the polynomial make up the ﬁrst line. Proceed as usual.
6. 6. Simplify Polynomials using Synthetic Substitution• Recall Synthetic Division can be used to divide a polynomial by a binomial.• Synthetic Division can also be used to evaluate a polynomial at a speciﬁc value.• Such as given f ( x ) = x − x + 5 , ﬁnd f ( −2 ) 2• The x value goes in the box and the coefﬁcients of the polynomial make up the ﬁrst line. Proceed as usual. −2 1 −1 5 −2 6 1 −3 11
7. 7. Simplify Polynomials using Synthetic Substitution• Recall Synthetic Division can be used to divide a polynomial by a binomial.• Synthetic Division can also be used to evaluate a polynomial at a speciﬁc value.• Such as given f ( x ) = x − x + 5 , ﬁnd f ( −2 ) 2• The x value goes in the box and the coefﬁcients of the polynomial make up the ﬁrst line. Proceed as usual.• The remainder is the value. −2 1 −1 5 −2 6 1 −3 11
8. 8. Simplify Polynomials using Synthetic Substitution• Recall Synthetic Division can be used to divide a polynomial by a binomial.• Synthetic Division can also be used to evaluate a polynomial at a speciﬁc value.• Such as given f ( x ) = x − x + 5 , ﬁnd f ( −2 ) 2• The x value goes in the box and the coefﬁcients of the polynomial make up the ﬁrst line. Proceed as usual.• The remainder is the value. −2 1 −1 5• So −2 6 f ( −2 ) = 11 1 −3 11
9. 9. Try it: Simplify using Synthetic SubstitutionFind g ( 5 ) for g ( x ) = −x + x − 7x + 8 3 2
10. 10. Try it: Simplify using Synthetic SubstitutionFind g ( 5 ) for g ( x ) = −x + x − 7x + 8 3 2 5 −1 1 −7 8 −5 −20 −135 −1 −4 −27 −127
11. 11. Try it: Simplify using Synthetic SubstitutionFind g ( 5 ) for g ( x ) = −x + x − 7x + 8 3 2 5 −1 1 −7 8 −5 −20 −135 −1 −4 −27 −127 Therefore, g ( 5 ) = −127.
12. 12. Try it: Simplify using Synthetic SubstitutionFind f ( −3) for g ( x ) = x − 2x − 3x + 5x − 6 4 3 2
13. 13. Try it: Simplify using Synthetic SubstitutionFind f ( −3) for g ( x ) = x − 2x − 3x + 5x − 6 4 3 2 −3 1 −2 −3 5 −6 −3 15 −36 93 1 −5 12 −31 87
14. 14. Try it: Simplify using Synthetic SubstitutionFind f ( −3) for g ( x ) = x − 2x − 3x + 5x − 6 4 3 2 −3 1 −2 −3 5 −6 −3 15 −36 93 1 −5 12 −31 87 Therefore, f ( −3) = 87.
15. 15. Try it: Simplify using Synthetic SubstitutionFind h ( 2b ) for h ( x ) = x + 3x − 2x + 1 3 2Don’t be intimidated by the 2b. Use the sameprocess as before.
16. 16. Try it: Simplify using Synthetic SubstitutionFind h ( 2b ) for h ( x ) = x + 3x − 2x + 1 3 2Don’t be intimidated by the 2b. Use the sameprocess as before. 2b 1 3 −2 1 2b 4b 2 + 6b 8b 3 + 12b 2 − 4b 2 3 2 1 2b + 3 4b + 6b − 2 8b + 12b − 4b + 1
17. 17. Try it: Simplify using Synthetic SubstitutionFind h ( 2b ) for h ( x ) = x + 3x − 2x + 1 3 2Don’t be intimidated by the 2b. Use the sameprocess as before. 2b 1 3 −2 1 2b 4b 2 + 6b 8b 3 + 12b 2 − 4b 2 3 2 1 2b + 3 4b + 6b − 2 8b + 12b − 4b + 1 Therefore, h ( 2b ) = 8b 3 + 12b 2 − 4b + 1.
18. 18. Find Factors given a Polynomial & one factor• Synthetic Division can be useful in factoring polynomials that cannot be factored using traditional methods.
19. 19. Find Factors given a Polynomial & one factor• Synthetic Division can be useful in factoring polynomials that cannot be factored using traditional methods.• It’s a great way to ‘break down’ the big, nasty looking polynomials.
20. 20. Find Factors given a Polynomial & one factor• Synthetic Division can be useful in factoring polynomials that cannot be factored using traditional methods.• It’s a great way to ‘break down’ the big, nasty looking polynomials.• Care to factor this polynomial? 3 2 x − 3x − 13x + 15
21. 21. Factoring when know 1 factor 3 2 x − 3x − 13x + 15• Say we know one factor of this nasty polynomial is (x + 3) or the zero of the function is x = -3. We can use synthetic division to help us ﬁnd the remaining factors.
22. 22. Factoring when know 1 factor 3 2 x − 3x − 13x + 15• Say we know one factor of this nasty polynomial is (x + 3) or the zero of the function is x = -3. We can use synthetic division to help us ﬁnd the remaining factors.• First step is to divide by the factor you know.
23. 23. Factoring when know 1 factor 3 2 x − 3x − 13x + 15• Say we know one factor of this nasty polynomial is (x + 3) or the zero of the function is x = -3. We can use synthetic division to help us ﬁnd the remaining factors.• First step is to divide by the factor you know. −3 1 −3 −13 15 −3 18 −15 1 −6 5 0
24. 24. Factoring when know 1 factor 3 2 x − 3x − 13x + 15• Say we know one factor of this nasty polynomial is (x + 3) or the zero of the function is x = -3. We can use synthetic division to help us ﬁnd the remaining factors.• First step is to divide by the factor you know. −3 1 −3 −13 15 −3 18 −15 1 −6 5 0• Remainder is zero so this indicates that (x + 3) is indeed a zero of the polynomial.
25. 25. Continued... −3 1 −3 −13 15 −3 18 −15 1 −6 5 0• Write the new polynomial from your synthetic division.• Look, we now have a nice quadratic. Use any method you’ve learned to factor the quadratic.
26. 26. Continued... −3 1 −3 −13 15 −3 18 −15 1 −6 5 0• Write the new polynomial from your synthetic division. 2 x − 6x + 5• Look, we now have a nice quadratic. Use any method you’ve learned to factor the quadratic.
27. 27. Continued... −3 1 −3 −13 15 −3 18 −15 1 −6 5 0• Write the new polynomial from your synthetic division. 2 x − 6x + 5• Look, we now have a nice quadratic. Use any method you’ve learned to factor the quadratic.
28. 28. Continued... −3 1 −3 −13 15 −3 18 −15 1 −6 5 0• Write the new polynomial from your synthetic division. 2 x − 6x + 5• Look, we now have a nice quadratic. Use any method you’ve learned to factor the quadratic. ( x − 5 ) ( x − 1)
29. 29. Continued... 3 2• So the factorization of x − 3x − 13x + 15 is ( x + 3) ( x − 5 ) ( x − 1)
30. 30. Continued... 3 2• So the factorization of x − 3x − 13x + 15 is ( x + 3) ( x − 5 ) ( x − 1)• When writing the factorization, be sure to include the ﬁrst factor you were given because without it, you do not have the original polynomial!
31. 31. Continued... 3 2Now find the zeros of x − 3x − 13x + 15 = 0.
32. 32. Continued... 3 2Now find the zeros of x − 3x − 13x + 15 = 0.Remember the factors are ... ( x + 3) ( x − 5 ) ( x − 1)
33. 33. Continued... 3 2Now find the zeros of x − 3x − 13x + 15 = 0.Remember the factors are ... ( x + 3) ( x − 5 ) ( x − 1)Set each factor to 0 and solve.
34. 34. Continued... 3 2Now find the zeros of x − 3x − 13x + 15 = 0.Remember the factors are ... ( x + 3) ( x − 5 ) ( x − 1)Set each factor to 0 and solve. x+3= 0 x−5=0 x −1= 0
35. 35. Continued... 3 2Now find the zeros of x − 3x − 13x + 15 = 0.Remember the factors are ... ( x + 3) ( x − 5 ) ( x − 1)Set each factor to 0 and solve. x+3= 0 x−5=0 x −1= 0 x = −3
36. 36. Continued... 3 2Now find the zeros of x − 3x − 13x + 15 = 0.Remember the factors are ... ( x + 3) ( x − 5 ) ( x − 1)Set each factor to 0 and solve. x+3= 0 x−5=0 x −1= 0 x = −3 x=5
37. 37. Continued... 3 2Now find the zeros of x − 3x − 13x + 15 = 0.Remember the factors are ... ( x + 3) ( x − 5 ) ( x − 1)Set each factor to 0 and solve. x+3= 0 x−5=0 x −1= 0 x = −3 x=5 x =1
38. 38. Continued... 3 2Now find the zeros of x − 3x − 13x + 15 = 0.Remember the factors are ... ( x + 3) ( x − 5 ) ( x − 1)Set each factor to 0 and solve. x+3= 0 x−5=0 x −1= 0 x = −3 x=5 x =1 The zeros are {-3, 1, 5}.
39. 39. Your turn...Show ( x − 3) is a factor of x + 4x − 15x − 18 3 2
40. 40. Your turn...Show ( x − 3) is a factor of x + 4x − 15x − 18 3 2 3 1 4 −15 −18 3 21 18 1 7 6 0 Remainder is 0, therefor x - 3 is a factor.
41. 41. Now take the last problem a little further... 3 2Factor x + 4x − 15x − 18Remember you were given a factor.
42. 42. Now take the last problem a little further... 3 2Factor x + 4x − 15x − 18Remember you were given a factor. 3 1 4 −15 −18 3 21 18 1 7 6 0
43. 43. Now take the last problem a little further... 3 2Factor x + 4x − 15x − 18Remember you were given a factor. 3 1 4 −15 −18 3 21 18 1 7 6 0 Write the new polynomial.
44. 44. Now take the last problem a little further... 3 2Factor x + 4x − 15x − 18Remember you were given a factor. 3 1 4 −15 −18 3 21 18 1 7 6 0 2 Write the new polynomial. x + 7x + 6
45. 45. Now take the last problem a little further... 3 2Factor x + 4x − 15x − 18Remember you were given a factor. 3 1 4 −15 −18 3 21 18 1 7 6 0 2 Write the new polynomial. x + 7x + 6 It’s a quadratic. Factor using any method you like.
46. 46. Now take the last problem a little further... 3 2Factor x + 4x − 15x − 18Remember you were given a factor. 3 1 4 −15 −18 3 21 18 1 7 6 0 2 Write the new polynomial. x + 7x + 6 It’s a quadratic. Factor using any method you like. ( x + 6 ) ( x + 1)
47. 47. Now take the last problem a little further... 3 2Factor x + 4x − 15x − 18Remember you were given a factor. 3 1 4 −15 −18 3 21 18 1 7 6 0 2 Write the new polynomial. x + 7x + 6 It’s a quadratic. Factor using any method you like. ( x + 6 ) ( x + 1)x + 4x − 15x − 18 = ( x − 3) ( x + 6 ) ( x + 1) 3 2
48. 48. And a little further... 3 2Find the zeros of x + 4x − 15x − 18
49. 49. And a little further... 3 2Find the zeros of x + 4x − 15x − 18Remember ... x + 4x − 15x − 18 = ( x − 3) ( x + 6 ) ( x + 1) 3 2
50. 50. And a little further... 3 2Find the zeros of x + 4x − 15x − 18Remember ... x + 4x − 15x − 18 = ( x − 3) ( x + 6 ) ( x + 1) 3 2Set each factor to 0 and solve.
51. 51. And a little further... 3 2Find the zeros of x + 4x − 15x − 18Remember ... x + 4x − 15x − 18 = ( x − 3) ( x + 6 ) ( x + 1) 3 2Set each factor to 0 and solve. x−3= 0 x+6=0 x +1= 0
52. 52. And a little further... 3 2Find the zeros of x + 4x − 15x − 18Remember ... x + 4x − 15x − 18 = ( x − 3) ( x + 6 ) ( x + 1) 3 2Set each factor to 0 and solve. x−3= 0 x+6=0 x +1= 0 x=3
53. 53. And a little further... 3 2Find the zeros of x + 4x − 15x − 18Remember ... x + 4x − 15x − 18 = ( x − 3) ( x + 6 ) ( x + 1) 3 2Set each factor to 0 and solve. x−3= 0 x+6=0 x +1= 0 x=3 x = −6
54. 54. And a little further... 3 2Find the zeros of x + 4x − 15x − 18Remember ... x + 4x − 15x − 18 = ( x − 3) ( x + 6 ) ( x + 1) 3 2Set each factor to 0 and solve. x−3= 0 x+6=0 x +1= 0 x=3 x = −6 x = −1
55. 55. And a little further... 3 2Find the zeros of x + 4x − 15x − 18Remember ... x + 4x − 15x − 18 = ( x − 3) ( x + 6 ) ( x + 1) 3 2Set each factor to 0 and solve. x−3= 0 x+6=0 x +1= 0 x=3 x = −6 x = −1 The zeros are {-6, -1, 3}.
56. 56. Practice some more...• Follow this link to practice using synthetic division to ﬁnd the zeros of the polynomials.
57. 57. Rational Roots Theorem• Watch this video to learn more about the Rational Roots Theorem and how to use it to ﬁnd any possible rational roots of a polynomial.• Aquí está una versión española del vídeo racional del teorema de las raíces.
58. 58. Factoring when you don’t know a factor
59. 59. Factoring when you don’t know a factor• The previous problems weren’t too bad because you knew a factor.
60. 60. Factoring when you don’t know a factor• The previous problems weren’t too bad because you knew a factor.• The video you just watched showed you how to determine if a polynomial has a rational root.
61. 61. Factoring when you don’t know a factor• The previous problems weren’t too bad because you knew a factor.• The video you just watched showed you how to determine if a polynomial has a rational root.• Now we will use this test to ﬁnd all factors of a polynomial.
62. 62. Factoring when you don’t know a factor• The previous problems weren’t too bad because you knew a factor.• The video you just watched showed you how to determine if a polynomial has a rational root.• Now we will use this test to ﬁnd all factors of a polynomial.• Watch this video to see how the Rational Roots Theorem is applied to ﬁnd the solutions of a polynomial.• Aquí está una versión española para aplicar el teorema racional de las raíces para encontrar las soluciones de un polinomio.
63. 63. Try one:Solve using the Rational Roots Theorem 3 2 x + 6x + 3x − 10 = 0
64. 64. Try one:Solve using the Rational Roots Theorem 3 2 x + 6x + 3x − 10 = 0Find factors of leading coefﬁcient.
65. 65. Try one:Solve using the Rational Roots Theorem 3 2 x + 6x + 3x − 10 = 0Find factors of leading coefﬁcient. ±1
66. 66. Try one:Solve using the Rational Roots Theorem 3 2 x + 6x + 3x − 10 = 0Find factors of leading coefﬁcient. ±1Find factors of constant term.
67. 67. Try one:Solve using the Rational Roots Theorem 3 2 x + 6x + 3x − 10 = 0Find factors of leading coefﬁcient. ±1Find factors of constant term. ±1, ±2, ±5, ±10
68. 68. Try one:Solve using the Rational Roots Theorem 3 2 x + 6x + 3x − 10 = 0Find factors of leading coefﬁcient. ±1Find factors of constant term. ±1, ±2, ±5, ±10Possible roots are made up of theconstant term factors over thefactors of the leading coefﬁcient.No need to repeat numbers andalways simplify.
69. 69. Try one:Solve using the Rational Roots Theorem 3 2 x + 6x + 3x − 10 = 0Find factors of leading coefﬁcient. ±1Find factors of constant term. ±1, ±2, ±5, ±10Possible roots are made up of theconstant term factors over the ±1, ±2, ±5, ±10factors of the leading coefﬁcient.No need to repeat numbers andalways simplify.
70. 70. Try one:Solve using the Rational Roots Theorem 3 2 x + 6x + 3x − 10 = 0Find factors of leading coefﬁcient. ±1Find factors of constant term. ±1, ±2, ±5, ±10Possible roots are made up of theconstant term factors over the ±1, ±2, ±5, ±10factors of the leading coefﬁcient.No need to repeat numbers andalways simplify.Lots from which to choose butlets start with an easy one.
71. 71. Try one:Solve using the Rational Roots Theorem 3 2 x + 6x + 3x − 10 = 0Find factors of leading coefﬁcient. ±1Find factors of constant term. ±1, ±2, ±5, ±10Possible roots are made up of theconstant term factors over the ±1, ±2, ±5, ±10factors of the leading coefﬁcient.No need to repeat numbers andalways simplify. f ( x ) = x 3 + 6x 2 + 3x − 10Lots from which to choose butlets start with an easy one. 3 2 f (1) = (1) + 6 (1) + 3(1) − 10 = 0
72. 72. Continued...Solution was 0 so we lucked out on our ﬁrst try.Use Synthetic Division to simplify the polynomial.
73. 73. Continued...Solution was 0 so we lucked out on our ﬁrst try.Use Synthetic Division to simplify the polynomial. 1 1 6 3 −10 1 7 10 1 7 10 0
74. 74. Continued...Solution was 0 so we lucked out on our ﬁrst try.Use Synthetic Division to simplify the polynomial. 1 1 6 3 −10 1 7 10 1 7 10 0Write the new polynomial.
75. 75. Continued...Solution was 0 so we lucked out on our ﬁrst try.Use Synthetic Division to simplify the polynomial. 1 1 6 3 −10 1 7 10 1 7 10 0Write the new polynomial. 2 x + 7x + 10 = 0
76. 76. Continued...Solution was 0 so we lucked out on our ﬁrst try.Use Synthetic Division to simplify the polynomial. 1 1 6 3 −10 1 7 10 1 7 10 0Write the new polynomial. 2 x + 7x + 10 = 0Look at the trinomial. Can it be factored?
77. 77. Continued... 2 x + 7x + 10 = 0Yes! It’s a factorable quadratic so use any methodyou would like to factor.
78. 78. Continued... 2 x + 7x + 10 = 0Yes! It’s a factorable quadratic so use any methodyou would like to factor. ( x + 5 )( x + 2 ) = 0
79. 79. Continued... 2 x + 7x + 10 = 0Yes! It’s a factorable quadratic so use any methodyou would like to factor. ( x + 5 )( x + 2 ) = 0Now set each factor to 0 and solve.
80. 80. Continued... 2 x + 7x + 10 = 0Yes! It’s a factorable quadratic so use any methodyou would like to factor. ( x + 5 )( x + 2 ) = 0Now set each factor to 0 and solve. ( x + 5) = 0 x = −5
81. 81. Continued... 2 x + 7x + 10 = 0Yes! It’s a factorable quadratic so use any methodyou would like to factor. ( x + 5 )( x + 2 ) = 0Now set each factor to 0 and solve. ( x + 5) = 0 ( x + 2) = 0 x = −5 x = −2
82. 82. Continued... 2 x + 7x + 10 = 0Yes! It’s a factorable quadratic so use any methodyou would like to factor. ( x + 5 )( x + 2 ) = 0Now set each factor to 0 and solve. ( x + 5) = 0 ( x + 2) = 0 x = −5 x = −2 Solutions: {−5, −2,1}
83. 83. Try another:Solve using the Rational Roots Theorem 3 2 5x − 29x + 55x − 28 = 0
84. 84. Try another:Solve using the Rational Roots Theorem 3 2 5x − 29x + 55x − 28 = 0Find factors of leading coefﬁcient.
85. 85. Try another:Solve using the Rational Roots Theorem 3 2 5x − 29x + 55x − 28 = 0Find factors of leading coefﬁcient. ±1, ±5
86. 86. Try another:Solve using the Rational Roots Theorem 3 2 5x − 29x + 55x − 28 = 0Find factors of leading coefﬁcient. ±1, ±5Find factors of constant term.
87. 87. Try another:Solve using the Rational Roots Theorem 3 2 5x − 29x + 55x − 28 = 0Find factors of leading coefﬁcient. ±1, ±5Find factors of constant term. ±1, ±2, ±4, ±7, ±14, ±28
88. 88. Try another:Solve using the Rational Roots Theorem 3 2 5x − 29x + 55x − 28 = 0Find factors of leading coefﬁcient. ±1, ±5Find factors of constant term. ±1, ±2, ±4, ±7, ±14, ±28Possible roots are leading coefﬁcient factors overconstant term factors without repeats.
89. 89. Try another:Solve using the Rational Roots Theorem 3 2 5x − 29x + 55x − 28 = 0Find factors of leading coefﬁcient. ±1, ±5Find factors of constant term. ±1, ±2, ±4, ±7, ±14, ±28Possible roots are leading coefﬁcient factors overconstant term factors without repeats. ±1, ±2, ±4, ±7, ±14, ±28 1 2 4 7 14 28 ± ,± ,± ,± ,± ,± 5 5 5 5 5 5
90. 90. Try another:Solve using the Rational Roots Theorem 3 2 5x − 29x + 55x − 28 = 0Find factors of leading coefﬁcient. ±1, ±5Find factors of constant term. ±1, ±2, ±4, ±7, ±14, ±28Possible roots are leading coefﬁcient factors overconstant term factors without repeats. ±1, ±2, ±4, ±7, ±14, ±28 1 2 4 7 14 28 ± ,± ,± ,± ,± ,± 5 5 5 5 5 5Wow, that’s a lot! Not to worry. Start with the smallestpositive integer and look for a pattern to help ﬁnd a root.
91. 91. Continued... ±1, ±2, ±4, ±7, ±14, ±28 1 2 4 7 14 28 f ( x ) = 5x 3 − 29x 2 + 55x − 28 ± ,± ,± ,± ,± ,± 5 5 5 5 5 5Start with 1.
92. 92. Continued... ±1, ±2, ±4, ±7, ±14, ±28 1 2 4 7 14 28 f ( x ) = 5x 3 − 29x 2 + 55x − 28 ± ,± ,± ,± ,± ,± 5 5 5 5 5 5Start with 1. 3 2 f (1) = 5 (1) − 29 (1) + 55 (1) − 28 = 3
93. 93. Continued... ±1, ±2, ±4, ±7, ±14, ±28 1 2 4 7 14 28 f ( x ) = 5x 3 − 29x 2 + 55x − 28 ± ,± ,± ,± ,± ,± 5 5 5 5 5 5Start with 1. 3 2 f (1) = 5 (1) − 29 (1) + 55 (1) − 28 = 3Didn’t work so try the next highest integer.
94. 94. Continued... ±1, ±2, ±4, ±7, ±14, ±28 1 2 4 7 14 28 f ( x ) = 5x 3 − 29x 2 + 55x − 28 ± ,± ,± ,± ,± ,± 5 5 5 5 5 5Start with 1. 3 2 f (1) = 5 (1) − 29 (1) + 55 (1) − 28 = 3Didn’t work so try the next highest integer. 3 2 f ( 2 ) = 5 ( 2 ) − 29 ( 2 ) + 55 ( 2 ) − 28 = 6
95. 95. Continued... ±1, ±2, ±4, ±7, ±14, ±28 1 2 4 7 14 28 f ( x ) = 5x 3 − 29x 2 + 55x − 28 ± ,± ,± ,± ,± ,± 5 5 5 5 5 5Start with 1. 3 2 f (1) = 5 (1) − 29 (1) + 55 (1) − 28 = 3Didn’t work so try the next highest integer. 3 2 f ( 2 ) = 5 ( 2 ) − 29 ( 2 ) + 55 ( 2 ) − 28 = 6No good again because but notice our remainder is gettinglarger. This indicates we are moving away from 0 so we aregoing in the wrong direction. Try a negative number.
96. 96. Continued... ±1, ±2, ±4, ±7, ±14, ±28 1 2 4 7 14 28 f ( x ) = 5x 3 − 29x 2 + 55x − 28 ± ,± ,± ,± ,± ,± 5 5 5 5 5 5Start with 1. 3 2 f (1) = 5 (1) − 29 (1) + 55 (1) − 28 = 3Didn’t work so try the next highest integer. 3 2 f ( 2 ) = 5 ( 2 ) − 29 ( 2 ) + 55 ( 2 ) − 28 = 6No good again because but notice our remainder is gettinglarger. This indicates we are moving away from 0 so we aregoing in the wrong direction. Try a negative number. 3 2 f ( −1) = 5 ( −1) − 29 ( −1) + 55 ( −1) − 28 = −117
97. 97. Continued... ±1, ±2, ±4, ±7, ±14, ±28 1 2 4 7 14 28 f ( x ) = 5x 3 − 29x 2 + 55x − 28 ± ,± ,± ,± ,± ,± 5 5 5 5 5 5Start with 1. 3 2 f (1) = 5 (1) − 29 (1) + 55 (1) − 28 = 3Didn’t work so try the next highest integer. 3 2 f ( 2 ) = 5 ( 2 ) − 29 ( 2 ) + 55 ( 2 ) − 28 = 6No good again because but notice our remainder is gettinglarger. This indicates we are moving away from 0 so we aregoing in the wrong direction. Try a negative number. 3 2 f ( −1) = 5 ( −1) − 29 ( −1) + 55 ( −1) − 28 = −117Now it’s way too small. Remember we are try to get 0.
98. 98. Continued... ±1, ±2, ±4, ±7, ±14, ±28 1 2 4 7 14 28 f ( x ) = 5x 3 − 29x 2 + 55x − 28 ± ,± ,± ,± ,± ,± 5 5 5 5 5 5 3 2 f (1) = 5 (1) − 29 (1) + 55 (1) − 28 = 3 3 2f ( −1) = 5 ( −1) − 29 ( −1) + 55 ( −1) − 28 = −117Because 0 falls between 3 and -117, we know theroot must fall between -1 and 1.
99. 99. Continued... ±1, ±2, ±4, ±7, ±14, ±28 1 2 4 7 14 28 f ( x ) = 5x 3 − 29x 2 + 55x − 28 ± ,± ,± ,± ,± ,± 5 5 5 5 5 5 3 2 f (1) = 5 (1) − 29 (1) + 55 (1) − 28 = 3 3 2 f ( −1) = 5 ( −1) − 29 ( −1) + 55 ( −1) − 28 = −117Because 0 falls between 3 and -117, we know theroot must fall between -1 and 1.Yes, that means we now get to try a fraction!
100. 100. Continued... ±1, ±2, ±4, ±7, ±14, ±28 1 2 4 7 14 28 f ( x ) = 5x 3 − 29x 2 + 55x − 28 ± ,± ,± ,± ,± ,± 5 5 5 5 5 5 3 2 f (1) = 5 (1) − 29 (1) + 55 (1) − 28 = 3 3 2 f ( −1) = 5 ( −1) − 29 ( −1) + 55 ( −1) − 28 = −117Because 0 falls between 3 and -117, we know theroot must fall between -1 and 1.Yes, that means we now get to try a fraction! 3 2  1  1  1  1 f   = 5   − 29   + 55   − 28 = −243 25  5  5  5  5
101. 101. Continued... ±1, ±2, ±4, ±7, ±14, ±28 1 2 4 7 14 28 f ( x ) = 5x 3 − 29x 2 + 55x − 28 ± ,± ,± ,± ,± ,± 5 5 5 5 5 5 3 2 f (1) = 5 (1) − 29 (1) + 55 (1) − 28 = 3 3 2 f ( −1) = 5 ( −1) − 29 ( −1) + 55 ( −1) − 28 = −117Because 0 falls between 3 and -117, we know theroot must fall between -1 and 1.Yes, that means we now get to try a fraction! 3 2  1  1  1  1 f   = 5   − 29   + 55   − 28 = −243 25  5  5  5  5No good but it’s negative. This means the root is between1/5 and 1.
102. 102. Continued... ±1, ±2, ±4, ±7, ±14, ±28 1 2 4 7 14 28f ( x ) = 5x 3 − 29x 2 + 55x − 28 ± ,± ,± ,± ,± ,± 5 5 5 5 5 5 3 2  4  4  4  4 f   = 5   − 29   + 55   − 28 = 0  5  5  5  5
103. 103. Continued... ±1, ±2, ±4, ±7, ±14, ±28 1 2 4 7 14 28 f ( x ) = 5x 3 − 29x 2 + 55x − 28 ± ,± ,± ,± ,± ,± 5 5 5 5 5 5 3 2  4  4  4  4 f   = 5   − 29   + 55   − 28 = 0  5  5  5  5Found it! So now use Synthetic Division to simplify thepolynomial.
104. 104. Continued... ±1, ±2, ±4, ±7, ±14, ±28 1 2 4 7 14 28 f ( x ) = 5x 3 − 29x 2 + 55x − 28 ± ,± ,± ,± ,± ,± 5 5 5 5 5 5 3 2  4  4  4  4 f   = 5   − 29   + 55   − 28 = 0  5  5  5  5Found it! So now use Synthetic Division to simplify thepolynomial. 4 5 5 −29 55 −28 4 −20 28 5 −25 35 0
105. 105. Continued... ±1, ±2, ±4, ±7, ±14, ±28 1 2 4 7 14 28 f ( x ) = 5x 3 − 29x 2 + 55x − 28 ± ,± ,± ,± ,± ,± 5 5 5 5 5 5 3 2  4  4  4  4 f   = 5   − 29   + 55   − 28 = 0  5  5  5  5Found it! So now use Synthetic Division to simplify thepolynomial. 4 5 5 −29 55 −28 4 −20 28 5 −25 35 0Write the resulting polynomial.
106. 106. Continued... ±1, ±2, ±4, ±7, ±14, ±28 1 2 4 7 14 28 f ( x ) = 5x 3 − 29x 2 + 55x − 28 ± ,± ,± ,± ,± ,± 5 5 5 5 5 5 3 2  4  4  4  4 f   = 5   − 29   + 55   − 28 = 0  5  5  5  5Found it! So now use Synthetic Division to simplify thepolynomial. 4 5 5 −29 55 −28 4 −20 28 5 −25 35 0Write the resulting polynomial. 5x 2 − 25x + 35 = 0
107. 107. Continued... 5x 2 − 25x + 35 = 0Factor the trinomial using any method but always look for aGCF ﬁrst!
108. 108. Continued... 5x 2 − 25x + 35 = 0Factor the trinomial using any method but always look for aGCF ﬁrst! 5 ( x 2 − 5x + 7 ) = 0
109. 109. Continued... 5x 2 − 25x + 35 = 0Factor the trinomial using any method but always look for aGCF ﬁrst! 5 ( x 2 − 5x + 7 ) = 0Because this is an equation, divide by 5 to simplify.
110. 110. Continued... 5x 2 − 25x + 35 = 0Factor the trinomial using any method but always look for aGCF ﬁrst! 5 ( x 2 − 5x + 7 ) = 0Because this is an equation, divide by 5 to simplify. 2 x − 5x + 7 = 0
111. 111. Continued... 5x 2 − 25x + 35 = 0Factor the trinomial using any method but always look for aGCF ﬁrst! 5 ( x 2 − 5x + 7 ) = 0Because this is an equation, divide by 5 to simplify. 2 x − 5x + 7 = 0The trinomial isn’t easily factored so use the quadraticformula or complete the square to ﬁnd the solution.
112. 112. Continued... x 2 − 5x + 7 = 0The quadratic formula will be the better optionbecause the middle term is odd and will result in afraction when completing the square.
113. 113. Continued... x 2 − 5x + 7 = 0The quadratic formula will be the better optionbecause the middle term is odd and will result in afraction when completing the square. a = 1; b = −5; c = 7 2 2 −b ± b − 4ac 5 ± ( −5 ) − 4 ⋅1⋅ 7 5 ± 25 − 28 x= = = 2a 2 ⋅1 2 5 ± −3 5 ± i 3 x= = 2 2
114. 114. Continued... 5x 3 − 29x 2 + 55x − 28 = 0Our original polynomial was degree 3, whichindicates 3 solutions. We found 3 solutionsalthough only 1 solutions was real.
115. 115. Continued... 5x 3 − 29x 2 + 55x − 28 = 0Our original polynomial was degree 3, whichindicates 3 solutions. We found 3 solutionsalthough only 1 solutions was real.The solutions to the equation are...
116. 116. Continued... 5x 3 − 29x 2 + 55x − 28 = 0Our original polynomial was degree 3, whichindicates 3 solutions. We found 3 solutionsalthough only 1 solutions was real.The solutions to the equation are...  4 5 + i 3 5 − i 3    5, ,    2 2  
117. 117. The end.