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- 1. Solving Polynomials Using Synthetic Division
- 2. Simplify Polynomials using Synthetic Substitution• Recall Synthetic Division can be used to divide a polynomial by a binomial.
- 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. 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. 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. 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. 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. 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. Try it: Simplify using Synthetic SubstitutionFind g ( 5 ) for g ( x ) = −x + x − 7x + 8 3 2
- 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. 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. Try it: Simplify using Synthetic SubstitutionFind f ( −3) for g ( x ) = x − 2x − 3x + 5x − 6 4 3 2
- 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. 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. 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. 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. 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. Find Factors given a Polynomial & one factor• Synthetic Division can be useful in factoring polynomials that cannot be factored using traditional methods.
- 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. Continued... 3 2• So the factorization of x − 3x − 13x + 15 is ( x + 3) ( x − 5 ) ( x − 1)
- 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. Continued... 3 2Now find the zeros of x − 3x − 13x + 15 = 0.
- 32. Continued... 3 2Now find the zeros of x − 3x − 13x + 15 = 0.Remember the factors are ... ( x + 3) ( x − 5 ) ( x − 1)
- 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. 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. 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. 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. 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. 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. Your turn...Show ( x − 3) is a factor of x + 4x − 15x − 18 3 2
- 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. Now take the last problem a little further... 3 2Factor x + 4x − 15x − 18Remember you were given a factor.
- 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. 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. 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. 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. 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. 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. And a little further... 3 2Find the zeros of x + 4x − 15x − 18
- 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. 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. 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. 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. 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. 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. 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. Practice some more...• Follow this link to practice using synthetic division to ﬁnd the zeros of the polynomials.
- 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. Factoring when you don’t know a factor
- 59. Factoring when you don’t know a factor• The previous problems weren’t too bad because you knew a factor.
- 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. 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. 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. Try one:Solve using the Rational Roots Theorem 3 2 x + 6x + 3x − 10 = 0
- 64. Try one:Solve using the Rational Roots Theorem 3 2 x + 6x + 3x − 10 = 0Find factors of leading coefﬁcient.
- 65. Try one:Solve using the Rational Roots Theorem 3 2 x + 6x + 3x − 10 = 0Find factors of leading coefﬁcient. ±1
- 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. 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. 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. 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. 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. 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. Continued...Solution was 0 so we lucked out on our ﬁrst try.Use Synthetic Division to simplify the polynomial.
- 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. 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. 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. 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. Continued... 2 x + 7x + 10 = 0Yes! It’s a factorable quadratic so use any methodyou would like to factor.
- 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. 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. 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. 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. 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. Try another:Solve using the Rational Roots Theorem 3 2 5x − 29x + 55x − 28 = 0
- 84. Try another:Solve using the Rational Roots Theorem 3 2 5x − 29x + 55x − 28 = 0Find factors of leading coefﬁcient.
- 85. Try another:Solve using the Rational Roots Theorem 3 2 5x − 29x + 55x − 28 = 0Find factors of leading coefﬁcient. ±1, ±5
- 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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 5No good but it’s negative. This means the root is between1/5 and 1.
- 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. 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 5Found it! So now use Synthetic Division to simplify thepolynomial.
- 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 5Found it! So now use Synthetic Division to simplify thepolynomial. 4 5 5 −29 55 −28 4 −20 28 5 −25 35 0
- 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 5Found 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. 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 5Found 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. Continued... 5x 2 − 25x + 35 = 0Factor the trinomial using any method but always look for aGCF ﬁrst!
- 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. 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. 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. 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. 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. 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. 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. 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. 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. The end.

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