This document discusses using synthetic division to evaluate polynomials at specific values and factor polynomials. It provides examples of using synthetic division to: 1) Evaluate polynomials like f(x) = x^2 - x + 5 at specific values such as f(-2) 2) Factor polynomials when one factor is known, such as factoring x^3 - 3x^2 - 13x + 15 after determining (x + 3) is a factor 3) Find all zeros of a polynomial by setting each factor equal to 0 after factoring Step-by-step instructions and additional examples are provided to illustrate the process.

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 Substitution
Find g ( 5 ) for g ( x ) = −x + x − 7x + 8
3 2

10. Try it: Simplify using Synthetic Substitution
Find 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 Substitution
Find 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 Substitution
Find f ( −3) for g ( x ) = x − 2x − 3x + 5x − 6
4 3 2

13. Try it: Simplify using Synthetic Substitution
Find 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 Substitution
Find 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 Substitution
Find h ( 2b ) for h ( x ) = x + 3x − 2x + 1
3 2
Don’t be intimidated by the 2b. Use the same
process as before.

16. Try it: Simplify using Synthetic Substitution
Find h ( 2b ) for h ( x ) = x + 3x − 2x + 1
3 2
Don’t be intimidated by the 2b. Use the same
process 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 Substitution
Find h ( 2b ) for h ( x ) = x + 3x − 2x + 1
3 2
Don’t be intimidated by the 2b. Use the same
process 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.

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

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 2
Now find the zeros of x − 3x − 13x + 15 = 0.

32. Continued...
3 2
Now find the zeros of x − 3x − 13x + 15 = 0.
Remember the factors are ...
( x + 3) ( x − 5 ) ( x − 1)

33. Continued...
3 2
Now 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 2
Now 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 2
Now 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 2
Now 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 2
Now 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 2
Now 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 2
Factor x + 4x − 15x − 18
Remember you were given a factor.

42. Now take the last problem a little further...
3 2
Factor x + 4x − 15x − 18
Remember 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 2
Factor x + 4x − 15x − 18
Remember 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 2
Factor x + 4x − 15x − 18
Remember 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 2
Factor x + 4x − 15x − 18
Remember 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 2
Factor x + 4x − 15x − 18
Remember 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 2
Factor x + 4x − 15x − 18
Remember 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 2
Find the zeros of x + 4x − 15x − 18

49. And a little further...
3 2
Find the zeros of x + 4x − 15x − 18
Remember ...
x + 4x − 15x − 18 = ( x − 3) ( x + 6 ) ( x + 1)
3 2

50. And a little further...
3 2
Find the zeros of x + 4x − 15x − 18
Remember ...
x + 4x − 15x − 18 = ( x − 3) ( x + 6 ) ( x + 1)
3 2
Set each factor to 0 and solve.

51. And a little further...
3 2
Find the zeros of x + 4x − 15x − 18
Remember ...
x + 4x − 15x − 18 = ( x − 3) ( x + 6 ) ( x + 1)
3 2
Set each factor to 0 and solve.
x−3= 0 x+6=0 x +1= 0

52. And a little further...
3 2
Find the zeros of x + 4x − 15x − 18
Remember ...
x + 4x − 15x − 18 = ( x − 3) ( x + 6 ) ( x + 1)
3 2
Set each factor to 0 and solve.
x−3= 0 x+6=0 x +1= 0
x=3

53. And a little further...
3 2
Find the zeros of x + 4x − 15x − 18
Remember ...
x + 4x − 15x − 18 = ( x − 3) ( x + 6 ) ( x + 1)
3 2
Set 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 2
Find the zeros of x + 4x − 15x − 18
Remember ...
x + 4x − 15x − 18 = ( x − 3) ( x + 6 ) ( x + 1)
3 2
Set 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 2
Find the zeros of x + 4x − 15x − 18
Remember ...
x + 4x − 15x − 18 = ( x − 3) ( x + 6 ) ( x + 1)
3 2
Set 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 = 0
Find factors of leading coefficient.

65. Try one:
Solve using the Rational Roots Theorem
3 2
x + 6x + 3x − 10 = 0
Find factors of leading coefficient. ±1

66. Try one:
Solve using the Rational Roots Theorem
3 2
x + 6x + 3x − 10 = 0
Find factors of leading coefficient. ±1
Find factors of constant term.

67. Try one:
Solve using the Rational Roots Theorem
3 2
x + 6x + 3x − 10 = 0
Find factors of leading coefficient. ±1
Find factors of constant term. ±1, ±2, ±5, ±10

68. Try one:
Solve using the Rational Roots Theorem
3 2
x + 6x + 3x − 10 = 0
Find factors of leading coefficient. ±1
Find factors of constant term. ±1, ±2, ±5, ±10
Possible roots are made up of the
constant term factors over the
factors of the leading coefficient.
No need to repeat numbers and
always simplify.

69. Try one:
Solve using the Rational Roots Theorem
3 2
x + 6x + 3x − 10 = 0
Find factors of leading coefficient. ±1
Find factors of constant term. ±1, ±2, ±5, ±10
Possible roots are made up of the
constant term factors over the ±1, ±2, ±5, ±10
factors of the leading coefficient.
No need to repeat numbers and
always simplify.

70. Try one:
Solve using the Rational Roots Theorem
3 2
x + 6x + 3x − 10 = 0
Find factors of leading coefficient. ±1
Find factors of constant term. ±1, ±2, ±5, ±10
Possible roots are made up of the
constant term factors over the ±1, ±2, ±5, ±10
factors of the leading coefficient.
No need to repeat numbers and
always simplify.
Lots from which to choose but
lets start with an easy one.

71. Try one:
Solve using the Rational Roots Theorem
3 2
x + 6x + 3x − 10 = 0
Find factors of leading coefficient. ±1
Find factors of constant term. ±1, ±2, ±5, ±10
Possible roots are made up of the
constant term factors over the ±1, ±2, ±5, ±10
factors of the leading coefficient.
No need to repeat numbers and
always simplify.
f ( x ) = x 3 + 6x 2 + 3x − 10
Lots from which to choose but
lets 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 0
Write 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 0
Write 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 0
Write the new polynomial.
2
x + 7x + 10 = 0
Look at the trinomial. Can it be factored?

77. Continued...
2
x + 7x + 10 = 0
Yes! It’s a factorable quadratic so use any method
you would like to factor.

78. Continued...
2
x + 7x + 10 = 0
Yes! It’s a factorable quadratic so use any method
you would like to factor.
( x + 5 )( x + 2 ) = 0

79. Continued...
2
x + 7x + 10 = 0
Yes! It’s a factorable quadratic so use any method
you would like to factor.
( x + 5 )( x + 2 ) = 0
Now set each factor to 0 and solve.

80. Continued...
2
x + 7x + 10 = 0
Yes! It’s a factorable quadratic so use any method
you would like to factor.
( x + 5 )( x + 2 ) = 0
Now set each factor to 0 and solve.
( x + 5) = 0
x = −5

81. Continued...
2
x + 7x + 10 = 0
Yes! It’s a factorable quadratic so use any method
you would like to factor.
( x + 5 )( x + 2 ) = 0
Now set each factor to 0 and solve.
( x + 5) = 0 ( x + 2) = 0
x = −5 x = −2

82. Continued...
2
x + 7x + 10 = 0
Yes! It’s a factorable quadratic so use any method
you would like to factor.
( x + 5 )( x + 2 ) = 0
Now 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 = 0
Find factors of leading coefficient.

85. Try another:
Solve using the Rational Roots Theorem
3 2
5x − 29x + 55x − 28 = 0
Find factors of leading coefficient. ±1, ±5

86. Try another:
Solve using the Rational Roots Theorem
3 2
5x − 29x + 55x − 28 = 0
Find factors of leading coefficient. ±1, ±5
Find factors of constant term.

87. Try another:
Solve using the Rational Roots Theorem
3 2
5x − 29x + 55x − 28 = 0
Find factors of leading coefficient. ±1, ±5
Find 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 = 0
Find factors of leading coefficient. ±1, ±5
Find factors of constant term. ±1, ±2, ±4, ±7, ±14, ±28
Possible roots are leading coefficient factors over
constant term factors without repeats.

89. Try another:
Solve using the Rational Roots Theorem
3 2
5x − 29x + 55x − 28 = 0
Find factors of leading coefficient. ±1, ±5
Find factors of constant term. ±1, ±2, ±4, ±7, ±14, ±28
Possible roots are leading coefficient factors over
constant 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 = 0
Find factors of leading coefficient. ±1, ±5
Find factors of constant term. ±1, ±2, ±4, ±7, ±14, ±28
Possible roots are leading coefficient factors over
constant term factors without repeats.
±1, ±2, ±4, ±7, ±14, ±28
1 2 4 7 14 28
± ,± ,± ,± ,± ,±
5 5 5 5 5 5
Wow, that’s a lot! Not to worry. Start with the smallest
positive 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 5
Start 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 5
Start 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 5
Start with 1.
3 2
f (1) = 5 (1) − 29 (1) + 55 (1) − 28 = 3
Didn’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 5
Start with 1.
3 2
f (1) = 5 (1) − 29 (1) + 55 (1) − 28 = 3
Didn’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 5
Start with 1.
3 2
f (1) = 5 (1) − 29 (1) + 55 (1) − 28 = 3
Didn’t work so try the next highest integer.
3 2
f ( 2 ) = 5 ( 2 ) − 29 ( 2 ) + 55 ( 2 ) − 28 = 6
No good again because but notice our remainder is getting
larger. This indicates we are moving away from 0 so we are
going 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 5
Start with 1.
3 2
f (1) = 5 (1) − 29 (1) + 55 (1) − 28 = 3
Didn’t work so try the next highest integer.
3 2
f ( 2 ) = 5 ( 2 ) − 29 ( 2 ) + 55 ( 2 ) − 28 = 6
No good again because but notice our remainder is getting
larger. This indicates we are moving away from 0 so we are
going 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 5
Start with 1.
3 2
f (1) = 5 (1) − 29 (1) + 55 (1) − 28 = 3
Didn’t work so try the next highest integer.
3 2
f ( 2 ) = 5 ( 2 ) − 29 ( 2 ) + 55 ( 2 ) − 28 = 6
No good again because but notice our remainder is getting
larger. This indicates we are moving away from 0 so we are
going in the wrong direction. Try a negative number.
3 2
f ( −1) = 5 ( −1) − 29 ( −1) + 55 ( −1) − 28 = −117
Now 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 2
f ( −1) = 5 ( −1) − 29 ( −1) + 55 ( −1) − 28 = −117
Because 0 falls between 3 and -117, we know the
root 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 = −117
Because 0 falls between 3 and -117, we know the
root 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 = −117
Because 0 falls between 3 and -117, we know the
root 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 = −117
Because 0 falls between 3 and -117, we know the
root 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 between
1/5 and 1.

102. 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

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 the
polynomial.

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 the
polynomial.
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  5
Found it! So now use Synthetic Division to simplify the
polynomial.
4
5 5 −29 55 −28
4 −20 28
5 −25 35 0
Write 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  5
Found it! So now use Synthetic Division to simplify the
polynomial.
4
5 5 −29 55 −28
4 −20 28
5 −25 35 0
Write the resulting polynomial.
5x 2 − 25x + 35 = 0

107. Continued...
5x 2 − 25x + 35 = 0
Factor the trinomial using any method but always look for a
GCF first!

108. Continued...
5x 2 − 25x + 35 = 0
Factor the trinomial using any method but always look for a
GCF first!
5 ( x 2 − 5x + 7 ) = 0

109. Continued...
5x 2 − 25x + 35 = 0
Factor the trinomial using any method but always look for a
GCF first!
5 ( x 2 − 5x + 7 ) = 0
Because this is an equation, divide by 5 to simplify.

110. Continued...
5x 2 − 25x + 35 = 0
Factor the trinomial using any method but always look for a
GCF first!
5 ( x 2 − 5x + 7 ) = 0
Because this is an equation, divide by 5 to simplify.
2
x − 5x + 7 = 0

111. Continued...
5x 2 − 25x + 35 = 0
Factor the trinomial using any method but always look for a
GCF first!
5 ( x 2 − 5x + 7 ) = 0
Because this is an equation, divide by 5 to simplify.
2
x − 5x + 7 = 0
The trinomial isn’t easily factored so use the quadratic
formula or complete the square to find the solution.

112. Continued... x 2 − 5x + 7 = 0
The quadratic formula will be the better option
because the middle term is odd and will result in a
fraction when completing the square.

113. Continued... x 2 − 5x + 7 = 0
The quadratic formula will be the better option
because the middle term is odd and will result in a
fraction 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 = 0
Our original polynomial was degree 3, which
indicates 3 solutions. We found 3 solutions
although only 1 solutions was real.

115. Continued... 5x 3 − 29x 2 + 55x − 28 = 0
Our original polynomial was degree 3, which
indicates 3 solutions. We found 3 solutions
although only 1 solutions was real.
The solutions to the equation are...

116. Continued... 5x 3 − 29x 2 + 55x − 28 = 0
Our original polynomial was degree 3, which
indicates 3 solutions. We found 3 solutions
although only 1 solutions was real.
The solutions to the equation are...

4 5 + i 3 5 − i 3 

 5, , 

 2 2  

117. The end.