The document provides information about solving polynomial equations. There are three main ways to solve polynomial equations: 1) Using factoring and the zero product property, 2) Using a graphing calculator to graph the equation, and 3) Using synthetic division. The maximum number of solutions a polynomial equation can have is equal to the degree of the polynomial. Examples are provided to demonstrate solving polynomial equations by factoring.
2. Solving Polynomial Equations
★ Solving a polynomial equation is the same as solving a
quadratic equation, except that the quadratic might be
replaced by a different kind of polynomial (such as a
cubic or a quartic).
3. Solving Polynomial Equations
★ Solving a polynomial equation is the same as solving a
quadratic equation, except that the quadratic might be
replaced by a different kind of polynomial (such as a
cubic or a quartic).
★ There are 3 ways to solve Polynomial Equations
4. Solving Polynomial Equations
★ Solving a polynomial equation is the same as solving a
quadratic equation, except that the quadratic might be
replaced by a different kind of polynomial (such as a
cubic or a quartic).
★ There are 3 ways to solve Polynomial Equations
(1) Using factoring and the zero product property
5. Solving Polynomial Equations
★ Solving a polynomial equation is the same as solving a
quadratic equation, except that the quadratic might be
replaced by a different kind of polynomial (such as a
cubic or a quartic).
★ There are 3 ways to solve Polynomial Equations
(1) Using factoring and the zero product property
(2) Using the graphing calculator to graph
6. Solving Polynomial Equations
★ Solving a polynomial equation is the same as solving a
quadratic equation, except that the quadratic might be
replaced by a different kind of polynomial (such as a
cubic or a quartic).
★ There are 3 ways to solve Polynomial Equations
(1) Using factoring and the zero product property
(2) Using the graphing calculator to graph
(3) Using Synthetic Division (separate notes)
7. Total Number of Solutions
★ The maximum number of solutions that a polynomial
equation can have is equal to the degree of the
polynomial.
★ It is possible for a polynomial equation to have fewer
solutions (or none at all).
★ The degree of the polynomial gives you the maximum
number of solutions that are theoretically possible.
★ Some solutions may be irrational or even imaginary.
8. Total Number of Solutions
★ The maximum number of solutions that a polynomial
equation can have is equal to the degree of the
polynomial. ( ) 3
f x = x + x − x +12
★ It is possible for a polynomial equation to have fewer
solutions (or none at all).
★ The degree of the polynomial gives you the maximum
number of solutions that are theoretically possible.
★ Some solutions may be irrational or even imaginary.
9. Total Number of Solutions
★ The maximum number of solutions that a polynomial
equation can have is equal to the degree of the
polynomial. ( ) 3
f x = x + x − x +12
Degree is 3. Therefore 3 or fewer solutions possible.
10. Total Number of Solutions
★ The maximum number of solutions that a polynomial
equation can have is equal to the degree of the
polynomial. ( ) 3
f x = x + x − x +12
Degree is 3. Therefore 3 or fewer solutions possible.
★ It is possible for a polynomial equation to have fewer
solutions (or none at all).
11. Total Number of Solutions
★ The maximum number of solutions that a polynomial
equation can have is equal to the degree of the
polynomial. ( ) 3
f x = x + x − x +12
Degree is 3. Therefore 3 or fewer solutions possible.
★ It is possible for a polynomial equation to have fewer
solutions (or none at all).
★ The degree of the polynomial gives you the maximum
number of solutions that are theoretically possible.
12. Total Number of Solutions
★ The maximum number of solutions that a polynomial
equation can have is equal to the degree of the
polynomial. ( ) 3
f x = x + x − x +12
Degree is 3. Therefore 3 or fewer solutions possible.
★ It is possible for a polynomial equation to have fewer
solutions (or none at all).
★ The degree of the polynomial gives you the maximum
number of solutions that are theoretically possible.
★ Some solutions may be irrational or even imaginary.
13. Solving by Factoring
★ When using factoring, remember to always look for a
GCF (Greatest Common Factor) first. This will help
make the remaining factoring easier.
★ Factor completely!
★ Set each factor equal to 0 and solve.
14. Solving by Factoring
★ When using factoring, remember to always look for a
GCF (Greatest Common Factor) first. This will help
make the remaining factoring easier. 3x 3 − 12x = 0
GCF ( )
3x x 2 − 4 = 0
★ Factor completely!
★ Set each factor equal to 0 and solve.
15. Solving by Factoring
★ When using factoring, remember to always look for a
GCF (Greatest Common Factor) first. This will help
make the remaining factoring easier. 3x 3 − 12x = 0
GCF ( )
3x x 2 − 4 = 0
★ Factor completely!
16. Solving by Factoring
★ When using factoring, remember to always look for a
GCF (Greatest Common Factor) first. This will help
make the remaining factoring easier. 3x 3 − 12x = 0
GCF ( )
3x x 2 − 4 = 0
★ Factor completely! Keep going here.
17. Solving by Factoring
★ When using factoring, remember to always look for a
GCF (Greatest Common Factor) first. This will help
make the remaining factoring easier. 3x 3 − 12x = 0
GCF ( )
3x x 2 − 4 = 0
★ Factor completely! Keep going here.
3x ( x − 2 ) ( x + 2 ) = 0
18. Solving by Factoring
★ When using factoring, remember to always look for a
GCF (Greatest Common Factor) first. This will help
make the remaining factoring easier. 3x 3 − 12x = 0
GCF ( )
3x x 2 − 4 = 0
★ Factor completely! Keep going here.
3x ( x − 2 ) ( x + 2 ) = 0
★ Set each factor equal to 0 and solve.
19. Solving by Factoring
★ When using factoring, remember to always look for a
GCF (Greatest Common Factor) first. This will help
make the remaining factoring easier. 3x 3 − 12x = 0
GCF ( )
3x x 2 − 4 = 0
★ Factor completely! Keep going here.
3x ( x − 2 ) ( x + 2 ) = 0
★ Set each factor equal to 0 and solve.
3x = 0 ( x − 2) = 0 ( x + 2) = 0
20. Solving by Factoring
★ When using factoring, remember to always look for a
GCF (Greatest Common Factor) first. This will help
make the remaining factoring easier. 3x 3 − 12x = 0
GCF ( )
3x x 2 − 4 = 0
★ Factor completely! Keep going here.
3x ( x − 2 ) ( x + 2 ) = 0
★ Set each factor equal to 0 and solve.
3x = 0 ( x − 2) = 0 ( x + 2) = 0
x=0 x=2 x = −2
22. Example: Solve by Factoring
4 2
x − 6x = 27 Degree is 4 so can up
to 4 solutions.
23. Example: Solve by Factoring
4 2
x − 6x = 27 Degree is 4 so can up
4 2 to 4 solutions.
x − 6x − 27 = 0
24. Example: Solve by Factoring
4 2
x − 6x = 27 Degree is 4 so can up
4 2 to 4 solutions.
x − 6x − 27 = 0
(x 2
)(
−9 x +3 =0 2
)
25. Example: Solve by Factoring
4 2
x − 6x = 27 Degree is 4 so can up
4 2 to 4 solutions.
x − 6x − 27 = 0
(x 2
)(
−9 x +3 =0 2
)
2
x −9=0
26. Example: Solve by Factoring
4 2
x − 6x = 27 Degree is 4 so can up
4 2 to 4 solutions.
x − 6x − 27 = 0
(x 2
)(
−9 x +3 =0 2
)
2 2
x −9=0 x +3= 0
27. Example: Solve by Factoring
4 2
x − 6x = 27 Degree is 4 so can up
4 2 to 4 solutions.
x − 6x − 27 = 0
(x 2
)(
−9 x +3 =0 2
)
2 2
x −9=0 x +3= 0
2
x =9
28. Example: Solve by Factoring
4 2
x − 6x = 27 Degree is 4 so can up
4 2 to 4 solutions.
x − 6x − 27 = 0
(x 2
)(
−9 x +3 =0 2
)
2 2
x −9=0 x +3= 0
2
x =9
x=± 9
29. Example: Solve by Factoring
4 2
x − 6x = 27 Degree is 4 so can up
4 2 to 4 solutions.
x − 6x − 27 = 0
(x 2
)(
−9 x +3 =0 2
)
2 2
x −9=0 x +3= 0
2
x =9
x=± 9
x = ±3
30. Example: Solve by Factoring
4 2
x − 6x = 27 Degree is 4 so can up
4 2 to 4 solutions.
x − 6x − 27 = 0
(x 2
)(
−9 x +3 =0 2
)
2 2
x −9=0 x +3= 0
2 2
x =9 x = −3
x=± 9
x = ±3
31. Example: Solve by Factoring
4 2
x − 6x = 27 Degree is 4 so can up
4 2 to 4 solutions.
x − 6x − 27 = 0
(x 2
)(
−9 x +3 =0 2
)
2 2
x −9=0 x +3= 0
2 2
x =9 x = −3
x=± 9 x = ± −3
x = ±3
32. Example: Solve by Factoring
4 2
x − 6x = 27 Degree is 4 so can up
4 2 to 4 solutions.
x − 6x − 27 = 0
(x 2
)(
−9 x +3 =0 2
)
2 2
x −9=0 x +3= 0
2 2
x =9 x = −3
x=± 9 x = ± −3
x = ±3 x = ±i 3
33. Example: Solve by Factoring
4 2
x − 6x = 27 Degree is 4 so can up
4 2 to 4 solutions.
x − 6x − 27 = 0
(x 2
)(
−9 x +3 =0 2
) Solutions:
2
x −9=0 2
x +3= 0
{±3, ±i 3}
2 2
x =9 x = −3
x=± 9 x = ± −3
x = ±3 x = ±i 3
35. Try this: Solve by Factoring
3 2
x + 3x − x − 3 = 0 Degree is 3 so can up
to 3 solutions.
36. Try this: Solve by Factoring
3 2
x + 3x − x − 3 = 0 Degree is 3 so can up
to 3 solutions.
( )
x + 3x + ( −x − 3) = 0
3 2
37. Try this: Solve by Factoring
3 2
x + 3x − x − 3 = 0 Degree is 3 so can up
to 3 solutions.
( )
x + 3x + ( −x − 3) = 0
3 2
x ( x + 3) − 1( x + 3) = 0
2
38. Try this: Solve by Factoring
3 2
x + 3x − x − 3 = 0 Degree is 3 so can up
to 3 solutions.
( )
x + 3x + ( −x − 3) = 0
3 2
x ( x + 3) − 1( x + 3) = 0
2
( )
x − 1 ( x + 3) = 0
2
39. Try this: Solve by Factoring
3 2
x + 3x − x − 3 = 0 Degree is 3 so can up
to 3 solutions.
( )
x + 3x + ( −x − 3) = 0
3 2
x ( x + 3) − 1( x + 3) = 0
2
( )
x − 1 ( x + 3) = 0
2
2
x −1= 0
40. Try this: Solve by Factoring
3 2
x + 3x − x − 3 = 0 Degree is 3 so can up
to 3 solutions.
( )
x + 3x + ( −x − 3) = 0
3 2
x ( x + 3) − 1( x + 3) = 0
2
( )
x − 1 ( x + 3) = 0
2
2
x −1= 0 x+3= 0
41. Try this: Solve by Factoring
3 2
x + 3x − x − 3 = 0 Degree is 3 so can up
to 3 solutions.
( )
x + 3x + ( −x − 3) = 0
3 2
x ( x + 3) − 1( x + 3) = 0
2
( )
x − 1 ( x + 3) = 0
2
2
x −1= 0 x+3= 0
x = −3
42. Try this: Solve by Factoring
3 2
x + 3x − x − 3 = 0 Degree is 3 so can up
to 3 solutions.
( )
x + 3x + ( −x − 3) = 0
3 2
x ( x + 3) − 1( x + 3) = 0
2
( )
x − 1 ( x + 3) = 0
2
2
x −1= 0 x+3= 0
2
x =1 x = −3
43. Try this: Solve by Factoring
3 2
x + 3x − x − 3 = 0 Degree is 3 so can up
to 3 solutions.
( )
x + 3x + ( −x − 3) = 0
3 2
x ( x + 3) − 1( x + 3) = 0
2
( )
x − 1 ( x + 3) = 0
2
2
x −1= 0 x+3= 0
2
x =1 x = −3
x = ±1
44. Try this: Solve by Factoring
3 2
x + 3x − x − 3 = 0 Degree is 3 so can up
to 3 solutions.
( )
x + 3x + ( −x − 3) = 0
3 2
x ( x + 3) − 1( x + 3) = 0
2
Solutions:
( )
x − 1 ( x + 3) = 0
2
{1, −1, −3}
2
x −1= 0 x+3= 0
2
x =1 x = −3
x = ±1
46. Try this: Solve by Factoring
3 2
x + x − 4x = 0 Degree is 3 so can up
to 3 solutions.
47. Try this: Solve by Factoring
3 2
x + x − 4x = 0 Degree is 3 so can up
to 3 solutions.
( 2
x x +x−4 =0 )
48. Try this: Solve by Factoring
3 2
x + x − 4x = 0 Degree is 3 so can up
to 3 solutions.
(
x x +x−4 =02
)
x=0
49. Try this: Solve by Factoring
3
x + x − 4x = 0 2 Degree is 3 so can up
to 3 solutions.
(
x x +x−4 =02
)
2
x=0 x +x−4=0
50. Try this: Solve by Factoring
3
x + x − 4x = 0 2 Degree is 3 so can up
to 3 solutions.
(
x x +x−4 =02
)
2 Not factorable so use
x=0 x +x−4=0 the quadratic formula
51. Try this: Solve by Factoring
3
x + x − 4x = 0 2 Degree is 3 so can up
to 3 solutions.
(
x x +x−4 =02
)
2 Not factorable so use
x=0 x +x−4=0 the quadratic formula
a = 1; b = 1; c = −4
−1 ± 12 − 4 (1) ( −4 )
x=
2 (1)
52. Try this: Solve by Factoring
3
x + x − 4x = 0 2 Degree is 3 so can up
to 3 solutions.
(
x x +x−4 =02
)
2 Not factorable so use
x=0 x +x−4=0 the quadratic formula
a = 1; b = 1; c = −4
−1 ± 12 − 4 (1) ( −4 )
x=
2 (1)
−1 ± 17
x=
2
53. Try this: Solve by Factoring
3
x + x − 4x = 0 2 Degree is 3 so can up
to 3 solutions.
(
x x +x−4 =02
)
2 Not factorable so use
x=0 x +x−4=0 the quadratic formula
a = 1; b = 1; c = −4
−1 ± 12 − 4 (1) ( −4 )
x=
2 (1) Solutions:
−1 ± 17 −1 ± 17
x= 0,
2
2
55. Try this: Solve by Factoring
f ( x ) = x + 64
3
Degree is 3 so can up to 3 solutions.
56. Try this: Solve by Factoring
f ( x ) = x + 64
3
Degree is 3 so can up to 3 solutions.
3
0 = x + (4)
3
57. Try this: Solve by Factoring
f ( x ) = x + 64
3
Degree is 3 so can up to 3 solutions.
3
0 = x + (4)
3
Sum of cubes. Apply the formula.
58. Try this: Solve by Factoring
f ( x ) = x + 64
3
Degree is 3 so can up to 3 solutions.
3
0 = x + ( 4 ) Sum of cubes. Apply the formula.
3
0 = ( x + 4 ) ( x − 4x + 16 )
2
59. Try this: Solve by Factoring
f ( x ) = x + 64
3
Degree is 3 so can up to 3 solutions.
3
0 = x + ( 4 ) Sum of cubes. Apply the formula.
3
0 = ( x + 4 ) ( x − 4x + 16 )
2
x+4=0
60. Try this: Solve by Factoring
f ( x ) = x + 64
3
Degree is 3 so can up to 3 solutions.
3
0 = x + ( 4 ) Sum of cubes. Apply the formula.
3
0 = ( x + 4 ) ( x − 4x + 16 )
2
2
x+4=0 x − 4x + 16 = 0
61. Try this: Solve by Factoring
f ( x ) = x + 64
3
Degree is 3 so can up to 3 solutions.
3
0 = x + ( 4 ) Sum of cubes. Apply the formula.
3
0 = ( x + 4 ) ( x − 4x + 16 )
2
2
x+4=0 x − 4x + 16 = 0
−4 = x
62. Try this: Solve by Factoring
f ( x ) = x + 64
3
Degree is 3 so can up to 3 solutions.
3
0 = x + ( 4 ) Sum of cubes. Apply the formula.
3
0 = ( x + 4 ) ( x − 4x + 16 )
2
Not factorable so
2
x+4=0 x − 4x + 16 = 0 use completing
the square.
−4 = x
63. Try this: Solve by Factoring
f ( x ) = x + 64
3
Degree is 3 so can up to 3 solutions.
3
0 = x + ( 4 ) Sum of cubes. Apply the formula.
3
0 = ( x + 4 ) ( x − 4x + 16 )
2
Not factorable so
2
x+4=0 x − 4x + 16 = 0 use completing
the square.
2
−4 = x x − 4x + ( −2 ) = −16 + 4
2
64. Try this: Solve by Factoring
f ( x ) = x + 64
3
Degree is 3 so can up to 3 solutions.
3
0 = x + ( 4 ) Sum of cubes. Apply the formula.
3
0 = ( x + 4 ) ( x − 4x + 16 )
2
Not factorable so
2
x+4=0 x − 4x + 16 = 0 use completing
the square.
2
−4 = x x − 4x + ( −2 ) = −16 + 4
2
2
( x − 2 ) = −12
65. Try this: Solve by Factoring
f ( x ) = x + 64
3
Degree is 3 so can up to 3 solutions.
3
0 = x + ( 4 ) Sum of cubes. Apply the formula.
3
0 = ( x + 4 ) ( x − 4x + 16 )
2
Not factorable so
2
x+4=0 x − 4x + 16 = 0 use completing
the square.
2
−4 = x x − 4x + ( −2 ) = −16 + 4
2
2
( x − 2 ) = −12
x − 2 = ± −12
66. Try this: Solve by Factoring
f ( x ) = x + 64
3
Degree is 3 so can up to 3 solutions.
3
0 = x + ( 4 ) Sum of cubes. Apply the formula.
3
0 = ( x + 4 ) ( x − 4x + 16 )
2
Not factorable so
2
x+4=0 x − 4x + 16 = 0 use completing
the square.
2
−4 = x x − 4x + ( −2 ) = −16 + 4
2
2
( x − 2 ) = −12
x − 2 = ± −12
x = 2 ± 2i 3
67. Try this: Solve by Factoring
f ( x ) = x + 64
3
Degree is 3 so can up to 3 solutions.
3
0 = x + ( 4 ) Sum of cubes. Apply the formula.
3
0 = ( x + 4 ) ( x − 4x + 16 )
2
Not factorable so
2
x+4=0 x − 4x + 16 = 0 use completing
the square.
2
−4 = x x − 4x + ( −2 ) = −16 + 4
2
2
( x − 2 ) = −12
Solutions: x − 2 = ± −12
{−4, 2 ± 2i 3} x = 2 ± 2i 3
68. Practice Time!
★ Follow this link to practice solving polynomial equations
using Factoring.
69. Solutions by Observing the Graph
★ The degree of the function tells you the maximum
number of solutions possible.
70. Solutions by Observing the Graph
★ The degree of the function tells you the maximum
number of solutions possible.
★ The real solutions are where the function crosses or
touches the x-axis.
71. Solutions by Observing the Graph
★ The degree of the function tells you the maximum
number of solutions possible.
★ The real solutions are where the function crosses or
touches the x-axis.
★ The graph below has 4 solutions because it crosses the
x-axis in 4 places. Notice 2 are positive real numbers
and 2 are negative real numbers.
82. You try: Find all real zeros on the graph.
★ The real zeros for the graph below are {−1, 2, 5}
83. Practice Time!
★ Follow this link to practice solving polynomial equations
using Factoring.
84. Solving by Graphing in Calculator
★ Graph the left side of the equation in Y1.
85. Solving by Graphing in Calculator
★ Graph the left side of the equation in Y1.
★ Graph the right side of the equation in Y2.
86. Solving by Graphing in Calculator
★ Graph the left side of the equation in Y1.
★ Graph the right side of the equation in Y2.
★ Find all the points the two graphs intersect. The x-
coordinate is the solution.
87. Solving by Graphing in Calculator
★ Graph the left side of the equation in Y1.
★ Graph the right side of the equation in Y2.
★ Find all the points the two graphs intersect. The x-
coordinate is the solution.
★ If you are given a function such as f(x) = x2 - 1, use zero
for f(x). So Y1 = 0 and Y2 = x2 - 1. The find all the
intersections.
89. Example: Solve by Graphing
2
f ( x ) = 0.25 ( x + 2 ) ( x − 1)
Y1 = 0
2
Y 2 = 0.25 ( x + 2 ) ( x − 1)
90. Example: Solve by Graphing
2
f ( x ) = 0.25 ( x + 2 ) ( x − 1)
Y1 = 0
2
Y 2 = 0.25 ( x + 2 ) ( x − 1)
91. Example: Solve by Graphing
2
f ( x ) = 0.25 ( x + 2 ) ( x − 1)
Y1 = 0
2
Y 2 = 0.25 ( x + 2 ) ( x − 1)
92. Example: Solve by Graphing
2
f ( x ) = 0.25 ( x + 2 ) ( x − 1)
Y1 = 0
2
Y 2 = 0.25 ( x + 2 ) ( x − 1)
93. Example: Solve by Graphing
2
f ( x ) = 0.25 ( x + 2 ) ( x − 1)
Y1 = 0
2
Y 2 = 0.25 ( x + 2 ) ( x − 1)
Solutions:
{−2,1}
94. Example: Solve by Graphing
2
f ( x ) = 0.25 ( x + 2 ) ( x − 1)
Y1 = 0
2
Y 2 = 0.25 ( x + 2 ) ( x − 1)
Solutions:
{−2,1}
Notice if you used the zero product
property, x = 1 would have occurred
twice. We say 1 has multiplicity of 2.
