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.
Factor Theorem and Remainder Theorem. Mathematics10 Project under Mrs. Marissa De Ocampo. Prepared by Danielle Diva, Ronalie Mejos, Rafael Vallejos and Mark Lenon Dacir of 10- Einstein. CNSTHS.
* Find zeros of polynomial functions
* Use the Fundamental Theorem of Algebra to find a function that satisfies given conditions
* Find all zeros of a polynomial function
"Harmonic and Other Sequences" presentation includes a brief historical background, problems and solutions to the simplest problems which you may face in your Mathematics.
Factor Theorem and Remainder Theorem. Mathematics10 Project under Mrs. Marissa De Ocampo. Prepared by Danielle Diva, Ronalie Mejos, Rafael Vallejos and Mark Lenon Dacir of 10- Einstein. CNSTHS.
* Find zeros of polynomial functions
* Use the Fundamental Theorem of Algebra to find a function that satisfies given conditions
* Find all zeros of a polynomial function
"Harmonic and Other Sequences" presentation includes a brief historical background, problems and solutions to the simplest problems which you may face in your Mathematics.
Unit 8 - Information and Communication Technology (Paper I).pdfThiyagu K
This slides describes the basic concepts of ICT, basics of Email, Emerging Technology and Digital Initiatives in Education. This presentations aligns with the UGC Paper I syllabus.
2024.06.01 Introducing a competency framework for languag learning materials ...Sandy Millin
http://sandymillin.wordpress.com/iateflwebinar2024
Published classroom materials form the basis of syllabuses, drive teacher professional development, and have a potentially huge influence on learners, teachers and education systems. All teachers also create their own materials, whether a few sentences on a blackboard, a highly-structured fully-realised online course, or anything in between. Despite this, the knowledge and skills needed to create effective language learning materials are rarely part of teacher training, and are mostly learnt by trial and error.
Knowledge and skills frameworks, generally called competency frameworks, for ELT teachers, trainers and managers have existed for a few years now. However, until I created one for my MA dissertation, there wasn’t one drawing together what we need to know and do to be able to effectively produce language learning materials.
This webinar will introduce you to my framework, highlighting the key competencies I identified from my research. It will also show how anybody involved in language teaching (any language, not just English!), teacher training, managing schools or developing language learning materials can benefit from using the framework.
Embracing GenAI - A Strategic ImperativePeter Windle
Artificial Intelligence (AI) technologies such as Generative AI, Image Generators and Large Language Models have had a dramatic impact on teaching, learning and assessment over the past 18 months. The most immediate threat AI posed was to Academic Integrity with Higher Education Institutes (HEIs) focusing their efforts on combating the use of GenAI in assessment. Guidelines were developed for staff and students, policies put in place too. Innovative educators have forged paths in the use of Generative AI for teaching, learning and assessments leading to pockets of transformation springing up across HEIs, often with little or no top-down guidance, support or direction.
This Gasta posits a strategic approach to integrating AI into HEIs to prepare staff, students and the curriculum for an evolving world and workplace. We will highlight the advantages of working with these technologies beyond the realm of teaching, learning and assessment by considering prompt engineering skills, industry impact, curriculum changes, and the need for staff upskilling. In contrast, not engaging strategically with Generative AI poses risks, including falling behind peers, missed opportunities and failing to ensure our graduates remain employable. The rapid evolution of AI technologies necessitates a proactive and strategic approach if we are to remain relevant.
A Strategic Approach: GenAI in EducationPeter Windle
Artificial Intelligence (AI) technologies such as Generative AI, Image Generators and Large Language Models have had a dramatic impact on teaching, learning and assessment over the past 18 months. The most immediate threat AI posed was to Academic Integrity with Higher Education Institutes (HEIs) focusing their efforts on combating the use of GenAI in assessment. Guidelines were developed for staff and students, policies put in place too. Innovative educators have forged paths in the use of Generative AI for teaching, learning and assessments leading to pockets of transformation springing up across HEIs, often with little or no top-down guidance, support or direction.
This Gasta posits a strategic approach to integrating AI into HEIs to prepare staff, students and the curriculum for an evolving world and workplace. We will highlight the advantages of working with these technologies beyond the realm of teaching, learning and assessment by considering prompt engineering skills, industry impact, curriculum changes, and the need for staff upskilling. In contrast, not engaging strategically with Generative AI poses risks, including falling behind peers, missed opportunities and failing to ensure our graduates remain employable. The rapid evolution of AI technologies necessitates a proactive and strategic approach if we are to remain relevant.
Instructions for Submissions thorugh G- Classroom.pptxJheel Barad
This presentation provides a briefing on how to upload submissions and documents in Google Classroom. It was prepared as part of an orientation for new Sainik School in-service teacher trainees. As a training officer, my goal is to ensure that you are comfortable and proficient with this essential tool for managing assignments and fostering student engagement.
Operation “Blue Star” is the only event in the history of Independent India where the state went into war with its own people. Even after about 40 years it is not clear if it was culmination of states anger over people of the region, a political game of power or start of dictatorial chapter in the democratic setup.
The people of Punjab felt alienated from main stream due to denial of their just demands during a long democratic struggle since independence. As it happen all over the word, it led to militant struggle with great loss of lives of military, police and civilian personnel. Killing of Indira Gandhi and massacre of innocent Sikhs in Delhi and other India cities was also associated with this movement.
Acetabularia Information For Class 9 .docxvaibhavrinwa19
Acetabularia acetabulum is a single-celled green alga that in its vegetative state is morphologically differentiated into a basal rhizoid and an axially elongated stalk, which bears whorls of branching hairs. The single diploid nucleus resides in the rhizoid.
Macroeconomics- Movie Location
This will be used as part of your Personal Professional Portfolio once graded.
Objective:
Prepare a presentation or a paper using research, basic comparative analysis, data organization and application of economic information. You will make an informed assessment of an economic climate outside of the United States to accomplish an entertainment industry objective.
Synthetic Fiber Construction in lab .pptxPavel ( NSTU)
Synthetic fiber production is a fascinating and complex field that blends chemistry, engineering, and environmental science. By understanding these aspects, students can gain a comprehensive view of synthetic fiber production, its impact on society and the environment, and the potential for future innovations. Synthetic fibers play a crucial role in modern society, impacting various aspects of daily life, industry, and the environment. ynthetic fibers are integral to modern life, offering a range of benefits from cost-effectiveness and versatility to innovative applications and performance characteristics. While they pose environmental challenges, ongoing research and development aim to create more sustainable and eco-friendly alternatives. Understanding the importance of synthetic fibers helps in appreciating their role in the economy, industry, and daily life, while also emphasizing the need for sustainable practices and innovation.
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.