This document discusses solving polynomial equations by factoring polynomials. It begins with essential questions and vocabulary about factoring polynomials and solving polynomial equations by factoring. It then provides the number of terms in a polynomial and the corresponding factoring technique that can be used. Examples of factoring various polynomials are also provided. The document aims to teach students how to factor polynomials and solve polynomial equations by factoring.
This presentation explains the basic information about Polynomial Function and Synthetic Division. Examples were given about easy ways to divide polynomial function using synthetic division. It also contains the steps on how to perform the division method of polynomial functions.
This presentation explains the basic information about Polynomial Function and Synthetic Division. Examples were given about easy ways to divide polynomial function using synthetic division. It also contains the steps on how to perform the division method of polynomial functions.
This PPT explains the concept of polynomial in detail. It describes the meaning of polynomials with the help of different examples.Furthermore different types of polynomials on the basis of degree and number of terms.This will be helpful for students and for teachers.
This PPT explains the concept of polynomial in detail. It describes the meaning of polynomials with the help of different examples.Furthermore different types of polynomials on the basis of degree and number of terms.This will be helpful for students and for teachers.
We all have good and bad thoughts from time to time and situation to situation. We are bombarded daily with spiraling thoughts(both negative and positive) creating all-consuming feel , making us difficult to manage with associated suffering. Good thoughts are like our Mob Signal (Positive thought) amidst noise(negative thought) in the atmosphere. Negative thoughts like noise outweigh positive thoughts. These thoughts often create unwanted confusion, trouble, stress and frustration in our mind as well as chaos in our physical world. Negative thoughts are also known as “distorted thinking”.
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.
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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.
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.
Students, digital devices and success - Andreas Schleicher - 27 May 2024..pptxEduSkills OECD
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How to Create Map Views in the Odoo 17 ERPCeline George
The map views are useful for providing a geographical representation of data. They allow users to visualize and analyze the data in a more intuitive manner.
5. Vocabulary
1. Prime Polynomials: Polynomials that cannot be
factored
2. Quadratic Form: Rewriting a polynomial so that it
fits the pattern of a standard form quadratic;
occurs with the degree of the first term is twice
that of the second, with a constant third term
6. Number
of Terms
Factoring Technique General Case
Any
Greatest Common Factor
(GCF)
2 Difference of Squares
2 Sum of Cubes
2 Difference of Cubes
3 Perfect Square Trinomial
3 General Trinomials
4+ Grouping
7. Number
of Terms
Factoring Technique General Case
Any
Greatest Common Factor
(GCF)
2 Difference of Squares
2 Sum of Cubes
2 Difference of Cubes
3 Perfect Square Trinomial
3 General Trinomials
4+ Grouping
4a5
b7
+12ab = 4ab(a4
b6
+ 3)
8. Number
of Terms
Factoring Technique General Case
Any
Greatest Common Factor
(GCF)
2 Difference of Squares
2 Sum of Cubes
2 Difference of Cubes
3 Perfect Square Trinomial
3 General Trinomials
4+ Grouping
4a5
b7
+12ab = 4ab(a4
b6
+ 3)
a2
− b2
= (a + b)(a − b)
9. Number
of Terms
Factoring Technique General Case
Any
Greatest Common Factor
(GCF)
2 Difference of Squares
2 Sum of Cubes
2 Difference of Cubes
3 Perfect Square Trinomial
3 General Trinomials
4+ Grouping
4a5
b7
+12ab = 4ab(a4
b6
+ 3)
a2
− b2
= (a + b)(a − b)
a3
+ b3
= (a + b)(a2
− ab + b2
)
10. Number
of Terms
Factoring Technique General Case
Any
Greatest Common Factor
(GCF)
2 Difference of Squares
2 Sum of Cubes
2 Difference of Cubes
3 Perfect Square Trinomial
3 General Trinomials
4+ Grouping
4a5
b7
+12ab = 4ab(a4
b6
+ 3)
a2
− b2
= (a + b)(a − b)
a3
+ b3
= (a + b)(a2
− ab + b2
)
a3
− b3
= (a − b)(a2
+ ab + b2
)
11. Number
of Terms
Factoring Technique General Case
Any
Greatest Common Factor
(GCF)
2 Difference of Squares
2 Sum of Cubes
2 Difference of Cubes
3 Perfect Square Trinomial
3 General Trinomials
4+ Grouping
4a5
b7
+12ab = 4ab(a4
b6
+ 3)
a2
− b2
= (a + b)(a − b)
a3
+ b3
= (a + b)(a2
− ab + b2
)
a3
− b3
= (a − b)(a2
+ ab + b2
)
a2
+ 2ab + b2
= (a + b)2
12. Number
of Terms
Factoring Technique General Case
Any
Greatest Common Factor
(GCF)
2 Difference of Squares
2 Sum of Cubes
2 Difference of Cubes
3 Perfect Square Trinomial
3 General Trinomials
4+ Grouping
4a5
b7
+12ab = 4ab(a4
b6
+ 3)
a2
− b2
= (a + b)(a − b)
a3
+ b3
= (a + b)(a2
− ab + b2
)
a3
− b3
= (a − b)(a2
+ ab + b2
)
a2
+ 2ab + b2
= (a + b)2
a2
− 2ab + b2
= (a − b)2
13. Number
of Terms
Factoring Technique General Case
Any
Greatest Common Factor
(GCF)
2 Difference of Squares
2 Sum of Cubes
2 Difference of Cubes
3 Perfect Square Trinomial
3 General Trinomials
4+ Grouping
4a5
b7
+12ab = 4ab(a4
b6
+ 3)
a2
− b2
= (a + b)(a − b)
a3
+ b3
= (a + b)(a2
− ab + b2
)
a3
− b3
= (a − b)(a2
+ ab + b2
)
a2
+ 2ab + b2
= (a + b)2
a2
− 2ab + b2
= (a − b)2
acx 2
+ (ad + bc)x + bd
= (ax + b)(cx + d )
14. Number
of Terms
Factoring Technique General Case
Any
Greatest Common Factor
(GCF)
2 Difference of Squares
2 Sum of Cubes
2 Difference of Cubes
3 Perfect Square Trinomial
3 General Trinomials
4+ Grouping
4a5
b7
+12ab = 4ab(a4
b6
+ 3)
a2
− b2
= (a + b)(a − b)
a3
+ b3
= (a + b)(a2
− ab + b2
)
a3
− b3
= (a − b)(a2
+ ab + b2
)
a2
+ 2ab + b2
= (a + b)2
a2
− 2ab + b2
= (a − b)2
acx 2
+ (ad + bc)x + bd
= (ax + b)(cx + d )
ax + bx + ay + by
= (a + b)(x + y )
15. Example 1
Factor each polynomial. If the polynomial cannot be
factored, write prime.
a. x 3
−1000
b. 24x 5
+ 3x 2
y 3
c. x 3
+ 400
16. Example 1
Factor each polynomial. If the polynomial cannot be
factored, write prime.
a. x 3
−1000
(x −10)(x 2
+10x +100)
b. 24x 5
+ 3x 2
y 3
c. x 3
+ 400
17. Example 1
Factor each polynomial. If the polynomial cannot be
factored, write prime.
a. x 3
−1000
(x −10)(x 2
+10x +100)
b. 24x 5
+ 3x 2
y 3
3x 2
(8x 3
+ y 3
)
c. x 3
+ 400
18. Example 1
Factor each polynomial. If the polynomial cannot be
factored, write prime.
a. x 3
−1000
(x −10)(x 2
+10x +100)
b. 24x 5
+ 3x 2
y 3
3x 2
(8x 3
+ y 3
)
3x 2
(2x + y )(4x 2
− 2xy + y 2
)
c. x 3
+ 400
19. Example 1
Factor each polynomial. If the polynomial cannot be
factored, write prime.
a. x 3
−1000
(x −10)(x 2
+10x +100)
b. 24x 5
+ 3x 2
y 3
3x 2
(8x 3
+ y 3
)
3x 2
(2x + y )(4x 2
− 2xy + y 2
)
c. x 3
+ 400
Prime
20. Example 2
Factor each polynomial. If the polynomial cannot be
factored, write prime.
a. x 3
+ 5x 2
− 2x −10
21. Example 2
Factor each polynomial. If the polynomial cannot be
factored, write prime.
a. x 3
+ 5x 2
− 2x −10
(x 3
+ 5x 2
)+ (−2x −10)
22. Example 2
Factor each polynomial. If the polynomial cannot be
factored, write prime.
a. x 3
+ 5x 2
− 2x −10
(x 3
+ 5x 2
)+ (−2x −10)
x 2
(x + 5)− 2(x + 5)
23. Example 2
Factor each polynomial. If the polynomial cannot be
factored, write prime.
a. x 3
+ 5x 2
− 2x −10
(x 3
+ 5x 2
)+ (−2x −10)
x 2
(x + 5)− 2(x + 5)
(x + 5)(x 2
− 2)
24. Example 2
Factor each polynomial. If the polynomial cannot be
factored, write prime.
b. a2
+ 3ay + 2ay 2
+ 6y 3
25. Example 2
Factor each polynomial. If the polynomial cannot be
factored, write prime.
b. a2
+ 3ay + 2ay 2
+ 6y 3
(a2
+ 3ay )+ (2ay 2
+ 6y 3
)
26. Example 2
Factor each polynomial. If the polynomial cannot be
factored, write prime.
b. a2
+ 3ay + 2ay 2
+ 6y 3
(a2
+ 3ay )+ (2ay 2
+ 6y 3
)
a(a + 3y )+ 2y 2
(a + 3y )
27. Example 2
Factor each polynomial. If the polynomial cannot be
factored, write prime.
b. a2
+ 3ay + 2ay 2
+ 6y 3
(a2
+ 3ay )+ (2ay 2
+ 6y 3
)
a(a + 3y )+ 2y 2
(a + 3y )
(a + 3y )(a + 2y 2
)
28. Example 3
Factor each polynomial. If the polynomial cannot be
factored, write prime.
a. x 2
y 3
− 3xy 3
+ 2y 3
+ x 2
z3
− 3xz3
+ 2z3
29. Example 3
Factor each polynomial. If the polynomial cannot be
factored, write prime.
a. x 2
y 3
− 3xy 3
+ 2y 3
+ x 2
z3
− 3xz3
+ 2z3
(x 2
y 3
− 3xy 3
+ 2y 3
)+ (x 2
z3
− 3xz3
+ 2z3
)
30. Example 3
Factor each polynomial. If the polynomial cannot be
factored, write prime.
a. x 2
y 3
− 3xy 3
+ 2y 3
+ x 2
z3
− 3xz3
+ 2z3
(x 2
y 3
− 3xy 3
+ 2y 3
)+ (x 2
z3
− 3xz3
+ 2z3
)
y 3
(x 2
− 3x + 2)+ z3
(x 2
− 3x + 2)
31. Example 3
Factor each polynomial. If the polynomial cannot be
factored, write prime.
a. x 2
y 3
− 3xy 3
+ 2y 3
+ x 2
z3
− 3xz3
+ 2z3
(x 2
y 3
− 3xy 3
+ 2y 3
)+ (x 2
z3
− 3xz3
+ 2z3
)
y 3
(x 2
− 3x + 2)+ z3
(x 2
− 3x + 2)
(x 2
− 3x + 2)(y 3
+ z3
)
32. Example 3
Factor each polynomial. If the polynomial cannot be
factored, write prime.
a. x 2
y 3
− 3xy 3
+ 2y 3
+ x 2
z3
− 3xz3
+ 2z3
(x 2
y 3
− 3xy 3
+ 2y 3
)+ (x 2
z3
− 3xz3
+ 2z3
)
y 3
(x 2
− 3x + 2)+ z3
(x 2
− 3x + 2)
(x 2
− 3x + 2)(y 3
+ z3
)
(x − 2)(x −1)(y + z)(y 2
− yz + z2
)
33. Example 3
Factor each polynomial. If the polynomial cannot be
factored, write prime.
b. 64x 6
− y 6
34. Example 3
Factor each polynomial. If the polynomial cannot be
factored, write prime.
b. 64x 6
− y 6
(8x 3
+ y 3
)(8x 3
− y 3
)
35. Example 3
Factor each polynomial. If the polynomial cannot be
factored, write prime.
b. 64x 6
− y 6
(8x 3
+ y 3
)(8x 3
− y 3
)
(2x + y )(4x 2
− 2xy + y 2
)(2x − y )(4x 2
+ 2xy + y 2
)
36. Example 4
A small cube is cut from a larger cube.
Determine the dimensions of the cubes if the
length of the smaller cube is is one half the
length of the larger cube, and the volume of the
object is 23,625 cubic centimeters.
37. Example 4
A small cube is cut from a larger cube.
Determine the dimensions of the cubes if the
length of the smaller cube is is one half the
length of the larger cube, and the volume of the
object is 23,625 cubic centimeters.
x
x
x
38. Example 4
A small cube is cut from a larger cube.
Determine the dimensions of the cubes if the
length of the smaller cube is is one half the
length of the larger cube, and the volume of the
object is 23,625 cubic centimeters.
x 3
−
x
2
⎛
⎝⎜
⎞
⎠⎟
3
= 23625
x
x
x
39. Example 4
A small cube is cut from a larger cube.
Determine the dimensions of the cubes if the
length of the smaller cube is is one half the
length of the larger cube, and the volume of the
object is 23,625 cubic centimeters.
x 3
−
x
2
⎛
⎝⎜
⎞
⎠⎟
3
= 23625
x
x
x
x 3
−
1
8
x 3
= 23625
40. Example 4
A small cube is cut from a larger cube.
Determine the dimensions of the cubes if the
length of the smaller cube is is one half the
length of the larger cube, and the volume of the
object is 23,625 cubic centimeters.
x 3
−
x
2
⎛
⎝⎜
⎞
⎠⎟
3
= 23625
x
x
x
x 3
−
1
8
x 3
= 23625
7
8
x 3
= 23625
41. Example 4
A small cube is cut from a larger cube.
Determine the dimensions of the cubes if the
length of the smaller cube is is one half the
length of the larger cube, and the volume of the
object is 23,625 cubic centimeters.
x 3
−
x
2
⎛
⎝⎜
⎞
⎠⎟
3
= 23625
x
x
x
x 3
−
1
8
x 3
= 23625
7
8
x 3
= 23625
x 3
= 27000
42. Example 4
A small cube is cut from a larger cube.
Determine the dimensions of the cubes if the
length of the smaller cube is is one half the
length of the larger cube, and the volume of the
object is 23,625 cubic centimeters.
x 3
−
x
2
⎛
⎝⎜
⎞
⎠⎟
3
= 23625
x
x
x
x 3
−
1
8
x 3
= 23625
7
8
x 3
= 23625
x 3
= 27000
x 33
= 270003
43. Example 4
A small cube is cut from a larger cube.
Determine the dimensions of the cubes if the
length of the smaller cube is is one half the
length of the larger cube, and the volume of the
object is 23,625 cubic centimeters.
x 3
−
x
2
⎛
⎝⎜
⎞
⎠⎟
3
= 23625
x
x
x
x 3
−
1
8
x 3
= 23625
7
8
x 3
= 23625
x 3
= 27000
x 33
= 270003
x = 30
44. Example 4
A small cube is cut from a larger cube.
Determine the dimensions of the cubes if the
length of the smaller cube is is one half the
length of the larger cube, and the volume of the
object is 23,625 cubic centimeters.
x 3
−
x
2
⎛
⎝⎜
⎞
⎠⎟
3
= 23625
x
x
x
x 3
−
1
8
x 3
= 23625
7
8
x 3
= 23625
x 3
= 27000
x 33
= 270003
x = 30
The larger cube has a length of 30cm.
The smaller cube has a length of 15 cm.
45. Example 5
Write each expression in quadratic form, if possible.
a. 2x 6
− x 3
+ 9 b. x 4
+ 2x 3
−1
46. Example 5
Write each expression in quadratic form, if possible.
a. 2x 6
− x 3
+ 9
2(x 3
)2
− (x 3
)+ 9
b. x 4
+ 2x 3
−1
47. Example 5
Write each expression in quadratic form, if possible.
a. 2x 6
− x 3
+ 9
2(x 3
)2
− (x 3
)+ 9
Let u = x 3
b. x 4
+ 2x 3
−1
48. Example 5
Write each expression in quadratic form, if possible.
a. 2x 6
− x 3
+ 9
2(x 3
)2
− (x 3
)+ 9
2u2
−u + 9
Let u = x 3
b. x 4
+ 2x 3
−1
49. Example 5
Write each expression in quadratic form, if possible.
a. 2x 6
− x 3
+ 9
2(x 3
)2
− (x 3
)+ 9
2u2
−u + 9
Let u = x 3
b. x 4
+ 2x 3
−1
Not possible