8

Quarter 1
Lesson Exemplar Lesson

for Mathematics 4

PILOT IMPLEMENTATION OF THE MATATAG K TO 10 CURRICULUM

Lesson Exemplar for Mathematics Grade 8
Quarter 1: Lesson 4 (Week 4)
SY 2024-2025

This material is intended exclusively for the use of teachers participating in the pilot implementation of the MATATAG K to 10 Curriculum during the
School Year 2024-2025. It aims to assist in delivering the curriculum content, standards, and lesson competencies. Any unauthorized reproduction, distribution,
modification, or utilization of this material beyond the designated scope is strictly prohibited and may result in appropriate legal actions and disciplinary measures.

Borrowed content included in this material are owned by their respective copyright holders. Every effort has been made to locate and obtain permission
to use these materials from their respective copyright owners. The publisher and development team do not represent nor claim ownership over them.

Development Team
Writer:
• Argiel L. Agapay (Liliw National High School)

Validators:
• Roldan S. Cardona (Philippine Normal University – North Luzon)
• PNU – RITQ Development Team

Management Team
Philippine Normal University
Research Institute for Teacher Quality
SiMERR National Research Centre

Every care has been taken to ensure the accuracy of the information provided in this material. For inquiries or feedback, please write or call the Office
of the Director of the Bureau of Learning Resources via telephone numbers (02) 8634-1072 and 8631-6922 or by email at blr.od@deped.gov.ph.
MATHEMATICS / QUARTER 1 / GRADE 8

I. CURRICULUM CONTENT, STANDARDS, AND LESSON COMPETENCIES

A. Content The learners should have knowledge and understanding of special products for binomials, and factorization of
Standards polynomials.

B. Performance
By the end of the quarter, the learners are able to obtain special binomial products.
Standards

C. Learning Learning Competency

Competencies 1. The learners are able to use special product patterns to multiply binomials.
and Objectives Learning Objectives
By the end of the lesson, the learners are expected to:
1. solve for the square of binomial;
2. solve for the product of sum and difference of two terms;
3. solve for the cube of binomial; and
4. solve for the square of a trinomial.

D. Content Special Products

1. Square of Binomial
2. Product of Sum and Difference of Two Terms
3. Cube of Binomial
4. Square of Trinomial

E. Integration

II. LEARNING RESOURCES

Marecek, L., Anthony-Smith, M., Mathis, A. H., (2020, April 22). 6.4 Special Products - Elementary Algebra 2E | OpenStax.
https://openstax.org/books/elementary-algebra-2e/pages/6-4-special-products
“Ex.1: Multiplying binomial to get difference of squares”. Khan Academy.
https://www.khanacademy.org/math/algebra/x2f8bb11595b61c86:quadratics-multiplying-factoring/x2f8bb11595b61c86:special-
product-binomials/v/special-polynomials-products-1
“Ex.2: Finding the square of binomial with one variable”. Khan Academy.
https://www.khanacademy.org/math/algebra/x2f8bb11595b61c86:quadratics-multiplying-factoring/x2f8bb11595b61c86:special-
product-binomials/v/square-a-binomial

1
III. TEACHING AND LEARNING PROCEDURE NOTES TO TEACHERS

A. Activating Prior DAY 1 This activity is intended to recall

Knowledge 1. Short Review the concept of multiplication and
Activity 1: Do You Remember? division of monomials and
Instruction: Let the learners multiply/divide the given polynomials. multinomials. Allot enough time
a. (𝑥 + 3)(𝑥 − 4) for the learners to solve each
b. (𝑥 + 3)(𝑥 2 − 2𝑥 + 3) problem on their own. After that,
c.
12𝑥 4 −16𝑥 3 +8𝑥 2 you may call learners to show
4𝑥 2 and explain their work on the
board.
2. Feedback (Optional)

B. Establishing 1. Lesson Purpose

Lesson Purpose Activity 2: Find the Area!
Instruction: Let the learners read and analyze the situation and answer the Activity 2 is intended to give the
questions that follow. learner an idea regarding one of
the special products which is the
A contractor wants to extend a house on square of binomial. You may also
two sides. The figure below shows the add other questions if necessary.
layout of the plan.
Guide Questions: Answer Key:
1. What is the original area of the house? 1. 2500 sq. ft
2. Represent the area of the house after it 2. (x + 50)2= (x2+100x+2500)
is extended and solve for the product. sq. ft
3. If x = 10, what is the area of the 3. 3600 sq. ft
extension of the house?

2. Unlocking Content Vocabulary

Special Products is a Mathematical term in which factors are combined to form
products. It is called "special" because they do not need long solutions.

C. Developing and SUB-TOPIC 1: Square of Binomial You may allot 1 day for each
Deepening 1. Explicitation subtopic of this lesson.
Understanding For the square of binomial, we have the following equation:
(𝑎 ± 𝑏)2 = 𝑎2 ± 2𝑎𝑏 + 𝑏 2
where a is the first term and b is the second term of the binomial. The result of
squaring a binomial is called a perfect square trinomial.

2
2. Worked Example
Example 1: From Activity 2, after extending the house by x ft, the area is
represented as the square of (50 + 𝑥) or (50 + 𝑥)2 and we can solve this by using
the patter for the square of a binomial.
Solution: Since the first term is 50 and the second term is x.
Square the first term (50)2 = 2500
Multiply the product of the first and
2(50)(x) = 100x
second term by 2
Square the second term (x)2 = x 2
Hence, the answer is 2500 + 100𝑥 + 𝑥 or 𝑥 + 100𝑥 + 2500 𝑠𝑞. 𝑓𝑡.
2 2

Example 2: (𝑥 − 8)2
Solution: Since the first term is x and the second term is -8.
Square the first term (x)2 = x 2
Multiply the product of the first and
2(x)(−8) = −16x
second term by 2
Square the second term (−8)2 = 64
Hence, the answer is 𝑥 − 16𝑥 + 64.
2

Example 3: (−3𝑎 + 7)2

Solution: Since the first term is -3a and the second term is 7.
Square the first term (−3a)2 = 9a2
Multiply the product of the first and
2(−3a)(7) = −42a
second term by 2
Square the second term (7)2 = 49
Hence, the answer is 9𝑎 − 42𝑎 + 49.
2

Example 4: (5𝑥𝑦 − 3)2

Solution: Since the first term is 5𝑥𝑦 and the second term is -3.
Square the first term (5xy)2 = 25x 2 y 2
Multiply the product of the first and
2(5xy)(−3) = −30xy
second term by 2
Square the second term (−3)2 = 9 You may add more examples if
Hence, the answer is 25𝑥 𝑦 − 30𝑥𝑦 + 9.
2 2 needed.

3
3. Lesson Activity Provide enough time for the
Activity 3A: Square It to Solve It! learners to accomplish this
Instruction: Let the learners use the concept of square of binomial in solving the activity.
following.
1. (𝑥 + 3)2 6. (−2𝑥 − 7)2 You may adjust the indicated
2. (𝑦 − 2)2
7. (𝑎𝑏 + 3)2 time in the worksheet for this
3. (3𝑢 − 1) 2
8. (𝑥𝑦 − 7)2 activity if necessary.
4. (4𝑘 + 5)2 9. (4𝑥 − 3𝑦)2
5. (−9𝑛 + 1) 2
10. (5𝑎 + 7𝑏)2 Activity 3A Answer Key:
1. 𝑥 2 + 6𝑥 + 9
DAY 2 2. 𝑦 2 − 4𝑥 + 4
SUB-TOPIC 2: Product of Sum and Difference of Two Terms 3. 9𝑢2 − 6𝑢 + 1
1. Explicitation 4. 16𝑘 2 + 40𝑘 + 25
In finding the product of the sum and difference of two terms, we can use: 5. 81𝑛2 − 18𝑛 + 1
(𝑎 + 𝑏)(𝑎 − 𝑏) = 𝑎2 − 𝑏 2 6. 4𝑥 2 + 28𝑥 + 49
where a is the first term and b is the second term. The result of getting the 7. 𝑎2 𝑏 2 + 6𝑎𝑏 + 9
product of the sum and difference of two terms is a difference of two squares. 8. 𝑥 2 𝑦 2 − 14𝑥𝑦 + 49
9. 16𝑥 2 − 24𝑥𝑦 + 9𝑦 2
2. Worked Example 10. 25𝑎2 + 70𝑎𝑏 + 49𝑏 2
Example 1: (𝑥 + 3)(𝑥 − 3)
Solution: Since the first term is x and the second term is 3
(𝑥 + 3)(𝑥 − 3) = (𝑥)2 − (3)2
Hence the answer is 𝑥 − 9
2

Example 2: (2𝑦 − 5)(2𝑦 + 5)

Solution: Since the first term is 2y and the second term is 5
(2𝑦 + 5)(2𝑦 − 5) = (2𝑦)2 − (5)2
Hence the answer is 4𝑦 − 25
2

1 1
Example 3: (3 + 4𝑥) (3 − 4𝑥)
1
Solution: Since the first term is 3
and the second term is 4x
1 1 1 2
(3 + 4𝑥) (3 − 4𝑥) = (3) − (4𝑥)2
1
Hence the answer is 9 − 16𝑥 2

4
 Example 4: (−𝑎2 − 𝑥𝑦)(−𝑎2 + 𝑥𝑦) You may add more examples if
Solution: Since the first term is -a2 and the second term is xy needed.
(−𝑎2 − 𝑥𝑦)(−𝑎2 + 𝑥𝑦) = (−𝑎2 )2 − (𝑥𝑦)2
Hence the answer is 𝑎4 − 𝑥 2 𝑦 2

3. Lesson Activity
Activity 3B: Finding Errors!
Instruction: Let the learners analyze the given problem and its solution. If the Provide enough time for the
solution and answer is correct, shade the smiley face emoji. Otherwise, shade learners to accomplish this
the sad face emoji and write the correct solution and answer in the space activity.
provided. You may adjust the indicated
= (𝑡)2 − (7)2 time in the worksheet for this
1. (𝑡 + 7)(𝑡 − 7) activity if necessary.
= 𝑡 2 − 49
= (𝑚)2 − (6)2 Activity 3B Answer Key:
2. (𝑚 + 6)(𝑚 − 6)
= 𝑚2 − 12
1.
= (4𝑥)2 − (1)2
3. (4𝑥 − 1)(4𝑥 + 1) = (𝑚)2 − (6)2
= 4𝑥 2 − 1 2.
= 𝑚2 − 36
= (4)2 − (2𝑏)2 = (4𝑥)2 − (1)2
4. (4 − 2𝑏)(4 + 2𝑏) 3.
= 16 − 4𝑏 2 = 16𝑥 2 − 1
= (3𝑘)2 − (5)2 4.
5. (3𝑘 + 5)(3𝑘 − 5)
= 9𝑘 2 − 25
5.
= (𝑝)2 − (10𝑞)2
6. (𝑝 + 10𝑞)(𝑝 − 10𝑞)
= 𝑝2 − 100𝑞2 6.
2 2
= (7𝑚) − (8𝑛) = (7𝑚)2 − (8𝑛)2
7. (7𝑚 + 8𝑛)(7𝑚 − 8𝑛) 7.
= 49𝑚 − 64𝑛 = 49𝑚2 − 64𝑛2
1 2 1 2
2
= (𝑐) − ( ) = (𝑐)2 − (4)
8. (𝑐
1
+ 4) (𝑐 −
1
) 4 8. 1
4 2 = 𝑐 2 − 16
= 𝑐2 −
4
9.
= (−𝑦)2 − (4)2
9. (−𝑦 + 4)(−𝑦 − 4) = (3𝑎𝑏)2 − (𝑥𝑦)2
= 𝑦 2 − 16 10.
= 9𝑎2 𝑏2 − 𝑥 2 𝑦 2
= (3𝑎𝑏)2 − (𝑥𝑦)2
10. (3𝑎𝑏 + 𝑥𝑦)(3𝑎𝑏 − 𝑥𝑦)
= 9𝑎𝑏 2 − 𝑥𝑦 2

5
DAY 3
SUB-TOPIC 3: Cube of Binomial
1. Explicitation
For the cube of binomial, we have the following equation:
(𝑎 + 𝑏)3 = 𝑎3 + 3𝑎2 𝑏 + 3𝑎𝑏 2 + 𝑏 3
(𝑎 − 𝑏)3 = 𝑎3 − 3𝑎2 𝑏 − 3𝑎𝑏 2 − 𝑏 3
where a is the first term and b is the second term.

2. Worked Example
Example 1: (𝑥 + 2)3
Solution: Since the first term is x and the second term is 2
(𝑥 + 2)3 = (𝑥)3 + 3(𝑥)2 (2) + 3(𝑥)(2)2 + (2)3 You may add more examples if
Hence the answer is 𝑥 3 + 6𝑥 2 + 12𝑥 + 8 needed.
Example 2: (2𝑦 − 1)3
Solution: Since the first term is 2y and the second term is 1
(2𝑦 − 1)3 = (2𝑦)3 − 3(2𝑦)2 (1) + 3(2𝑦)(1)2 − (1)3
Hence the answer is 8𝑦 3 − 12𝑦 2 + 6𝑦 − 1 Provide enough time for the
Example 3: (3𝑥 + 4𝑦) 3 learners to accomplish this
Solution: Since the first term is 3x and the second term is 4y activity.
(3𝑥 + 4𝑦)3 = (3𝑥)3 + 3(3𝑥)2 (4𝑦) + 3(3𝑥)(4𝑦)2 + (4𝑦)3
Hence the answer is 27𝑥 3 + 108𝑥 2 𝑦 + 144𝑥𝑦 2 + 64𝑦 3 You may adjust the indicated
time in the worksheet for this
Example 4: (𝑏 2 − 3)3 activity if necessary.
Solution: Since the first term is b2 and the second term is 3 Activity 3C Answer Key:
(𝑏 2 − 3)3 = (𝑏 2 )3 − 3(𝑏 2 )2 (3) + 3(𝑏 2 )(3)2 − (3)3 1. 𝑥 3 − 15𝑥 2 + 75𝑥 − 125
Hence the answer is 𝑏 6 − 9𝑏 4 + 27𝑏 2 − 27 2. 𝑦 3 + 9𝑦 2 + 27𝑦 + 27
3. 𝑚3 + 3𝑚2 + 3𝑚 + 1
3. Lesson Activity 4. 27𝑥 3 + 189𝑥 2 + 441𝑥 + 343
Activity 3C: What’s the Product? 5. 8𝑎3 − 24𝑎2 + 24𝑎 − 8
Instruction: Let the learners find the product of the following by using the 6. 125𝑥 3 − 450𝑥 2 + 540𝑥 − 216
concept of cube of binomial. 7. 𝑥 3 + 3𝑥 2 𝑦 + 3𝑥𝑦 2 + 𝑦 3
1. (𝑥 − 5)3 6. (5𝑥 − 6)3 8. 27𝑝3 − 108𝑝2 𝑞 + 144𝑝𝑞2 −
2. (𝑦 + 3)3 7. (𝑥 + 𝑦)3 64𝑞 3
3. (𝑚 + 1) 3
8. (3𝑝 − 4𝑞)3 9. 125𝑥 3 + 225𝑥 2 𝑦 + 135𝑥𝑦 2 +
4. (3𝑥 + 7) 3
9. (5𝑥 + 3𝑦)3 27𝑦 3
1 1
5. (2𝑎 − 2)3 10. (3𝑐 2 − 3)3 10. 27𝑐 6 − 9𝑐 4 + 𝑐 2 − 27

6
DAY 4
SUB-TOPIC 4: Square of Trinomial
1. Explicitation
In finding the square of a trinomial, we can use:
(𝑎 + 𝑏 + 𝑐)2 = 𝑎2 + 𝑏 2 + 𝑐 2 + 2𝑎𝑏 + 2𝑎𝑐 + 2𝑏𝑐
where a is the first term, b is the second term and c is the third term.

2. Worked Example
Example 1: (𝑥 + 𝑦 + 𝑧)2
Solution: Since the first term is x, the second term is y and the third term is z
Square the first term (x)2 = x 2
Square of the second term (y)2 = y 2
Square the third term (z)2 = z 2
Multiply the product of the first and
2(x)(y) = 2xy
second term by 2
Multiply the product of the first and
2(x)(z) = 2xz
third term by 2
Multiply the product of the second and
2(y)(z) = 2yz
third term by 2
Hence, the answer is 𝑥 2 + 𝑦 2 + 𝑧 2 + 2𝑥𝑦 + 2𝑥𝑧 + 2𝑦𝑧.

Example 2: (𝑎 + 2𝑏 + 3𝑐)2
Solution: Since the first term is a, the second term is 2b, the third term is 3c
Square the first term (a)2 = a2
Square of the second term (2b)2 = 4b2
Square the third term (3c)2 = 9c 2
Multiply the product of the first and
2(a)(2b) = 4ab
second term by 2
Multiply the product of the first and
2(a)(3c) = 6ac
third term by 2
Multiply the product of the second and
2(2b)(3c) = 12bc
third term by 2
Hence, the answer is 𝑎2 + 4𝑏 2 + 9𝑐 2 + 4𝑎𝑏 + 6𝑎𝑐 + 12𝑏𝑐.

Example 3: (2𝑥 − 3𝑦 − 4𝑧)2

Solution: Since the first term is 2x, the second term is -3y, the third term is -4z
Square the first term (2x)2 = 4x 2

7
 Square of the second term (−3y)2 = 9y 2
Square the third term (−4z)2 = 16z 2
Multiply the product of the first and
2(2x)(−3y) = −12xy
second term by 2
Multiply the product of the first and
2(2x)(−4z) = −16xz
third term by 2
Multiply the product of the second and You may add more examples if
2(−3y)(−4z) = 24yz
third term by 2 needed.
Hence, the answer is 4𝑥 2 + 9𝑦 2 + 16𝑧 2 − 12𝑥𝑦 − 16𝑥𝑧 + 24𝑦𝑧.

3. Lesson Activity Provide enough time for the

Activity 3D: Complete It! learners to accomplish this
Instruction: Let the learners complete the solution for each square of trinomial. activity.
1. (𝑥 + 3𝑦 + 5𝑧)2
Square the first term _______________ You may adjust the indicated
Square of the second term _______________ time in the worksheet for this
Square the third term _______________ activity if necessary.
Multiply the product of the first and
_______________ Activity 3D Answer Key:
second term by 2
Multiply the product of the first and third 1. (𝑥)2 = 𝑥 2
_______________
term by 2 (3𝑦)2 = 9𝑦 2
Multiply the product of the second and (5𝑧)2 = 25𝑧 2
_______________
third term by 2 2(𝑥)(3𝑦) = 6𝑥𝑦
Answer: _________________________________________ 2(𝑥)(5𝑧) = 10𝑥𝑧
2(3𝑦)(5𝑧) = 30𝑦𝑧
2. (4𝑎 − 7𝑏 − 2𝑐)2 𝑥 2 + 9𝑦 2 + 25𝑧 2 + 6𝑥𝑦 + 10𝑥𝑧
Square the first term _______________ + 30𝑦𝑧
Square of the second term _______________ 2. (4𝑎)2 = 16𝑎2
Square the third term _______________ (−7𝑏)2 = 49𝑏2
Multiply the product of the first and (−2𝑐)2 = 4𝑐 2
_______________
second term by 2 2(4𝑎)(−7𝑏) = −56𝑎𝑏
Multiply the product of the first and third 2(4𝑎)(−2𝑐) = −16𝑎𝑐
_______________
term by 2 2(−7𝑏)(−2𝑐) = 28𝑏𝑐
Multiply the product of the second and 16𝑎2 + 49𝑏 2 + 4𝑐 2 − 56𝑎𝑏
_______________
third term by 2 − 16𝑎𝑐 + 28𝑏𝑐
Answer: _________________________________________

8
 3. (−2𝑝 + 4𝑞 + 5)2 3. (−2𝑝)2 = 4𝑝2
Square the first term _______________ (4𝑞)2 = 16𝑞2
Square of the second term _______________ (5)2 = 25
Square the third term _______________ 2(−2𝑝)(4𝑞) = −16𝑝𝑞
Multiply the product of the first and 2(−2𝑝)(5) = −20𝑝
_______________
second term by 2 2(4𝑞)(5) = 40𝑞
Multiply the product of the first and third 4𝑝2 + 16𝑞 2 − 16𝑝𝑞 − 20𝑝 + 40𝑞
_______________
term by 2 + 25
Multiply the product of the second and
_______________
third term by 2 Note: For no.3 the square of the
Answer: _________________________________________ third term should be the last term
since it is a constant.

D. Making DAY 4 The activity is intended to

Generalizations Learners’ Takeaways and Reflection on Learning determine what the learners
Activity 4: Closing the Loop! have learned as well as to give
Instruction: Let the learners answer the following questions. feedback to their experiences
1. What are the key concepts of our lesson? during the lesson. Allot enough
2. Which part of the lesson is the easiest for you? Why? time to listen and process the
3. Which part of the lesson is the hardest for you? Why? responses of your learners. You
4. How are we as a class today? may do this activity at the end of
each subtopic. You may also add
questions if needed.

IV. EVALUATING LEARNING: FORMATIVE ASSESSMENT AND TEACHER’S REFLECTION NOTES TO TEACHERS

A. Evaluating 1. Formative Assessment Answer Key:

Learning Activity 5: Let’s Solve It! 1. 9𝑥 2 + 30𝑥 + 25
Instruction: Let the learners use special products in solving the following. 2. 25𝑚2 − 1
1. (3𝑥 + 5)2 3. 8𝑝3 − 36𝑝2 + 54𝑝 − 27
2. (5𝑚2 + 1)(5𝑚2 − 1) 4. 𝑥 6 + 3𝑥 4 𝑥 2 + 3𝑥 2 𝑦 4 + 𝑦 6
3. (2𝑝 − 3)3 5. 𝑎2 + 4𝑏 2 + 9𝑐 2 + 4𝑎𝑏 − 12𝑏𝑐 − 6𝑎𝑐
4. (𝑥 2 + 𝑦 2 )3
5. (𝑎 + 2𝑏 − 3𝑐)2

2. Homework (Optional)

9
B. Teacher’s Note observations on any The teacher may take note of
Effective Practices Problems Encountered
Remarks of the following areas: some observations related to the
effective practices and problems
strategies explored encountered after utilizing the
different strategies, materials
materials used used, learner engagement, and
other related stuff.
learner engagement/
interaction
Teachers may also suggest ways
others to improve the different activities
explored/lesson exemplar.

C. Teacher’s Reflection guide or prompt can be on: Teacher’s reflection in every

Reflection • principles behind the teaching lesson conducted/facilitated is
What principles and beliefs informed my lesson? essential and necessary to
Why did I teach the lesson the way I did? improve practice. You may also
consider this as an input for the
• students LAC/Collab sessions.
What roles did my students play in my lesson?
What did my students learn? How did they learn?

• ways forward
What could I have done differently?
What can I explore in the next lesson?

10