This is the Grade 9 Mathematics self-learning module for Quarter 1, Module 3, “Solving Quadratic Equation by Factoring,” set for Week 1 under learning code M9AL-Ia-2.1. It is for Junior High School learners who have already solved quadratic equations by extracting square roots and are now learning a second method. After the module, learners should be able to factor a quadratic equation, use the Zero Product Property, and find its solutions.
On this page 8 sections
What learners will learn
The module states one learning competency. Learners will be able to:
- solve quadratic equations by: (b) factoring (M9AL-Ia-2.1)
Review: factoring quadratic expressions
The “What’s In” part reviews the factoring learners did in Grade 8, because solving by factoring depends on it. It covers three kinds of expressions:
| Type | Module example | Factored form |
|---|---|---|
| Factoring by GCF | 2x² − 4x | 2x(x − 2) |
| x² + bx + c | x² + 4x − 12 | (x − 2)(x + 6) |
| ax² + bx + c | 2x² + 3x − 2 | (x + 2)(2x − 1) |
- By GCF: Write each term as a product of factors (2x² = 2 · x · x and 4x = 2 · 2 · x). The common factor is 2x. Then use the distributive property.
- x² + bx + c: Look for two factors of c whose sum is b. For x² + 4x − 12, the factors of −12 that add up to 4 are −2 and 6.
- ax² + bx + c: Try factors of a and factors of c until the outer and inner products add up to b. Check your answer by multiplying the binomials: First, Outer, Inner, Last.
The Zero Product Property
This property is the key idea of the module. If the product of two quantities is zero, then one or both of them must be zero. In symbols, if ab = 0, then a = 0 or b = 0. For example, 5(0) = 0 and 0(7) = 0.
The module shows how to use it with three worked examples:
- 4x² = 6x: Subtract 6x to get 4x² − 6x = 0. Factor by GCF: 2x(2x − 3) = 0. So 2x = 0 or 2x − 3 = 0, which gives x = 0 or x = 3/2.
- x² − 3x = 18: Subtract 18 to get x² − 3x − 18 = 0. Factor: (x + 3)(x − 6) = 0. The solutions are −3 and 6.
- 2x² − 3 = −x: Add x to get 2x² + x − 3 = 0. Factor: (2x + 3)(x − 1) = 0. The solutions are −3/2 and 1.
Steps in solving a quadratic equation by factoring
The “What I Have Learned” part sums up the method in six steps:
- Move all the terms to the left side if needed, so that the right side is zero.
- Combine similar terms on the left side.
- Factor the left side.
- Set each factor that contains the variable equal to zero.
- Solve each of these equations.
- Check the answers by substituting them into the original equation.
Real-life problems in the module
The module uses word problems to show where factoring helps:
- Picture Frame (What’s New): A rectangular glass with an area of 168 in² is 2 inches longer than it is wide. Learners draw a diagram, write expressions for its length, width, and area, and write the equation for a larger glass with an area of 960 in².
- Rectangle problem (Assessment): A rectangle’s length is 5 inches more than its width, and its area is 24 square inches. Learners find its dimensions.
- Throwing of Ball (Problem-Based Worksheet): Using h(t) = −16t² + v₀t + h₀, learners model a ball Linda throws from 5 ft with an initial velocity of 16 ft/s, then solve by factoring to find how long she has before it hits the ground.
How to use this module
The module follows this order:
- What I Know: A 10-item multiple-choice pretest. If you get an honest perfect score, the module says you may skip it.
- What’s In: Factoring review with five “Let’s Try!!” items.
- What’s New: The Picture Frame activity.
- What Is It: The Zero Product Property and the worked examples.
- What’s More: Five equations to solve, showing your solution.
- What I Have Learned: The six steps.
- What I Can Do: Ten equations to solve, showing your complete solution and encircling the final answer.
- Assessment: A 10-item test. If your answer is not among the choices, write E with your final answer.
- Additional Activities: A graphic organizer of the steps and a reflection on whether problems in your community also have “factors.”
- Problem-Based Worksheet: The Throwing of Ball problem.
For learners: Write your answers on a separate answer sheet. After the pretest, note the items you missed and look for the right answers as you go through the module.
For parents: Encourage your child to show every step of the solution, since most activities ask for a complete solution, not just the final answer.
For teachers: The module ends with an answer key for What I Know, What’s In, What’s More, What I Can Do, the Assessment, and the Problem-Based Worksheet. Review it before checking work, as a few items may need a second look.
Frequently asked questions
What grade and quarter is this module for?
It is for Grade 9 Mathematics, Quarter 1, Module 3, scheduled for Week 1. Its learning code is M9AL-Ia-2.1.
What is the Zero Product Property?
It states that if the product of two quantities is zero, then one or both of them are zero: if ab = 0, then a = 0 or b = 0. It lets you split a factored quadratic equation into two simple equations.
What should learners know before using this module?
The module expects learners to have solved quadratic equations by extracting square roots and to know factoring from Grade 8. The What’s In part reviews factoring by GCF, x² + bx + c, and ax² + bx + c.
Does the module have an answer key?
Yes. The last page has answers for the pretest, the factoring review, What’s More, What I Can Do, the Assessment, and the Problem-Based Worksheet.
Can a learner skip this module?
The module says that a learner who gets an honest perfect score of 10 on the What I Know pretest may skip it.
Full text of the module
The text below is a transcription of the PDF above, as printed.
Mathematics Quarter 1-Module 3 Solving Quadratic Equation By Factoring
Week 1 Learning Code – M9AL-Ia-2.1
[illustration: teacher and learners in an online learning setting, with books, laptops, and a chat bubble]
GRADE 9 Learning Module for Junior High School Mathematics
Quarter 1 – Module 3 – New Normal Math for G9 First Edition 2020 Copyright © 2020
Republic Act 8293, section 176 states that: No copyright shall subsist in any work of the Government of the Philippines. However, prior approval of the government agency or office wherein the work is created shall be necessary for exploitation of such work for profit. Such agency or office may, among other things, impose as a condition the payment of royalties.
Borrowed materials (i.e. songs, stories, poems, pictures, photos, brand names, trademarks, etc.) included in this module are owned by their respective copyright holders. Every effort had been exerted to locate and seek permission to use these materials from their respective copyright owners. The publisher and authors do not represent nor claim ownership over them.
Published by the Department of Education Secretary: Leonor Magtolis Briones Undersecretary: Diosdado M. San Antonio
Development Team of the Module
Writers: Jhon Edward M. Valera – T1; Rowena F. Reyes – T1; Analyn M. Argel – MTII; Roderick F. Borja – TIII
Editor: Sally C. Caleja – Head Teacher VI; Chuci R. Rocero – Head Teacher VI; Mio Mata – Head Teacher III
Validators: Remylinda T. Soriano, EPS, Math; Angelita Z. Modesto, PSDS; George B. Borromeo, PSDS
Illustrator: Writers
Layout Artist: Writers
Management Team: Malcolm S. Garma, Regional Director; Genia V. Santos, CLMD Chief; Dennis M. Mendoza, Regional EPS in Charge of LRMS and Regional ADM Coordinator; Maria Magdalena M. Lim, CESO V, Schools Division Superintendent; Aida H. Rondilla, Chief-CID; Lucky S. Carpio, Division EPS in Charge of LRMS and Division ADM Coordinator
GRADE 9 Learning Module for Junior High School Mathematics
Learning Competency
The learners will be able to:
- solve quadratic equations by: (b) factoring; M9AL-Ia-2.1
Solving Quadratic Equation by Factoring
You already acquired how to solve quadratic equation by extracting square roots. In this topic, you will learn another approach in solving quadratic equation by factoring.
What I Know
Find out how much you already know about solving quadratic equations by factoring. Write the letter that you think is the best answer to each question on your answer sheet. Answer all items. After taking and checking this short test, take note of the items that you were not able to answer correctly and look for the right answer as you go through this module.
If ab = 0, then a = 0 or b = ______.
a. 0
b. 1
c. -1
d. 2The factored form of x² – 25 is _________.
a. (x – 5) (x – 5)
b. (x – 5) (x + 5)
c. (x + 5) (x + 5)
d. (x + 1) (x – 5)Which of the following is the GCF of 4x² and 12x?
a. 3x
b. 2x
c. 4x
d. 5xIf xy = 0, then x = 0 or y = 0 where both x and y are real numbers is called ___________.
a. Square Root Property
b. Addition Property
c. Zero Product Property
d. Multiplication PropertySolve for y in the equation y² – 3y = 0
a. y = 1, 2
b. y = 0, 3
c. y = 2, 3
d. y = 1, 3Which of the following quadratic equations cannot be factored?
a. x² + 3x + 2 = 0
b. x² – 4x + 4 = 0
c. x² + 3x – 18 = 0
d. x² + 2x – 1 = 0What are the solutions of x² + 5x + 4 = 0?
a. -4, -1
b. 4, -1
c. -4, 1
d. 5, 4
GRADE 9 Learning Module for Junior High School Mathematics
Find the solution set of x² – 7x = −12.
a. {-2, 4}
b. {-4, 1}
c. {4, -4}
d. {3, 4}Using factoring, solve for x: 6x² + 7x = 3.
a. x = −3/2, 1/3
b. x = 3, 1/3
c. x = −3/2, 1
d. x = −3/2, 2/3Which of the following is NOT a factor of 12x²?
a. 3 and x
b. 12 and x²
c. 4 and x³
d. 12 and x
*** If you got an honest 10 points (perfect score), you may skip this module.
What’s In
The processes in factoring were already tackled when you were in Grade 8. In this part, you will be focusing on factoring quadratic expressions which will be needed in finding the solutions of quadratic equations.
Factoring Quadratic Expressions
A. Factoring by GCF
Factor this polynomial: 2x² − 4x
Aligning the factor of each term will help you identify the GCF.
2x² = 2 · x · x 4x = 2 · 2 · x
Find the GCF.
The GCF of 2x² and 4x is 2x.
x(2x) − 2(2x) Write terms as products using the GCF as factor.
2x(x − 2) Use the Distributive property to factor.
Therefore, the factored form is 2x(x − 2).
GRADE 9 Learning Module for Junior High School Mathematics
The factored form is (x + 2)(2x − 1).
Let’s Try!!
A. Factor each quadratic expression. Show your COMPLETE SOLUTION then ENCIRCLE your final answer.
- 3x² + 5x
- 4x² + 16x
- x² + 5x + 6
- x² − 5x + 4
- 2x² + 11x + 5
B. FACTORING: x² + bx + c
Factor this polynomial: x² + 4x − 12
(x + ■)(x + ■)
In this form, b = 4 and c = -12; look for the factors of -12 with a sum of 4.
Guess and Check
| Factors of -12 | Sum |
|---|---|
| -1 and 12 | 11 ✖ |
| 1 and -12 | -11 ✖ |
| -2 and 6 | 4 ✓ |
The factors needed are -2 and 6.
(x − 2)(x + 6)
The factored form is (x − 2)(x + 6).
C. FACTORING: ax² + bx + c
Factor this polynomial: 2x² + 3x − 2
(■x + ■)(■x + ■)
Guess and Check
| a = 2: Factors of 2 | c = -2: Factors of -2 | b = 3: Outer + Inner |
|---|---|---|
| 1 and 2 | 1 and -2 | 1(-2) + 2(1)=0 ✖ |
| 1 and 2 | -1 and 2 | 1(2) + 2(-1)=0 ✖ |
| 1 and 2 | 2 and -1 | 1(-1) + 2(2)=3 ✓ |
(x + 2) (2x − 1) = 2x² + 3x − 2
The factored form is (x + 2)(2x − 1).
When checking, remember these when multiplying binomials: First Terms, Outer Terms, Inner Terms, Last Terms.
GRADE 9 Learning Module for Junior High School Mathematics
[transcriber’s note: blank answer lines follow the Let’s Try!! items on the page]
What’s New
Collaboration, Creativity and Critical Thinking
Picture Frame
Directions: Use the situation below to answer the questions that follow.
[photo: hands placing a rectangular glass into a wooden picture frame, measured with a ruler]
[photo: a rectangular glass frame being marked with a measuring tape and pencil]
A rectangular glass with an area of 168in² is placed on a picture frame. The length of the glass should be 2 inches longer than its width.
Draw a diagram that represents this situation.
IT’S YOUR TURN!
- How will you represent the length, width and area of the glass?
- What expression would represent the area of the glass?
- If the picture frame is enlarged and the glass needed should have an area of 960in², what equation would describe its area?
- How will you solve for the length and width of the glass?
GRADE 9 Learning Module for Junior High School Mathematics
What Is It
In order to find the length and width of the problem above, the property and examples may help you.
Zero Product Property
This states that if ab=0, then either a=0 or b=0 (or both). This property is useful when solving quadratic equations.
For all real number a and b,
| WORDS | NUMBERS | ALGEBRA |
|---|---|---|
| If the product of two quantities is equal to zero, then either or both two quantities are equal to zero. | 5(0) = 0 0(7) = 0 | If mn = 0, then m = 0 or n=0 |
Example 1: Solve using factoring.
A. 4x² = 6x
Solution:
4x² = 6x
4x² − 6x = 6x − 6x Subtract 6x from both sides
4x² − 6x = 0
2x(2x − 3) = 0 Factor by GCF (GCF: 2x)
2x = 0 or 2x − 3 = 0 Use the Zero Product Property
x = 0 or x = 3/2 Solve each equation
The solutions are 0 and 3/2.
Example 2: A. x² − 3x = 18
Solution:
x² − 3x = 18
x² − 3x − 18 = 18 − 18 Subtract 18 from both sides
x² − 3x − 18 = 0
(x + 3)(x − 6) = 0 Factor the trinomial
x + 3 = 0 or x − 6 = 0 Use the Zero Product Property
x = −3 or x = 6 Solve each equation
The solutions are -3 and 6.
GRADE 9 Learning Module for Junior High School Mathematics
Example 3: A. 2x² − 3 = −x
Solution:
2x² − 3 = −x
2x² − 3 + x = −x + x Add x to both sides
2x² + x − 3 = 0
(2x + 3)(x − 1) = 0 Factor the trinomial
2x + 3 = 0 or x − 1 = 0 Use the Zero Product Property
x = −3/2 or x = 1 Solve each equation
The solutions are −3/2 and 1.
What’s More
Solve the following quadratic equations by factoring: (SHOW YOUR SOLUTION)
- x² − 4x = 0
- x² + 2x = 15
- x² − 8x + 16 = 0
- x² − 5x = 2x² − 5x
- 20x² − 9x = 20
What I Have Learned
Steps on how to solve a quadratic equation by factoring:
- Move all the terms of the equation in the left side if necessary. In this case, the other side must be zero.
- Combine the similar terms in the left side.
- Factor the left side of the equation.
- Equate each factor that holds the unknown variable to zero.
- Solve the equated form.
- Check the answers by evaluating it to the original equation.
WHAT I CAN DO
Solve each quadratic equation by factoring. Show your COMPLETE SOLUTION then ENCIRCLE your final answer.
| 1) x² + 5x = 0 | 6) x² + 4x − 21 = 0 |
| 2) 7x² + 21x = 0 | 7) x² − 14x = -24 |
| 3) 16x² = 12x | 8) 2x² − 12 = 2x |
| 4) −4x² = 6x | 9) 2x² + 5x + 2 = 0 |
| 5) x² + 8x + 12 = 0 | 10) 3x² − 5x = 12 |
ASSESSMENT
Write the letter of the correct answer on your answer sheet. If your answer is not among the choices, write E together with your final answer.
For any real number a and b, if ab = 0, then either a= 0 or b = 0 is called _____.
A. Zero Product Property
B. Square Root Property
C. Addition Property
D. Subtraction PropertyThe (x + 3)(x – 3) is a factored form of equation ______.
A. x² – 3 = 0
B. 2x² = 9
C. x² – 9 = 0
D. x² + 32 = 0What is the factored form of x² + 10x + 25 = 0?
A. (x − 5)²
B. (x + 5)²
C. (x + 2) (x + 5)
D. (x − 2) (x + 5)Solve for the value of x in equation 4x² + 8x = 0.
A. x = 0, −2
B. x = 2, −2
C. x = 1, 2
D. x = −1, −2In question number 4, find the GCF of 4x² and 8x.
A. 2x
B. x
C. 3x
D. 4xWhat is the solution set of 6x² – 30x = −36?
A. { 1,2}
B. {2,3}
C. {3,4}
D. {4,5}Find the roots of 3x² – 4x = 0.
A. 0, 4/3
B. 4, 1
C. 3, 4
D. 1/2, 2/3Which of the following are factors of 4 – 8x + 3x²?
A. (2 – x) (2 + 4x)
B. (2 – x) (2 – 3x)
C. (1 – x) (4 – 4x)
D. (4 + x) (1 – 3x)What are the roots in question number 7?
A. x = −2, −2/3
B. x = −3, 1/3
C. x = 2/3, 1
D. x = 3/4, −3/2A rectangle has a length of five inches more than its width. Its area is 24 square inches. Find the dimensions of the rectangle.
A. width= 3 in., length= 4 in
B. width =3 in, length = 8 in.
C. width = 4 in, length = 5 in
D. width = 8 in, length = 2 in.
ADDITIONAL ACTIVITIES
Communication, Collaboration, and Character Building
A.
Copy and complete the graphic organizer. In each box, write a step used in solving quadratic equation by factoring.
[chart: graphic organizer; Solving Quadratic Equation by Factoring, with boxes “Step 1”, “Step 2”, and “Step ….”]
B. REFLECT
Solving quadratic equation through factoring requires the factors of the given quadratic equation before breaking them down into two linear equations. This is helpful in solving for the values of the unknown.
Do you think the problems within your community also have factors? Do you think it’s easier to solve for these factors first before solving the main problem? Why?
E-Search
You may also check the following link for your reference and further learnings about this module:
- https://www.youtube.com/watch?v=sLIaI7lQ7AM
- https://www.youtube.com/watch?v=6PTt9BoIfHg&t=94s
REFERENCES
Morgan, F. M., & Paige, B. L. (n.d.). Algebra 2. pp 98-102 .America: Henry Holt and Company.
Oronce, O. A., Santos, G. C., & Ona, M. I. (n.d.). *Interactive Mathematics III (Concepts,Structures, adn Metods for High School.*Manila: Rex Book Store
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https://build-basic.com/wp-content/uploads/2015/03/DIY-Photo-Frame-by-Build-Basic-Step-7-copy-300×300.png
https://www.stcs.org/view/11836.pdf
https://www.stcs.org/view/11835.pdf
file:///C:/Users/samsung-pc/Desktop/math_9_lm_draft_3.24.2014.pdf
PROBLEM – BASED WORKSHEET
Throwing of Ball
Any ball thrown into the air is a projectile. The general function that approximates the height in feet of a projectile on Earth after t seconds is given by the function: h(t) = -16t² + vot + ho. This model has a limitation because it does not account for air resistance, wind, and other factors.
Suppose Linda throws a ball into the air to a height of 5 ft with an initial velocity of 16ft/s.
[illustration: a girl throwing a ball into the air along a curved path]
Write an equation that can be used to model the height of the ball in feet seconds after the ball is thrown.
Write an equation of the model when it reaches the ground.
Solve the equation by factoring.
How long does Linda have to catch the ball before it hits the ground?
[transcriber’s note: this page is printed upside down; it is transcribed below in reading order after rotation]
ANSWER KEY
WHAT’S IN
- x (3x + 5)
- 4x (x + 4)
- (x + 3) (x + 2)
- (x – 1) (x – 4)
- (x + 5) (2x + 1)
WHAT I CAN DO
- −5 and 0
- −3 and 0
- 0 and 4/3 [?]
- −3/2 [?] and 0
- −6 and −2
- −7 and 3
- 2 and 12
- −2 and 3
- −2 and −1/2
- −4/3 and 3
WHAT’S MORE
- 0, 4
- −5, 3
- 4
- 0
- 5/4 and −4/5 [?]
WHAT I KNOW
- A
- B
- C
- C
- B
- D
- A
- D
- A
- C
ASSESSMENT
- A
- C
- B
- A
- D
- B
- A
- B
- A
- B
PROBLEM – BASED WORKSHEET
- h(t) = −16t² + 16t + 5
- −16t² + 16t + 5 = 0
- (−4t + 5)(4t + 1) = 0
- 1.25 seconds
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