How to learn from this module

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Module 16

All About “F”

This module will focus on using special products and factoring in solving problems involving number relations and areas of polygons such as rectangles, squares and triangles.

This module contains the following lessons:

Lesson 1 Using Special Products and Factoring in Solving Problems Involving Numbers

Lesson 2 Using Special Products and Factoring in Solving Problems Involving Squares

Lesson 3 Using Special Products and Factoring in Solving Problems Involving Rectangles

Lesson 4 Using Special Products and Factoring in Solving Problems Involving Triangles

After going through this module, you are expected to use special products and factoring to solve problems on

 number relations

 areas and dimensions of rectangles, squares and triangles

What this module is all about

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This is your guide for the proper use of the module:

1. Read the items in the module carefully.

2. Follow the directions as you read the materials.

3. Answer all the questions that you encounter. As you go through the module, you will find help to answer these questions. Sometimes, the answers are found at the end of the module for immediate feedback.

4. To be successful in undertaking this module, you must be patient and industrious in doing the suggested tasks.

5. Take your time to study and learn. Happy learning!

The following flowchart serves as your quick guide in using this module.

How to learn from this module

Start

Take the Pretest

Check your paper and count your correct answers.

Is your score 80% or above?

Yes Scan the items you missed.

No Study this module

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In order to learn from this module, you need to recall the special product and factoring formulas, the formulas for areas and dimensions of rectangle, square, and triangle: and the properties of equality.

You refer to the previous module for the special product and factoring formulas.

Let’s recall the properties of equality.

Exploration 1

Take the following pretest.

Multiple Choice. Choose the letter of the correct answer.

1. The square of a positive number is 47 less than twice the square of the next consecutive positive number. What are the numbers?

a. 5 and 6 b. 6 and 7 c. 7 and 8 d. 8 and 9

2. In a school audio-visual room, the number of seats in each row is 4 more than the number of rows. The audio-visual room sits 60 persons. How many seats are there in each row?

a. 16 b. 14 c. 12 d. 10

3. The length of a rectangular field is 12 m longer than its width. The area of the field is 108 sq m. What is its width?

a. 4 m b. 5 m c. 6 m d. 7 m

PROPERTIES OF EQUALITY

For any real numbers a, b and c  Reflexive: a = a

 Symmetric: If a=b, then b=a

 Transitive: If a=b and b=c, then a=c

 Addition Property of Equality (APE): If a = b, then a + c = b + c

 Multiplication Property of Equality (MPE): If a = b, then ac = bc

 Zero Property for Multiplication: ab = 0, then a = 0 or b = 0

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4. The height of a triangle is one more than thrice its base. What is the height of the triangle if the area is 15 sq cm.?

a. 7 cm b. 10 cm c. 13 cm d. 16 cm

5. The square of a number decreased by 64 is equal to 17. What is the number?

a. 7 b. 8 c. 9 d. 10

Lesson 1

Using Special Products and Factoring in Solving Problems

Involving Numbers

In this lesson, you will learn to solve problems about number relations. It is worthwhile to recall the elements of the set of counting numbers, whole numbers, and integers.

Counting numbers= {1, 2, 3, . . .} Whole numbers= {0, 1, 2, 3, . . .}

Integers= { . . ., -3, -2, -1, 0, 1, 2, 3, . . .}

Consider the problem below. Read the problem carefully and understand the method of solving it.

Problem 1. The square of a counting number is 98 less than twice the square of the next consecutive counting number. What are the numbers?

To solve the problem, determine the following:

Unknown: 2 consecutive counting numbers

Exploration

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Known: the square of the first is 98 less than twice the square of the second consecutive counting number

Representation: Let x = the first counting number x + 1 = the second counting number x2 = the square of the first

(x + 1)2 = the square of the second

2(x + 1)2 – 98 = 98 less than twice the square of the second

Equation: =

x2 = 2(x + 1)2 – 98

Solve:

x2 = 2(x + 1)2 – 98

x2 = 2(x2 + 2x + 1) – 98 → Square of a binomial x2 = 2x2 + 4x + 2 – 98 → Distributive Property x2 = 2x2 + 4x – 96 → Combine 2 and - 98 2x2 + 4x – 96 = x2 → Symmetric Property

2x2 + (-x2) + 4x – 96 = x2 + (-x2) → Addition Property of Equality x2 + 4x – 96 = 0 → Inverse Property for Addition

(x – 8)(x + 12) = 0 → Factoring trinomial ax2 + bx + c; a = 1 x – 8 = 0; x + 12 = 0 → Zero Property for Multiplication

x = 8; x = -12

x =- 12 is rejected because counting numbers do not include negative number x = 8 → the first counting number

x + 1 = 9 → the next counting number

Therefore, the counting numbers are 8 and 9.

Check: Is the square of the first counting number(8) equals 98 less than twice the square of the next counting number (9)?

x2 ? 2(x + 1)2 – 98

82 ?2(9)2 – 98 64 ?2 (81) – 98

64 ?162 – 9

64 =64 It’s Correct!

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Read the problem below. This time you will be required to answer questions as you read and solve the problem.

Problem 2. The square of a number decreased by 36 is equal to 64. Find the numbers.

What are the unknown and the known in the problem?

Unknown_________

Known___________

Did you answer. . .

Unknown: a number

Known: the square of the number decreased by 36 is 64

Then you are right!

How do you represent the unknown in this problem?

Let x = _____________________

Check your answers with mine.

My answers:

Representation: Let x = the number

x2 = the square of the number

What is the equation?_____________

Did you answer x2 – 36 = 64? Very good!

Now you read and supply the reason for each of the steps in solving this problem.

Equation: x2 – 36 = 64 Solve: x2 – 36 = 64

x2 – 36 – 64 = 0 → Addition Property of Equality

x2 – 100 = 0 → (1) _________________________

x2 – 102 = 0 → Express 100 in exponential form

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x – 10 = 0 ; x + 10 = 0 → (2)___________________________

x = 10 ; x = -10

Therefore, the numbers are 10 and -10.

Compare your answers with mine.

My answers:

(1) Combine 36 and - 64

(2) Zero Property for Multiplication

Now check if your answers are correct.

Check: x2 – 36 = 64 x2 – 36 = 64 for x = 10; (10)2 – 36 ?64 for x –10; (-10)2 – 36 ?64 100 – 36 ?64 100 – 36?64

64 = 64 64 = 64

You are now ready to solve the next problem.

Problem 3. Find two positive numbers whose difference is 2 and whose product is 24.

Show your solution below.

Compare your answers with mine.

My answers are found in the following solution:

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Known: The difference of the unknown positive numbers is 2 and their product is 24

Representation: Let x = the smaller number x+2 = the larger number

Equation: x(x+2) = 24

Solve: x(x+2) = 24

x2 + 2x = 24 Distributive property x2 + 2x – 24= 0 APE

(x+6) (x – 4) = 0 Factoring

x = -6 and x = 4

x = -6 is rejected since you are looking for positive numbers.

Therefore, the numbers are x= 4 and x +2 = 6.

Check: Is their difference 2? 6-4 = 2 YES Is their product 24?

6(4) = 24 YES

Hence, the answers are correct since they satisfied the conditions in the problem.

Solve the following problems.

1. One number is 4 more than the other. If the sum of their squares is 40, find the numbers.

2. The square of a number decreased by 25 is 24. Find the numbers.

Compare your answers with those in the Answer key. If you got all the answers correct, CONGRATULATIONS! You did well in the lesson and can proceed to the next lesson. If not, you need to go over the same lesson and take note of the mistakes committed. It pays to have a second look.

Self-check 1

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Lesson 2

Using Special Products and Factoring in Solving Problems

Involving Squares

In this lesson, you will learn to solve problems involving area and dimensions of a square.

Let us recall the definition, parts and formula to find the area of a square.

Read and analyze the problem below.

Problem 1. The area of a square lot is 144 sq m. Find the length of one side.

Unknown: length of one side of the square Known: area is 144 sq m

Representation: Let x = the length of one side

Equation: 144 = x2 A = s2

Solve: 144 = x2

x2 = 144 → Symmetric Property of Equality x2 – 144 = 0 → Addition Property of Equality

x2 – 122 = 0 → Express 144 in exponential form (x –12)(x + 12) = 0 → a2 – b2 = (a – b) (a + b)

x – 12 = 0 ; x + 12 = 0 → Zero Property for Multiplication

x = 12 ; x = -12

x = -12 is rejected for there is no negative linear measurement.

 A square is a four-sided polygon with four sides congruent and having four right angles.

s

 Area of a Square

Area = square of the side: A = s2

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Therefore, the side of the square lot is 12 m.

Check : Is the area of the square lot 144 sq m if the side is 12 m? A = s2

= (12)2

= 144 sq m It’s correct!

Read and understand the problem below. You need to fill up the missing parts in the solution.

Problem 2. The side of a square is increased by 3. What is its area in terms of x?

x x=3

What is the unknown and known in the problem?

Unknown:___________________

Known:_____________________

Compare your answer with mine.

My answers:

Unknown: area of the square in terms of x

Known: the side of the square is increased by 3

How do you represent the unknown in this problem?

Let x=__________________

___________________

Did you answer . . .

Representation: Let x = the side of a square

x+3 = the side of the square increased by 3

Then you are right!

What is the equation? _____________

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Did you answer A= (x+3)2? You’re correct!

Now we solve the equation to answer the question.

Solve: A = (x+3)2 using the formula A = s2

A = x2 + 4x + 4 (a+ b)2 = a2 + 2ab + b2

1. What is the side of a square bulletin board if its area is 64 sq ft?

2. How long is the side of a square field having an area of 169 sq m?

Lesson 3 Using Special Products and Factoring in Solving Problems Involving Rectangles

In this lesson, you will learn to solve problems involving rectangles.

Let us recall the definition, parts and formula to find the area of a rectangle.

Self-check 2

Exploration

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Consider the problem below. Read the problem carefully and follow the method of solving it.

Problem 1. A photograph is 8 cm wide and 12cm long. The length and the width are increased by an equal amount in order to double the area of the photograph. What are the dimensions of the new photograph?

To solve the problem, determine the

Unknown: dimension of the new paragraph Known: The width of the photograph is 8cm. Its length is 12cm.

Area = lw

Substitute = 8cm (12), thus 12 + x

Area = 96 sq m

12 cm

8 cm 8+ x

Representation: Let x = the equal amount of increase 8 + x = the width of the new photograph 2 + x = the length of the new photograph 2(96) = the area of the new photograph

Equation: (8 + x) (12 + x) = 2(96) A = LW

96 + 20x + x2 = 192 → FOIL Method

x2 + 20x + 96 = 192 → Commutative Property x2 + 20x + 96 – 192 = 0 → Inverse Property

x2 + 20x – 96 = 0 → Combined 96 - 192 A = 96 sq cm

A = 2(96)

 A rectangle is a four-sided polygon with two pairs of opposite sides congruent and having four right angles.

l w

 Area of a Rectangle

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(x + 24)(x – 4) = 0 → Factoring trinomial ax2 + bx + c, a = 1 x + 24 = 0 ; x – 4 = 0 → Zero Property for Multiplication

x = -24 ; x = 4

x = -24 is rejected

If x = 4: 8 + x = 8 + 4 = 12 cm → the width of the new photograph 12 + x = 12 + 4 = 16 cm → the length of the new photograph

Therefore, the dimensions of the new photograph are 12 cm by 16 cm.

Check: Are the dimensions of the new photograph double the area of the given photograph?

12(16) ? 2(96)

192 = 192 It’s correct!

Now you will solve the problem below completely.

Problem 2. The length of the floor of the classroom is 2 m longer than the width. The sum of their squares is 130. Find the length and the width.

x+ 2

x

My answer:

Unknown: length and width of the floor of the classroom Known: the sum of their squares is 130.

Representation: Let x = the width x+ 2 = the length

x2 = the square of the width (x + 2)2 = the square of the length

= 130

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Equation: x2 + (x + 2)2 = 130

x2 + (x2 + 4x + 4) = 130 → Square of a binomial

2x2 + 4x + 4 – 130 = 0 → Addition Property of Equality 2x2 + 4x – 126 = 0 → Combine 4 – 130

2 (x2 + 2x – 63) = 0 → Factor out common monomial factor 2 (x + 9) (x – 7) = 0 → Factor trinomial ax2 + bx + c; a = 1

x + 9 = 0; x – 7 = 0 → Zero Property for Multiplication x = -9 rejected; x = 7

x = 7 m is the width

x + 2 = 7 + 2 = 9 m is the length

Therefore, the floor of the classroom measures 7 m by 9 m.

Check : Is the length 2 m longer than the width? 9 - 7 = 2 YES

Is the sum of the squares of the length and the width equal to 130? 72 + 92 ?130

49 + 81 ?130

130 = 130 It’s correct!

You have noticed that negative solutions for problems involving linear measurement are rejected because negative value does not make any sense as far as linear measurement is concern.

1. The length of a rectangle is 3 m longer than the width. The area is 40 sq m . Find the length and the width.

2. The length of a corridor is twice its width. The width is doubled and the length is increased by 5 m in order to triple the area of the corridors. Find the dimensions of the corridor.

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Lesson 4

Using Special Products and Factoring in Solving Problems

Involving Triangles

In this lesson, you will learn to solve problems involving triangles.

Let us recall the definition, parts and formula to find the area of a triangle.

Read and understand the procedure of solving the problem that follows.

 A triangle is a three-sided polygon. It can be a scalene triangle

( no sides congruent), an isosceles triangle( 2 sides congruent), or an equilateral triangle( 3 sides congruent).

h

b

 Area of a triangle

Area = one-half the product of its base and height: A bh

2 1



 Area of a Rectangle

Area = length times width: A= lw

Exploration

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Problem 1. The height of a triangular flag is twice its base. The area of the flag is 81 sq cm. Find the base and the height.

Unknown: the base and height of the triangle Known: the height is twice the base

Area is 81 sq m.

Representation: Let x = the base 2x = the height

Equation: 81 = ½ (x)(2x) A bh

2 1



Solve: 81 = x2 x2 = 81 x2 – 81 = 0 (x – 9)(x + 9) = 0

x – 9 = 0 ; x + 9 = 0 x = 9 ; x = -9

x = 9 cm is the base

2x = 2(9) = 18 cm is the height

Therefore, the base and height of the flag are 9 cm and 18 cm respectively.

Check: Is the height twice the base? 9(2) = 18 YES Is the area equals 81 sq cm ?

A = (9)18 81 2

1 _

YES

Now you are ready to solve the problem below completely. Try it.

Problem 2. The height of a triangular garden is equal to its base. The area covered by the garden is 50 sq m. How long is the base of the garden?

2x

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My answer:

Unknown: the base of the triangular garden Known: 50 sq m is the area

the base is equal to the height

Representation: Let x = the base x = the height

Equation: 50 = ½ (x)(x) A bh

2 1



Solve: 50 = ½ (x)(x) 50 = ½ x2 2(50) = 2(1/2x2)

100 = x2 x2 – 100 = 0 (x – 10)(x + 10) = 0

x – 10 = 0 ; x + 10 = 0

x = 10; x = -10 is rejected

x = 10 m is the base x

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Therefore, the base of the triangle is 10 m long.

Check: Is the area equals 50 sq m if the base = height = 10 m?

50 ? ½ (10)(10)

50

2 100 ?

50 = 50

1. The height of a triangle is thrice the base. The area of the triangle is 24 sq units. Find the base and height of the triangle.

2. The base of triangular pond is one-half the height. The area of the pond is 25 sq m. Find the base and the height of the pond.

Look back!

Special products and factoring techniques are important tools for solving equations. These mathematical methods provide you with patterns that can be used in solving practical problems. Number relations and finding dimensions of enclosed figures are just few of the many problem situations that need mathematical manipulation.

Self-check 4

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Take the following posttest.

Multiple Choice. Choose the letter of the correct answer.

1. A certain number decreased by 7 is multiplied by the sum of the number and 7. The product is 51. What is the number?

a. 9 c. 11

b. 10 d. 12

2. The length of a corridor is four times its width. Its area is 100 sq m. How long is the corridor?

a. 5m c. 20m

b. 10m d. 100m

3. The area of a square glass top is 4m2. What is the length of one side of the glass top?

a. 1m c. 3m

b. 2m d. 4m

4. One number is 3 less than the other. The sum of the number multiplied by the larger equals 35. What is the smaller number?

a. 5 c. 3

b. 4 d. 2

5. The height of a triangle is three more than twice the base. The area of the triangle is 45 sq units. What is the height of the triangle?

a. 15 c. 3

b. 12 d. 2

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Pretest page 3

1. a 4. b

2. d 5. c

3. c

Lesson 1 Exploration1 andSelf-Check 1 on page 8

1. x2 + (x + 4)2 = 40 x2 + x2 +8x + 16 = 40 2x2 + 8x + 16 – 40 = 0 2x2 + 8x – 24 = 0 2(x2 + 4x – 12) = 0 2(x + 6)(x – 2) = 0

2 = 0; x + 6 = 0; x – 2 = 0 x = -6; x = 2

If x = -6

x+ 4 = -6 + 4 = -2 If x = 2

x+ 4 = 2 + 4 = 6

Therefore the numbers are -6 and -2 and 4 and 6.

2. x2 – 25 = 24 x2 – 25 – 24 = 0 x2 – 49 = 0 x2 – 72 = 0

(x – 7)(x + 7) = 0

x – 7 = 0; x + 7 = 0 x = 7; x = -7

Therefore, the numbers are 7 and -7.

Lesson 2 Exploration 2 andSelf-Check 2 page 11

1. 64 = x2 x2 = 64 x2 – 64 = 0 x2 – 82 = 0 (x – 8)(x + 8) = 0

x – 8 = 0; x + 8 = 0 x = 8 ; x = -8 rejected

Therefore, the side of the bulletin board is 8 ft.

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2. 169 = x2 x2 = 169 x2 – 169 = 0 x2 – 132 = 0 (x – 13)(x + 13) = 0

x – 13 = 0; x + 13 = 0 x = 13 ; x = -13 rejected

Therefore, the field has a side of 13 m.

Lesson 3 Exploration 3 and Self-Check 3 page 14

1. x(x + 3) = 40 x2 + 3x – 40 = 0 ( x + 8)(x – 5) = 0

x + 8 = 0 ; x – 5 = 0 x = -8; x = 5

x = -8 is rejected x = 5 m is the width x + 3 = 8 m is the length

Therefore, the rectangle measures 5 m by 8 m.

2. 2x(2x + 5) = 3x(2x) 4x2 + 10x = 6x2

6x2 = 4x2 + 10x 6x2 – 4x2 – 10x = 0

2x2 – 10x = 0 2x(x – 5) = 0

2x = 0 ; x – 5 = 0 x = 0; x = 5

x = 0 is rejected x = 5 m is the width 2x = 2 x 5 = 10 m is the length

Therefore, the corridor is 5 m wide and 10 metres long.

Lesson 4 Exploration 4 and Self-Check 4 page 18

1. ½(x)(3x) = 24 3x2/2 = 24

2(3x2/2) = 2(24) 3x2 = 48 3x2 – 48 = 0 3 (x2 – 16) = 0 3(x – 4)(x + 4) = 0

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x = 4 units is the base

3x = 3(4) = 12 units is the height

Therefore, the base and height of the triangle are 4 and 12 respectively.

2. ½ (1/2 x )(x) = 25 x2/4 = 25

4(x2/4) = 4(25) x2 = 100

x2 – 100 = 0 (x – 10)(x + 10) = 0

x – 10 = 0; x + 10 = 0 x = 10; x = -10 is rejected x = 10 m is the height

½ x = ½(10) = 5 m is the base

Therefore, the base and height of the pond are 5 m and 10 m respectively.

Posttest page 19

1. b 4. d

2. c 5. a

3. b

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BIBLIOGRAPHY

Fuller, Gordon. (1977). College algebra. 4th ed. New York, NY: Van Nostrand Company.

Leithold, Louis. (1989). College algebra. USA: Addison-Wesley Publishing Company, Inc.