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STUDY GUIDE GRADE 8 | UNIT 1 Factoring Table of Contents Introduction...................................................

STUDY GUIDE GRADE 8 | UNIT 1 Factoring Table of Contents Introduction.................................................................................................................................... 3 Test Your Prerequisite Skills........................................................................................................ 4 Objectives...................................................................................................................................... 5 Lesson 1: Factoring Polynomials with a Common Monomial Factor - Warm Up!........................................................................................................................... 5 - Learn about It!................................................................................................................... 6 - Let’s Practice!..................................................................................................................... 7 - Check Your Understanding!............................................................................................ 12 Lesson 2: Factoring the Difference of Two Squares - Warm Up!......................................................................................................................... 13 - Learn about It!................................................................................................................. 15 - Let’s Practice!................................................................................................................... 17 - Check Your Understanding!............................................................................................ 22 Lesson 3: Factoring the Sum or Difference of Two Cubes - Warm Up!......................................................................................................................... 23 - Learn about It!................................................................................................................. 24 - Let’s Practice!................................................................................................................... 25 - Check Your Understanding!............................................................................................ 30 Lesson 4: Factoring Perfect Square Trinomials - Warm Up!......................................................................................................................... 31 - Learn about It!................................................................................................................. 32 - Let’s Practice!................................................................................................................... 34 - Check Your Understanding!............................................................................................ 39 1  STUDY GUIDE Lesson 5: Factoring General Trinomials - Warm Up!......................................................................................................................... 40 - Learn about It!................................................................................................................. 41 - Let’s Practice!................................................................................................................... 42 - Check Your Understanding!............................................................................................ 49 Challenge Yourself!..................................................................................................................... 49 Performance Task....................................................................................................................... 50 Wrap-up....................................................................................................................................... 54 Key to Let’s Practice!.................................................................................................................... 55 References................................................................................................................................... 56 2  STUDY GUIDE Click Home icon to go back to Table of Contents GRADE 8 | MATHEMATICS UNIT 1 Factoring Mathematics relies on logic and creativity, and it is studied both for its theoretical and practical applications. For some people, not just the professionals, the pursuit to understand the essence of mathematics lies in the study of patterns and relationships. For grade 8 students, the pursuit to study mathematics starts with the understanding of the concepts of polynomials. For some, it is not outright clear when or where the application of the study polynomials can be readily applied. However, even the simplest tasks or scenarios can be modeled using polynomials. Have you ever wondered how architects maximize and design a certain floor plan of a house or a building? …or how an engineer computes for the right mixture of cement and gravel to be used for the foundations of skyscrapers and other infrastructures? …or simply how a carpenter builds a cabinet or drawer using limited supplies? In this unit, you will learn about the different factoring techniques that will be useful in simplifying polynomials as well as their applications in the real world. 3  STUDY GUIDE Test Your Prerequisite Skills Computing the area and volume of two- and three-dimensional figures Performing basic operations on polynomials Finding special products Before you get started, answer the following items to help you assess your prior knowledge and practice some skills that you will need in studying the lessons in this unit. 1. List all the positive integer factors of the given number. a. 9 b. 16 c. 51 2. Perform the indicated operation. a. (2𝑥 5 + 4𝑥 4 − 3𝑥 2 + 7) + (7𝑥 4 − 3𝑥 3 + 𝑥 2 − 9) b. (3 − 𝑚𝑥 − 3𝑥 2 ) − (−𝑚𝑥 − 2𝑥 2 − 4) c. (2𝑥 + 3)(𝑥 − 2) d. (4𝑥 + 1)(16𝑥 2 − 4𝑥 + 1) e. (−156𝑥 2 𝑦 4 𝑧 3 ) ÷ (26𝑥 2 𝑦 2 𝑧 2 ) f. (7𝑥 + 12 + 𝑥 2 ) ÷ (𝑥 + 3) 3. Expand the following expressions using special product formulas. a. (3𝑥 − 2)2 c. (5 − 4𝑥)(5 + 4𝑥) b. (9 + 2𝑥)3 d. (𝑥 + 2𝑦 − 7)2 4. Give the formulas for the following: a. area of a rectangle b. area of a square c. area of a triangle d. area of a circle e. volume of a cube f. volume of a rectangular prism 4 STUDY GUIDE Objectives At the end of this unit, you should be able to factor completely different types of polynomials (polynomials with common monomial factor, difference of two squares, sum and difference of two cubes, perfect square trinomials, and general trinomials; and solving problems involving factors of polynomials. Lesson 1: Factoring Polynomials with a Common Monomial Factor Warm Up! Something Common Materials Needed: fishbowl, pen and paper Instructions: 1. This activity may be played by the whole class. 2. Your teacher has prepared a fishbowl with rolled small pieces of papers having monomials written on them. 3. Each one of you is to pick a piece of rolled paper without looking at it. You must also not show the content of your paper to any of your classmates just yet. 4. After each of your classmates has taken his or her pick, each one of you may open the rolled paper and take a look at the monomial you got. 5. Each one of you must then go around the classroom and find two other persons with monomials that have a common factor (except 1) with your monomial. The common factor may be numerical (numeral) or literal (variable) or a combination of both. You are free to talk while in search for your partners that would form a triad. 5  STUDY GUIDE Examples of monomials with common factors: 12𝑦 and 9𝑥 (Both have 3 as a common numerical factor.) 11𝑥 2 and 5𝑥 2 (Both have 𝑥 2 as a common literal factor.) 4𝑤 2 and 8𝑤 (Both have 4𝑤 as a common factor.) 6. Once a triad is formed, the three must run to the teacher and show their monomials for verification. 7. The first three triads who get verified by the teacher get a prize. Learn about It! In factoring polynomials, the first factors taken out are the greatest common monomial factors. For example, let us factor 6𝑎𝑏 + 12𝑏𝑐. One may opt to list down the common factors of both terms of the polynomial and select the greatest factor. Nevertheless, the following steps may be helpful and easier to perform. Step 1: Find the GCF of the numerical coefficients and the GCF of the variables by prime factorization. Numerical coefficients: Variables: 6=3×2 ab = a ⋅ b 12 = 3 × 2 × 2 bc = b ⋅ c GCF = 3 × 2 = 6 GCF = b 6  STUDY GUIDE Step 2: Get the product of the GCFs for the numerical coefficients and the variables. This will be the greatest common monomial factor of the polynomial. 6 ⋅ 𝑏 = 6𝑏 Definition 1.1: The greatest common monomial factor is the product of the greatest common numerical factor and a second component made up of the common variable factors, each with the highest power common to each term. Step 3: Factor out the greatest common monomial factor and divide each term by this to find the other factor. 6𝑎𝑏 + 12𝑏𝑐 6𝑎𝑏 12𝑏𝑐 = + = 𝑎 + 2𝑐 6𝑏 6𝑏 6𝑏 The other factor is 𝑎 + 2𝑐. Therefore, 6𝑎𝑏 + 12𝑏𝑐 = 6𝑏(𝑎 + 2𝑐). Let’s Practice! Example 1: Completely factor the polynomial 45𝑎4 + 36𝑎3. Solution: Step 1: Find the GCF of the numerical coefficients and the GCF of the variables by prime factorization. 7  STUDY GUIDE Numerical coefficients: Variables: 45 = 3 × 3 × 5 a4 = a ⋅ a ⋅ a ⋅ a 36 = 3 × 3 × 4 a3 = a ⋅ a ⋅ a GCF = 3 × 3 = 9 GCF = a ⋅ a ⋅ a = a3 Step 2: Get the product of the GCFs for the numerical coefficients and the variables. This will be the greatest common monomial factor of the polynomial. 9 ⋅ 𝑎3 = 9𝑎3 Step 3: Factor out the greatest common monomial factor and divide each term by this to find the other factor. 45𝑎4 + 36𝑎3 45𝑎4 36𝑎3 = + = 5𝑎 + 4 9𝑎3 9𝑎3 9𝑎3 The other factor is 5𝑎 + 4. Therefore, 45𝑎4 + 36𝑎3 = 9𝑎3 (5𝑎 + 4). Try It Yourself! Completely factor the polynomial 48ℎ5 – 32ℎ3. Example 2: Find the greatest common monomial factor of 24𝑥 3 𝑦 2 + 18𝑥 2 𝑦 2 − 30𝑥𝑦 3 to factor it completely. Solution: Step 1: Find the GCF of the numerical coefficients and the GCF of the variables. Besides prime factorization, you may also use continuous division. 8  STUDY GUIDE The GCF of the numerical coefficients is 6. The GCF of the variables is 𝑥𝑦 2. Step 2: Get the product of the GCFs for the numerical coefficients and the variables. This will be the greatest common monomial factor of the polynomial. 6 ⋅ 𝑥𝑦 2 = 6𝑥𝑦 2 Step 3: Factor out the greatest common monomial factor and divide each term by this to find the other factor. 24𝑥 3 𝑦 2 + 18𝑥 2 𝑦 2 − 30𝑥𝑦 3 = 4𝑥 2 + 3𝑥 − 5𝑦 6𝑥𝑦 2 The other factor is 4𝑥 2 + 3𝑥 − 5𝑦. Therefore, 24𝑥 3 𝑦 2 + 18𝑥 2 𝑦 2 − 30𝑥𝑦 3 = 6𝑥𝑦 2 (4𝑥 2 + 3𝑥 − 5𝑦). 9  STUDY GUIDE Try It Yourself! Completely factor the polynomial 42𝑝7 𝑟 8 𝑠 4 − 63𝑝5 𝑟 4 𝑠 5 + 105𝑝4 𝑟 4 𝑠 7. Example 3: Given that 75𝑚5 + 15𝑚3 + 45𝑚2 represents a positive integer, prove that it is a composite number. Solution: 75𝑚5 + 15𝑚3 + 45𝑚2 is composite if we can factor it. Step 1: Find the GCF of the numerical coefficients and the GCF of the variables. Besides prime factorization, you may also use continuous division. The GCF of 75, 15, and 45 is 15. The GCF of 𝑚5 , 𝑚3 , and 𝑚2 is 𝑚2. Step 2: Get the product of the GCFs for the numerical coefficients and the variables. This will be the greatest common monomial factor of the polynomial. 15 ⋅ 𝑚2 = 15𝑚2 Step 3: Factor out the greatest common monomial factor and divide each term by this to find the other factor. 75𝑚5 + 15𝑚3 + 45𝑚2 2 = 5𝑚3 + 𝑚 + 3 15𝑚 The other factor is 5𝑚3 + 𝑚 + 3. Written in factored form, 75𝑚5 + 15𝑚3 + 45𝑚2 = 15𝑚2 (5𝑚3 + 𝑚 + 3). Since we have factored 75𝑚5 + 15𝑚3 + 45𝑚2 , it is a composite number. 10  STUDY GUIDE Try It Yourself! Is the polynomial 85𝑤 9 𝑥 5 𝑦 6 + 68𝑤 7 𝑥 6 𝑦 7 − 153𝑤 5 𝑥 7 𝑦 4 a prime or composite number? Justify your answer. Real-World Problems Example 4: Mr. Santos plans to construct a circular dinner table with a square centerpiece in the center. The radius of the table should be the same measure as the side of the square centerpiece. Write an expression for the remaining area of the table not covered by the centerpiece in terms of the side 𝑠 of the square. Write your answer in factored form. Solution: Step 1: Draw a figure to illustrate the given problem. Step 2: Note that we are looking for the area not covered by the centerpiece. Hence, we use the formula for the area of the circle and that of the square. 𝐴𝑐𝑖𝑟𝑐𝑙𝑒 = 𝜋𝑟 2 𝐴𝑠𝑞𝑢𝑎𝑟𝑒 = 𝑠 2 11  STUDY GUIDE Since the radius should be the same measure as the side of the square, we write 𝐴𝑐𝑖𝑟𝑐𝑙𝑒 = 𝜋𝑟 2 = 𝜋𝑠 2. Step 3: Solving for the area not covered by the centerpiece, we have 𝐴 = (𝜋𝑠 2 ) − 𝑠 2. Step 4: Rewriting the equation in factored form, we have 𝐴 = 𝑠 2 (𝜋 − 1). Therefore the area of the table not covered by the centerpiece is given by 𝐴 = 𝑠 2 (𝜋 − 1). Try It Yourself! If the table in Example 4 is in the form of a square having a circular mantle and the radius of the mantle is equal to the side of the square 𝑠, what would be the expression for the area not covered by the mantle? Check Your Understanding! 1. Factor the following expressions completely. a. 10𝑥 4 − 5𝑥 2 b. 36𝑥 2 𝑦 2 − 30𝑥𝑦 3 c. 𝑢𝑣 2 𝑟 + 𝑢2 𝑣 3 𝑟 2 + 𝑢3 𝑣𝑟 3 d. 28𝑎𝑏 2 𝑐 − 14𝑎2 𝑏𝑐 3 + 7𝑎3 𝑏 2 𝑐 12 STUDY GUIDE 2. Write an expression (in factored form) for the area of the shaded region in terms of 𝑟 and 𝑠. Lesson 2: Factoring the Difference of Two Squares Warm Up! A Tale of Two Squares Materials Needed: colored papers, ruler, pencil, scissors, paper and pen, cartolina, markers Instructions: 1. Form groups of three. 2. Cut two square-shaped piece from the colored papers. One square should be larger than the other, and the squares must be of different colors. Label the side of the bigger square 𝑥 and the side of the smaller square 𝑦. 13 STUDY GUIDE 3. Trace the smaller square onto the bigger square at the corner and cut along the trace. With your groupmates, find an expression for the area of the remaining figure in terms of 𝑥 and 𝑦? 4. Draw a horizontal line segment dividing the remaining figure into two rectangles as shown and cut along this line. 5. Move the cut out region to the right as shown. 6. Find the expressions of the dimensions (width and length) of the rectangle in terms of 𝑥 and 𝑦. 14  STUDY GUIDE 7. Find the expression for the area of the rectangle using the dimensions you obtained in step 6. How does this compare to the area obtained in step 3? 8. Write your answers to the expressions required and your solutions on a cartolina and present your work in class. Learn about It! A constant is a perfect square if it is obtained by multiplying a number by itself. Observe the following examples: Number Square Perfect (x) (x2) Square 1 12 = 1 1 2 22 = 4 4 2 3 3 =9 9 2 5 5 = 25 25 10 102 = 100 100 A variable is a perfect square when its exponent is even, which is divisible by 2. Number Square Perfect (x) (x2) Square a a2 a2 b2 (b2)2 = b4 b4 c3 (c3)2 = c6 c6 d5 (d5)2 = d10 d10 e10 (e10)2 = e20 e20 15  STUDY GUIDE In Warm Up!, the area of the new figure in step 3, is in the form of a difference of two perfect squares. The difference of the squares of two numbers is equal to the product of their sum and their difference as shown by the length and width in step 5 of the activity in Warm Up!. That is, 𝒙𝟐 − 𝒚𝟐 = (𝒙 + 𝒚)(𝒙 − 𝒚) Let us take the case of 𝑛2 − 4. What are its factors? Check if the given involves difference of two squares: ✓ It is a binomial. ✓ The operation involved is subtraction. ✓ 𝑛2 is the square of 𝑛, while 4 is the square of 2. Following the format above, the factors of 𝑛2 − 4 are the sum and the difference of the numbers being squared, 𝑛 and 2. Thus, 𝑛2 − 4 = (𝑛 + 2)(𝑛 − 2). 16  STUDY GUIDE Let’s Practice! Example 1: Factor 4𝑥 2 − 25. Solution: Step 1: Check if there is a common monomial factor. 4x2 – 25 have no common monomial factor. Step 2: Check if the given polynomial involves a difference of two squares. Step 3: Identify the square root of each of the terms. The square root is the number being multiplied by itself to obtain another number. 4𝑥 2 = (𝟐𝒙)2 25 = 𝟓2 Step 4: Write the factors as the sum and the difference of the answers in step 3. 4𝑥 2 − 25 = (2𝑥 + 5)(2𝑥 − 5) 17  STUDY GUIDE Try It Yourself! Factor 9𝑚4 − 49. Example 2: Factor 45𝑛3 − 5𝑛. Solution: Step 1: Check if there is a common monomial factor. The GCF of the numerical coefficients is 5 and the GCF of the variables is 𝑛. Therefore, the greatest common monomial factor is 5𝑛. We factor this out. 45𝑛3 − 5𝑛 = 5𝑛(9𝑛2 − 1) Step 2: Check if any of the factors involves a difference of two squares. Step 3: Identify the square roots of each of the terms. 9𝑛2 = (𝟑𝒏)2 1 = 12 18  STUDY GUIDE Step 4: Write the factors from steps 1 and 3. 45𝑛3 − 5𝑛 = 5𝑛(9𝑛2 − 1) = 5𝑛(3𝑛 + 1)(3𝑛 − 1) Thus, 45𝑛3 − 5𝑛 is factored as 5𝑛(3𝑛 + 1)(3𝑛 − 1). Try It Yourself! Factor completely: 16𝑛4 − 1. Example 3: Simplify (222 − 212 ) without using a calculator. Solution: Step 1: Check if there is a common monomial factor. The terms of the given expression has no common monomial factor. Step 2: Check if any of the factors involves a difference of two squares. Notice that the given involves the difference of two squares since 22 and 21 are squared. Step 3: Identify the square roots of each of the terms. 222 = (22)2 212 = (21)𝟐 19  STUDY GUIDE Step 4: Write the factors from step 3. (222 − 212 ) = (22 + 21)(22 − 21) = (43)(1) = 43 Thus, (222 − 212 ) is equal to 43. Try It Yourself! Simplify (1002 − 992 ) without using a calculator. Real-World Problems Example 4: The difference of the areas of two limited edition square stamps is 128 cm2. One side of the bigger stamp is thrice the length of one side of the smaller. Find the lengths of the sides of the two stamps. Solution: Step 1: Identify the given information. Area of bigger square – Area of smaller square = 128 cm2. 20 STUDY GUIDE Step 2: Write the equation. Remember that the area of a square is equal to the square of the length of its side. (3𝑠)2 − 𝑠 2 = 128 Step 3: Solve for the value of s by factoring. (3𝑠)2 − 𝑠 2 = 128 9𝑠 − 𝑠2 2 = 128 Square 3s. 8𝑠 2 = 128 Subtract s2 from 9s2. 8𝑠 2 − 128 = 0 Subtract 128 from both sides of the equation. 2 8(𝑠 − 16) = 0 Factor out the greatest common monomial factor, 8. 2 𝑠 − 16 = 0 Divide both sides of the equation by 8. (𝑠 + 4)(𝑠 − 4) = 0 Factor the difference of two squares. Remember that a product can only be zero if one of the factors is zero. In this case, either 𝑠 + 4 is zero or 𝑠 − 4 is zero. 𝑠 + 4 = 0 ⇒ 𝑠 = −4 or 𝑠−4=0 ⇒𝑠 = 4 The measurement of length cannot be a negative value. Thus, 𝑠 = 4. The length of a side of the smaller stamp is 4 cm while the length of one side of the larger stamp is 12 cm. 21 STUDY GUIDE Try It Yourself! What will be the length of the sides of the two stamps in Example 4, if the difference of their areas is 512 cm2? Check Your Understanding! 1. Factor the following expressions completely: a. 𝑠 2 − 81 b. 625𝑎4 – 100 c. 0.09𝑥 2 𝑦 4 – 121 d. 16𝑥 2𝑛 − 1 2. Without raising any number to a power, evaluate the following: a. 10002 − 9992 b. 802 − 702 c. 20182 − 20172 3. A photo is surrounded by a frame of uniform width. The length of the outer side of the frame is 4 inches more than the length of a side of the photo. If the area covered by the frame alone is 80 square inches, what is the length of a side of the photo? 22 STUDY GUIDE Lesson 3: Factoring the Sum or Difference of Two Cubes Warm Up! Special Delivery! Materials Needed: paper and pen, stopwatch or timer Instructions: 1. This activity may be played by the entire class with the members of the class divided into groups of five and a person assigned as timer and facilitator. 2. The facilitator will stand in front of the class and choose one expression to be simplified from the given table below. 3. One representative from each group will move to the back of the classroom to answer or give the product of the expressions flashed by the facilitator. 4. The members of each group are to take turns in answering. 5. The fastest representative to reach the facilitator with the correct answer will a get a point. 6. The group with the highest score out of the 5 items in the table will be considered the winner. Questions (𝑥 + 𝑦)(𝑥 2 − 𝑥𝑦 + 𝑦 2 ) (𝑥 − 𝑦)(𝑥 2 + 𝑥𝑦 + 𝑦 2 ) (𝑥 + 4)(𝑥 2 − 4𝑥 + 16) (𝑥 − 3)(𝑥 2 + 3𝑥 + 9) (𝑥 + 2)(𝑥 2 − 2𝑥 + 4) 23  STUDY GUIDE Learn about It! Recall first what a perfect cube is. A constant is a perfect cube if it is obtained by using a number as a factor three times. Observe the following examples: Number Cube Perfect 3 (x) (x ) Cube 1 13 = 1 1 3 33 = 27 27 5 53 = 125 125 8 83 = 512 512 3 10 10 = 1 000 1 000 A variable is a perfect cube when its exponent is divisible by 3. Variable Cube Perfect 3 (x) (x ) Cube a a3 a3 b2 (b2)3 = b6 b6 c3 (c3)3 = c9 c9 d5 (d5)3 = d15 d15 e10 (e10)3 = e30 e30 The expression given in Warm Up! result in either a sum or a difference of two cubes which can be factored as 24  STUDY GUIDE 𝒙𝟑 + 𝒚𝟑 = (𝒙 + 𝒚)(𝒙𝟐 − 𝒙𝒚 + 𝒚𝟐 ) 𝒙𝟑 − 𝒚𝟑 = (𝒙 − 𝒚)(𝒙𝟐 + 𝒙𝒚 + 𝒚𝟐 ) Let us take the case of 𝑘 3 − 125. What are its factors? Check if the given expression involves a sum or difference of two cubes: ✓ It is a binomial. ✓ The operation involved is either addition or subtraction. In this case, it is subtraction. ✓ 𝑘 3 is the cube of 𝑘, while 125 is the cube of 5. Following the format above for difference of two cubes, we have 𝒙 = 𝒌 and 𝒚 = 𝟓. So, 𝑥 3 − 𝑦 3 = (𝑥 − 𝑦)(𝑥 2 − 𝑥𝑦 + 𝑦 2 ) 𝒌3 − 𝟓3 = (𝒌 − 𝟓)(𝒌2 + 𝒌(𝟓) + 𝟓2 ) 𝑘 3 − 125 = (𝑘 − 5)(𝑘 2 + 5𝑘 + 25) Let’s Practice! Example 1: Find the factors of 8b3 + 1 Solution: Step 1: Check if there is a common monomial factor. 8b3 and 1 have no common monomial factor. Step 2: Check if the given polynomial involves the sum or the difference of two cubes. 25  STUDY GUIDE ✓ 8b3 + 1 is a binomial ✓ The operation involved is addition. ✓ 8b3 is the cube of 2𝑏, while 1 is the cube of 1. Step 3: Identify the cube root of each of the terms. 8𝑏 3 = (2𝑏)3 , so √(2𝑏)3 = 2𝑏 3 1 = (1)3, so √(1)3 = 1 3 Step 4: Form the factors using the formula 𝑥 3 + 𝑦 3 = (𝑥 + 𝑦)(𝑥 2 − 𝑥𝑦 + 𝑦 2 ) with 𝑥 = 2𝑏 and 𝑦 = 1. 8𝑏 3 + 1 = (2𝑏 + 1)[(2𝑏)2 − (2𝑏)(1) + 12 ] = (2𝑏 + 1)(4𝑏 2 − 2𝑏 + 1) Therefore, the factors of 8𝑏 3 + 1 are (2𝑏 + 1) and (4𝑏 2 − 2𝑏 + 1). Try It Yourself! Find the factors of 𝑚9 + 64𝑛6. Example 2: Factor 250𝑥 3 𝑧 4 − 54𝑦 6 𝑧 4 Solution: Step 1: Check if there is a common monomial factor. 250𝑥 3 𝑧 4 − 54𝑦 6 𝑧 4 = 2𝑧 4 (125𝑥 3 − 27𝑦 6 ) Step 2: Check if any of the factors involves the sum or the difference of two cubes. 26  STUDY GUIDE ✓ 125𝑥 3 − 27𝑦 6 is a binomial. ✓ The operation involved is subtraction. ✓ 125𝑥 3 is the cube of 5𝑥, while 27𝑦 6 is the cube of 3𝑦 2. Step 3: Identify the cube root of each of the terms. 125𝑥 3 = (5𝑥)3 , so √(5𝑥)3 = 5𝑥 3 27𝑦 6 = (3𝑦 2 )3, so √(3𝑦 2 )3 = 3𝑦 2 3 Step 4: Write the factors from steps 1 and 3. 125𝑥 3 − 27𝑦 6 = (5𝑥 + 3𝑦)(25𝑥 2 − 15𝑥𝑦 + 9𝑦 2 ) 250𝑥 3 𝑧 4 − 54𝑦 6 𝑧 4 = 2𝑧 4 (125𝑥 3 − 27𝑦 6 ) = 2𝑧 4 (5𝑥 − 3𝑦 2 )(25𝑥 2 + 15𝑥𝑦 + 9𝑦 4 ) Therefore, 𝟐𝟓𝟎𝒙𝟑 𝒛𝟒 − 𝟓𝟒𝒚𝟔 𝒛𝟒 = 𝟐𝒛𝟒 (𝟓𝒙 − 𝟑𝒚𝟐 )(𝟐𝟓𝒙𝟐 + 𝟏𝟓𝒙𝒚 + 𝟗𝒚𝟒 ). Try It Yourself! Factor completely: 135𝑟 3 𝑤 5 − 1715𝑠 6 𝑡 3 𝑤 5. Example 3: Factor completely: 𝑓 6 − 1. Solution: Step 1: Check if there is a common monomial factor. 𝑓 6 and 1 have no common monomial factor. 27 STUDY GUIDE Step 2: Check if the given is a difference of two squares. 𝑓 6 is the square of 𝑓 3 , and 1 is the square of 1. Using 𝑥 2 − 𝑦 2 = (𝑥 + 𝑦)(𝑥 − 𝑦), we have 𝑓 6 − 1 = (𝑓 3 + 1)(𝑓 3 − 1) Step 3: Check if any of the factors involves the sum or the difference of two cubes. ✓ (𝑓 3 + 1) and (𝑓 3 − 1) are binomials ✓ The operations involved are addition and subtraction. ✓ 𝑓 3 is the cube of 𝑓, while 1 is the cube of 1. Step 4: Identify the cube root of each of the terms. 𝑓 3 = (𝑓)3 , so √(𝑓)3 = 𝑓 3 1 = (1)3, so √(1)3 = 1 3 Step 5: Write the factors from steps 1 and 3. Using the formula for factoring the sum of two cubes, we have 𝑓 3 + 1 = (𝑓 + 1)(𝑓 2 − 𝑓 + 1) 𝑓 6 − 1 = (𝑓 3 + 1)(𝑓 3 − 1) = (𝑓 + 1)(𝑓 2 − 𝑓 + 1)(𝑓 − 1)(𝑓 2 + 𝑓 + 1) = (𝑓 + 1)(𝑓 − 1)(𝑓 2 − 𝑓 + 1)(𝑓 2 + 𝑓 + 1) Therefore, in factored form, 𝑓 6 − 1 = (𝑓 + 1)(𝑓 − 1)(𝑓 2 − 𝑓 + 1)(𝑓 2 + 𝑓 + 1). 28  STUDY GUIDE Try It Yourself! Factor completely: 𝑎6 + 1. Real-World Problems Example 4: The volume of a Rubik’s cube with side n is equal to n3. Another Rubik’s cube whose side measures 8 cm is placed on top of the first cube. If the sum of their volumes is 1 843 cm3, how long is one side of the other cube? Solution: Step 1: Identify the given and illustrate the given problem. Sum of the volumes = 1843 cm3. Step 2: Write the equation. Remember that the volume of a cube is equal to the cube of the length of one side. 𝑛3 + 83 = 1 843 29  STUDY GUIDE Step 3: Solve for the value of 𝑛 by factoring. 𝑛3 + 83 = 1 843 𝑛3 + 512 = 1 843 Evaluate 83. 𝑛3 − 1 331 = 0 Subtract 1 843 from both sides of the equation. 2 (𝑛 − 11)(𝑛 + 11𝑛 + 121) = 0 Factor the difference of two cubes. Remember that a product can only be zero if one of the factors is zero. In this case, either 𝑛 − 11 = 0 or 𝑛2 + 11𝑛 + 121 = 0. But 𝑛2 + 11𝑛 + 121 is no longer factorable and definitely would not equal to zero for any positive value of n, so we have 𝑛 − 11 = 0 or 𝑛 = 11. Therefore, the length of one side of the bigger cube is 11 cm. Try It Yourself! What will be the length of a side of the bigger cube in Example 4, if the smaller cube has a side whose length 5 cm and the sum of their volumes is 341 cm3? Check Your Understanding! 1. Identify if the following mathematical statement is true or false: (𝑚9 − 1) = (𝑚3 + 1)(𝑚3 − 1). 30  STUDY GUIDE 2. Factor the following expressions completely. a. 𝑡 3 − 27 b. 64𝑎3 + 125 64 c. − 𝑦3 1000 d. (𝑎 + 2)3 + (𝑎 − 2)3 3. A wooden block is in the shape of a cube which is hollow in the inside. The hollow space inside is also in the shape of a cube. The outside length of a side of the block is four cm longer than a side of the cube-shaped hollow space inside. If the total volume occupied by the wooden part of the block is 448 cm3, find the side length of the cube and the side length of the cube-shaped hollow space inside. Lesson 4: Factoring Perfect Square Trinomials Warm Up! Tile Four-mation Materials Needed: cartolina, pen, and scissors Instructions: 1. This activity may be done in groups of three. 2. Each group will be assigned a number. 3. You and your groupmates shall cut out from cartolinas the following: 4 big squares measuring 4” × 4” 8 rectangular tiles measuring 4” × 1” 16 small squares measuring 1” × 1” 31  STUDY GUIDE 4. Each of the groups shall form the following for each round. 1 big square tile, 2 rectangular tiles, and 1 small square. 1 big square tile, 4 rectangular tiles, and 4 small squares. 1 big square tile, 6 rectangular tiles, and 9 small squares. 1 big square tile, 8 rectangular tiles, and 16 small squares. 4 big square tiles, 4 rectangular tiles, and 1 small square. 4 big square tiles, 8 rectangular tiles, and 4 small squares. 5. Once your group is done forming a square with the indicated pieces, shout out your group number. 6. The teacher shall then verify if a square was formed and the pieces are correct. 7. The group that is first to get the pieces right shall receive a point. 8. The group with the most number of points after 6 rounds wins. Learn about It! In Warm Up!, you were able to create squares of varying dimensions using the tiles. Consider the square formed using 1 big square tile, 8 rectangular tiles, and 16 small squares tiles. If the side of the 4” × 4” square is 𝑥, the shorter side of the 4” × 1” rectangle is 1 unit, and the side of the 1” × 1” square is 1 unit, the area covered by the square can be modeled by the polynomial: 𝑥 2 + 8𝑥 + 16. Can you try to find the measure of one of its sides? To solve this kind of problem, remember that the area of a square is equal to the square of the length of its side. So, if the length of one side of a square is x, then the area is x2. Similarly, if the length is 𝑎 + 𝑏, then the area is (𝑎 + 𝑏)2 = 𝑎2 + 2𝑎𝑏 + 𝑏 2. Moreover, if the length is a – b, then the area is (𝑎 − 𝑏)2 = 𝑎2 − 2𝑎𝑏 + 𝑏 2. 32  STUDY GUIDE Such expression for the area of a square is called a perfect square. Moreover, since the square of a + b results in a polynomial with three terms, it is called a perfect square trinomial. Definition 4.1: A perfect square trinomial is an algebraic expression that can be written in the form 𝒂𝟐 ± 𝟐𝒂𝒃 + 𝒃𝟐 In the case of the square formed by 1 big square tile, 8 rectangular tiles, and 16 small square tiles, the side is of length 𝑥 + 4. That means the area is (𝑥 + 4)2. Thus, equating the two areas, we have 𝑥 2 + 8𝑥 + 16 = (𝑥 + 4)2 and found a way to factor the perfect square trinomial 𝑥 2 + 8𝑥 + 16. Generally for a perfect square trinomial 𝑎2 + 2𝑎𝑏 + 𝑏 2 , by working backwards, we have 𝒂𝟐 ± 𝟐𝒂𝒃 + 𝒃𝟐 = (𝒂 ± 𝒃)𝟐 33  STUDY GUIDE Note that the first and last terms are squares of 𝑎 and 𝑏 respectively (which should always be positive), and the middle term is positive or negative twice the product of 𝑎 and 𝑏. Hence, in the factored form, the operation involved depends on the sign of the middle term. Let us check if the given area (in square meters) in the problem, 𝑥 2 + 8𝑥 + 16. Does it follow the pattern for a perfect square trinomial? In 𝑥 2 + 8𝑥 + 16, observe that 𝑥 2 = (𝒙)2 and 16 = (4)2. To check if it is a perfect square trinomial, the middle term must be equal to twice the product of 𝒙 and 𝟒. Since 2(𝑥)(4) = 8𝑥 (the middle term), the given expression for the area is a perfect square trinomial. Factoring the expression, we have 𝑥 2 + 8𝑥 + 16 = (𝑥 + 4)2. Hence, we also find by factoring that the length of one side of the square is 𝑥 + 4 meters. Let’s Practice! Example 1: Write 𝑛2 − 14𝑛 + 49 as the square of a binomial. Solution: Step 1: Check if there is a common monomial factor. 𝑛2 − 14𝑛 + 49 has no common monomial factor. Step 2: Identify if the given polynomial is a perfect square trinomial. 34  STUDY GUIDE ✓ The first and last terms are both perfect squares. 𝑛2 = (𝑛)2 and 49 = (7)2 ✓ The middle term is negative two times the product of the square roots of the first and last terms. – 2(𝑛)(7) = – 14𝑛 Step 3: Use the square root of the first term of the trinomial as the first term of the factor and the square root of the last term of the trinomial as the second term of the factor. Square the binomial formed. (Note that the operation involved in the factor depends on the sign of the middle term.) √𝑛2 = 𝑛 and √49 = 7 (𝑛 + 7)2 Hence, 𝑛2 − 14𝑛 + 49 = (𝑛 − 7)2. Try It Yourself! Write 4𝑝2 + 36𝑝 + 81 as the square of a binomial. Example 2: Factor completely: 18𝑥 3 + 60𝑥 2 + 50𝑥. Solution: Step 1: Check if there is a common monomial factor. 18𝑥 3 + 60𝑥 2 + 50𝑥 = 𝟐𝒙(9𝑥 2 + 30𝑥 + 25) 35 STUDY GUIDE Step 2: Identify if the trinomial factor is a perfect square. ✓ The first and last terms are both perfect squares. 9𝑥 2 = (3𝑥)2 and 25 = (5)2 ✓ The middle term is twice the product of the square roots of the first and last terms. 2(3𝑥)(5) = 30𝑥 Step 3: Use the square root of the first term of the trinomial as the first term of the factor; use the square root of the last term of the trinomial as the second term of the factor. Square the binomial formed. Note that the middle term is positive (addition) so the operation to be used in the factor is addition. √9𝑥 2 = 3𝑥 and √25 = 5 18𝑥 3 + 60𝑥 2 + 50𝑥 = 2𝑥 (9𝑥 2 + 30𝑥 + 25) = 2𝑥(3𝑥 + 5)2 Hence, 18𝑥 3 + 60𝑥 2 + 50𝑥 = 2𝑥(3𝑥 + 5)2. Try It Yourself! Factor completely: 108𝑚2 𝑝2 − 72𝑚2 𝑝 + 12𝑚2. 36  STUDY GUIDE Example 3: The square of a number is (𝑛 + 3)2 + 4(𝑛 + 3) + 4. What is the number if 𝑛 = 9? Solution: Step 1: List down the given information. Square of the number = (𝑛 + 3)2 + 4(𝑛 + 3) + 4 𝑛=9 Step 2: Find the square root of (𝑛 + 3)2 + 4(𝑛 + 3) + 4 by factoring. We first check if (𝑛 + 3)2 + 4(𝑛 + 3) + 4 is a perfect square trinomial. Notice that (𝑛 + 3)2 may be taken as the first term and is the square of 𝑛 + 3 4 may be taken as the last term and is the square of 2 4(𝑛 + 3) may be taken as the middle term and is twice the product of 𝑛 + 3 and 2 Therefore, (𝑛 + 3)2 + 4(𝑛 + 3) + 4 is in the form of 𝑎2 + 2𝑎𝑏 + 𝑏 2 where 𝑎 = 𝑛 + 3 and 𝑏 = 2. Following the format for the factors of a perfect square trinomial, we have (𝑛 + 3)2 + 4(𝑛 + 3) + 4 = [(𝑛 + 3) + 2]2 = (𝑛 + 5)2 Hence, the number is equivalent to 𝑛 + 5. Step 3: Substitute 9 for n. 𝑛+5=9+5 = 14 Therefore, the number is 14. 37  STUDY GUIDE Try It Yourself! How long is one side of a square whose area is 36𝑟 4 + 60𝑟 2 + 25 cm2? Real-World Problems Example 4: Mang Berting is planting palay on a square piece of land. The land area is given by the expression 100 − 20𝑚 + 𝑚2 square meters. What is the length of the side of the land if 𝑚 = 4? Solution: Step 1: List down the given information. Area of the square piece of land = 100 − 20𝑚 + 𝑚2 𝑚=4 Step 2: Find the square root of 100 − 20𝑚 + 𝑚2 by factoring. 100 − 20𝑚 + 𝑚2 is a perfect square trinomial in the form 𝑎2 − 2𝑎𝑏 + 𝑏 2 with 𝑎 = 10 and 𝑏 = 𝑚 because 100 is a perfect square; it is the square of 10. 𝑚2 is a perfect square; it is the square of 𝑚. −20𝑚 is equal to −2(10)(𝑚). Following the formula for factoring a perfect square trinomial, we have 100 − 20𝑚 + 𝑚2 = (10 − 𝑚)2 Hence, the side of the square is equivalent to (10 − 𝑚). 38 STUDY GUIDE Step 3: Substitute the value of 𝑚. (10 − 𝑚) = 10 − 4 =6 Therefore, the measure of the side of the square is 6 m. Try It Yourself! The area of a square coaster is given by the expression 16 − 8𝑎𝑏 + 𝑎2 𝑏 2 cm2. What is the length of a side of the coaster if 𝑎 = 4 and 𝑏 = 2? Check Your Understanding! 1. Factor completely. a. 64 + 16𝑝 + 𝑝2 b. 25𝑥 2 + 30𝑥𝑦 + 9𝑦 2 c. 144𝑎2 − 72𝑎 + 9 2. Complete each expression to form a perfect square trinomial. a. 25𝑥 2 + 10𝑥 + _____ b. 16𝑥 2 + ______ + 1 c. _____ − 30𝑎4 𝑏 3 + 25𝑏 6 3. A wall clock has a shape of a square. Its area is given by the algebraic expression 4𝑤 2 − 12𝑤 + 9 cm2. Find the length of a side of the clock if 𝑤 = 14. 39 STUDY GUIDE Lesson 5: Factoring General Trinomials Warm Up! Cards on the Table Materials Needed: flaglet for each group, standard deck of cards (only the spade and heart suits are needed) Instructions: 1. This activity may be played by the whole class. One of your classmates shall assume the role of game master. Divide yourselves in groups of five. 2. The game master shall take hold of the cards and shuffle them. 3. Each card shall be assigned an integer: A = 0 7 = 7 A = 0 7 = –7 1 = 1 8 = 8 1 = –1 8 = –8 2 = 2 9 = 9 2 = –2 9 = –9 3 = 3 10 = 10 3 = –3 10 = –10 4 = 4 J = 11 4 = –4 J = –11 5 = 5 Q = 12 5 = –5 Q = –12 6 = 6 K = 13 6 = –6 K = –13 4. The game master shall then pick two cards at random, remember them, and have them face down on the table. 5. Using the table above, the game master shall then compute the sum and the product of the corresponding integers of the cards and tell these to the class. 40  STUDY GUIDE 6. The groups are then supposed to try to correctly guess the card as fast they can. 7. Once a group has the answers, they should raise their flaglet to be recognized by the game master. 8. The group that correctly guesses the two cards (with the correct number and suit) gets a point. 9. The group that gets the most points wins. Learn about It! Suppose the product of two numbers is 𝑥 2 + 5𝑥 + 6. What are the numbers? To factor out a general trinomial, you can use the technique called the AC method, which is basically the “inverse” of the FOIL method. First, identify the coefficients 𝐴 and 𝐶 of the trinomial in the form 𝐴𝑥 2 + 𝐵𝑥 + 𝐶, and then find their product. 𝑥 2 + 5𝑥 + 6 = 𝟏𝑥 2 + 5𝑥 + 𝟔 𝐴=1 𝐶=6 𝐴𝐶 = 6 Then find the factors of 𝐴𝐶 = 6 that add up to the middle term’s coefficient, 𝐵. In this case, 𝐵 = 5. Factors of 6 Sum 1, 6 7 2, 3 5 The factors of 6 that have a sum of 5 are 2 and 3. Next, rewrite the trinomial by expanding the middle term, 5𝑥, into 2𝑥 + 3𝑥: 41  STUDY GUIDE 𝑥 2 + 𝟓𝒙 + 6 = 𝑥 2 + 𝟐𝒙 + 𝟑𝒙 + 6 Regroup the polynomial into two sets of binomials. Factor each binomial using their greatest common factor. Then we factor the common binomial factor from each set to form the two factors. 𝑥 2 + 5𝑥 + 6 = (𝑥 2 + 2𝑥) + (3𝑥 + 6) = 𝒙(𝑥 + 2) + 𝟑(𝑥 + 2) = (𝒙 + 𝟑)(𝒙 + 𝟐) Thus, written in factored form, 𝒙𝟐 + 𝟓𝒙 + 𝟔 = (𝒙 + 𝟑)(𝒙 + 𝟐). Let’s Practice! Example 1: Factor completely: 𝑥 2 + 7𝑥 + 10. Solution: Step 1: Check if there is a common monomial factor and/or if the given trinomial is a perfect square. If so, factor accordingly. The terms of 𝑥 2 + 7𝑥 + 10 have no common monomial factor. 𝑥 2 + 7𝑥 + 10 is not a perfect square trinomial. Step 2: Find 𝐴𝐶. Given 𝑥 2 + 7𝑥 + 10, 𝐴 = 1 and 𝐶 = 10. 𝐴𝐶 = 1 × 10 = 𝟏𝟎 42 STUDY GUIDE Step 3: Find the factors of 10 whose sum is the coefficient of the middle term, 7. Factors of 10 Sum 1, 10 11 2, 5 7 Note: When the coefficients of the middle and last terms are both positive, the factors of 𝐴𝐶 must be both positive as well. Step 4: Rewrite the middle term of the original trinomial using the factors obtained in step 3. 𝑥 2 + 𝟕𝒙 + 10 = 𝑥 2 + 𝟐𝒙 + 𝟓𝒙 + 10 Step 5: Regroup the resulting polynomial and factor each group using their greatest common factor. Then factor out the common binomial factor from set. 𝑥 2 + 7𝑥 + 10 = (𝑥 2 + 2𝑥) + (5𝑥 + 10) = 𝒙(𝑥 + 2) + 𝟓(𝑥 + 2) = (𝑥 + 5)(𝑥 + 2) Therefore, 𝒙𝟐 + 𝟕𝒙 + 𝟏𝟎 = (𝒙 + 𝟓)(𝒙 + 𝟐). Try It Yourself! Factor 𝑥 2 + 9𝑥 + 18. 43  STUDY GUIDE Example 2: Factor completely: 10𝑥 2 + 15𝑥 − 70. Solution: Step 1: Check if there is a common monomial factor and/or if the given trinomial is a perfect square. If so, factor accordingly. 10𝑥 2 + 15𝑥 − 70 = 𝟓(2𝑥 2 + 3𝑥 − 14) The trinomial factor 2𝑥 2 + 3𝑥 − 14 is not a perfect square. Step 2: For the trinomial 2𝑥 2 + 3𝑥 − 14, find 𝐴𝐶. Given 2𝑥 2 + 3𝑥 − 14, 𝐴 = 2 and 𝐶 = −14. 𝐴𝐶 = 2 × (−14) = −𝟐𝟖 Step 3: Find the factors of –28 whose sum is the coefficient of the middle term, 3. Factors of –28 Sum 1, –28 –27 –1, 28 27 2, –14 –12 –2, 14 12 4, –7 –3 –4, 7 3 Note: When the last term is negative, one of the factors of 𝐴𝐶 must be positive while the other must be negative. 44  STUDY GUIDE Step 4: Rewrite the middle term of the original trinomial using the factors obtained in step 3. 2𝑥 2 + 𝟑𝒙 − 14 = 𝑥 2 − 𝟒𝒙 + 𝟕𝒙 − 14 Step 5: Regroup the resulting polynomial and factor each group using their greatest common factor. Then factor out the common binomial factor from each set. 2𝑥 2 + 3𝑥 − 14 = (2𝑥 2 − 4𝑥) + (7𝑥 − 14) = 𝟐𝒙(𝑥 − 2) + 𝟕(𝑥 − 2) = (2𝑥 + 7)(𝑥 – 2) Therefore, 10𝑥 2 + 15𝑥 − 70 = 5(2𝑥 + 7)(𝑥 − 2). Try It Yourself! Find the factors of 3𝑥 2 − 6𝑥 − 45. Example 3: The product of two positive integers is in the form 𝑥 2 − 8𝑥 + 12. Without evaluating the given expression itself, give two such integers if 𝑥 = 31. Solution: To get the numbers, we can find the factors of 𝑥 2 − 8𝑥 + 12. Step 1: Check if there is a common monomial factor and/or if the given trinomial is a perfect square. If so, factor accordingly. The terms of 𝑥 2 − 8𝑥 + 12 has no common monomial factor. 𝑥 2 − 8𝑥 + 12 is not a perfect square trinomial. 45 STUDY GUIDE Step 2: Find 𝐴𝐶. Given 𝑥 2 − 8𝑥 + 12, 𝐴 = 1 and 𝐶 = 12. 𝐴𝐶 = 𝟏 × 𝟏𝟐 = 𝟏𝟐 Step 3: Find the factors of 12 whose sum is the coefficient of the middle term, –8. Factors of –28 Sum –1, –12 –13 –2, –6 –8 –3, –4 –7 Note: When the last term is positive but the middle term is negative, the factors of 𝐴𝐶 must be both negative. Step 4: Rewrite the middle term of the original trinomial using the factors obtained in step 3. 𝑥 2 − 𝟖𝒙 + 12 = 𝑥 2 − 𝟐𝒙 − 𝟔𝒙 + 12 Step 5: Regroup the resulting polynomial and factor each group using their greatest common factor. Then factor out the common binomial factor from each set. 𝑥 2 − 8𝑥 + 12 = (𝑥 2 − 8𝑥) + (−6𝑥 + 12) = 𝒙(𝑥 − 2) − 𝟔(𝑥 − 2) = (𝑥 − 6)(𝑥 − 2) Step 6: Substitute x = 31 to find the values of the missing numbers. 𝑥 − 6 = 31 − 6 = 25 𝑥 − 2 = 31 − 2 = 29 Therefore, two such integers are 25 and 29. 46  STUDY GUIDE Try It Yourself! The product of two positive integers is in the form 𝑥 2 − 9𝑥 + 18. Without evaluating the given expression itself, give two such integers if 𝑥 = 25. Real-World Problems Example 4: The floor area of a chain of boutiques is given by 75 + 28𝑎 + 𝑎2 square meters. Without evaluating the given expression itself, find its dimensions if the length is 𝑎 + 25, and 𝑎 = 2. Solution: Step 1: Check if there is a common monomial factor and/or if the given trinomial is a perfect square. If so, factor accordingly. ✓ 75 + 28𝑎 + 𝑎2 has no common monomial factor. ✓ Also it is not a perfect square. Step 2: To factor 75 + 28𝑎 + 𝑎2 , find 𝐴𝐶. 75 + 28𝑎 + 𝑎2 written in standard form is 𝑎2 + 28𝑎 + 75. Thus, 𝐴 = 1 and 𝐶 = 75. 𝐴𝐶 = 1 × 75 = 75 Step 3: Find the factors of 75 whose sum is the coefficient of the middle term, 28. Factors of 75 Sum 1, 75 76 –1, –75 –76 3, 25 28 47 STUDY GUIDE –3, –25 –28 5, 15 20 -5, -15 -18 Step 4: Rewrite the middle term of the original trinomial using the factors obtained in step 3. 75 + 28𝑎 + 𝑎2 = 75 + 𝟐𝟓𝒂 + 𝟑𝒂 + 𝑎2 Step 5: Regroup the resulting polynomial and factor each group using their greatest common factor. Then factor out the common binomial factor from each set. 75 + 28𝑎 + 𝑎2 = 75 + 25𝑎 + 3𝑎 + 𝑎2 = 𝟐𝟓(3 + 𝑎) + 𝒂(3 + 𝑎) = (25 + 𝑎)(3 + 𝑎) Step 6: Evaluate the length and the width using 𝑎 = 2. Since the length is 𝑎 + 25 = 25 + 𝑎, its width would be 3 + 𝑎. Finally, if 𝑎 = 2, then the length is 27 m, while the width is 5m. Try It Yourself! What will be the dimensions of the floor area of the chain of boutiques in Example 4 if the area is 12 + 7𝑥 + 𝑥 2 square meters, the length is 𝑥 + 4, and 𝑥 = 4? Compute for the required dimensions without evaluating the given expression itself. 48 STUDY GUIDE Check Your Understanding! 1. Use a table to factor each of the following trinomials. a. 𝑥 2 + 5𝑥 + 4 b. 𝑥 2 − 9𝑥 + 20 c. 𝑥 2 − 3𝑥 − 28 d. 2𝑥 2 + 16𝑥 + 14 e. 10𝑥 2 − 13𝑥 − 3 2. Find all values of 𝑘 for which the given trinomials can be factored. a. 𝑥 2 + 𝑘𝑥 + 36 b. 𝑥 2 + 𝑘𝑥 + 45 c. 𝑥 2 − 𝑘𝑥 − 30 3. Without evaluating the expression for the area, find the dimensions of a smartphone if its area is given by 2𝑥 2 + 9𝑥 + 10, its length is 2𝑥 + 5 and 𝑥 = 5. Challenge Yourself! 1. Provide an explanation for errors in factoring of the given polynomials below. a. 𝑥 2 + 16 = (𝑥 + 4)(𝑥 + 4) b. 1.6𝑥 2 − 9 = (0.4𝑥 − 3)(0.4𝑥 + 3) c. 3𝑥 2 − 27 is prime (i.e. it is not factorable). 2. Is the GCF of any expression always a monomial? Give an example that supports your answer. 3. Relate the concepts in specials products to the factoring techniques that you have studied in this unit. Provide explanations on their relationship. 49  STUDY GUIDE Performance Task (This activity will showcase your learning in this unit. You and two other teammates will assume the role of packaging specialists hired by a cosmetics company.) The company is launching their new products, and as the head of a team hired by the company, you are tapped to provide your expertise in formulating and answering inquiries regarding the packaging of the company’s new products. You will present the models and explain your computations to the chief executive officer (CEO) of the company and to the head of operations and logistics department. Specifically, you and your team are to make different prismatic (in the form of a prism) bundles by stacking products of different sizes. The products are packaged in boxes with the following dimensions: The first part of the task is purely computational. In this part, you are to answer some inquiries about the dimensions of the bundles for specific. The solutions shall not yet use the construction of the actual boxes as well as diagrams since they may be solved purely by operations on polynomials and factoring. You are to answer all inquiries for all of the following bundles: 50  STUDY GUIDE Bundle 1: Form a prismatic bundle by stacking 1 blue box, 2 green boxes and 1 red box. a) Can you compute for the total base area covered by if the height of the bundle is (𝑥 + 𝑦) cm? b) Can you provide its length and width in terms of 𝑥 and 𝑦? c) If 𝑥 = 8 cm and 𝑦 = 4 cm, provide the total base area covered as well as the length and width. Bundle 2: Form a prismatic bundle by stacking 2 blue boxes and 4 green boxes. a) Can you compute for the total base area covered by if the height of the bundle is 2𝑥 cm? b) Can you provide its length and width in terms of 𝑥 and 𝑦? c) If 𝑥 = 10 cm and 𝑦 = 5 cm, provide the total base area covered as well as the length and width. Bundle 3: Form a prismatic bundle with a volume equal to 2𝑥 3 + 3𝑥 2 𝑦 + 𝑥𝑦 2. a) What are the dimensions (length, width, height) of this prismatic bundle in terms of 𝑥 and 𝑦? b) If 𝑥 = 8 cm and 𝑦 = 5 cm, compute for the length, width, and height of the bundle. c) What products will fit into this bundle and what are their dimensions? Bundle 4: Form a prismatic bundle with a volume equal to 𝑥 3 + 4𝑥 2 𝑦 + 4𝑥𝑦 2. a) What are the dimensions (length, width, height) of this prismatic bundle in terms of 𝑥 and 𝑦? b) If 𝑥 = 10 cm and 𝑦 = 4 cm, compute for the length, width, and height of the bundle. c) What products will fit into this bundle and what are their dimensions? After accomplishing the computational part of task, the second part of the task is to create the model of a bundle. Your team is to create actual models of these boxes placed stacked to form one of these prismatic bundles. You shall choose to model only one (1) of the bundles stated previously. The aim of the actual model is to confirm if your 51  STUDY GUIDE computations for the dimensions asked in the inquiries agree with the boxes in your prismatic bundle models. You are free to use any material and to create your own patterns and designs in making the boxes. The third part of the task is the presentation of your computations and models to the class. You may refer to the given format below as a guide for your presentation. I. Product Name Project Leader Members Date Submitted II. Computations and Explanations (State each problem and present your complete and detailed solutions for each problem. Do not forget to specify the factoring concept or formula you used in solving the problem.) III. Prismatic Bundle Model (Indicate the bundle you chose to model. Present a diagram of the chosen bundle with dimensions. Present to the class the actual stacking of the boxes to form the prismatic bundle.) The presentation will be evaluated according to the following: explanation of the proposal, accuracy of computations, and appropriateness of the design and model. 52  STUDY GUIDE Performance Task Rubric Below Needs Successful Exemplary Criteria Expectation Improvement Performance Performance (0–49%) (50–74%) (75–99%) (99+%) Explanations and Explanations and presentation of presentation of Explanations and Explanations and Explanation of the layout is the layout is a presentation of presentation of the Design difficult to little difficult to the layout is the layout is Proposal understand and understand but clear. detailed and clear. is missing several includes critical components. components. The The The computations computations The computations computations done are done are done are accurate done are Accuracy of erroneous and accurate and and show an erroneous and Computations do not show wise shows moderate extensive use of show little use of use of the use of the the concepts of the concepts of concepts of concepts of factoring. factoring. factoring. factoring. The model is well- The model is well- crafted and crafted and useful The model is not The model is useful for for understanding useful in somewhat useful Appropriateness understanding the design understanding in understanding of the Designs the design proposal. It the design the design proposal. It showcases the proposal. proposal. showcases the desired layout and desired layout. is accurately done. 53 STUDY GUIDE Wrap-up Key Concept Description The greatest common monomial factor is the product of the GCF of the numerical coefficients and the GCF of the variables. To factor an expression using the greatest common monomial factor, follow these steps: Greatest Common Step 1: Factor out the greatest common monomial factor. Monomial Factor Step 2: Divide each term of the expression by the greatest comomon monomial factor to get the other factor. Note: Finding the greatest common monomial factor is the first step to be done in factoring any expression. Difference of Two 𝒂𝟐 − 𝒃𝟐 = (𝑎 + 𝑏)(𝑎 − 𝑏) Squares Sum and Difference 𝒂𝟑 + 𝒃𝟑 = (𝑎 + 𝑏)(𝑎2 − 𝑎𝑏 + 𝑏 2 ) of Two Cubes 𝒂𝟑 − 𝒃𝟑 = (𝑎 − 𝑏)(𝑎2 + 𝑎𝑏 + 𝑏 2 ) Perfect Square 𝒂𝟐 + 𝟐𝒂𝒃 + 𝒃𝟐 = (𝑎 + 𝑏)2 Trinomial 𝒂𝟐 − 𝟐𝒂𝒃 + 𝒃𝟐 = (𝑎 − 𝑏)2 To find the factors of a trinomial in the form 𝑨𝒙𝟐 + 𝑩𝒙 + 𝑪, follow these steps: Step 1: Find 𝐴𝐶. Step 2: Find factors of 𝐴𝐶 whose sum is 𝐵. Step 3: Rewrite the middle term using the values found in General Trinomial Step 2. Step 4: Regroup the terms and take greatest common monomial factor in each group. Step 5: Factor out the common binomial factor from each group to find the factors. 54