PREFACE. Synergy for Success in Mathematics 8 is designed for Grade 8 students. The textbook contains

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1 Synergy for Success in Mathematics 8 is designed for Grade 8 students. The textbook contains all the required learning competencies and is supplemented with some additional topics for enrichment. Lessons are presented using effective Singapore Math strategies that are intended for easy understanding and grasp of ideas for its target readers. Various exercises are also provided to help the learners acquire the necessary skills needed. The book is organized with the following recurring features in every chapter: PREFACE Learning Goals Introduction Historical Note Method/Exam Notes Examples Enhancing Skills Linking Together Chapter Test Chapter Project Making Connection This gives the specific objectives that are intended to be achieved in the end. The reader is given a bird's eye view of the contents. A brief historical account of a related topic is included giving the reader an awareness of some important contributions of some great mathematicians or even stories of great achievements related to mathematics. These additional tools help students recall important information, formula, and shortcuts, needed in working out solutions. Step-by-step and detailed demonstrations of how a specific concept or technique is applied in solving problems. These are practice exercises found after every lesson, that will consolidate and reinforce what the students have learned. This visual tool can help the students realize the connection of all the ideas presented in the chapter. This is a summative test given at the end in preparation for the expected actual classroom examination containing the topics included in the chapter. A challenging task is designed for the learner giving him an opportunity to use what he/she has learned in the chapter. This may be a manipulative type of activity that is specifically chosen to enhance understanding of the concepts learned in the chapter. The students are exposed to facts and information that connect mathematics and culture. This is for the purpose of letting the learners appreciate the subject because of tangible or true-to-life stories that show how useful and relevant mathematics is. Every effort has been made in order for all the discussions in this book to be clear, simple, and straightforward. This book also gives opportunities for the readers to see the beauty of mathematics as an essential tool in understanding the world we live in. With this in mind, appreciation of mathematics goes beyond seeing; realizing its critical application to decision making in life completes the purpose of knowing and understanding mathematics.

2 Table of C ntents CHAPTER 1 SPECIAL PRODUCTS AND FACTORING Introduction...1 Historical Note Special Products... The Product of a Monomial and a Polynomial... The Product of Two Binomials...4 The Product of the Sum and the Difference of Two Terms...5 The Square of a Binomial...6 The Cube of a Binomial...7 The Product Giving the Sum or the Difference of Two Cubes...9 Combining Special Products Factoring...16 Common Monomial Factoring...17 Factoring by Grouping...1 Factoring the Difference of Two Squares... Factoring a Perfect Square Trinomial...6 Factoring General Trinomials...8 Factoring the Cube of a Binomial... Factoring the Sum or the Difference of Two Cubes...5 Linking Together...41 Chapter Test...4 Chapter Project...45 Making Connection...46 CHAPTER INTEGRAL EXPONENTS AND RATIONAL EXPRESSIONS Introduction...47 Historical Note Integral Exponents...49

3 Positive Integer Exponents...49 Laws of Exponents...51 Zero and Negative Integer Exponents...55 Scientific Notation...61 Converting from Scientific Notation to Decimal Notation...61 Converting from Decimal Notation to Scientific Notation...6. Rational Expressions...67 Polynomials...67 Rational Expressions...67 Simplifying Rational Expressions...68 Equivalent Rational Expressions Operations on Rational Expressions...77 Multiplication and Division of Rational Expressions...77 Addition and Subtraction of Rational Expressions...8 Rational Expressions with Different Denominators...85 Complex Rational Expressions Equations Involving Rational Expressions...96 Solutions to Equations Involving Rational Expressions...96 Word Problems Using Rational Equations Linking Together Chapter Test Chapter Project...11 Making Connection CHAPTER LINEAR EQUATIONS AND LINEAR FUNCTIONS Introduction Historical Note Rectangular Coordinate System Relations and Functions...14 Relations...14 Equations as Relations...17 Graphs of Linear Equations...19

4 Functions...15 The Vertical Line Test...18 Linear Function Evaluating a Linear Function Slope and Equation of a Line The Slope of a Line The Equation of a Line Applications of Linear Functions Linking Together Chapter Test Chapter Project Making Connection...18 CHAPTER 4 SYSTEMS OF LINEAR EQUATIONS AND INEQUALITIES IN TWO VARIABLES Introduction...18 Historical Note Systems of Linear Equations in Two Variables Solving Systems of Two Linear Equations in Two Variables Applications of Systems of Linear Equations in Two Variables...08 Number and Geometric Relations Problems...08 Investment Problems...11 Mixture Problems...1 Uniform Motion and Work Problems Linear Inequalities and Systems of Linear Inequalities in Two Variables... Graphing a Linear Inequality in Two Variables...5 Systems of Linear Inequalities...8 Application of Linear Inequalities... Linking Together...45 Chapter Test...46 Chapter Project...49 Making Connection...50

5 CHAPTER 5 REASONING AND PROVING Introduction...51 Historical Note Conditional Statements...5 Truth Value and Counterexamples...55 Negation...55 Converse, Inverse, and Contrapositive of a Conditional Statement Inductive and Deductive Reasoning...67 Inductive Reasoning...67 Deductive Reasoning Formal Proofs...77 Direct Proof...77 Indirect Proof...81 Linking Together...88 Chapter Test...89 Chapter Project...91 Making Connection...9 CHAPTER 6 TRIANGLE CONGRUENCE AND ITS APPLICATIONS Introduction...9 Historical Note Triangle Congruence...95 Corresponding Parts...95 Proving Triangle Congruence...97 Using Triangle Congruence Using Congruent Triangles...16 Corresponding Parts of Congruent Triangles...16 Proving the Congruence of Sides and Angles...0 Segments Related to Triangles On Triangle Inequalities...46 Linking Together...61

6 Chapter Test...6 Chapter Project...67 Making Connection...68 CHAPTER 7 STATISTICS Introduction...69 Historical Note Measures of Central Tendency...71 Measures of Central Tendency for Ungrouped Data...7 Measures of Central Tendency for Grouped Data Measures of Variability Measures of Variability of Ungrouped Data Measures of Variability of Grouped Data Linking Together...4 Chapter Test...4 Chapter Project...47 Making Connection...48 CHAPTER 8 PROBABILITY Introduction...49 Historical Note Introduction to Probability...41 Types of Probability...4 Theoretical Probability...4 Probability of Two or More Events...45 Counting Outcomes Experimental Probability Linking Together Chapter Test Chapter Project Making Connection Glossary Index Bibliography Photo Credits...47

7 1 SPECIAL PRODUCTS AND FACTORING Learning Goals At the end of the chapter, you should be able to: Suppose you are asked in a quiz bee to evaluate the following expressions ; 4 (4)(5) + 5 How can you evaluate these expressions accurately within a limited time? Recall that these expressions follow certain patterns called special products. In this chapter, you will learn how to make fast algebraic calculations using some techniques. These techniques are helpful in solving real-life problems. The review on special products is necessary in the introduction of factoring polynomials with the use of special product formulas, one can multiply special polynomials fast. Also, factoring techniques are very helpful in dividing polynomials and simplifying rational expressions. Review multiplication of polynomials and derive special products Identify polynomials which are special products and factor different types of polynomials completely using special product formulas, grouping, and other techniques

8 Historical Note Blaise Pascal (June 19, 16 August 19, 166) was a French mathematician, physicist, inventor, writer, and religious philosopher. He was a child prodigy. When he was 18, he made the mechanical calculator called Pascaline, which could be used to add, subtract, multiply, and divide numbers. Also, he devised the Pascal s triangle (also called arithmetical triangle), which gives the coefficients of the terms in the expansion of n ( x + y). He contributed significantly in various mathematical areas like projective geometry and probability theory.

9 Synergy for Success in Mathematics Chapter Special Products Special products are simplified products of two or more polynomial factors. This lesson is a review of the special products. Knowledge of the special products is necessary in factoring polynomials. The Product of a Monomial and a Polynomial One of the basic special products involves multiplying a polynomial by a monomial. This technique is simply based on the distributive property of real numbers. The Product of a Monomial and a Polynomial The product of a polynomial and a monomial is the sum of the products of each term of the polynomial and the monomial. ax ( + y+ z)= ax + ay + az Example 1 Multiply the following expressions. (a) 1 xy 5x + xy+ y (b) 6aa+ 5b (c) x yz xyz 10xy z xyz (a) 1 xy 5x + xy+ y = + 1 xy 5x xy xy + xy y = 15x y x y 6xy (b) 6aa+ 5b = 6aa + 6a( 5b) = 6a + 0ab ( )

10 (c) x yz xyz 10xy z xyz x yz xyz x yz 10xyz x yz = = 6x y z 0x yz x y z ( xyz ) The Product of Two Binomials Recall the FOIL method in finding the product of two binomials. FOIL stands for the product of the First terms, the product of the Outer terms, the product of the Inner terms, and the product of the Last terms. F L F O I L ( x+ a) ( x+ b)= x( x)+( x)( b)+( a)( x)+( a)( b) I O = x + bx+ ax+ ab = x + a+ bx+ab F O I L x+ a ( x b)= x + bx+ ax+ ab + Example Multiply the following pairs of binomials. ( ) ( x y) ( x+ y) (a) 5x+ x 5 (b) 5 6 ( ) =( 5x)( x)+( 5x) ( 5)+( )( x)+( ) ( 5) (a) 5x+ x 5 = 10x 5x+ 6x 15 = 10x 19x 15 ( + ) =( x)( 5x)+( x)( 6y)+ ( y)( 5x)+ ( y)( 6y) (b) x y 5x 6 y = 15x + 18xy 5xy 6y = 15x + 1xy 6 y 4

11 Synergy for Success in Mathematics Chapter 1 Example Use the FOIL method in multiplying the following binomials. ( + ) (a) a+ 6 a ( + ) (b) ab 1 4ab 5 ( + ) (a) a+ 6 a = a( a)+ a()+ 6 ( a)+ 6 () = a + a+ 18a + 18 = a + 1a+ 18 ( + ) =+()+ + (b) ab 1 4ab 5 () ab 4ab ab 5 1 4ab = 8ab + 10ab 4ab 5 4 = 8ab + 6ab 5 The Product of the Sum and the Difference of Two Terms The product of the sum and the difference of two terms is simply the square of the first term minus the square of the second term. The Product of the Sum and the Difference of Two Terms The product of the sum and the difference of two terms is the difference of the squares of those terms. x+ y ( x y)= x y Example 4 Find the products. 1 1 (a) 7x y 7x y + ( )( ) (b) 5xy + 4z 5xy 4z a a (c) 7 + 5c 7 5c b b 5

12 1 1 (a) 7x y 7x y + 1 =( 7x) y = 49x 1 9 y ( ) (b) 5xy + 4z 5xy 4z (c) =( 5xy ) ( 4z ) = 5x y 16z 4 4 7a 7a + 5c 5c b b 7a = 5c b 49a = b 5c The Square of a Binomial is a trinomial The square of the sum of two terms x+ y that is the square of the first term of the binomial plus twice the product of the first and second term plus the square of the last term of the binomial. = + + x+ y x xy y On the other hand, the square of the difference of two terms ( x y) is a trinomial that is the square of the first term of the binomial minus twice the product of the first and second term plus the square of the last term of the binomial. = + x y x xy y The Square of a Binomial The square of a binomial is the square of its first term plus or minus twice the product of the two terms plus the square of its second term. = + + = + x+ y x xy y x y x xy y ( x+ y) x + y Exam Note 6

13 Synergy for Success in Mathematics Chapter 1 Example 5 Find the square of each binomial. (a) a+ 8b (b) mn -pq (c) 8m + n n = + + (a) a+ 8b a a 8b 8b (b) mn pq = 4a + ab+ 64b =( mn ) ( mn )( pq )+( pq ) = 4mn 1mn pq + 9p q = 4mn 1mn pq + 9pq 4 6 (c) 8m + n n 8m 8 m = n n n + n 64m = + 48m+ 9n n + The Cube of a Binomial is the cube of the first The cube of a binomial sum x+ y term plus thrice the product of the square of the first term and the second term plus thrice the product of the first term and square of the second term plus the cube of the second term. = x+ y x x y xy y The cube of a binomial difference x y with alternating signs. = + x y x x y xy y is similar, except 7

14 The Cube of a Binomial The cube of a binomial is the cube of its first term plus (or minus) thrice the product of the square of its first term and second term plus thrice the product of its first term and the square of its second terms plus (or minus) the cube of its second term. = x+ y x x y xy y = + x y x x y xy y ( x+ y) x + y Exam Note Example 6 Find the special products. (a) x + ( a b) (b) 5 (a) x + (b) 5a =( x) + ( x) ()+ ( x)()+( ) 8x 4x x + = + ()+ 9 7 = 8x + 6x + 54x + 7 ( b) =( 5a ) ( 5a ) ( b)+ ( 5a )( b) ( b) a 5a b = + 5 = 15a 75a b+ 15a b b 6 4 a b b Example 7 Find the cube of the binomial 1 a a = a a a ()+ () ( ) = a 6 1 a a = a a + 6a 8 8 8

15 The Product Giving the Sum or the Difference of Two Cubes Another special product technique involves multiplying a binomial and a trinomial that gives a result of either the sum or the difference of two cubes. A binomial and a trinomial factor results in either the sum or the difference of two cubes if the following conditions are satisfied: The first term of the trinomial is the square of the first term of the binomial. The second term of the trinomial is the additive inverse of the product of the first and second terms of the binomial. The third term of the trinomial is the square of the second term of the binomial. Synergy for Success in Mathematics Chapter 1 The Product Giving the Sum or the Difference of Two Cubes and the trinomial The product of the binomial x+ y ( x xy+ y ) is the cube of the first term plus the cube of the second term of the binomial. x+ y ( x xy y )= x + y + and the trinomial The product of the binomial x y ( x + xy+ y ) is the cube of the first term minus the cube of the second term of the binomial. x y ( x xy y )= x y + + Example 8 Multiply the following expressions. ( + ) ( ab c) ( a 4 b + a bc + c ) (a) x+ y x xy 4 y (b) (a) x+ y x xy 4 y = x +( y) = x + 8 y 9

16 4 ( + + ) (b) ab 5c 9a b 15a bc 5c =( ab) ( 5c) = 7ab 15c 6 Method Note Let ab be the first term and 5c be the second term. The special product is the difference of the cube of those two terms. Example 9 Multiply a a 1 + 9a a 4 1 =( a ) + a 6 = 7a + 1 7a 1 + 9a a 4 1 a+ 9a 1 a+ 9a. Combining Special Products There are some instances in which special products cannot be used right away to get the product of the given polynomials. Multiplying those polynomials then becomes time consuming. In such cases, the terms of the given are grouped; and then the special products are applied. Example 10 and Group the terms of the expression w+ x y z solve it using special products. ( w+ x y z) = ( w+ x) ( y+ z) = ( w+ x) ( w+ x) ( y+ z)+ ( y+ z) = ( w + wx + x ) ( wy+ wz+ xy+ xz)+ y + yz + z = w + wx+ x wy wz xy xz + y + yz+ z = w + x + y + z + wx wy wz xy xz + yz 10

17 Example 11 Find for the products of the following expressions. (a) ( x y+ 5) (b) a+ 5 6b+ 7 a 5 6b 7 ( + + ) (a) The given polymomial does not comply with any of the special products previously discussed. However, the terms in the factor can be grouped, and we can still apply special products. Group the first two terms of x y+ 5. Then apply the square of a binomial to find the product. ( x y+ 5) = ( x y)+ 5 = ( x y) + () ( x y)() 5 5 = ( x y) + 10( x y)+ 5 = ( x y) + ( 0x 0 y)+ 5 = x y 0x + 0 y + 5 +() = ( x) +( x) ( y)+ ( y) 0x 0y = 4x 1xy+ 9y + 0x 0y+ 5 Synergy for Success in Mathematics Chapter 1 ( + + ) (b) a+ 5 6b+ 7 a 5 6b 7 = ( a+ 5) ( 6b 7) ( a+ 5)+ ( 6b 7) = a+ 5 6b 7 ( ) = ( a) + ( a)()+( 5 5) 6b 6b 7 + ( )+ 7 = ( 9a + 0a+ 5) ( 6b 84b+ 49) = 9a + 0a 6b + 84b = 9a + 0a 6b + 84b 4 = 9a 6b + 0a+ 84b 4 11

18 Example 1 Find each product. ( ) (a) 9x+ 5 9x 5 (b) n ( ) (c) 4. a a (a) ( 9x+ 5) ( 9x 5)= 9x 5 = 81x 5 (b) n 9 n n = n 414n 81 = n 69n 81 = (c) ( 4. a ) ( 4. a )=( 4. a) ( ) = a

19 Synergy for Success in Mathematics Chapter 1 ENHANCING SKILLS Find the special products. (1) x x 5x + 1 (6) 4r 5s () ab5ab+ 11ab 5b + + (7) x 9x 6x 4 ( + ) () 5y 7z 8y 5z + + (8) x 9x 6x 4 ( ) (4) 8ab+ 7c 8ab 7c ( ) (9) p +5q p 5q (5) x + 5 ( )( + ) (10) m 4n m 4n 1

20 ( )( ) (11) 5ax + by 5ax by ( )( ) (16) 6x y z 5x y 9z ( )( + ) (1) 5ab 8c 5ab 8c (17) m 6n (1) ( 6ab 7) (18) 4t + t (14) 11x y+ xy + (19) v+ 4z 9v 1vz 16z ( + ) (15) 5rs 11 rs 8 (0) 5x x 5 14

21 ( )( ) (1) 7a b 49a 1a b 9b Synergy for Success in Mathematics Chapter 1 ( )( ) (6) 7x y 49x 14x y 4 y 1 1 () 5x + y x y 5 ( + + ) (7) a 5b+ a 5b () m 1 + n 4 (8) x y (4) x y + 6 ( + ) (9) 5a b 5a b (5) 6ab 7ab ( + ) (0) x y x x y x 15

22 1. Factoring Factoring is the process of getting the factors of a given polynomial. It is the reverse of finding the product of polynomials; factoring the product results in the original polynomials that were multiplied together to get the product. Consider the problem below. A man pours cement on the floor of his garage. The garage floor has an area of 15 square meters. Its dimensions are prime numbers. Can you determine the length and the width of the garage? What procedure will you follow to find those dimensions? You will most likely express 15 as a factor of two numbers. To factor a number means to write it as a product of two or more numbers. = =() The factors of 15 are 1,, 5, and 15. Note that and 5 are the prime numbers among the factors. Hence, the dimensions of the garage floor are meters and 5 meters. Also, a polynomial can be expressed as a product of prime factors. This process is called factoring. Method Note Recall that a prime number has only two positive integral factors, 1 and itself. 16

23 Synergy for Success in Mathematics Chapter 1 Factoring a polynomial means writing it as a product of prime factors. A polynomial is prime or irreducible if it cannot be expressed as a product of at least two polynomials. The factors of a prime or irreducible polynomial are only 1 and itself or -1 and itself. For example, the binomial x + can only be expressed as a product of the factors ( x + ) and 1. Hence, the polynomial x + is prime or irreducible since it cannot be expressed as a product of two polynomials. On the other hand, the binomial 4x + 10xcan be expressed as the product of x and x + 5. Hence, 4x + 10x is factorable. Below are examples of prime polynomials: (a) x + (b) ab-c (c) x+ y+ z (d) x + y A polynomial is said to be factored completely if it is written as a product consisting only of prime or irreducible factors. Common Monomial Factoring The first step in factoring a polynomial is to determine if there is a monomial factor that is common to all terms of the polynomial. It is advisable to look for the greatest common factor (GCF) of the terms. The greatest common factor of a polynomial with more than one term is a monomial with the greatest coefficient and degree that is common to all the terms. For example, to find the greatest common factor of the polynomial 1x + 0x, calculate first the GCF of the numerical coefficients 1 and 0. The GCF of 1 and 0 is 6. Next, look for the GCF of the literal coefficients x and x. Their GCF is x. Hence, the GCF of the terms of 1x + 0x is 6x. 17

24 Finding the Greatest Common Monomial Factor Step 1 Step Find the greatest common factor of the numerical coefficients. Find the GCF of the literal coefficients. The product of the GCF of the numerical coefficients and the GCF of the literal coefficients is the greatest common monomial factor of the polynomial. Once the greatest common monomial factor of a polynomial has been determined, the polynomial can then be factored. The GCF of the terms of 1x + 0x is 6x. This is the first factor. The second factor is solved by dividing each term of 1x + 0x by 6x. 1x + 0x 6x + () 6x x 6x 5 = 6x 6x x+ 5 = 6x = x + 5 Thus, the factors of 1x + 0x are 6x and x + 5. The polynomial 1x + 0x written in factored form is 1x + 0x = 6x ( x+ 5). Greatest Common Monomial Factoring The following steps show the process of finding the greatest common monomial factor of a polynomial: Step 1 Find the GCF of the terms of the polynomial to get the first factor. Step Find the second factor by dividing the given polynomial by the GCF. polynomial =( GCF) polynomial GCF Method Note This factoring uses the distributive property ax + ay = ax ( + y). 18

25 Example 1 Find the common monomial factor of each polynomial and write it in factored form. (a) x + 18 (b) 9rs+ 15rs (c) x -10x -6x (d) 5t + 0t 4 5 (a) Common monomial factor: The other factor: x +6 Factored form: x + 6 (b) Common monomial factor: -rs The other factor: r-5s Factored form: rs r 5s (c) Common monomial factor: x The other factor: x -5x- Factored form: x x 5x (d) Common monomial factor: 5t The other factor: t + 6 Factored form: 5t t + 6 Synergy for Success in Mathematics Chapter 1 Example Factor the polynomials completely. (a) 0x y + 8x y (b) 18abc -6a bc (c) 1rs + 4rs 60rs (a) The GCF of the numerical coefficients 0x 4 y 5 + 8x y 6 is 4, while the GCF of the literal coefficients is x y 5. Thus, the greatest common monomial factor is 4x y 5. 19

26 The other factor is obtained by dividing 0x 4 y 5 + 8x y 6 by 4x y 5. This yields 0x y + 8x y 5 4x y x y 5x 4x y y = 5 4x y 5 4x y 5x+ y = 5 4x y = 5x+ y Hence, 0x 4 y 5 + 8x y 6 when factored completely is 5 4x y ( 5x+ y). (b) The greatest common monomial factor of the given polynomial is 6abc. 4 Use it as a divisor to get the other factor abc 6a bc 4 6abc 4 4 ac 6abc 6a bc = 4 6abc 4 6abc a c 1 = 4 6abc 1 = ac 6 4 Thus, the factored form of 18abc - 6a bc is 6abc 4 ( a c 1). (c) Find the GCF of the terms. Since the first term is negative, you may place a negative sign beside the GCF to make it negative, Then, get the other factor by dividing the given polynomial with the GCF. 1rs + 4rs 60rs 1rs = 1rs rs r + 5s 1rs = rs r+5 s. Method Note Factoring out the negative of the GCF is usually desirable when the terms in the remaining factor are not in descending order. Thus, when factored completely, 1rs + 4rs 60rs is equal to 1rs rs r+ 5s. 0

27 Synergy for Success in Mathematics Chapter 1 Example Identify the greatest common factor of 15x y z + 75x yz 5x y z and factor the polynomial completely. The greatest common factor of 15x y z + 75x yz 5x y z is 5x y z. Get the other factor by finding the quotient of the polynomial and the greatest common factor. 15x y z + 75x yz 5x y z 5x y z 5x y z 5x+ y z = 5x y z = 5x+ y z Hence, 15x y z + 75x yz 5x y z when factored completely is 5x y z ( 5x+ y z). Factoring by Grouping There are instances in which there is no common factor among the terms of a polynomial. Examine the polynomial below. x+ y+ xz + yz The terms of the given polynomial does not have a common factor. However, notice that the first two terms both have a numerical coefficient of and the last two terms a literal have coefficient of z. The terms can then be grouped so that each group has a common factor. x+ y+ xz + yz = ( x+ y)+ xz + yz ( + ) = ( x+ y)+ z x+ y = x+ y z Exam Note If you cannot see that x+ y is a common factor, try replacing the binomial with another variable. Method Note Pair up the two terms with the common factor in one group and the other two terms with the common factor z in another group. Method Note Factor each group. In this case, the common factor of the two groups is ( x+ y). 1

28 Factoring by Grouping The following steps show the process of factoring by grouping. Step 1 Find any greatest monomial factor that is common among all terms of the given polynomial. Step Group pairs of terms so that the terms in each group have a common factor. Step Find the GCF in each group. If there is a common binomial factor, then factor the polynomial. Method Note If there is no common binomial factor, try to interchange and regroup the terms. Step 4 Write the polynomial as a product of the common factor and the other remaining factors. Example 4 Factor the polynomials completely. (a) rv + sv rw sw (b) 5xy 10xz + y z (c) x 10x 4x + 0 (d) 10ab 10ab + 5ab 5ab (e) 4rs t+ 8rs + 1rt + 4rt (a) The first two terms have the common monomial factor v, and the other two terms have the common factor w. Grouping the first two terms and the other two terms yields rv + sv rw sw = ( rv + sv) ( rw + sw) = v( r+ s) w( r+ s) Factoring out the common binomial factor r+ s gives the factored form ( r+ s) ( v w). (b) 5xy 10xz + y z = ( 5xy 10xz)+ y z ( + ) = 5x( y z)+ y z = y z 5x 1

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