Page 1 FACTORISATION 217 14.1 Introduction 14.1.1 Factors of natural numbers Y ou will remember what you learnt about factors in Class VI. Let us take a natural number , say 30, and write it as a product of other natural numbers, say 30 = 2 × 15 = 3 × 10 = 5 × 6 Thus, 1, 2, 3, 5, 6, 10, 15 and 30 are the factors of 30. Of these, 2, 3 and 5 are the prime factors of 30 (Why?) A number written as a product of prime factors is said to be in the prime factor form; for example, 30 written as 2 × 3 × 5 is in the prime factor form. The prime factor form of 70 is 2 × 5 × 7. The prime factor form of 90 is 2 × 3 × 3 × 5, and so on. Similarly, we can express algebraic expressions as products of their factors. This is what we shall learn to do in this chapter. 14.1.2 Factors of algebraic expressions W e have seen in Class VII that in algebraic expressions, terms are formed as products of factors. For example, in the algebraic expression 5xy + 3x the term 5xy has been formed by the factors 5, x and y, i.e., 5xy = y x× × 5 Observe that the factors 5, x and y of 5xy cannot further be expressed as a product of factors. We may say that 5, x and y are ‘prime’ factors of 5xy. In algebraic expressions, we use the word ‘irreducible’ in place of ‘prime’. W e say that 5 × x × y is the irreducible form of 5xy. Note 5 × (xy) is not an irreducible form of 5xy, since the factor xy can be further expressed as a product of x and y, i.e., xy = x × y. Factorisation CHAPTER 14 Note 1 is a factor of 5xy, since 5xy = y x× × × 5 1 In fact, 1 is a factor of every term. As in the case of natural numbers, unless it is specially required, we do not show 1 as a separate factor of any term. We know that 30 can also be written as 30 = 1 × 30 Thus, 1 and 30 are also factors of 30. You will notice that 1 is a factor of any number. For example, 101 = 1 × 101. However, when we write a number as a product of factors, we shall not write 1 as a factor, unless it is specially required. 2019-20 Page 2 FACTORISATION 217 14.1 Introduction 14.1.1 Factors of natural numbers Y ou will remember what you learnt about factors in Class VI. Let us take a natural number , say 30, and write it as a product of other natural numbers, say 30 = 2 × 15 = 3 × 10 = 5 × 6 Thus, 1, 2, 3, 5, 6, 10, 15 and 30 are the factors of 30. Of these, 2, 3 and 5 are the prime factors of 30 (Why?) A number written as a product of prime factors is said to be in the prime factor form; for example, 30 written as 2 × 3 × 5 is in the prime factor form. The prime factor form of 70 is 2 × 5 × 7. The prime factor form of 90 is 2 × 3 × 3 × 5, and so on. Similarly, we can express algebraic expressions as products of their factors. This is what we shall learn to do in this chapter. 14.1.2 Factors of algebraic expressions W e have seen in Class VII that in algebraic expressions, terms are formed as products of factors. For example, in the algebraic expression 5xy + 3x the term 5xy has been formed by the factors 5, x and y, i.e., 5xy = y x× × 5 Observe that the factors 5, x and y of 5xy cannot further be expressed as a product of factors. We may say that 5, x and y are ‘prime’ factors of 5xy. In algebraic expressions, we use the word ‘irreducible’ in place of ‘prime’. W e say that 5 × x × y is the irreducible form of 5xy. Note 5 × (xy) is not an irreducible form of 5xy, since the factor xy can be further expressed as a product of x and y, i.e., xy = x × y. Factorisation CHAPTER 14 Note 1 is a factor of 5xy, since 5xy = y x× × × 5 1 In fact, 1 is a factor of every term. As in the case of natural numbers, unless it is specially required, we do not show 1 as a separate factor of any term. We know that 30 can also be written as 30 = 1 × 30 Thus, 1 and 30 are also factors of 30. You will notice that 1 is a factor of any number. For example, 101 = 1 × 101. However, when we write a number as a product of factors, we shall not write 1 as a factor, unless it is specially required. 2019-20 218 MATHEMATICS Next consider the expression 3x (x + 2). It can be written as a product of factors. 3, x and (x + 2) 3x(x + 2) = ( ) 2 3 + × × x x The factors 3, x and (x +2) are irreducible factors of 3x (x + 2). Similarly, the expression 10x (x + 2) (y + 3) is expressed in its irreducible factor form as 10x (x + 2) (y + 3) = ( ) ( ) 2 5 2 3 x x y × × × + × + . 14.2 What is Factorisation? When we factorise an algebraic expression, we write it as a product of factors. These factors may be numbers, algebraic variables or algebraic expressions. Expressions like 3xy, y x 2 5 , 2x (y + 2), 5 (y + 1) (x + 2) are already in factor form. Their factors can be just read off from them, as we already know. On the other hand consider expressions like 2x + 4, 3x + 3y, x 2 + 5x, x 2 + 5x + 6. It is not obvious what their factors are. W e need to develop systematic methods to factorise these expressions, i.e., to find their factors. This is what we shall do now. 14.2.1 Method of common factors • We begin with a simple example: Factorise 2x + 4. W e shall write each term as a product of irreducible factors; 2x = 2 × x 4 = 2 × 2 Hence 2x + 4 = (2 × x) + (2 × 2) Notice that factor 2 is common to both the terms. Observe, by distributive law 2 × (x + 2) = (2 × x) + (2 × 2) Therefore, we can write 2x + 4 = 2 × (x + 2) = 2 (x + 2) Thus, the expression 2x + 4 is the same as 2 (x + 2). Now we can read off its factors: they are 2 and (x + 2). These factors are irreducible. Next, factorise 5xy + 10x. The irreducible factor forms of 5xy and 10x are respectively, 5xy = 5 × x × y 10x = 2 × 5 × x Observe that the two terms have 5 and x as common factors. Now, 5xy + 10x = (5 × x × y) + (5 × x × 2) = (5x × y) + (5x × 2) W e combine the two terms using the distributive law, (5x× y) + (5x× 2) = 5x × ( y + 2) Therefore, 5xy + 10x = 5 x (y + 2). (This is the desired factor form.) 2019-20 Page 3 FACTORISATION 217 14.1 Introduction 14.1.1 Factors of natural numbers Y ou will remember what you learnt about factors in Class VI. Let us take a natural number , say 30, and write it as a product of other natural numbers, say 30 = 2 × 15 = 3 × 10 = 5 × 6 Thus, 1, 2, 3, 5, 6, 10, 15 and 30 are the factors of 30. Of these, 2, 3 and 5 are the prime factors of 30 (Why?) A number written as a product of prime factors is said to be in the prime factor form; for example, 30 written as 2 × 3 × 5 is in the prime factor form. The prime factor form of 70 is 2 × 5 × 7. The prime factor form of 90 is 2 × 3 × 3 × 5, and so on. Similarly, we can express algebraic expressions as products of their factors. This is what we shall learn to do in this chapter. 14.1.2 Factors of algebraic expressions W e have seen in Class VII that in algebraic expressions, terms are formed as products of factors. For example, in the algebraic expression 5xy + 3x the term 5xy has been formed by the factors 5, x and y, i.e., 5xy = y x× × 5 Observe that the factors 5, x and y of 5xy cannot further be expressed as a product of factors. We may say that 5, x and y are ‘prime’ factors of 5xy. In algebraic expressions, we use the word ‘irreducible’ in place of ‘prime’. W e say that 5 × x × y is the irreducible form of 5xy. Note 5 × (xy) is not an irreducible form of 5xy, since the factor xy can be further expressed as a product of x and y, i.e., xy = x × y. Factorisation CHAPTER 14 Note 1 is a factor of 5xy, since 5xy = y x× × × 5 1 In fact, 1 is a factor of every term. As in the case of natural numbers, unless it is specially required, we do not show 1 as a separate factor of any term. We know that 30 can also be written as 30 = 1 × 30 Thus, 1 and 30 are also factors of 30. You will notice that 1 is a factor of any number. For example, 101 = 1 × 101. However, when we write a number as a product of factors, we shall not write 1 as a factor, unless it is specially required. 2019-20 218 MATHEMATICS Next consider the expression 3x (x + 2). It can be written as a product of factors. 3, x and (x + 2) 3x(x + 2) = ( ) 2 3 + × × x x The factors 3, x and (x +2) are irreducible factors of 3x (x + 2). Similarly, the expression 10x (x + 2) (y + 3) is expressed in its irreducible factor form as 10x (x + 2) (y + 3) = ( ) ( ) 2 5 2 3 x x y × × × + × + . 14.2 What is Factorisation? When we factorise an algebraic expression, we write it as a product of factors. These factors may be numbers, algebraic variables or algebraic expressions. Expressions like 3xy, y x 2 5 , 2x (y + 2), 5 (y + 1) (x + 2) are already in factor form. Their factors can be just read off from them, as we already know. On the other hand consider expressions like 2x + 4, 3x + 3y, x 2 + 5x, x 2 + 5x + 6. It is not obvious what their factors are. W e need to develop systematic methods to factorise these expressions, i.e., to find their factors. This is what we shall do now. 14.2.1 Method of common factors • We begin with a simple example: Factorise 2x + 4. W e shall write each term as a product of irreducible factors; 2x = 2 × x 4 = 2 × 2 Hence 2x + 4 = (2 × x) + (2 × 2) Notice that factor 2 is common to both the terms. Observe, by distributive law 2 × (x + 2) = (2 × x) + (2 × 2) Therefore, we can write 2x + 4 = 2 × (x + 2) = 2 (x + 2) Thus, the expression 2x + 4 is the same as 2 (x + 2). Now we can read off its factors: they are 2 and (x + 2). These factors are irreducible. Next, factorise 5xy + 10x. The irreducible factor forms of 5xy and 10x are respectively, 5xy = 5 × x × y 10x = 2 × 5 × x Observe that the two terms have 5 and x as common factors. Now, 5xy + 10x = (5 × x × y) + (5 × x × 2) = (5x × y) + (5x × 2) W e combine the two terms using the distributive law, (5x× y) + (5x× 2) = 5x × ( y + 2) Therefore, 5xy + 10x = 5 x (y + 2). (This is the desired factor form.) 2019-20 FACTORISATION 219 TRY THESE Example 1: Factorise 12a 2 b + 15ab 2 Solution: We have 12a 2 b = 2 × 2 × 3 × a × a × b 15ab 2 = 3 × 5 × a × b × b The two terms have 3, a and b as common factors. Therefore, 12a 2 b + 15ab 2 = (3 × a × b × 2 × 2 × a) + (3 × a × b × 5 × b) = 3 × a × b × [(2 × 2 × a) + (5 × b)] = 3ab × (4a + 5b) = 3ab (4a + 5b) (required factor form) Example 2: Factorise 10x 2 – 18x 3 + 14x 4 Solution: 10x 2 = 2 × 5 × x × x 18x 3 = 2 × 3 × 3 × x × x × x 14x 4 = 2 × 7 × x × x × x × x The common factors of the three terms are 2, x and x. Therefore, 10x 2 – 18x 3 + 14x 4 = (2 × x × x × 5) – (2 × x × x × 3 × 3 × x) + (2 × x × x × 7 × x × x) = 2 × x × x ×[(5 – (3 × 3 × x) + (7 × x × x)] = 2x 2 × (5 – 9x + 7x 2 ) = 2 2 2 (7 9 5) x x x - + Factorise: (i) 12x + 36 (ii) 22y – 33z (iii) 14pq + 35pqr 14.2.2 Factorisation by regrouping terms Look at the expression 2xy + 2y + 3x + 3. Y ou will notice that the first two terms have common factors 2 and y and the last two terms have a common factor 3. But there is no single factor common to all the terms. How shall we proceed? Let us write (2xy + 2y) in the factor form: 2xy + 2y = (2 × x × y) + (2 × y) = (2 × y × x) + (2 × y × 1) = (2y × x) + (2y × 1) = 2y (x + 1) Similarly , 3x + 3 = (3 × x) + (3 × 1) = 3 × (x + 1) = 3 ( x + 1) Hence, 2xy + 2y + 3x + 3 = 2y (x + 1) + 3 (x +1) Observe, now we have a common factor (x + 1) in both the terms on the right hand side. Combining the two terms, 2xy + 2y + 3x + 3 = 2y (x + 1) + 3 (x + 1) = (x + 1) (2y + 3) The expression 2xy + 2y + 3x + 3 is now in the form of a product of factors. Its factors are (x + 1) and (2y + 3). Note, these factors are irreducible. Note, we need to show1 as a factor here. Why? Do you notice that the factor form of an expression has only one term? (combining the three terms) (combining the terms) 2019-20 Page 4 FACTORISATION 217 14.1 Introduction 14.1.1 Factors of natural numbers Y ou will remember what you learnt about factors in Class VI. Let us take a natural number , say 30, and write it as a product of other natural numbers, say 30 = 2 × 15 = 3 × 10 = 5 × 6 Thus, 1, 2, 3, 5, 6, 10, 15 and 30 are the factors of 30. Of these, 2, 3 and 5 are the prime factors of 30 (Why?) A number written as a product of prime factors is said to be in the prime factor form; for example, 30 written as 2 × 3 × 5 is in the prime factor form. The prime factor form of 70 is 2 × 5 × 7. The prime factor form of 90 is 2 × 3 × 3 × 5, and so on. Similarly, we can express algebraic expressions as products of their factors. This is what we shall learn to do in this chapter. 14.1.2 Factors of algebraic expressions W e have seen in Class VII that in algebraic expressions, terms are formed as products of factors. For example, in the algebraic expression 5xy + 3x the term 5xy has been formed by the factors 5, x and y, i.e., 5xy = y x× × 5 Observe that the factors 5, x and y of 5xy cannot further be expressed as a product of factors. We may say that 5, x and y are ‘prime’ factors of 5xy. In algebraic expressions, we use the word ‘irreducible’ in place of ‘prime’. W e say that 5 × x × y is the irreducible form of 5xy. Note 5 × (xy) is not an irreducible form of 5xy, since the factor xy can be further expressed as a product of x and y, i.e., xy = x × y. Factorisation CHAPTER 14 Note 1 is a factor of 5xy, since 5xy = y x× × × 5 1 In fact, 1 is a factor of every term. As in the case of natural numbers, unless it is specially required, we do not show 1 as a separate factor of any term. We know that 30 can also be written as 30 = 1 × 30 Thus, 1 and 30 are also factors of 30. You will notice that 1 is a factor of any number. For example, 101 = 1 × 101. However, when we write a number as a product of factors, we shall not write 1 as a factor, unless it is specially required. 2019-20 218 MATHEMATICS Next consider the expression 3x (x + 2). It can be written as a product of factors. 3, x and (x + 2) 3x(x + 2) = ( ) 2 3 + × × x x The factors 3, x and (x +2) are irreducible factors of 3x (x + 2). Similarly, the expression 10x (x + 2) (y + 3) is expressed in its irreducible factor form as 10x (x + 2) (y + 3) = ( ) ( ) 2 5 2 3 x x y × × × + × + . 14.2 What is Factorisation? When we factorise an algebraic expression, we write it as a product of factors. These factors may be numbers, algebraic variables or algebraic expressions. Expressions like 3xy, y x 2 5 , 2x (y + 2), 5 (y + 1) (x + 2) are already in factor form. Their factors can be just read off from them, as we already know. On the other hand consider expressions like 2x + 4, 3x + 3y, x 2 + 5x, x 2 + 5x + 6. It is not obvious what their factors are. W e need to develop systematic methods to factorise these expressions, i.e., to find their factors. This is what we shall do now. 14.2.1 Method of common factors • We begin with a simple example: Factorise 2x + 4. W e shall write each term as a product of irreducible factors; 2x = 2 × x 4 = 2 × 2 Hence 2x + 4 = (2 × x) + (2 × 2) Notice that factor 2 is common to both the terms. Observe, by distributive law 2 × (x + 2) = (2 × x) + (2 × 2) Therefore, we can write 2x + 4 = 2 × (x + 2) = 2 (x + 2) Thus, the expression 2x + 4 is the same as 2 (x + 2). Now we can read off its factors: they are 2 and (x + 2). These factors are irreducible. Next, factorise 5xy + 10x. The irreducible factor forms of 5xy and 10x are respectively, 5xy = 5 × x × y 10x = 2 × 5 × x Observe that the two terms have 5 and x as common factors. Now, 5xy + 10x = (5 × x × y) + (5 × x × 2) = (5x × y) + (5x × 2) W e combine the two terms using the distributive law, (5x× y) + (5x× 2) = 5x × ( y + 2) Therefore, 5xy + 10x = 5 x (y + 2). (This is the desired factor form.) 2019-20 FACTORISATION 219 TRY THESE Example 1: Factorise 12a 2 b + 15ab 2 Solution: We have 12a 2 b = 2 × 2 × 3 × a × a × b 15ab 2 = 3 × 5 × a × b × b The two terms have 3, a and b as common factors. Therefore, 12a 2 b + 15ab 2 = (3 × a × b × 2 × 2 × a) + (3 × a × b × 5 × b) = 3 × a × b × [(2 × 2 × a) + (5 × b)] = 3ab × (4a + 5b) = 3ab (4a + 5b) (required factor form) Example 2: Factorise 10x 2 – 18x 3 + 14x 4 Solution: 10x 2 = 2 × 5 × x × x 18x 3 = 2 × 3 × 3 × x × x × x 14x 4 = 2 × 7 × x × x × x × x The common factors of the three terms are 2, x and x. Therefore, 10x 2 – 18x 3 + 14x 4 = (2 × x × x × 5) – (2 × x × x × 3 × 3 × x) + (2 × x × x × 7 × x × x) = 2 × x × x ×[(5 – (3 × 3 × x) + (7 × x × x)] = 2x 2 × (5 – 9x + 7x 2 ) = 2 2 2 (7 9 5) x x x - + Factorise: (i) 12x + 36 (ii) 22y – 33z (iii) 14pq + 35pqr 14.2.2 Factorisation by regrouping terms Look at the expression 2xy + 2y + 3x + 3. Y ou will notice that the first two terms have common factors 2 and y and the last two terms have a common factor 3. But there is no single factor common to all the terms. How shall we proceed? Let us write (2xy + 2y) in the factor form: 2xy + 2y = (2 × x × y) + (2 × y) = (2 × y × x) + (2 × y × 1) = (2y × x) + (2y × 1) = 2y (x + 1) Similarly , 3x + 3 = (3 × x) + (3 × 1) = 3 × (x + 1) = 3 ( x + 1) Hence, 2xy + 2y + 3x + 3 = 2y (x + 1) + 3 (x +1) Observe, now we have a common factor (x + 1) in both the terms on the right hand side. Combining the two terms, 2xy + 2y + 3x + 3 = 2y (x + 1) + 3 (x + 1) = (x + 1) (2y + 3) The expression 2xy + 2y + 3x + 3 is now in the form of a product of factors. Its factors are (x + 1) and (2y + 3). Note, these factors are irreducible. Note, we need to show1 as a factor here. Why? Do you notice that the factor form of an expression has only one term? (combining the three terms) (combining the terms) 2019-20 220 MATHEMATICS What is regrouping? Suppose, the above expression was given as 2xy + 3 + 2y + 3x; then it will not be easy to see the factorisation. Rearranging the expression, as 2xy + 2y + 3x + 3, allows us to form groups (2xy + 2y) and (3x + 3) leading to factorisation. This is regrouping. Regrouping may be possible in more than one ways. Suppose, we regroup the expression as: 2xy + 3x + 2y + 3. This will also lead to factors. Let us try: 2xy + 3x + 2y + 3 = 2 × x × y + 3 × x + 2 × y + 3 = x × (2y + 3) + 1 × (2y + 3) = (2y + 3) (x + 1) The factors are the same (as they have to be), although they appear in different order. Example 3: Factorise 6xy – 4y + 6 – 9x. Solution: Step 1 Check if there is a common factor among all terms. There is none. Step 2 Think of grouping. Notice that first two terms have a common factor 2y; 6xy – 4y = 2y (3x – 2) (a) What about the last two terms? Observe them. If you change their order to – 9x + 6, the factor ( 3x – 2) will come out; –9x + 6 = –3 (3x) + 3 (2) = – 3 (3x – 2) (b) Step 3 Putting (a) and (b) together, 6xy – 4y + 6 – 9x = 6xy – 4y – 9x + 6 = 2y (3x – 2) – 3 (3x – 2) = (3x – 2) (2y – 3) The factors of (6xy – 4y + 6 – 9 x) are (3x – 2) and (2y – 3). EXERCISE 14.1 1. Find the common factors of the given terms. (i) 12x, 36 (ii) 2y, 22xy (iii) 14 pq, 28p 2 q 2 (iv) 2x, 3x 2 , 4 (v) 6 abc, 24ab 2 , 12 a 2 b (vi) 16 x 3 , – 4x 2 , 32x (vii) 10 pq, 20qr, 30rp (viii) 3x 2 y 3 , 10x 3 y 2 ,6 x 2 y 2 z 2. Factorise the following expressions. (i) 7x – 42 (ii) 6p – 12q (iii) 7a 2 + 14a (iv) – 16 z + 20 z 3 (v) 20 l 2 m + 30 a l m (vi) 5 x 2 y – 15 xy 2 (vii) 10 a 2 – 15 b 2 + 20 c 2 (viii) – 4 a 2 + 4 ab – 4 ca (ix) x 2 y z + x y 2 z + x y z 2 (x) a x 2 y + b x y 2 + c x y z 3. Factorise. (i) x 2 + x y + 8x + 8y (ii) 15 xy – 6x + 5y – 2 2019-20 Page 5 FACTORISATION 217 14.1 Introduction 14.1.1 Factors of natural numbers Y ou will remember what you learnt about factors in Class VI. Let us take a natural number , say 30, and write it as a product of other natural numbers, say 30 = 2 × 15 = 3 × 10 = 5 × 6 Thus, 1, 2, 3, 5, 6, 10, 15 and 30 are the factors of 30. Of these, 2, 3 and 5 are the prime factors of 30 (Why?) A number written as a product of prime factors is said to be in the prime factor form; for example, 30 written as 2 × 3 × 5 is in the prime factor form. The prime factor form of 70 is 2 × 5 × 7. The prime factor form of 90 is 2 × 3 × 3 × 5, and so on. Similarly, we can express algebraic expressions as products of their factors. This is what we shall learn to do in this chapter. 14.1.2 Factors of algebraic expressions W e have seen in Class VII that in algebraic expressions, terms are formed as products of factors. For example, in the algebraic expression 5xy + 3x the term 5xy has been formed by the factors 5, x and y, i.e., 5xy = y x× × 5 Observe that the factors 5, x and y of 5xy cannot further be expressed as a product of factors. We may say that 5, x and y are ‘prime’ factors of 5xy. In algebraic expressions, we use the word ‘irreducible’ in place of ‘prime’. W e say that 5 × x × y is the irreducible form of 5xy. Note 5 × (xy) is not an irreducible form of 5xy, since the factor xy can be further expressed as a product of x and y, i.e., xy = x × y. Factorisation CHAPTER 14 Note 1 is a factor of 5xy, since 5xy = y x× × × 5 1 In fact, 1 is a factor of every term. As in the case of natural numbers, unless it is specially required, we do not show 1 as a separate factor of any term. We know that 30 can also be written as 30 = 1 × 30 Thus, 1 and 30 are also factors of 30. You will notice that 1 is a factor of any number. For example, 101 = 1 × 101. However, when we write a number as a product of factors, we shall not write 1 as a factor, unless it is specially required. 2019-20 218 MATHEMATICS Next consider the expression 3x (x + 2). It can be written as a product of factors. 3, x and (x + 2) 3x(x + 2) = ( ) 2 3 + × × x x The factors 3, x and (x +2) are irreducible factors of 3x (x + 2). Similarly, the expression 10x (x + 2) (y + 3) is expressed in its irreducible factor form as 10x (x + 2) (y + 3) = ( ) ( ) 2 5 2 3 x x y × × × + × + . 14.2 What is Factorisation? When we factorise an algebraic expression, we write it as a product of factors. These factors may be numbers, algebraic variables or algebraic expressions. Expressions like 3xy, y x 2 5 , 2x (y + 2), 5 (y + 1) (x + 2) are already in factor form. Their factors can be just read off from them, as we already know. On the other hand consider expressions like 2x + 4, 3x + 3y, x 2 + 5x, x 2 + 5x + 6. It is not obvious what their factors are. W e need to develop systematic methods to factorise these expressions, i.e., to find their factors. This is what we shall do now. 14.2.1 Method of common factors • We begin with a simple example: Factorise 2x + 4. W e shall write each term as a product of irreducible factors; 2x = 2 × x 4 = 2 × 2 Hence 2x + 4 = (2 × x) + (2 × 2) Notice that factor 2 is common to both the terms. Observe, by distributive law 2 × (x + 2) = (2 × x) + (2 × 2) Therefore, we can write 2x + 4 = 2 × (x + 2) = 2 (x + 2) Thus, the expression 2x + 4 is the same as 2 (x + 2). Now we can read off its factors: they are 2 and (x + 2). These factors are irreducible. Next, factorise 5xy + 10x. The irreducible factor forms of 5xy and 10x are respectively, 5xy = 5 × x × y 10x = 2 × 5 × x Observe that the two terms have 5 and x as common factors. Now, 5xy + 10x = (5 × x × y) + (5 × x × 2) = (5x × y) + (5x × 2) W e combine the two terms using the distributive law, (5x× y) + (5x× 2) = 5x × ( y + 2) Therefore, 5xy + 10x = 5 x (y + 2). (This is the desired factor form.) 2019-20 FACTORISATION 219 TRY THESE Example 1: Factorise 12a 2 b + 15ab 2 Solution: We have 12a 2 b = 2 × 2 × 3 × a × a × b 15ab 2 = 3 × 5 × a × b × b The two terms have 3, a and b as common factors. Therefore, 12a 2 b + 15ab 2 = (3 × a × b × 2 × 2 × a) + (3 × a × b × 5 × b) = 3 × a × b × [(2 × 2 × a) + (5 × b)] = 3ab × (4a + 5b) = 3ab (4a + 5b) (required factor form) Example 2: Factorise 10x 2 – 18x 3 + 14x 4 Solution: 10x 2 = 2 × 5 × x × x 18x 3 = 2 × 3 × 3 × x × x × x 14x 4 = 2 × 7 × x × x × x × x The common factors of the three terms are 2, x and x. Therefore, 10x 2 – 18x 3 + 14x 4 = (2 × x × x × 5) – (2 × x × x × 3 × 3 × x) + (2 × x × x × 7 × x × x) = 2 × x × x ×[(5 – (3 × 3 × x) + (7 × x × x)] = 2x 2 × (5 – 9x + 7x 2 ) = 2 2 2 (7 9 5) x x x - + Factorise: (i) 12x + 36 (ii) 22y – 33z (iii) 14pq + 35pqr 14.2.2 Factorisation by regrouping terms Look at the expression 2xy + 2y + 3x + 3. Y ou will notice that the first two terms have common factors 2 and y and the last two terms have a common factor 3. But there is no single factor common to all the terms. How shall we proceed? Let us write (2xy + 2y) in the factor form: 2xy + 2y = (2 × x × y) + (2 × y) = (2 × y × x) + (2 × y × 1) = (2y × x) + (2y × 1) = 2y (x + 1) Similarly , 3x + 3 = (3 × x) + (3 × 1) = 3 × (x + 1) = 3 ( x + 1) Hence, 2xy + 2y + 3x + 3 = 2y (x + 1) + 3 (x +1) Observe, now we have a common factor (x + 1) in both the terms on the right hand side. Combining the two terms, 2xy + 2y + 3x + 3 = 2y (x + 1) + 3 (x + 1) = (x + 1) (2y + 3) The expression 2xy + 2y + 3x + 3 is now in the form of a product of factors. Its factors are (x + 1) and (2y + 3). Note, these factors are irreducible. Note, we need to show1 as a factor here. Why? Do you notice that the factor form of an expression has only one term? (combining the three terms) (combining the terms) 2019-20 220 MATHEMATICS What is regrouping? Suppose, the above expression was given as 2xy + 3 + 2y + 3x; then it will not be easy to see the factorisation. Rearranging the expression, as 2xy + 2y + 3x + 3, allows us to form groups (2xy + 2y) and (3x + 3) leading to factorisation. This is regrouping. Regrouping may be possible in more than one ways. Suppose, we regroup the expression as: 2xy + 3x + 2y + 3. This will also lead to factors. Let us try: 2xy + 3x + 2y + 3 = 2 × x × y + 3 × x + 2 × y + 3 = x × (2y + 3) + 1 × (2y + 3) = (2y + 3) (x + 1) The factors are the same (as they have to be), although they appear in different order. Example 3: Factorise 6xy – 4y + 6 – 9x. Solution: Step 1 Check if there is a common factor among all terms. There is none. Step 2 Think of grouping. Notice that first two terms have a common factor 2y; 6xy – 4y = 2y (3x – 2) (a) What about the last two terms? Observe them. If you change their order to – 9x + 6, the factor ( 3x – 2) will come out; –9x + 6 = –3 (3x) + 3 (2) = – 3 (3x – 2) (b) Step 3 Putting (a) and (b) together, 6xy – 4y + 6 – 9x = 6xy – 4y – 9x + 6 = 2y (3x – 2) – 3 (3x – 2) = (3x – 2) (2y – 3) The factors of (6xy – 4y + 6 – 9 x) are (3x – 2) and (2y – 3). EXERCISE 14.1 1. Find the common factors of the given terms. (i) 12x, 36 (ii) 2y, 22xy (iii) 14 pq, 28p 2 q 2 (iv) 2x, 3x 2 , 4 (v) 6 abc, 24ab 2 , 12 a 2 b (vi) 16 x 3 , – 4x 2 , 32x (vii) 10 pq, 20qr, 30rp (viii) 3x 2 y 3 , 10x 3 y 2 ,6 x 2 y 2 z 2. Factorise the following expressions. (i) 7x – 42 (ii) 6p – 12q (iii) 7a 2 + 14a (iv) – 16 z + 20 z 3 (v) 20 l 2 m + 30 a l m (vi) 5 x 2 y – 15 xy 2 (vii) 10 a 2 – 15 b 2 + 20 c 2 (viii) – 4 a 2 + 4 ab – 4 ca (ix) x 2 y z + x y 2 z + x y z 2 (x) a x 2 y + b x y 2 + c x y z 3. Factorise. (i) x 2 + x y + 8x + 8y (ii) 15 xy – 6x + 5y – 2 2019-20 FACTORISATION 221 (iii) ax + bx – ay – by (iv) 15 pq + 15 + 9q + 25p (v) z – 7 + 7 x y – x y z 14.2.3 Factorisation using identities We know that (a + b) 2 = a 2 + 2ab + b 2 (I) (a – b) 2 = a 2 – 2ab + b 2 (II) (a + b) (a – b) = a 2 – b 2 (III) The following solved examples illustrate how to use these identities for factorisation. What we do is to observe the given expression. If it has a form that fits the right hand side of one of the identities, then the expression corresponding to the left hand side of the identity gives the desired factorisation. Example 4: Factorise x 2 + 8x + 16 Solution: Observe the expression; it has three terms. Therefore, it does not fit Identity III. Also, it’s first and third terms are perfect squares with a positive sign before the middle term. So, it is of the form a 2 + 2ab + b 2 where a = x and b = 4 such that a 2 + 2ab + b 2 = x 2 + 2 (x) (4) + 4 2 = x 2 + 8x + 16 Since a 2 + 2ab + b 2 = (a + b) 2 , by comparison x 2 + 8x + 16 = ( x + 4) 2 (the required factorisation) Example 5: Factorise 4y 2 – 12y + 9 Solution: Observe 4y 2 = (2y) 2 , 9 = 3 2 and 12y = 2 × 3 × (2y) Therefore, 4y 2 – 12y + 9 = (2y) 2 – 2 × 3 × (2y) + (3) 2 = ( 2y – 3) 2 (required factorisation) Example 6: Factorise 49p 2 – 36 Solution: There are two terms; both are squares and the second is negative. The expression is of the form (a 2 – b 2 ). Identity III is applicable here; 49p 2 – 36 = (7p) 2 – ( 6 ) 2 = (7p – 6 ) ( 7p + 6) (required factorisation) Example 7: Factorise a 2 – 2ab + b 2 – c 2 Solution: The first three terms of the given expression form (a – b) 2 . The fourth term is a square. So the expression can be reduced to a difference of two squares. Thus, a 2 – 2ab + b 2 – c 2 = (a – b) 2 – c 2 (Applying Identity II) = [(a – b) – c) ((a – b) + c)] (Applying Identity III) = (a – b – c) (a – b + c) (required factorisation) Notice, how we applied two identities one after the other to obtain the required factorisation. Example 8: Factorise m 4 – 256 Solution: We note m 4 = (m 2 ) 2 and 256 = (16) 2 Observe here the given expression is of the form a 2 – 2ab + b 2 . Where a = 2y, and b = 3 with 2ab = 2 × 2y × 3 = 12y. 2019-20Read More

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