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 â€˜primeâ€™ factors of 5xy. In algebraic expressions, we use the word â€˜irreducibleâ€™ in place of â€˜primeâ€™. 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. 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 â€˜primeâ€™ factors of 5xy. In algebraic expressions, we use the word â€˜irreducibleâ€™ in place of â€˜primeâ€™. 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. 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) = ( ) ( ) 25 2 3 xx 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 â€¢ W e 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.)Read More

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### NCERT Solutions(Part- 1)- Factorisation

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### Factorisation using Identities: Part 1

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