Factorisation

Welcome dear students! Today we are going to learn about Factorisation from Class 8 Maths.

Let us begin with factors of natural numbers. You will remember what you learnt about factors in Class Six. Let us take a natural number, say thirty, 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. 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. Similarly, we can express algebraic expressions as products of their factors. This is what we shall learn to do in this chapter.

We have seen in Class Seven 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, that is, 5xy = 5 × x × y. 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. We 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, that is, xy = x × y. 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. Note 1 is a factor of 5xy, since 5xy = 1 × 5 × x × y. 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.

Next consider the expression 3x(x + 2). It can be written as a product of factors 3, x and (x + 2). 3x(x + 2) = 3 × x × (x + 2). 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 × x × (x + 2) × (y + 3).

[CHECKPOINT]

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, 5x²y, 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, x² + 5x, x² + 5x + 6. It is not obvious what their factors are. We need to develop systematic methods to factorise these expressions, that is, to find their factors. This is what we shall do now.

We begin with a simple example: Factorise 2x + 4. We 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). We combine the two terms using the distributive law, (5x × y) + (5x × 2) = 5x × (y + 2). Therefore, 5xy + 10x = 5x(y + 2). This is the desired factor form.

[CHECKPOINT]

Let us look at Example one. Factorise 12a²b + 15ab². Solution: We have 12a²b = 2 × 2 × 3 × a × a × b. 15ab² = 3 × 5 × a × b × b. The two terms have 3, a and b as common factors. Therefore, 12a²b + 15ab² = (3 × a × b × 2 × 2 × a) + (3 × a × b × 5 × b) = 3 × a × b × [(2 × 2 × a) + (5 × b)] by combining the terms. This equals 3ab × (4a + 5b) which is 3ab(4a + 5b). This is the required factor form.

Example two: Factorise 10x² – 18x³ + 14x⁴. Solution: 10x² = 2 × 5 × x × x. 18x³ = 2 × 3 × 3 × x × x × x. 14x⁴ = 2 × 7 × x × x × x × x. The common factors of the three terms are 2, x and x. Therefore, 10x² – 18x³ + 14x⁴ = (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)] by combining the three terms. This equals 2x² × (5 – 9x + 7x²) which is 2x²(7x² – 9x + 5).

Now, let us try these. Factorise: (i) 12x + 36. We write 12x as 2 × 2 × 3 × x, and 36 as 2 × 2 × 3 × 3. The common factors are 2, 2 and 3, which multiply to 12. So 12x + 36 = 12(x + 3). (ii) 22y – 33z. 22y is 2 × 11 × y. 33z is 3 × 11 × z. The common factor is 11. So 22y – 33z = 11(2y – 3z). (iii) 14pq + 35pqr. 14pq is 2 × 7 × p × q. 35pqr is 5 × 7 × p × q × r. The common factors are 7, p and q. So 14pq + 35pqr = 7pq(2 + 5r).

[CHECKPOINT]

Look at the expression 2xy + 2y + 3x + 3. You 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). Note, we need to show 1 as a factor here. Why? Because when we factor out 2y, we are left with x in the first term and one in the second term. 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.

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, although they appear in different order.

[CHECKPOINT]

Example three: Factorise 6xy – 4y + 6 – 9x. Solution: Step one: Check if there is a common factor among all terms. There is none. Step two: Think of grouping. Notice that first two terms have a common factor 2y. 6xy – 4y = 2y(3x – 2). 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). Step three: Putting these 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 – 9x) are (3x – 2) and (2y – 3).

Now let us solve Exercise twelve point one. Question one: Find the common factors of the given terms. (i) 12x, 36. Prime factors of 12x are 2, 2, 3, x. Prime factors of 36 are 2, 2, 3, 3. Common factors are 2, 2, 3. So common factor is 12. (ii) 2y, 22xy. 2y is 2 × y. 22xy is 2 × 11 × x × y. Common factors are 2, y. So common factor is 2y. (iii) 14pq, 28p²q². 14pq is 2 × 7 × p × q. 28p²q² is 2 × 2 × 7 × p × p × q × q. Common factors are 2, 7, p, q. So common factor is 14pq. (iv) 2x, 3x², 4. 2x is 2 × x. 3x² is 3 × x × x. 4 is 2 × 2. Only common factor is 1. (v) 6abc, 24ab², 12a²b. 6abc is 2 × 3 × a × b × c. 24ab² is 2 × 2 × 2 × 3 × a × b × b. 12a²b is 2 × 2 × 3 × a × a × b. Common factors are 2, 3, a, b. So common factor is 6ab. (vi) 16x³, –4x², 32x. 16x³ is 2⁴ × x³. –4x² is –2² × x². 32x is 2⁵ × x. Common factors are 2, 2, x. So common factor is 4x. (vii) 10pq, 20qr, 30rp. 10pq is 2 × 5 × p × q. 20qr is 2 × 2 × 5 × q × r. 30rp is 2 × 3 × 5 × r × p. Common factors are 2, 5. So common factor is 10. (viii) 3x²y³, 10x³y², 6x²y²z. 3x²y³ is 3 × x² × y³. 10x³y² is 2 × 5 × x³ × y². 6x²y²z is 2 × 3 × x² × y² × z. Common factors are x², y². So common factor is x²y².

[CHECKPOINT]

Question two: Factorise the following expressions. (i) 7x – 42. 7x – 7 × 6 = 7(x – 6). (ii) 6p – 12q. 6p – 6 × 2q = 6(p – 2q). (iii) 7a² + 14a. 7a² + 7a × 2 = 7a(a + 2). (iv) –16z + 20z³. Factor out 4z. 4z(–4 + 5z²) or 4z(5z² – 4). (v) 20l²m + 30alm. Common factor is 10lm. 10lm(2l + 3a). (vi) 5x²y – 15xy². Common factor is 5xy. 5xy(x – 3y). (vii) 10a² – 15b² + 20c². Common factor is 5. 5(2a² – 3b² + 4c²). (viii) –4a² + 4ab – 4ca. Common factor is –4a. –4a(a – b + c). (ix) x²yz + xy²z + xyz². Common factor is xyz. xyz(x + y + z). (x) ax²y + bxy² + cxyz. Common factor is xy. xy(ax + by + cz).

Question three: Factorise. (i) x² + xy + 8x + 8y. Group as (x² + xy) + (8x + 8y) = x(x + y) + 8(x + y) = (x + y)(x + 8). (ii) 15xy – 6x + 5y – 2. Group as (15xy – 6x) + (5y – 2) = 3x(5y – 2) + 1(5y – 2) = (5y – 2)(3x + 1). (iii) ax + bx – ay – by. Group as (ax + bx) – (ay + by) = x(a + b) – y(a + b) = (a + b)(x – y). (iv) 15pq + 15 + 9q + 25p. Rearrange as 15pq + 25p + 9q + 15. Group as 5p(3q + 5) + 3(3q + 5) = (3q + 5)(5p + 3). (v) z – 7 + 7xy – xyz. Rearrange as z – xyz – 7 + 7xy. Group as z(1 – xy) – 7(1 – xy) = (1 – xy)(z – 7).

[CHECKPOINT]

We know that (a + b)² = a² + 2ab + b². This is Identity one. (a – b)² = a² – 2ab + b². This is Identity two. (a + b)(a – b) = a² – b². This is Identity three. 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 four: Factorise x² + 8x + 16. Solution: Observe the expression; it has three terms. Therefore, it does not fit Identity three. Also, its first and third terms are perfect squares with a positive sign before the middle term. So, it is of the form a² + 2ab + b² where a = x and b = 4 such that a² + 2ab + b² = x² + 2(x)(4) + 4² = x² + 8x + 16. Since a² + 2ab + b² = (a + b)², by comparison x² + 8x + 16 = (x + 4)². This is the required factorisation.

Example five: Factorise 4y² – 12y + 9. Solution: Observe 4y² = (2y)², 9 = 3² and 12y = 2 × 3 × (2y). Therefore, 4y² – 12y + 9 = (2y)² – 2 × 3 × (2y) + (3)² = (2y – 3)². This is the required factorisation.

Example six: Factorise 49p² – 36. Solution: There are two terms; both are squares and the second is negative. The expression is of the form (a² – b²). Identity three is applicable here. 49p² – 36 = (7p)² – (6)² = (7p – 6)(7p + 6). This is the required factorisation.

[CHECKPOINT]

Example seven: Factorise a² – 2ab + b² – c². 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, a² – 2ab + b² – c² = (a – b)² – c² by applying Identity two. This equals [(a – b) – c][(a – b) + c] by applying Identity three. This equals (a – b – c)(a – b + c). This is the required factorisation. Notice, how we applied two identities one after the other to obtain the required factorisation.

Example eight: Factorise m⁴ – 256. Solution: We note m⁴ = (m²)² and 256 = (16)². Thus, the given expression fits Identity three. Therefore, m⁴ – 256 = (m²)² – (16)² = (m² – 16)(m² + 16) using Identity three. Now, (m² + 16) cannot be factorised further, but (m² – 16) is factorisable again as per Identity three. m² – 16 = m² – 4² = (m – 4)(m + 4). Therefore, m⁴ – 256 = (m – 4)(m + 4)(m² + 16).

Let us now discuss how we can factorise expressions in one variable, like x² + 5x + 6, y² – 7y + 12, z² – 4z – 12, 3m² + 9m + 6. Observe that these expressions are not of the type (a + b)² or (a – b)², that is, they are not perfect squares. For example, in x² + 5x + 6, the term 6 is not a perfect square. These expressions obviously also do not fit the type (a² – b²) either. They, however, seem to be of the type x² + (a + b)x + ab. We may therefore, try to use Identity four studied in the last chapter to factorise these expressions: (x + a)(x + b) = x² + (a + b)x + ab. For that we have to look at the coefficients of x and the constant term.

[CHECKPOINT]

Example nine: Factorise x² + 5x + 6. Solution: If we compare the right hand side of Identity four with x² + 5x + 6, we find ab = 6, and a + b = 5. From this, we must obtain a and b. The factors then will be (x + a) and (x + b). If ab = 6, it means that a and b are factors of 6. Let us try a = 6, b = 1. For these values a + b = 7, and not 5, so this choice is not right. Let us try a = 2, b = 3. For this a + b = 5 exactly as required. The factorised form of this given expression is then (x + 2)(x + 3). In general, for factorising an algebraic expression of the type x² + px + q, we find two factors a and b of q, that is, the constant term, such that ab = q and a + b = p. Then, the expression becomes x² + (a + b)x + ab or x² + ax + bx + ab or x(x + a) + b(x + a) or (x + a)(x + b) which are the required factors.

Example ten: Find the factors of y² – 7y + 12. Solution: We note 12 = 3 × 4 and 3 + 4 = 7. Therefore, y² – 7y + 12 = y² – 3y – 4y + 12 = y(y – 3) – 4(y – 3) = (y – 3)(y – 4). Note, this time we did not compare the expression with that in Identity four to identify a and b. After sufficient practice you may not need to compare the given expressions for their factorisation with the expressions in the identities; instead you can proceed directly as we did above.

Example eleven: Obtain the factors of z² – 4z – 12. Solution: Here ab = –12; this means one of a and b is negative. Further, a + b = –4, this means the one with larger numerical value is negative. We try a = –4, b = 3; but this will not work, since a + b = –1. Next possible values are a = –6, b = 2, so that a + b = –4 as required. Hence, z² – 4z – 12 = z² – 6z + 2z – 12 = z(z – 6) + 2(z – 6) = (z – 6)(z + 2).

Example twelve: Find the factors of 3m² + 9m + 6. Solution: We notice that 3 is a common factor of all the terms. Therefore, 3m² + 9m + 6 = 3(m² + 3m + 2). Now, m² + 3m + 2 = m² + m + 2m + 2 (as 2 = 1 × 2) = m(m + 1) + 2(m + 1) = (m + 1)(m + 2). Therefore, 3m² + 9m + 6 = 3(m + 1)(m + 2).

[CHECKPOINT]

Now let us solve Exercise twelve point two. Question one: Factorise the following expressions. (i) a² + 8a + 16. This is a² + 2(a)(4) + 4², so it equals (a + 4)². (ii) p² – 10p + 25. This is p² – 2(p)(5) + 5², so it equals (p – 5)². (iii) 25m² + 30m + 9. This is (5m)² + 2(5m)(3) + 3², so it equals (5m + 3)². (iv) 49y² + 84yz + 36z². This is (7y)² + 2(7y)(6z) + (6z)², so it equals (7y + 6z)². (v) 4x² – 8x + 4. Factor out 4 first: 4(x² – 2x + 1) = 4(x – 1)². (vi) 121b² – 88bc + 16c². This is (11b)² – 2(11b)(4c) + (4c)², so it equals (11b – 4c)². (vii) (l + m)² – 4lm. Expand first: l² + 2lm + m² – 4lm = l² – 2lm + m², which is (l – m)². (viii) a⁴ + 2a²b² + b⁴. This is (a²)² + 2(a²)(b²) + (b²)², so it equals (a² + b²)².

Question two: Factorise. (i) 4p² – 9q². This is (2p)² – (3q)², so it equals (2p – 3q)(2p + 3q). (ii) 63a² – 112b². Factor out 7: 7(9a² – 16b²) = 7[(3a)² – (4b)²] = 7(3a – 4b)(3a + 4b). (iii) 49x² – 36. This is (7x)² – 6², so it equals (7x – 6)(7x + 6). (iv) 16x⁵ – 144x³. Factor out 16x³: 16x³(x² – 9) = 16x³(x – 3)(x + 3). (v) (l + m)² – (l – m)². Apply difference of squares: [(l + m) – (l – m)][(l + m) + (l – m)] = [l + m – l + m][l + m + l – m] = (2m)(2l) = 4lm. (vi) 9x²y² – 16. This is (3xy)² – 4², so it equals (3xy – 4)(3xy + 4). (vii) (x² – 2xy + y²) – z². First part is (x – y)². So it is (x – y)² – z², which equals (x – y – z)(x – y + z). (viii) 25a² – 4b² + 28bc – 49c². Group last three terms: 25a² – (4b² – 28bc + 49c²). The bracket is (2b – 7c)². So it is (5a)² – (2b – 7c)², which equals (5a – 2b + 7c)(5a + 2b – 7c).

[CHECKPOINT]

Question three: Factorise the expressions. (i) ax² + bx. Common factor is x. x(ax + b). (ii) 7p² + 21q². Common factor is 7. 7(p² + 3q²). (iii) 2x³ + 2xy² + 2xz². Common factor is 2x. 2x(x² + y² + z²). (iv) am² + bm² + bn² + an². Group: m²(a + b) + n²(b + a) = (a + b)(m² + n²). (v) (lm + l) + m + 1. Group: l(m + 1) + 1(m + 1) = (m + 1)(l + 1). (vi) y(y + z) + 9(y + z). Common factor is (y + z). (y + z)(y + 9). (vii) 5y² – 20y – 8z + 2yz. Rearrange: 5y² – 20y + 2yz – 8z. Group: 5y(y – 4) + 2z(y – 4) = (y – 4)(5y + 2z). (viii) 10ab + 4a + 5b + 2. Group: 2a(5b + 2) + 1(5b + 2) = (5b + 2)(2a + 1). (ix) 6xy – 4y + 6 – 9x. We already solved this in Example three. It equals (3x – 2)(2y – 3).

Question four: Factorise. (i) a⁴ – b⁴. This is (a²)² – (b²)² = (a² – b²)(a² + b²) = (a – b)(a + b)(a² + b²). (ii) p⁴ – 81. This is (p²)² – 9² = (p² – 9)(p² + 9) = (p – 3)(p + 3)(p² + 9). (iii) x⁴ – (y + z)⁴. This is [x²]² – [(y + z)²]² = [x² – (y + z)²][x² + (y + z)²] = [x – (y + z)][x + (y + z)][x² + (y + z)²] = (x – y – z)(x + y + z)(x² + (y + z)²). (iv) x⁴ – (x – z)⁴. This is [x²]² – [(x – z)²]² = [x² – (x – z)²][x² + (x – z)²]. First bracket: [x – (x – z)][x + (x – z)] = (z)(2x – z). So overall: z(2x – z)[x² + (x – z)²]. (v) a⁴ – 2a²b² + b⁴. This is (a² – b²)², which equals [(a – b)(a + b)]² = (a – b)²(a + b)².

[CHECKPOINT]

Question five: Factorise the following expressions. (i) p² + 6p + 8. Find two numbers whose product is 8 and sum is 6. They are 2 and 4. So it equals (p + 2)(p + 4). (ii) q² – 10q + 21. Find two numbers whose product is 21 and sum is –10. They are –3 and –7. So it equals (q – 3)(q – 7). (iii) p² + 6p – 16. Find two numbers whose product is –16 and sum is 6. They are 8 and –2. So it equals (p + 8)(p – 2).

We have learnt how to add and subtract algebraic expressions. We also know how to multiply two expressions. We have not however, looked at division of one algebraic expression by another. This is what we wish to do in this section. We recall that division is the inverse operation of multiplication. Thus, 7 × 8 = 56 gives 56 ÷ 8 = 7 or 56 ÷ 7 = 8. We may similarly follow the division of algebraic expressions. For example, 2x × 3x² = 6x³. Therefore, 6x³ ÷ 2x = 3x² and also, 6x³ ÷ 3x² = 2x. 5x(x + 4) = 5x² + 20x. Therefore, (5x² + 20x) ÷ 5x = x + 4 and also (5x² + 20x) ÷ (x + 4) = 5x. We shall now look closely at how the division of one expression by another can be carried out. To begin with we shall consider the division of a monomial by another monomial.

Consider 6x³ ÷ 2x. We may write 2x and 6x³ in irreducible factor forms. 2x = 2 × x. 6x³ = 2 × 3 × x × x × x. Now we group factors of 6x³ to separate 2x. 6x³ = 2 × x × (3 × x × x) = (2x) × (3x²). Therefore, 6x³ ÷ 2x = 3x². A shorter way to depict cancellation of common factors is as we do in division of numbers. 77 ÷ 7 = 77/7 = (7 × 11)/7 = 11. Similarly, 6x³ ÷ 2x = 6x³/2x = (2 × 3 × x × x × x)/(2 × x) = 3 × x × x = 3x².

[CHECKPOINT]

Example thirteen: Do the following divisions. (i) –20x⁴ ÷ 10x². Solution: –20x⁴ = –2 × 2 × 5 × x × x × x × x. 10x² = 2 × 5 × x × x. Therefore, (–20x⁴) ÷ 10x² = (–2 × 2 × 5 × x × x × x × x)/(2 × 5 × x × x) = –2 × x × x = –2x². (ii) 7x²y²z² ÷ 14xyz. Solution: 7x²y²z² ÷ 14xyz = (7 × x × x × y × y × z × z)/(2 × 7 × x × y × z) = (x × y × z)/2 = 1/2 xyz.

Now let us try these. Divide. (i) 24xy²z³ by 6yz². 24 divided by 6 is 4. x remains. y² over y is y. z³ over z² is z. So result is 4xyz. (ii) 63a²b⁴c⁶ by 7a²b²c³. 63 divided by 7 is 9. a² over a² is 1. b⁴ over b² is b². c⁶ over c³ is c³. So result is 9b²c³.

Let us consider the division of the trinomial 4y³ + 5y² + 6y by the monomial 2y. 4y³ + 5y² + 6y = (2 × 2 × y × y × y) + (5 × y × y) + (2 × 3 × y). Here, we expressed each term of the polynomial in factor form. We find that 2 × y is common in each term. Therefore, separating 2 × y from each term, we get 4y³ + 5y² + 6y = 2 × y × (2 × y × y) + 2 × y × (5/2 × y) + 2 × y × 3 = 2y(2y²) + 2y(5/2 y) + 2y(3) = 2y(2y² + 5/2 y + 3). The common factor 2y is shown separately. Therefore, (4y³ + 5y² + 6y) ÷ 2y = (4y³ + 5y² + 6y)/2y = (2y(2y² + 5/2 y + 3))/2y = 2y² + 5/2 y + 3. Alternatively, we could divide each term of the trinomial by the monomial using the cancellation method. (4y³ + 5y² + 6y) ÷ 2y = (4y³ + 5y² + 6y)/2y = 4y³/2y + 5y²/2y + 6y/2y = 2y² + 5/2 y + 3. Here, we divide each term of the polynomial in the numerator by the monomial in the denominator.

[CHECKPOINT]

Example fourteen: Divide 24(x²yz + xy²z + xyz²) by 8xyz using both the methods. Solution: 24(x²yz + xy²z + xyz²) = 2 × 2 × 2 × 3 × [(x × x × y × z) + (x × y × y × z) + (x × y × z × z)] = 2 × 2 × 2 × 3 × x × y × z × (x + y + z) = 8 × 3 × xyz × (x + y + z) by taking out the common factor. Therefore, 24(x²yz + xy²z + xyz²) ÷ 8xyz = (8 × 3 × xyz × (x + y + z))/(8 × xyz) = 3 × (x + y + z) = 3(x + y + z). Alternately, 24(x²yz + xy²z + xyz²) ÷ 8xyz = 24x²yz/8xyz + 24xy²z/8xyz + 24xyz²/8xyz = 3x + 3y + 3z = 3(x + y + z).

Consider (7x² + 14x) ÷ (x + 2). We shall factorise (7x² + 14x) first to check and match factors with the denominator: 7x² + 14x = (7 × x × x) + (2 × 7 × x) = 7 × x × (x + 2) = 7x(x + 2). Now (7x² + 14x) ÷ (x + 2) = (7x² + 14x)/(x + 2) = 7x(x + 2)/(x + 2) = 7x by cancelling the factor (x + 2).

Example fifteen: Divide 44(x⁴ – 5x³ – 24x²) by 11x(x – 8). Solution: Factorising 44(x⁴ – 5x³ – 24x²), we get 44(x⁴ – 5x³ – 24x²) = 2 × 2 × 11 × x²(x² – 5x – 24) by taking the common factor x² out of the bracket. This equals 2 × 2 × 11 × x²(x² – 8x + 3x – 24) = 2 × 2 × 11 × x²[x(x – 8) + 3(x – 8)] = 2 × 2 × 11 × x²(x + 3)(x – 8). Therefore, 44(x⁴ – 5x³ – 24x²) ÷ 11x(x – 8) = (2 × 2 × 11 × x × x × (x + 3) × (x – 8))/(11 × x × (x – 8)) = 2 × 2 × x(x + 3) = 4x(x + 3). We cancel the factors 11, x and (x – 8) common to both the numerator and denominator.

[CHECKPOINT]

Example sixteen: Divide z(5z² – 80) by 5z(z + 4). Solution: Dividend = z(5z² – 80) = z[(5 × z²) – (5 × 16)] = z × 5 × (z² – 16) = 5z × (z + 4)(z – 4) using the identity a² – b² = (a + b)(a – b). Thus, z(5z² – 80) ÷ 5z(z + 4) = 5z(z – 4)(z + 4)/5z(z + 4) = (z – 4).

Now let us solve Exercise twelve point three. Question one: Carry out the following divisions. (i) 28x⁴/56x. 28 over 56 is 1/2. x⁴ over x is x³. So 1/2 x³. (ii) –36y³/9y². –36 over 9 is –4. y³ over y² is y. So –4y. (iii) 66pq²r³/11qr². 66 over 11 is 6. p remains. q² over q is q. r³ over r² is r. So 6pqr. (iv) 34x³y³z³/51xy²z³. 34 over 51 is 2/3. x³ over x is x². y³ over y² is y. z³ over z³ is 1. So 2/3 x²y. (v) 12a⁸b⁸/(–6a⁶b⁴). 12 over –6 is –2. a⁸ over a⁶ is a². b⁸ over b⁴ is b⁴. So –2a²b⁴.

Question two: Divide the given polynomial by the given monomial. (i) (5x² – 6x)/3x. Divide each term: 5x²/3x is 5/3 x. –6x/3x is –2. So 5/3 x – 2. (ii) (3y⁸ – 4y⁶ + 5y⁴)/y⁴. Divide each term: 3y⁸/y⁴ is 3y⁴. –4y⁶/y⁴ is –4y². 5y⁴/y⁴ is 5. So 3y⁴ – 4y² + 5. (iii) 8(x³y²z² + x²y³z² + x²y²z³)/4x²y²z². Factor out 8/4 which is 2. Divide each term inside: x³/x² is x. y²/y² is 1. z²/z² is 1. So first term gives x. Second: x²/x² is 1, y³/y² is y, z²/z² is 1. Gives y. Third: x²/x² is 1, y²/y² is 1, z³/z² is z. Gives z. So 2(x + y + z). (iv) (x³ + 2x² + 3x)/2x. Divide each term: x³/2x is 1/2 x². 2x²/2x is x. 3x/2x is 3/2. So 1/2 x² + x + 3/2. (v) (p³q⁶ – p⁶q³)/p³q³. Divide each term: p³q⁶/p³q³ is q³. –p⁶q³/p³q³ is –p³. So q³ – p³.

[CHECKPOINT]

Question three: Work out the following divisions. (i) (10x – 25)/5. Factor numerator: 5(2x – 5)/5 = 2x – 5. (ii) (10x – 25)/(2x – 5). Factor numerator: 5(2x – 5)/(2x – 5) = 5. (iii) 10y(6y + 21)/5(2y + 7). Factor inside: 6y + 21 is 3(2y + 7). So 10y × 3(2y + 7)/5(2y + 7). Cancel (2y + 7). 30y/5 = 6y. (iv) 9x²y²(3z – 24)/27xy(z – 8). Factor inside: 3z – 24 is 3(z – 8). Numerator becomes 9x²y² × 3(z – 8) = 27x²y²(z – 8). Divide by 27xy(z – 8). Cancel 27, (z – 8), one x, one y. Result is xy. (v) 96abc(3a – 12)(5b – 30)/144(a – 4)(b – 6). Factor inside: 3a – 12 is 3(a – 4). 5b – 30 is 5(b – 6). Numerator becomes 96abc × 3(a – 4) × 5(b – 6) = 1440abc(a – 4)(b – 6). Divide by 144(a – 4)(b – 6). Cancel (a – 4), (b – 6). 1440 over 144 is 10. So 10abc.

Question four: Divide as directed. (i) 5(2x + 1)(3x + 5)/(2x + 1). Cancel (2x + 1). Result is 5(3x + 5). (ii) 26xy(x + 5)(y – 4)/13x(y – 4). Cancel 13x(y – 4). 26 over 13 is 2. Result is 2y(x + 5). (iii) 52pqr(p + q)(q + r)(r + p)/104pq(q + r)(r + p). Cancel pq(q + r)(r + p). 52 over 104 is 1/2. Result is 1/2 r(p + q). (iv) 20(y + 4)(y² + 5y + 3)/5(y + 4). Cancel 5(y + 4). 20 over 5 is 4. Result is 4(y² + 5y + 3). (v) x(x + 1)(x + 2)(x + 3)/x(x + 1). Cancel x(x + 1). Result is (x + 2)(x + 3).

[CHECKPOINT]

Question five: Factorise the expressions and divide them as directed. (i) (y² + 7y + 10)/(y + 5). Factor numerator: (y + 2)(y + 5). Divide by (y + 5) gives (y + 2). (ii) (m² – 14m – 32)/(m + 2). Factor numerator: (m – 16)(m + 2). Divide by (m + 2) gives (m – 16). (iii) (5p² – 25p + 20)/(p – 1). Factor out 5: 5(p² – 5p + 4) = 5(p – 1)(p – 4). Divide by (p – 1) gives 5(p – 4). (iv) 4yz(z² + 6z – 16)/2y(z + 8). Factor inside: z² + 6z – 16 is (z + 8)(z – 2). Numerator becomes 4yz(z + 8)(z – 2). Divide by 2y(z + 8). Cancel 2y(z + 8). Result is 2z(z – 2). (v) 5pq(p² – q²)/2p(p + q). Factor p² – q² as (p – q)(p + q). Numerator is 5pq(p – q)(p + q). Divide by 2p(p + q). Cancel p(p + q). Result is 5/2 q(p – q). (vi) 12xy(9x² – 16y²)/4xy(3x + 4y). Factor 9x² – 16y² as (3x – 4y)(3x + 4y). Numerator is 12xy(3x – 4y)(3x + 4y). Divide by 4xy(3x + 4y). Cancel 4xy(3x + 4y). Result is 3(3x – 4y). (vii) 39y³(50y² – 98)/26y²(5y + 7). Factor 50y² – 98 as 2(25y² – 49) = 2(5y – 7)(5y + 7). Numerator becomes 39y³ × 2(5y – 7)(5y + 7) = 78y³(5y – 7)(5y + 7). Divide by 26y²(5y + 7). Cancel 26y²(5y + 7). 78 over 26 is 3. y³ over y² is y. Result is 3y(5y – 7).

Let us review the key points from this chapter. One: When we factorise an expression, we write it as a product of factors. These factors may be numbers, algebraic variables or algebraic expressions. Two: An irreducible factor is a factor which cannot be expressed further as a product of factors. Three: A systematic way of factorising an expression is the common factor method. It consists of three steps: write each term of the expression as a product of irreducible factors, look for and separate the common factors, and combine the remaining factors in each term in accordance with the distributive law. Four: Sometimes, all the terms in a given expression do not have a common factor; but the terms can be grouped in such a way that all the terms in each group have a common factor. When we do this, there emerges a common factor across all the groups leading to the required factorisation of the expression. This is the method of regrouping. Five: In factorisation by regrouping, we should remember that any regrouping of the terms in the given expression may not lead to factorisation. We must observe the expression and come out with the desired regrouping by trial and error. Six: A number of expressions to be factorised are of the form or can be put into the form: a² + 2ab + b², a² – 2ab + b², a² – b² and x² + (a + b) + ab. These expressions can be easily factorised using Identities one, two, three and four. Seven: In expressions which have factors of the type (x + a)(x + b), remember the numerical term gives ab. Its factors, a and b, should be so chosen that their sum, with signs taken care of, is the coefficient of x. Eight: We know that in the case of numbers, division is the inverse of multiplication. This idea is applicable also to the division of algebraic expressions. Nine: In the case of division of a polynomial by a monomial, we may carry out the division either by dividing each term of the polynomial by the monomial or by the common factor method. Ten: In the case of division of a polynomial by a polynomial, we cannot proceed by dividing each term in the dividend polynomial by the divisor polynomial. Instead, we factorise both the polynomials and cancel their common factors. Eleven: In the case of divisions of algebraic expressions that we studied in this chapter, we have Dividend = Divisor × Quotient. In general, however, the relation is Dividend = Divisor × Quotient + Remainder. Thus, we have considered in the present chapter only those divisions in which the remainder is zero.

Thank you for listening! Keep revising and practicing. Goodbye! [CHAPTER_COMPLETE]