Expanding brackets and factorising

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1 CHAPTER 8 Epanding brackets and factorising 8 CHAPTER Epanding brackets and factorising 8.1 Epanding brackets There are three rows. Each row has n students. The number of students is 3 n 3n. Two students are added to each of the three rows, so students are added. There are now n 2 students in each row. n 2 n 2 n 2 Total number of students 3(n 2) 3 n 3 2 3n 6 The simplest way of writing this is 3(n 2) 3n 6 Writing 3(n 2) as 3n 6 is called epanding brackets.this is also known as multiplying out brackets. Eample 1 Epand 5(2 3). Solution 1 5(2 3) Multiply each term inside the bracket by the term outside the bracket. Eample 2 Epand p(p 3). Solution 2 p(p 3) p p p 3 p 2 3p Multiply each term inside the bracket by the term outside the bracket. Positive ( ) negative ( ) negative ( ). Eample 3 Epand and simplify 5p 3(p q). Solution 3 5p 3(p q) 5p 3 p 3 q 5p 3p 3q 2p 3q Multiply each term inside the bracket by the 3 outside the bracket. Negative ( 3) negative ( q) is positive ( 3q). Collect like terms. 106

2 8.2 Factorising by taking out common factors CHAPTER 8 Eample 4 Epand 2y(y 2 4y 3). Solution 4 2y(y 2 4y 3) 2y y 2 2y 4y 2y 3 2y 3 8y 2 6y Multiply each term inside the bracket by the term outside the bracket. Eercise 8A 1 Epand a 2(n 1) b 2(2 n) c 4(p 2) d 2(3 n) e 5(c d) f 2(c d) g 4(m n) h 2(2 3y) i 5(3g 4) j 2(3g 4h) k 7( 2y 3) l 3(p 2q 1) 2 Epand a n(n 1) b b(b 2 3b) c 2n(n 3) d a(4 a) e y(3y y 2 ) f 3(5n 4) g 3(2a 1) h 2(n 4) i (y 1) j 5(d 2 d) k a(b c) l 2( ) 3 Epand and simplify a 2(p 1) 5 b 2(t 3) t c 2(n 1) 3(n 2) d 2(2c 1) 3(3c 2) e 2(m 3) 3(2m 1) f 2(3d 4) 3(1 2d) g 2(2p 3q) 2(2p q) h 4p 2(p 1) i 5y 3(2y 1) j 3 (n 3) k 2q 3(q 1) l 4 3(2 1) 4 Epand and simplify a ( 1) 1( 1) b q(q 1) 3(q 1) c s(s 4) 2(s 4) d t(t 3) 2(t 5) e a(a 3) 2(a 1) f n(n 2) 4(n 2) 8.2 Factorising by taking out common factors Epanding 3(2b 5) gives 6b 15 Factorising is the reverse process to epanding brackets. So factorising 6b 15 gives 3(2b 5). 3 is the highest common factor of 6 and 15 The two factors of the epression 6b 15 are 3 and 2b 5 Eample 5 Factorise 8c 12s. Solution 5 8c 12s 4 2c 4 3s 4(2c 3s) The common factors of 8 and 12 are 1, 2 and 4 The HCF of 8 and 12 is 4 Write the HCF outside the bracket and the other factor inside the bracket. 107

3 CHAPTER 8 Epanding brackets and factorising Eample 6 Factorise ab ac. Solution 6 ab ac a b a c a(b c) The common factor of ab and ac is a. Write the a outside the bracket. Eample 7 Factorise y 2 y. Solution 7 y 2 y y y 1 y y y y 1 y(y 1) The common factor of y 2 and y is y. Re-writing 1 y as y 1 Write the y outside the bracket. Eample 8 Factorise completely 4a 8b 6c. Solution 8 4a 8b 6c 2 2a 2 4b 2 3c 2(2a 4b 3c) The common factor of 4a, 8b and 6c is 2. Write the 2 outside the bracket. In questions where you are asked to Factorise completely check that the terms of the epression inside the bracket do not have a common factor. Eample 9 Factorise completely 9a 2 b 12ab 2 c. Solution 9 9a 2 b 12ab 2 c 3 3 a a b 3 4 a b b c 3ab(3 a 4 b c) 3ab(3a 4bc) The common factor is 3 a b. Write the 3ab outside the bracket. Note that 3a and 4bc do not have a common factor. 108 Eercise 8B 1 Factorise these epressions. a 3y 6 b 2 2 c 5p 10 d 3d 6e e 6 2y f 3a 12b g 12p 6q h 4d 6e i 10c 6d j 8 12y k 10m 15n 5 l 8p 6q 4 m 9d 6e 12f n pq pr o y zy p ab 7b q db b r pq py p s y 2 yz t y 2 y u 4y 2 3y v y 3 2y w y 3 5y p 2 pq pr y ay 3 by 2 cy

4 8.3 Epanding the product of two brackets CHAPTER 8 2 Factorise completely a 2y 4 b 3pq 9qr 6pqr c 4y 8y 2 d 5ac abc e 4pq 6pqr f rt rst g 14y 4 7y 2 h 9d 3 6d 2 i ab 3 a 2 b j 10 2 y 15y 2 k 3y y 5 l 12pq 2 r 16qp 3 m 8a 2 b 16a 3 12ab 2 n 4a 4 b 2 6a 3 b 3 12a 2 b 2 o (p 2 q) 2 p 3 q 8.3 Epanding the product of two brackets Consider the areas of these two rectangles Area ( 2) Area 4( 2) Combining these two rectangles gives a single rectangle with length ( 4) and width ( 2) 4 2 Area ( 4)( 2) The area of the large rectangle is equal to the sum of the areas of the two smaller rectangles so ( 4)( 2) ( 2) 4( 2) To epand two brackets, multiply each term in the first bracket by the second bracket so ( p)( q) ( q) p( q) 2 q p pq Epanding ( p)( q) gives 2 q p pq. Eample 10 Epand and simplify ( 2)( 3). Solution 10 ( 2)( 3) ( 3) 2( 3) Multiply each term, and 2 in the first bracket by the second bracket, ( 3). Collect like terms. 109

5 CHAPTER 8 Epanding brackets and factorising Eample 11 Epand and simplify (n 2) 2. Solution 11 (n 2) 2 (n 2)(n 2) n(n 2) 2(n 2) (n 2) 2 (n 2) (n 2). Multiply each term in the first bracket by the second bracket. n 2 2n 2n 4 n 2 4n 4 Collect like terms. Eample 12 Epand and simplify (4p 5q)(3p 2q). Solution 12 (4p 5q)(3p 2q) 4p(3p 2q) 5q(3p 2q) Multiply each term in the first bracket by the second bracket. 12p 2 8pq 15qp 10q 2 12p 2 7pq 10q 2 Collecting like terms; note 15qp is the same as 15pq. Eercise 8C 1 Epand and simplify a ( 1)( 2) b (q 3)(q 1) c (r 1)(r 3) d ( 4)( 3) e ( 2)( 1) f ( 3)( 3) g (r 1)(r 3) h ( 2)( 4) i ( 2)( 1) j ( 3)( 4) 2 Epand and simplify a ( 1) 2 b ( 3) 2 c ( 7) 2 d ( 5) 2 e ( 8) 2 f ( 10) 2 3 Epand and simplify a ( y)( 2y) b (2 y)( y) c (p q)(3p q) d (3p q)(p q) e ( 3y)(2 5y) f (a b)(a b) g (p q)(p 3q) h (2 3y)(3 4y) i (3 2y)( y) j (4 5y)(2 3y) k ( y) 2 l (2 y) 2 m (3p 5q) 2 n (6a 5b)(6a 5b) 4 Epand a (a b)(c d) b (e f)(g h) 8.4 Factorising by grouping In Section 8.2, an epression was factorised by taking out the common factor of the terms in the epression. The common factor was a single term, for eample 3, 4a, ab. For some epressions the common factor can involve the sum or difference of terms, for eample 3 or 2a 4b. 110

6 8.4 Factorising by grouping CHAPTER 8 Eample 13 Factorise completely 12( 2) 2 9( 2). Solution 13 12( 2) 2 9( 2) 3 4 ( 2) ( 2) 3 3 ( 2) 3 ( 2) 4 ( 2) 3 ( 2) 3 3( 2)[4( 2) 3] 3( 2)[4 8 3] 3 is the HCF of 12 and 9 3 and ( 2) are both common factors. So write 3( 2) outside the square bracket. Simplify the terms inside the square bracket. 3( 2)(4 5) Epanding (a b)(c d) gives ac ad bc bd. Since factorising is the reverse process to epanding brackets, factorising ac ad bc bd gives (a b)(c d). In the epression ac ad bc bd there is no common factor of all four terms. Pairs of terms with a common factor can be grouped together and factorised to give a(c d) b(c d). These two terms have a common bracketed factor, (c d). Using this common factor gives (c d)(a b) which can also be written as (a b)(c d). This is called factorising by grouping. Eample 14 Factorise pr qs ps qr. Solution 14 pr qs ps qr pr ps qs qr p(r s) q(r s) (r s)(p q) Group the terms in pairs so that each pair has a common factor in this case p and q. Factorise each pair. The bracketed term must be the same. (r s) is a common factor. This answer could be written as (p q)(r s). Eercise 8D 1 Factorise completely a y( 3) 2( 3) b a( y) b( y) c p( 2y) q( 2y) d 2p( 4) 3q( 4) e ( 1) 2 4( 1) f ( y) 2 b( y) g ( 5) 3( 5) 2 h ( 3y) 2( 3y) 2 i 4( 2) 2 2( 2) j 6( y) 2 3( y) k 6( 4) 4( 4) 2 l 6y( 3y) 9( 3y) 2 2 Factorise completely a ab ac db dc b pq 2r pr 2q c 2 a 2 2a d ps qr pr qs e f g h

7 CHAPTER 8 Epanding brackets and factorising 8.5 Factorising epressions of the form 2 b c Epanding ( 4)( 2) gives Since factorising is the reverse process to epanding brackets, factorising gives ( 4)( 2). In the epression there is no common factor of all three terms. From Section 8.3 ( p)( q) 2 q p pq. So ( p)( q) 2 (p q) pq. The diagrams show the epansions of ( p)( q) and ( 4)( 2) p ( 4)( 2) ( 2) 4( 2) Taking out a common factor of just two of the three terms is not factorising the epression. An answer ( 6) 8 is incorrect. (p q) is the same as (p q). 4 2 p 2 4 q q pq For these epansions to be the same, obviously 2 2 but also pq 8 and p q that is, the product of p and the sum of p and q is 6, and q is 8 since p q (p q). To factorise find two numbers whose product is 8 and whose sum is 6. The two numbers are 2 and 4. So ( 2) 4( 2) ( 2)( 4) Eample 15 Factorise Solution Find two numbers whose product is 12 and whose sum is ( 4) 3( 4) ( 4)( 3) Write 7 as 4 3. Factorise by grouping. The bracketed term must be the same. ( 4) is a common factor. This answer could also be written as ( 3)( 4). 112

8 8.6 Factorising the difference of two squares CHAPTER 8 Eample 16 Factorise 2 6 Solution Find two numbers whose product is 6 and whose sum is ( 2) 3( 2) ( 2)( 3) Write as 2 3. Factorise by grouping. The bracketed term must be the same. ( 2) is a common factor. This answer could also be written as ( 3)( 2). Eercise 8E 1 Write down the two numbers a whose product is 10 and whose sum is 7 b whose product is 7 and whose sum is 8 c whose product is 24 and whose sum is 11 d whose product is 6 and whose sum is 5 e whose product is 8 and whose sum is 2 f whose product is 12 and whose sum is 1 g whose product is 12 and whose sum is 4 h whose product is 9 and whose sum is 6 i whose product is 20 and whose sum is 9 j whose product is 16 and whose sum is 0 2 Factorise these epressions. a b c d e f g h i j k l m n o p q Factorising the difference of two squares 2, 4, 4 2, 9, 1, p 2, A 2, B 2, (2t 1) 2 and 9(t 6) 2 are all squares A 2 B 2 1 p 2 All these epressions show the difference of two squares. (2t 1) 2 9(t 6) 2 An epression that is the difference of two squares can be factorised using the method in Section

9 CHAPTER 8 Epanding brackets and factorising Eample 17 Factorise 2 4 Solution Write 2 4 in the form 2 b c Find two numbers whose product is 4 and whose sum is ( 2) 2( 2) ( 2)( 2) Write 0 as 2 2. Factorise by grouping. The bracketed term must be the same. ( 2) is a common factor. This answer could also be written as ( 2) ( 2). Eample 18 Factorise 2 n 2. Solution 18 2 n n n 2 n n n 2 n n 0 2 n 2 2 n n n 2 ( n) n( n) ( n)( n) Find two numbers whose product is n 2 and whose sum is 0 Write 0 as n n. Factorise by grouping. The bracketed term must be the same. ( n) is a common factor. To factorise the difference of the squares of two terms, multiply the sum of the two terms by the difference of the two terms. So, for any two terms A and B, This answer could also be written as ( n)( n). A 2 B 2 (A B)(A B) Eample 19 Factorise Solution (2) (2 3)(2 3) 4 2 and 9 are squares since 4 2 (2) 2 and 9 3 2, so this epression is the difference of two squares. Write in the form A 2 B 2. Use A 2 B 2 (A B)(A B) with A 2 and B 3 114

10 8.6 Factorising the difference of two squares CHAPTER 8 Eample 20 Factorise y 2 Solution y 2 2(9 2 25y 2 ) 2[(3) 2 (5y) 2 ] 2[(3 5y)(3 5y)] Take out the common factor, 2, and 9 2 and 25y 2 are squares since 9 2 (3) 2 and 25y 2 (5y) 2 Write y 2 in the form A 2 B 2. Use A 2 B 2 (A B)(A B) with A 3 and B 5y. 2(3 5y)(3 5y) Eample 21 Factorise (2t 1) 2 9(t 6) 2 Solution 21 (2t 1) 2 9(t 6) 2 (2t 1) 2 [3(t 6)] 2 [(2t 1) 3(t 6)][(2t 1) 3(t 6)] [2t 1 3t 18][2t 1 3t 18] [5t 17][ t 19] Write (2t 1) 2 9(t 6) 2 in the form A 2 B 2. Use A 2 B 2 (A B)(A B) with A 2t 1 and B 3(t 6). Epand and simplify the terms in the squared brackets. (5t 17)( t 19) Eercise 8F 1 Factorise these epressions. a p 2 1 b y 2 9 c 2 36 d a e 144 b 2 f 1 p 2 g 4p 2 1 h i 25y 2 4 j ( 1) 2 25 k 49 (1 ) 2 2 a Factorise 2 p 2. b Hence find the value of i ii Find the value of Factorise these epressions completely. a 2p 2 32 b 27a 2 48 c 3y d 4a 2 64b 2 e 12p 2 27q 2 f 9(p 1) 2 4p 2 g 8( 2) 2 2( 1) 2 h 50(2 1) 2 18(1 ) 2 115

11 CHAPTER 8 Epanding brackets and factorising Chapter summary You should now be able to: epand (multiply out) brackets by multiplying each term inside the bracket by the term outside the bracket, for eample 3( 2) factorise by taking out a common factor, for eample 4c 6cs 2c(2 3s); factorising is the opposite process to epanding brackets epand two brackets by multiplying each term in the first bracket by the second bracket, foreample ( p)( q) ( q) p( q) 2 q p pq and then simplify if possible factorise by grouping factorise an epression of the form 2 b c factorise the difference of two squares using the general rule A 2 B 2 (A B)(A B). Chapter 8 review questions 1 a Multiply out 4(3 2) b Simplify 2(3 1) 3( 2) 2 Simplify a 3(a 2) 5 b 4(b 3) 3b c 2(3c d) 3(c d) 3 Simplify a 5(2n 3m) 3(n 4m) b 4(2n 2m 1) 3(2n 4p 3) 4 Epand and simplify 3(5 2) 2 (2 5) (1388 March 2002) 5 a Simplify i p 2 p 7 ii 8 3 iii y4 y 3 y b Epand t(3t 5 2 4) (1387 November 2003) 6 a Epand the brackets p(q p 2 ) b Epand and simplify 5(3p 2) 2(5p 3) (1387 November 2004) 7 Epand y(3y 2 5y) 8 Factorise a 2r 6 9 Factorise a 6a 12b 30 b 4s 10t b 8 12y 16z 10 a Epand and simplify 3(2 1) 2(2 3) b Factorise y 2 y (1387 November 2003) 11 Factorise k 2 k 12 Factorise a ab 2bc b 2 3a 3 116

12 Chapter 8 review questions CHAPTER 8 13 a Factorise 2 3 b Simplify k 5 k 2 (1387 June 2004) 14 Epand and simplify (y 5)(y 3) (1388 January 2005) 15 Epand and simplify ( 9)( 4) (1388 March 2005) 16 a Simplify 5p 4q 3p q b Simplify 2 c Factorise 4 6 d Multiply out and simplify ( 3)( 2) e Simplify (1386 November 2002) 7 17 a Epand and simplify ( 7)( 4) b Epand y(y 3 2y) c Factorise p 2 6p d Factorise completely 6 2 9y (1387 June 2005) 18 a Simplify k 5 k 2 b Epand and simplify i 4( 5) 3( 7) ii ( 3y)( 2y) c Factorise (p q) 2 5(p q) d Simplify (m 4 ) 2 e Simplify 2t 2 3r 3 t 4 (1387 June 2004) 19 Epand and simplify a 3b 1 4(b 2) b y(2y y 3 ) c ( 2)( 7) d 4 (m 1) 2 20 a Epand ( 5)( 8) b Factorise (1388 March 2004) 21 a Epand and simplify ( 1)( 7) b Factorise y 2 3y 10 (1388 November 2005) 22 a Factorise b Factorise c Factorise ab 2ay b 2y 23 a Factorise m 2 n 2 b Hence find the value of Factorise completely 6p 2 4p 3pq 2q (1385 June 1998) 25 Factorise completely a 3a 2 12b 2 b 8(n 1) 2 2(n 3) 2 26 a Factorise b Hence, factorise completely (3 4) 2 ( 2 4 4) 117

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