Chapter Factorisation Factorisation using the Highes - PDF document

57 Chapter 5 – Factorisation Factorisation using the Highest Common Factor (HCF) 5.1 We have already seen that we can expand an expression like 2(x + 2y) as follows: 2(x + 2y) = 2x + 4y by distributing the 2 in front of the () to each of the two term s inside the bracket. We can do the reverse and ‘undistribute the 2 from the two terms viz: 2x + 4y = 2(x + 2y). We call this process factorisation. Note the LHS has two terms but the factorised expression one the RHS has one term. E5.1 Factorise by ‘ undistributing’ the highest common factor (HCF) from each of these expressions: (i) 2x – 4y + 8z = (ii) 4a 2 b – 6ab + 12ab 2 = A5.1 (i) 2x – 4y + 8z = 2(x – 2y + 4z) (ii) 4a 2 b – 6ab + 12ab 2 = 2ab(2a – 3 + 6b) 5.2 When we convert an expression made up of two or more terms i nto an expression made up of 1 term with multiple factors we call the process FACTORISATION. E5.2 Identify the expressions which are already factorised: i. 3a 2 b 3 ii.3a 2 (m + 2n) iii. 2a + 4ab iv. 3a 2 (m + 2n) + 6a v. (a+b)(c+d) vi. (3(a+b)+ 4(c+d))(p+q) vii . (3(a+b)+ 4(c+d))(p+q) + r A5.2 i. Yes ii. No iii. No iv. No v. Yes vi. No vii. No 5.3 When we factorise two terms, pictorially we are doing something like this: □∆  □◊ = □(∆  ◊) Example: ax  ay = a(x  y) where the a is □, x is ∆ and c is ◊ E5.3 Factorise: ax + ay – az A5.3 ax + ay – az = a(x + y – z) 5.4 We can check our factorisation by expanding: 58 For example is the following factorisation correct? 2x 2 – 4xy + 8y 2 = 2xy(x – 2 + 4y) Expand the RHS: 2xy(x – 2 + 4y) = 2x 2 y – 4xy + 8 xy 2 ≠ 2x 2 – 4xy + 8y 2 so the answer is NO. E5.4 Check the following Factorisations: (i) 3x 2 + 6xy – 12y 2 = 3xy(x + 2 – 4y) (ii) - 2x + 4y = - 2(x + 2y) (iii) - 2x + 4y = 2( - x + 2y) (iv) - 2x + 4y = - 2(x + 2y) (v) - 2x + 4y = 2( - x – 2y) A5.4 (i) Incorrect (ii) incorrect (iii ) correct (iv) correct (v) correct 5.5 When we have an expression with a leading – ve sign e.g. - ax + ay – az, we have a choice between ‘ - a ’ as the HCF or ‘ a ’ as the HCF. If we choose - a as the HCF then our factorisation is: - ax + ay – az = - a(x – y + z ). Check (by expansion) - a(x – y + z) = - ax + ay – az OR - ax + ay – az = a( - x + y - z). Check (by expansion) a( - x + y - z) = - ax + ay – az E5.5 Factorise : - 3x + 6y using - 3 and then 3 as the HCF. - 3x + 6y = - 3( ) - 3x + 6y = 3( ) Chec k by expansion. A5.5 - 3x + 6y = - 3( x – 2y ) - 3x + 6y = 3( - x + 2y ) 5.6 Now consider the factorisation of: 3a 2 (m + 2n) + 6a 59 Note: This is made up two terms with HCF = 3a. So we get: 3a 2 (m + 2n) + 6a = 3a[a(m+2n) + 6] E5.6 Factorise: 2a 2 b(m - 2n) + 6ab A5.6 2a 2 b(m – 2n) + 6ab = 2ab[a(m – 2n) + 3] 5.7 Now Factorise a(m+n) + b(m+n) T his expression has two terms and the HCF is (m+n). So we get: a(m+n) + b(m+n) = (m+n)(a+b). E5.7 Factorise: (i) 2x(m+n) + 3y(m+n) = (ii) 2x(m+n) – 4y(m+n) = (iii) 2x 2 (m+n) – 4x(m+n) = A5.7 (i) 2x(m+n) + 3y(m+n) = (m + n)(2x + 3y) (ii) 2x(m+n) – 4y(m+n) = 2(m + n)(x – 2y) (iii) 2x 2 (m+n) – 4x(m+n) = 2x(m + n)(x – 2) 5.8 Now consider the factorisation of 4a 2 b 2 c 3 + 2ab 3 c 2 - 8a 3 b 3 c 3 The HCF is 2ab 2 c 2 , so we get: 4a 2 b 2 c 3 + 2ab 3 c 2 - 8a 3 b 3 c 3 = 2ab 2 c 2 (2ac + b - 4a 2 bc) Check by expanding the result: 2ab 2 c 2 (2ac + b - 4a 2 bc) = 4a 2 b 2 c 3 + 2ab 3 c 2 - 8a 3 b 3 c 3 E5.8 Factorise 3a 3 b 2 c - 6a 2 b 3 c 2 + 9a 3 b 3 c 3 and check by expansion. A5.8 3a 3 b 2 c - 6a 2 b 3 c 2 + 9a 3 b 3 c 3 = 3a 2 b 2 c(a – 2bc + 3abc 2 ) 5.9 Now consider the factorisation of 4ab + 8a 2 b. Here the HCF is 4ab. Remember that removing the factor 4ab from the first term leaves 1 not 0. It is a common mistake to take this as 0. 4ab + 8a 2 b = 4ab(1 + 2a). 60 E5.9 (i) 5ab + 10a 2 b 2 (ii) 7a 2 b - 14a 2 b 2 A5.9 (i) 5ab + 10a 2 b 2 (ii) 7a 2 b - 14a 2 b 2 =5ab(1 + 2ab) = 7a 2 b(1 – 2b) Two Stage Factorisation of a four term expression 5.10 Now consider the 4 term expression: 2am + 8an + 3bm + 12bn. The HCF is just the trivial 1. Note however that the 1 st two terms has the HCF 2a and the last two terms have the HCF 3b. Let’s factorise just the 1 st 2 terms and the last two terms separately. Then we get: 2am + 8an + 3bm + 12bn = 2a(m + 4n) + 3b(m+ 4n) Now we have gone from 4 ter ms to two terms. So we still haven’t factorised the expression but now we have only two terms and they both have the HCF of (m + 4n). So we can proceed to get: 2a(m + 4n) + 3b(m+ 4n) = (m + 4n)(2a + 3b) The fa ctorisation has been achieved in two stages by grouping the 4 term expression into two groups E5.10 Factorise the 4 term expression by grouping: (i) am + bm + na + nb (ii) 3m + 6n + am + 2an A5.10 (i) am + bm + na + nb = m(a + b) + n(a + b) = (a + b )(m + n) (ii) 3m + 6n + am + 2an = 3(m + 2n) + a(m + 2n) = (m + 2n)(3 + a) 5.11 Now with – ve signs, we get: 2am ─ 8an + 3bm ─ 12bn = 2a(m ─ 4n) + 3b(m ─ 4n) = (2a + 3b)(m ─ 4n) E5.11 Factorise: (i) am ─ bm + na ─ nb (ii) 3m ─ 6n + am ─ 2an A5.11 (i) am – bm + na – nb = m(a – b) + n(a – b) = (a – b)(m + n) (ii )3m – 6n + am – 2an = 3(m – 2n) + a(m – 2n) 61 = (m – 2n)(3 + a) 5.12 Another example with a leading negative in the second group : 2am – 8an – 3bm + 12bn = 2a(m – 4n) – 3b(m – 4n) = (m – 4n)(2a - 3b) E5.12 Factorise: (i) am ─ bm ─ na + nb (ii) 3m ─ 6n ─ am + 2an A5.12 (i) am – bm – na + nb = m(a – b) – n(a – b) = (a – b)(m – n) (ii)3m – 6n – am + 2an = 3(m – 2n) – a(m – 2n) = (m – 2n)(3 – a) Quadratic pairs 5.13 Firstly let’s find the numbers represented by the two ? in the following: (?)(?) = 10 and ? + ? = 7. i.e. find a pair of integer numbers whose product is 10 and whose sum is 7 . T he pair s (2,5) and (10,1) has a product of 10 The following pairs all sum to 7: (1,6), (2,5), (3,4) By trial and error we fi nd the quadratic pair to be 5 and 2 because 2+5 = 7; and (2)(5) = 10. E5.13 Find the the quadratic pair such that (i) (?)(?) = 15 and ? + ? = 8 (ii) (?)(?) = 15 and ? + ? = 16 A5.13 (i) (5)(3 ) = 15 and (1)(15) = 15 But 5 + 3 = 8 so the pair is 5 and 3. (ii) (3)(5 ) = 1 5 ; (15)(1) = 15 and 15 + 1 = 16 so the pair is 15 and 1. 5.1 4 Now let’s find (?)(?) = - 15 and ? + ? = - 2. Because (?)(?) is ─ ve we know that one ? is ─ ve and the other ? is +ve. Also because ? + ? is ─ ve we know that the number with a larger magnitude is – ve. So we can identify the pair as - 5 and +3. E5.1 4 Find the quadratic pairs: (i) (?)(?) = - 20 and ? + ? = - 8 62 (ii) (?)(?) = - 20 and ? + ? = 1 (iii) (?)(?) = - 20 and ? + ? = - 19 (iv) (?)(?) = - 20 and ? + ? = 19 (v) (?)(?) = - 20 and ? + ? = 8 (vi) (?)(?) = - 20 and ? + ? = - 1 A5.1 4 (i) - 10 , 2 (ii) 5 , - 4 (iii) - 20 , 1 (iv) 20 , - 1 (v) 10 , - 2 (vi) - 5 , 4 5.1 5 N ow let’s find (?)(?) = 15 and ?  ? = - 8. Because (?)(?) is + ve we know that both are ─ ve or both are +ve but because ? + ? is ─ ve we know that both the numbers are – ve. So we get from trial and error that ( - 5)( - 3) = 15 and - 5 + - 3 = - 8. So the quadratic pair is - 5 and - 3. E5.1 5 Find the quadratic pairs: (i) (?)(?) = 21 and ? + ? = - 10 (ii) (?)(?) = 21 and ? + ? = 10 (iii) (?)(?) = 21 and ? + ? = 22 (iv) (?)(?) = 21 and ? + ? = - 22 A5.1 5 (i) - 7, - 3 (ii) 7,3 (iii) 21 , 1 (iv) - 21, - 1 Factorising perfect Squares 5.1 6 Now we know the expansion: (x + 3) 2 = x 2 + 6x + 9. We say x 2 + 6x + 9 is a perfect square . Notice its structure: Of the 3 terms two are squares and the third is twice the cross. So we can factorise x 2 + 6x + 9 = (x + 3) 2 Note thi s a square of a sum. A n other exa mple is 100 + 140 + 49x 2 . Note 100 is the square of 10 i.e. 10 2 , 49x 2 is the square of 7x , A nd 140x is ‘ twice the cross ’ , 2( 10 )( 7x) = 140x. 63 So the perfect square is (10 + 7x) 2 You must learn to identify perfect squares. E5. 16 F actorise (i) x 2 + 10x + 25 (ii) x 2 + 2 x + 1 (iii) x 2 + 1 8 x + 81 (iv) 121 x 2 + 1 98 x + 81 (v) 144 + 120 x + 25x 2 A 5.1 6 (i) x 2 + 10x + 25 = (x + 5) 2 (ii) x 2 + 2 x + 1 = (x + 1) 2 (iii) x 2 + 1 8 x + 81 = (x + 9) 2 (iv) 121 x 2 + 1 98 x + 81 = (11x + 9) 2 (v) 144 + 120 x + 25x 2 = (12 + 5x) 2 5.1 7 Now consider an example where the coefficient of the cross term is negative: x 2 – 10x + 25 The per fect square is (x – 5) 2 . Note this the square of a difference . E5.17 F actorise (i) x 2 – 10x + 25 (ii) x 2 – 2 x + 1 (iii) x 2 – 1 8 x + 81 (iv) 121 x 2 – 1 98 x + 81 (v) 144 – 120 x + 25x 2 A5.17 (i) x 2 – 10x + 25 = (x – 5) 2 (ii) x 2 – 2 x + 1 = (x – 1) 2 (iii) x 2 – 1 8 x + 81 = (x – 9) 2 (iv) 121 x 2 – 1 98 x + 81 = (11x – 9) 2 (vi) 144 – 120 x + 25x 2 = (12 – 5x) 2 F actorising Quadratic functions 5.18 Now consider quadratic expressions which are not perfect squar es. Consider quadratics where the coefficient of x 2 is 1 . Consider x 2 + 8x + 12. This is called a quadratic expression. There are three terms : A n x 2 term, which we call the quadratic te rm , an x term, +8x, wh ich we call the linear term and the c onstant term , 1 2 . Can this be factorised? If we can find a quadratic pair such that : 64 (?)(?) = 12 ? + ? = 8, coefficient of the cross term then the answer is yes. The quadratic pair is 6,2 The factorised expression is (x + 6)(x + 2). Check this by expansion: (x + 6)(x + 2) = x 2 + 8x + 12 . E5.18 Complete the following for the expression: x 2 + 8x + 16 ? + ? = (?)(?) = The quadratic pair is …. And ….. Hence we can w rite: x 2 + 8x + 16 = (x + )(x + ) = (x + ) 2 A5.18 x 2 + 8x + 16 ( ? )( ? ) = 16 (?)(?) = 8 The quadratic pair is 4 And 4 Hence we can w rite: x 2 + 8x + 16 = (x + 4 )(x + 4 ) = (x + 4 ) 2 5.19 Now we can do this more succinctly as follows: x 2 + 1 5 x + 36 = (x + 12 ) (x + 3) by finding the quadratic pair 12 and 3 mentally. We can also check our answer by expanding: (x + 12 ) (x + 3) = x 2 + 12x + 3x + 36 = x 2 + 1 5 x + 36 E5.19 Comp lete the following: (i) x 2 + 6x + 8 = (x + ) (x + ) ; check: (x + ) (x + ) = x 2 + 6x + 8 (ii) x 2 + 7 x + 10 = (x + ) (x + ) ; check: (x + ) (x + ) = x 2 + 7 x + 10 A5.19 (i) x 2 + 6x + 8 = (x + 4 ) (x + 2 ) ; check: (x + 4 ) (x + 2 ) = x 2 + 6x + 8 (i) x 2 + 7 x + 10 = (x + 5 ) (x + 2 ) ; check: (x + 5 ) (x + 2 ) = x 2 + 7 x + 10 5.20 Now l et’s consider negative coefficients : x 2 ─ 4 x – 1 2 . The q uadratic pair is - 6, 2 so we factorise as follows: x 2 ─ 4 x – 1 2 = (x – 6)(x + 2) . 65 Another example is x 2 + x – 1 2 . The quadratic pair is 4 and - 3. Then we factorise: x 2 + x – 12 = (x + 4 )(x – 3) Another example: Another example is x 2 – 7 x + 1 2 . The quadratic pair is - 4 and - 3. Then we factorise: x 2 - 7 x + 12 = (x – 4 )(x – 3) E5.20 Complete the foll owing for the expression: x 2 ─ 8x + 7 ? + ? = (?)(?) = The quadratic pair is …. And ….. Hence we can write: x 2 ─ 8x + 7 = (x ─ )(x ─ ) A5.20 x 2 ─ 8 x + 7 ? + ? = - 8 (?)(?) = 7 The quadratic pair is - 7 a nd - 1 Hence we can write: x 2 ─ 8x – 7 = (x ─ 7 )(x ─ 1 ) Difference of two squares 5.2 1 Suppose we want to factorise: x 2 – 25. This is a quadratic in which the x term is missing because its coeficiceint is zero. We can write this as a quadratic as follows: x 2 + 0x – 25, then find the quadratic pair : (?)(?)= - 25 and ? + ? = 0 as the numbers +5 and - 5. Then we write : x 2 – 25 = (x + 5)(x – 5). E5.2 1 Factorise x 2 – 100. A5.21 We can write this as a quadratic as follows: x 2 + 0x – 100 , then find the quadratic pair : (?)(?)= - 100 and ? + ? = 0 as the numbers + 2 5 and - 2 5. 66 Then we write : x 2 – 100 = (x + 10 )(x – 10 ). 5.2 2 Here are some more done for you. Ca you see a pattern? x 2 – 16 = (x + 4)(x – 4) x 2 – 100 = (x – 10)(x + 10) x 2 – 1 = (x – 1)(x + 1) x 2 – 121 = (x + 11)(x – 11) The expression on the LHS is called the difference of two squares. Consider the first in the list: x 2 is the square of x. It has the root x. 16 is a square number with root 4. The two squares are being subtracted hence the word difference. E5.2 2 Factorise at sight, using difference of squares : i. x 2 – 16 = ii. x 2 – 81 = iii. x 2 – 49 = iv. x 2 – 36 = A5.2 2 i. x 2 – 16 = (x + 4 )(x – 4 ) ii. x 2 – 81 = (x + 9 )(x – 9 ) iii. x 2 – 49 = (x – 7 )(x + 7 ) iv. x 2 – 36 = (x + 6 )(x – 6 ) 5.2 3 Here are more examples: x 2 – y 2 = (x + y)(x – y) 4x 2 – y 2 = (2x + y)(2x – y) x 2 – 9y 2 = (x – 3y)(x + 3y) 4x 2 – 9y 2 = (2x + 3y)(2x – 3y) E5.2 3 Factorise the following: i. a 2 – b 2 = ii. 4x 2 – 9y 2 = iii. 16x 4 – 9y 2 = iv. 1 - 4x 2 = A5.2 3 i. a 2 – b 2 = (a – b)(a + b) ii. 4x 2 – 9y 2 = (2x + 3y)(2x – 3y) iii. 16x 4 – 9y 2 = (4x 2 – 3y )(4x 2 + 3y) iv. 1 - 4x 2 = (1 + 2x)(1 – 2x ) 5.2 4 Sometimes the factorisation can be multi - staged. Track the following factorisation: 16x 4 – 625 = (4x 2 + 25)(4x 2 – 25) =(4x 2 + 25)(2x + 5)(2x – 5) Note the second factor of the RHS is also a differerence of two squares. E5.2 4 Factorise ful ly: i. 16x 4 – y 4 ii. 81 – 16y 4 iii. 32 – 18y 2 67 A5.2 4 Factorise ful ly: i. 16x 4 – y 4 = (4x 2 – y 2 )(4x 2 + y 2 ) = ( 2 x – y ) (2x + y) (4x 2 + y 2 ) ii. 81 – 16y 4 = ( 9 – 4 y 2 )( 9 + 4 y 2 ) = ( 3 – 2 y ) ( 3 + 2 y) ( 9 + 4 y 2 ) iii. 32 – 18y 2 = 2 (16 – 9y 2 ) = 2 ( 4 – 3 y )(4 + 3 y ) 5.2 5 Now consider the factorisation of the quadratic function: 3x 2 + 8x + 4. Note that the coefficient of the x 2 term, the quadratic term is not 1 but 3. The coefficient of the x term, the linear term is 8 and the cons tant term is 4. Step 1: We find the quadratic pair: (?)(?) = (3)(4)= 12 and ? + ? = 8. Note the 12 is the product of the quadratic coefficient and the constant term i.e. 3(4) and the 8 is the coefficient of the linear term. The quadratic pair is 6 and 2. Ste p 2: Write our original 3 term quadratic expression as a 4 term quadratic expression by splitting the linear term into two terms using the quadratic pair as shown below: 3x 2 + 8x + 4 = 3x 2 + 6x + 2x + 4 Step 3:Factorise using our grouping method: 3x 2 + 6 x + 2x + 4 = 3x(x + 2) + 2(x + 2) = (x + 2)(3x + 2) Step 4: Check: (x + 2)(3x + 2) = 3x 2 + 2x + 6x + 4 = 3x 2 + 8x + 4 E5.2 5 Complete the following to factorise, 2x 2 + 7x + 6. Step 1: Find the quadratic pair: (?)(?) = 2(6) =12 and ? + ? = 6 . Step 2: Write our original 3 term quadratic expression as a 4 term 68 quadratic expression by splitting the linear term into two terms using the quadratic pair as shown below: = 2x 2  …x  ….x  4 Step 3: Factorise using our grouping m ethod: Step 4: Check: (…x  ….)(….x  ….) A5. 2 5 F actorise, 2x 2 + 7x + 6. Step 1: Find the quadratic pair: (?)(?) = 2(6) = 12 and ? + ? = 7. The quadratic pair is 4 , 3 Step 2: Write our original 3 term quadratic expression as a 4 term quadratic expression by splitting the linear term into two terms using the quadratic pair as shown below: 2x 2 + 7x + 6 = 2x 2 + 4 x + 3 x + 6 Step 3: Factorise using our grouping m ethod: 2x 2 + 4 x + 3 x + 6 =2 x (x +2) + 3(x+2) =(x + 2)( 2 x + 3) Step 4: Check: ( x + 2 )( 2 x + 3 ) = 2x 2 + 7x + 6 5.2 6 Now consider an example with – ve coefficients: Track the following: Factorise 5x 2 – 8x – 4 Step 1: Find the quadratic pair: (?)(?) = - 20 and ? + ? = - 8. The quadratic pair is - 10, +2. Step 2: W rite our original 3 term quadratic expression as a 4 term quadratic expression by splitting the linear term into two terms using the quadratic pair as shown below: = 5x 2 – 10x + 2x – 4 Step 3: Factorise using our grouping method: = 5x(x – 2) + 2(x – 2) 69 = (5x + 2)(x – 2) Step 4: Check (5x + 2)(x – 2) = 5x 2 – 10x + 2x – 4 = 5x 2 – 8x – 4 E5.2 6 Factorise: 7x 2 + 12x – 4 A5.2 6 F actorise, 7 x 2 + 12 x – 4 . Step 1: Find the quadratic pair: (?)(?) = 7( - 4) = - 28 and ? + ? = 12 . The quadratic pair is 1 4 , - 2 Step 2: Write our original 3 term quadratic expression as a 4 term quadratic expression by splitting the linear term into two terms using the quadratic pair as shown below: 7 x 2 + 12 x – 4 = 7 x 2 + 1 4 x – 2 x – 4 Step 3: Factorise using our grouping m ethod: 7 x 2 + 1 4 x – 2 x – 4 = 7 x (x +2) – 2 (x+2) =(x + 2)( 7 x – 2 ) Step 4: Check: ( x + 2 )( 7 x – 2 ) = 7 x 2 + 12 x – 4 70 Exercise 5 1. Are the follo wing factorisations correct? - 3a 3 b + 6a 2 b 2 – 9ab 3 = - 3ab(a 2 - 2ab + 3b 2 ) - 3a 3 b + 6a 2 b 2 – 9ab 3 = 3ab( - a 2 +2ab - 3b 2 ) 2. Factorise: 3a(m + n) + 5b(m + n) 3. Factorise by the grouping method: (i) am ─ bm + na ─ nb (ii) 3m ─ 6n ─ am + 2an 4 Factorise the following quadratic expressions: (i) x 2 + 7x + 10 (ii) x 2 – 7x + 12 (iii) x 2 – 2x – 8 (iv) x 2 + 2x – 63 5. Simplify the following: (Hint: Factori se numerator ) 6. Factorise fully where possible: a) 3x 2 y – 4xy + 5xy 2 b) - 3a 2 + 3b 2 – 18 c) x 2 – x – 2 d) 7x 2 y +5xy - x 2 y 2 +5x+5x 2 y 2 e ) x 2 – 1 f) 3(x – y) + 6(x + y) g ) x 2 - 16 y 2 h ) 4 – 6x – 7x 2 i ) 3a(a+b) 2 - 6b(a+b)