Factorising Algebraic Expressions

1 Factorising AlgebraicExpressions7175 Can you crack thecode, Mr. x?Chapter Contents7:01 Factorising using common factors7:02 Factorising by grouping in pairs7:03 Factorising using the difference oftwo squaresChallenge: The difference of two cubes7:04 Factorising quadratic trinomialsFun Spot: How much logic do you have?7:05 Factorising further quadratictrinomialsChallenge: Another Factorising methodfor harder trinomials7:06 Factorising : Miscellaneous typesFun Spot: What did the caterpillar saywhen it saw the butterfly?7:07 Simplifying Algebraic fractions :Multiplication and division7:08 Addition and subtraction of algebraicfractionsMathematical Terms, Diagnostic Test, Revision Assignment, Working MathematicallyLearning OutcomesStudents will be able to: Factorise using common factors. Factorise by grouping in pairs. Factorise using the difference of two squares. Factorise quadratic trinomials. Simplify Algebraic fractions by Factorising . Perform operations with Algebraic of InteractionApproaches to Learning (Knowledge Acquisition, Problem Solving, Logical Thinking, Reflection), Human Ingenuity176 INTERNATIONAL MATHEMATICS 4In Chapter 4, Algebraic Expressions , you were shown how to expand various Algebraic products that were written in a factorised form; that is, each product had to be rewritten without grouping example:3a(5 2a) 15a 6a2(a 2)(a + 7) a2 + 5a 14(x + 5)2 x2 + 10x + 25(m + 2)(m 2) m2 4 This chapter will show you how to reverse this process.

2 You will learn how to factorise various Algebraic :01 | Factorising Using Common FactorsTo factorise an Algebraic expression, we must determine the highest common factor (HCF) of the terms and insert grouping symbols, usually we expand the expression 5a(a 2) we obtain 5a2 10a. To factorise 5a2 10a we simply reverse the process. We notice that 5a is the HCF of 5a2 and 2, so 5a is written outside the brackets and the remainder is written inside the brackets: 5a2 10a = 5a(a 2). A factor of a given number is another number that will divide into the given number with no {1, 2, 3, 6, 9, 18} is the set of factors of (a 2)5a2 10a Gnidnapxe is thereverse of expanding .It s Factorising ,you dummy!worked examples15y + 15 = 5 y + 5 3 (HCF is 5)221x 24y = 3 7x 3 8y (HCF is 3)= 5(y + 3)= 3(7x 8y)312ab + 18a = 6a b + 6a 3 (HCF is 6a)45x2 30x = 5x x 5x 6 (HCF is 5x)= 6a(b + 3)= 5x(x 6)5 12x2 3x = 3x 4x 3x 1 (HCF is 3x)= 3x(4x + 1)63a2b 9ab2 + 15ab = 3ab a 3ab b + 3ab 5= 3ab(a b + 5)ab + ac = a(b + c)CHAPTER 7 Factorising Algebraic EXPRESSIONS177 Factorise the following + 15yb 3m m2c6xy 2xd15p 20qe15pq 20qf12st2 + 15stg 18xy 6xhat at2i7x2y + xyja2 + abFactorise each of the + ab + 3abxy 3x2 + 2xc12st 4t3 + 8td36 12ab + 18be3ab 9a2b + 12ab2 + a2b2f4m 8n 12mng3 + 5m 2nh 3n 5mn + 2n2i12x2 + 8x 4j12x2 y2 + 8xy2 4y2 Examine this examplex(x + 1) 2(x + 1) has a common factor of (x + 1) so it can be taken out as a common factor so x(x + 1) 2(x + 1) = (x + 1)(x 2)Now factorise these similar (x + 4) + y(x + 4)b4(a + 2) b(a + 2)cm(m 1) 3(m 1)

3 D2(s 3) + s(s 3)e2a(a 1) (a 1)f3m(9 2m) + 2(9 2m)gx(x 5) + 2(3x 15)hy(y + 5) + 2( y 5)ix(3 x) + 5(x 3)jab(9 a) 2(a 9)Factorise fully the following Algebraic + 6b10 + 15ac4m 6ndx2 + 7xe2a2 3af12y 6y2gab bxhst si4ab + 10bcj 4m + 6nk x2 3xl 15a + 5abm3x + x2 axnax + ay + azo4m 8n + 6pp2(a + x) + b(a + x)qx(3 + b) + 2(3 + b)ry(x 1) 3(x 1)s5ab 15ac + 10adtx2 7x + xyua(a + 3) (a + 3)Exercise 7:01 Common factors1 Complete the + 12 = 6(.. + 2) + 15ba2 6m 15b 2x2 + 4xFoundation Worksheet 7:01123 Note:(x + 1)(x 2) = (x 2)(x + 1)4178 INTERNATIONAL MATHEMATICS 47:02 | Factorising by Grouping in PairsFor some Algebraic Expressions , there may not be a factor common to every term. For example, there is no factor common to every term in the expression:3x + 3 + mx + mBut the first two terms have a common factor of 3 and the remaining terms have a common factor of m. So:3x + 3 + mx + m = 3(x + 1) + m(x + 1)Now it can be seen that (x + 1) is a common factor for each (x + 1) + m(x + 1) = (x + 1)(3 + m)Therefore:3x + 3 + mx + m = (x + 1)(3 + m)The original expression has been factorised by grouping the terms in these13a + 1825x + ax3pq 9bx5x2 2x6a3 + a279 3a8 5m 1099(a + 1) + x(a + 1)10x(x + y) 1(x + y)prepquiz7:02worked examples12x + 2y + ax + ay = 2(x + y) + a(x + y)2a2 + 3a + ax + 3x = a(a + 3) + x(a + 3)= (x + y)(2 + a)= (a + 3)(a + x)3ax bx + am bm = x(a b) + m(a b)4ab + b2 a b = b(a + b) 1(a + b)= (a b)(x + m)= (a + b)(b 1)55x + 2y + xy + 10 = 5x + 10 + 2y + xy= 5(x + 2) + y(2 + x)= (x + 2)(5 + y) Note.

4 Terms had to be rearranged to pair those with common + ac + bd + cd = a(b + c) + d(b + c)= (b + c)(a + d)CHAPTER 7 Factorising Algebraic EXPRESSIONS179 Complete the factorisation of each expression (a + b) + x(a + b)ba(x + 7) + p(x + 7)cm(x y) + n(x y)dx(m + n) y(m + n)ea2(2 x) + 7(2 x)fq(q 2) 2(q 2)g(x + y) + a(x + y)hx(1 3y) 2(1 3y)Factorise these + pb + qa + qbb3a + 3b + ax + bxcmn + 3np + 5m + 15pda2 + ab + ac + bce9x2 12x + 3xy 4yf12p2 16p + 3pq 4qgab + 3c + 3a + bchxy + y + 4x + 4ia3 + a2 + a + 1jpq + 5r + 5p + qrkxy x + y 1l8a 2 + 4ay ymmn + m + n + 1nx2 + my + xy + mxox2 xy + xw ywpx2 + yz + xz + xyq11a + 4c + 44 + acra3 a2 + a 1 Factorise the + xz wy wzbab + bc ad cdc5a + 15 ab 3bd6x 24 xy + 4ye11y + 22 xy 2xfax2 ax x + 1 Exercise 7:02 Grouping in pairs1 Complete the (x + 2) + a(x + 2) + 5a + 2m + 10 Foundation Worksheet 7:02123 This is an exercise you can sink your teeth into!

5 180 INTERNATIONAL MATHEMATICS 47:03 | Factorising Using the Difference of Two SquaresIf the expression we want to factorise is the difference of two squares, we can simply reverse the procedure seen in section 3 each of these 4ba2 16cm2 25dp2 81ey2 100fx2 121g9 x2h1 n2i49 y2ja2 b2kx2 a2ly2 a2m9a2 4n16x2 1o25p2 9p49 4a2q25p2 a2rm2 81n2s100a2 9b2t81x2 121y2 Factorise by first taking out a common 32b3x2 108c4a2 100d5y2 20e24a2 6b2f3x2 27y2g8y2 128h80p2 5q2i4x2 64j3x2 3k72p2 2l2 18x2m8a2 18m2n125 20a2o200x2 18y2p98m2 8n2 Simplify:123If x is positive, simplify:456 Expand and simplify:7(x 2)(x + 2)8(x + 5)(x 5)9(7 a)(7 + a)10(3m + 2n)(3m 2n)prepquiz7:031649121x29x264x2expandfac torise(x a)(x + a)x2 a2worked examples1x2 9 = x2 32225a2 b2 = (5a)2 b2= (x 3)(x + 3)= (5a b)(5a + b)3a4 64 = (a2)2 82436m2 49n2 = (6m)2 (7n)2= (a2 8)(a2 + 8)= (6m 7n)(6m + 7n)x2 y2 = (x y)(x + y) Note: (x y)(x + y) = (x + y)(x y)Exercise 7:0312worked example18x2 50 = 2(9x2 25)= 2([3x]2 52)= 2(3x 5)(3x + 5)CHAPTER 7 Factorising Algebraic EXPRESSIONS181 Challenge 7:03 | The difference of two cubes (Extension) The large cube has a volume of a3 cubic unitsIt is made up of four smaller parts (a cube and three rectangular prisms).

6 Our aim is to find an expression for thedifference of two cubes (a3 b3).1 Complete the table an expression for the volume of the large cube (a3) in terms of the volumes of the four smaller a3 = V1 + V2 + V3 + V43 Use your answer to question 2 to write an expression for a3 each volume as a product of its this to Algebraic Expressions , we could factorise a difference of two cubes:eg x3 8 = x3 23= (x 2)(x2 + 2x + 4)ExercisesFactorise these Expressions using the formula n32x3 y33a3 84m3 275x3 10006y3 125764 n3827 k398m3 271064x3 125y311125x3 8y31227m3 343n3challenge7:03aaaa bbbba ba b1234 Volume of part 1 Volume of part 2 Volume of part 3 Volume of part 4b b b Note: a3 b3 = (a b)(a2 + ab + b2)182 INTERNATIONAL MATHEMATICS 47:04 | Factorising Quadratic Trinomials An expression with three terms is called a trinomial. Expressions like x2 + 3x 4 are called quadratic trinomials. The highest power of the variable is 2.

7 Factorising is the reverse of expanding.(x + a)(x + b) = x2 + ax + bx + ab= x2 + (a + b)x + abUsing this result, to factorise x2 + 5x + 6we look for two values a and b, wherea + b = 5 and ab = numbers are 2 and 3, so:x2 + 5x + 6 = (x + 2)(x + 3)Expand:1(x + 2)(x + 3)2(a 1)(a + 3)3(m 7)(m 2)4(x + 5)25(a 2)2 Find two numbers a and b where:6a + b = 5 and ab = 67a + b = 9 and ab = 208a + b = 2 and ab = 159a + b = 3 and ab = 410a + b = 7 and ab = 18prepquiz7:04(x + a)(x + b)x2 + (a + b)x + ab2 and 3 add to give5 and multiply togive examplesFactorise:1x2 + 7x + 102m2 6m + 8 3y2 + y 124x2 9x 3653y2 + 15y 72 Solutions123 x2 + 7x + 10 m2 6m + 8 y2 + y 12= (x + 2)(x + 5)= (m 2)(m 4)= (y 3)(y + 4)453y2 + 15y 72= 3(y2 + 5y 24) x2 9x 36= (x + 3)(x 12) 3(y2 + 5y 24)= 3(y 3)(y + 8)If x2 + 7x + 10 = (x + a)(x + b) then a + b = 7 and ab = + 5 = 72 5 = 10( 2) + ( 4) = 6( 2) ( 4) = 8( 3) + 4 = 1( 3) 4 = 123 + ( 12) = 93 ( 12) = 36( 3) + 8 = 5( 3) 8 = 24 Step 1:Take out anycommon 7 Factorising Algebraic EXPRESSIONS183 Factorise each of these + 4x + 3bx2 + 3x + 2cx2 + 6x + 5dx2 + 7x + 6ex2 + 9x + 20fx2 + 10x + 25gx2 + 12x + 36hx2 + 10x + 21ix2 + 9x + 18jx2 + 14x + 40kx2 + 15x + 54lx2 + 13x + 36mx2 4x + 4nx2 12x + 36ox2 7x + 12px2 9x + 20qx2 + 2x 3rx2 + x 12sx2 + 4x 12tx2 + 7x 30ux2 x 2vx2 10x 24wx2 7x 30xx2 x 56 Factorise.

8 Aa2 + 6a + 8bm2 + 9m + 18cy2 + 13y + 42dp2 + 7p + 12ex2 + 12x + 20fn2 + 17n + 42gs2 + 21s + 54ha2 + 18a + 56ix2 3x 4ja2 2a 8kp2 5p 24ly2 + y 6mx2 + 7x 8nq2 + 5q 24om2 + 12m 45pa2 + 18a 63qy2 + 6y 55rx2 2x + 1sk2 5k + 6tx2 13x + 36ua2 22a + 72vp2 + 22p + 96wq2 12q 45xm2 4m 77 Factorise by first taking out a common factor (see example 5).a2x2 + 6x + 4b3x2 6x 9c5x2 10x 40d2x2 + 16x + 32e3x2 30x 33f3x2 + 21x + 36g4a2 12a 40h2n2 + 8n + 6i5x2 30x + 40j3x2 21x + 36k3a2 15a 108l5x2 + 15x 350 Fun Spot 7:04 | How much logic do you have?See if you can solve the three problems is the next letter in this sequence? O, T, T, F, F, S, S, ?2A man passing a beggar in the street exclaimed, I am that beggar s father! But the beggar was not the man s son. How can this be?3 Two guards are guarding two sacks. One guard always tells the truth, but theother guard always lies, but you do notknow which guard is which.

9 One of thesacks is full of gold; the other is full of peanuts. You are permitted to take one of the sacks but you are not sure which one contains the are also allowed to ask one of the guards just one question should you ask to ensure you get the sack of gold?Exercise 7:04 Factorising trinomials1 Which two integers:aadd to bmultiply togive 4?give 5?2 Factorise:am2 + 8m + 9bn2 3n + 2 Foundation Worksheet 7:04123funspot7:04184 INTERNATIONAL MATHEMATICS 47:05 | Factorising Further Quadratic Trinomials In all quadratic trinomials factorised so far, the coefficient ofx2 has been 1. We will now consider cases where the coefficientof x2 is not 1. To expand (5x 1)(x + 3) we can use a cross is the product of the two left terms. 3 is the product of the two right is the sum of the products along the cross, ie 15x + ( x). (5x 1)(x + 3) = 5x2 + 14x 3 One method used to factorise trinomials like 5x2 + 14x 3 is called the cross methodTo factorise 5x2 + 14x 3, we need to reverse the expanding need to choose two factors of 5x2 and two factors of 3 to write on the cross.

10 If (5x 3) and (x + 1) are the factors of 5x2 + 14x 3, then the products of numbers on the ends of each arm will have a sum of +14x. When we add the cross products here, we get:(5x) + ( 3x) = 2xThis does not give the correct middle term of 14x,so (5x 3) and (x + 1) are not factors. Vary the terms on the product = ( 15x) + (x) = 14x Cross product = (15x) + ( x) = 14x This must be the correct combination. 5x2 + 14x 3 = (5x 1)(x + 3)Examine the examples below. Make sure you understand the method. 3x2 + 7x + 9coefficient of x2 Remember5x2 3(5x 1)(x + 3) x+15x= 5x2 + 15x x 3= 5x2 + 14x 35xx 1+35xx 1+35xx+1 35xx 3+1 Try:5x 3andx+1 +1 Try:andx 3 5x 1 Try:andx+3 CHAPTER 7 Factorising Algebraic EXPRESSIONS185aWhich diagram will give the factors of 2x2 + 13x + 6?iiiiiiivbWhich diagram will give the factors of 9x2 9x 4?iiiiiiivExercise 7:05worked examplesFind the factors of:13x2 19x + 624x2 x 332x2 + 25x + 12 Solutions1 This cross product gives the correct middle term of 19x, so:3x2 19x + 6 = (3x 1)(x 6)Note: The factors of +6 had to be both negative to give a negative middle 4x2 x 3= (4x + 3)(x 1)3In practice, we would not draw a separatecross for each new set of factors.