Algebra: Factorising- Linear and Quadratics Skipton Girls’ High School

Factors Factors of 8: 1, 2, 4, 8 Factors of 2x: 1, 2, x, 2x What does the factor of a number mean? Numbers which divide the original number without a remainder. ? Factors of 8: 1, 2, 4, 8 ? Factors of 2x: ? 1, 2, x, 2x Factors of 2x2: 1, 2, x, 2x, x2, 2x2 ?

2x + 4 So 2x + 4 = 2(x + 2) Factorising Common factor = 2 ? ? Factorising is the reverse of expanding. When you have a sum of terms, just identify the common factor. i.e. Find the largest expression each of your terms is divisible by. Common factor = 2 ? 2x + 4 So 2x + 4 = 2(x + 2) ? (You could always check this by expanding out the brackets)

3x2 + 9x So 3x2 + 9x = 3x(x + 3) Factorising ? Common factor = 3x ? Factorising is the reverse of expanding. When you have a sum of terms, just identify the common factor. i.e. Find the largest expression each of your terms is divisible by. ? Common factor = 3x 3x2 + 9x We could have just ‘factored out’ the 3, but we wouldn’t have fully factorised because there’s also a factor of x. So 3x2 + 9x = 3x(x + 3) ?

Factorising xy + x = x(y + 1) ? ? 2xy + 4x = 2x(y + 2)

Exercise 1 1) 2𝑥 −4=2 𝑥−2 2) 𝑥𝑦+𝑦=𝑦(𝑥+1) 3) 𝑞𝑟−2𝑞=𝑞 𝑟−2 4) 6𝑥−3𝑦=3(2𝑥−𝑦) 5) 𝑥𝑦𝑧+𝑦𝑧=𝑦𝑧(𝑥+1) 6) 𝑥 2 𝑦+2𝑦𝑧=𝑦 𝑥 2 +2𝑧 7) 𝑥 3 𝑦+𝑥 𝑦 2 =𝑥𝑦 𝑥 2 +𝑦 8) 5𝑞𝑟+10𝑟=5𝑟 𝑞+2 9) 12𝑝 𝑤 2 −8 𝑤 2 𝑦=4 𝑤 2 3𝑝−2𝑦 10) 55 𝑝 3 +33 𝑝 2 =11 𝑝 2 5𝑝+3 11) 6𝑝 4 +8 𝑝 3 +10 𝑝 2 = 2𝑝 2 (3 𝑝 2 +4𝑝+5) 12) 10 𝑥 3 𝑦 2 +5 𝑥 2 𝑦 3 +15 𝑥 2 𝑦 2 =5 𝑥 2 𝑦 2 (2𝑥+𝑦+3) ? Extension Question: What integer (whole number) solutions are there to the equation xy+3x=15 Answer: 𝑥 𝑦+3 =15. So the two expressions we’re multiplying can be 1×15, 15×1, 3×5, 5×3, −1×−15, … This gives solutions (𝑥,𝑦) of 𝟏,𝟏𝟐 , 𝟑,𝟐 , 𝟓,𝟎 , 𝟏𝟓,−𝟐 , −𝟏,−𝟏𝟖 , −𝟑,−𝟖 , −𝟓,−𝟔 ,(−𝟏𝟓,−𝟒) ? ? ? ? ? ? ? ? ? ? ? ?

Exercise 2 ? ? ? ? ? ? ? ? 16 factors: 1, 2, x, y, z, 2x, 2y, 2z, xy, xz, yz, 2xy, 2xz, 2yz, xyz, 2xyz ? ? 36 factors ? 2 can either appear in the factor or not (2 possibilities) x can either appear 0 times, 1 time, up to a times (a + 1 possibilities) y similarly has b + 1 possibilities and z has c +1 possibilities. So 2(a + 1)(b + 1)(c + 1) possible factors.

Dealing with fractions When factorising, it’s convention to have any fractions outside the bracket. ? ? ? ? Bro Tip: Make sure the fractions have a common denominator.

Test Your Understanding ? ? ?

Exercise 3 ? ? ? ? ? ? ? ? ? ?

2𝑥2+4𝑥𝑧 2𝑥(𝑥+2𝑧) 𝑥2+3𝑥+2 (𝑥+1)(𝑥+2) 2x3 + 3x2 – 11x – 6 Factorising Overview Factorising means : To turn an expression into a product of factors. So what factors can we see here? Year 8 Factorisation Factorise 2𝑥2+4𝑥𝑧 2𝑥(𝑥+2𝑧) Year 9 Factorisation Factorise 𝑥2+3𝑥+2 (𝑥+1)(𝑥+2) A Level Factorisation Factorise 2x3 + 3x2 – 11x – 6 (2𝑥+1)(𝑥−2)(𝑥+3)

Six different types of factorisation 1. Factoring out a single term 2. 𝒙 𝟐 +𝒃𝒙+𝒄 2𝑥 2 +4𝑥=2𝑥 𝑥+2 ? ? 𝑥 2 +4𝑥−5= 𝑥+5 𝑥−1 3. Difference of two squares 4. 𝒂 𝒙 𝟐 +𝒃𝒙+𝒄 4 𝑥 2 −1= 2𝑥+1 2𝑥−1 ? 2 𝑥 2 +𝑥−3=(2𝑥+3)(𝑥−1) Strategy: either split the middle term, or ‘go commando’. ? 5. Pairwise 6. Intelligent Guesswork 𝑥 3 +2 𝑥 2 −𝑥−2= 𝑥 2 𝑥+2 −1 𝑥+2 = 𝑥 2 −1 𝑥+2 =… 𝑥 2 + 𝑦 2 +2𝑥𝑦+𝑥+𝑦 = 𝒙+𝒚+𝟏 𝒙+𝒚 ? ?

𝑥 2 −𝑥−30= 𝑥+5 𝑥−6 TYPE 2: 𝑥 2 +𝑏𝑥+𝑐 Expand: 𝑥+𝑎 𝑥+𝑏 = 𝒙 𝟐 + 𝒂+𝒃 𝒙+𝒂𝒃 𝑥+𝑎 𝑥+𝑏 = 𝒙 𝟐 + 𝒂+𝒃 𝒙+𝒂𝒃 How does this suggest we can factorise say 𝑥 2 +3𝑥+2? 𝑥 2 +3𝑥+2= 𝒙+𝟐 𝒙+𝟏 ? 𝑎 and 𝑏 add to give 3. 𝑎 and 𝑏 times to give 2. ? 𝑥 2 −𝑥−30= 𝑥+5 𝑥−6 ? Tip: Think of the factor pairs of 30. You want a pair where the sum or difference of the two numbers is the middle number (-1).

TYPE 2: 𝑥 2 +𝑏𝑥+𝑐 A few more examples: 𝑥 2 +6𝑥+5=(𝑥+5)(𝑥+1) ? 𝑥 2 +6𝑥+5=(𝑥+5)(𝑥+1) ? 𝑥 2 −12𝑥+35= 𝑥−7 𝑥−5 ? 𝑥 2 +5𝑥−14=(𝑥+7)(𝑥−2) ? 𝑥 2 +6𝑥+9= 𝑥+3 2 𝑥 2 −6𝑥+9= 𝑥−3 2 ?

Exercise 2 1 𝑥 2 +4𝑥+3= 𝑥+3 𝑥+1 𝑥 2 −8𝑥+7= 𝑥−1 𝑥−7 𝑥 2 +2𝑥−8= 𝑥+4 𝑥−2 𝑥 2 +16𝑥−36= 𝑥+18 𝑥−2 𝑦 2 −𝑦−56=(𝑦+7)(𝑦−8) 𝑧 2 +3𝑧−54= 𝑧+9 𝑧−6 𝑧 2 −3𝑧−54= 𝑧−6 𝑧+9 𝑧 2 +15𝑧+54= 𝑧+6 𝑧+9 𝑥 2 +4𝑥+4= 𝑥+2 2 𝑥 2 −14𝑥+49= 𝑥−7 2 𝑥 2 +10𝑥−39= 𝑥+13 𝑥−3 ? 11 𝑥 2 +4𝑥+4 = 𝑥+2 2 𝑥 2 −17𝑥+66 = 𝑥−11 𝑥−6 𝑎 2 −2𝑎−63 = 𝑎+7 𝑎−9 𝑦 2 −10𝑦+25 = 𝑦−5 2 2 ? ? 3 ? 12 𝝅 ? ? 4 ? 13 5 ? ? 6 ? 14 7 ? ? 8 ? 9 ? 10 ? Hardcore N4 ? N1 𝑥 4 +5 𝑥 2 +4= 𝑥 2 +1 𝑥 2 +4 𝑥 2 −2𝑎𝑥+ 𝑎 2 = 𝑥−𝑎 2 𝑥 4 −6𝑎𝑏 𝑥 2 +9 𝑎 2 𝑏 2 = 𝑥 2 −3𝑎𝑏 2 ? 𝑥 9 − 𝑥 8 −2 𝑥 7 = 𝑥 7 𝑥+2 𝑥−1 𝑥 11 +2 𝑥 9 + 𝑥 7 = 𝑥 7 𝑥 2 +1 2 N5 ? N2 ? N3 ?

Six different types of factorisation 1. Factoring out a single term 2. 𝑥 2 +𝑏𝑥+𝑐 2𝑥 2 +4𝑥=2𝑥 𝑥+2 𝑥 2 +4𝑥−5= 𝑥+5 𝑥−1 3. Difference of two squares 4. 𝑎 𝑥 2 +𝑏𝑥+𝑐 4 𝑥 2 −1= 2𝑥+1 2𝑥−1 ? 2 𝑥 2 +𝑥−3=(2𝑥+3)(𝑥−1) Strategy: either split the middle term, or ‘go commando’. ? 5. Pairwise 6. Intelligent Guesswork 𝑥 3 +2 𝑥 2 −𝑥−2= 𝑥 2 𝑥+2 −1 𝑥+2 = 𝑥 2 −1 𝑥+2 =… 𝑥 2 + 𝑦 2 +2𝑥𝑦+𝑥+𝑦 = 𝒙+𝒚+𝟏 𝒙+𝒚 ? ?

TYPE 3: Difference of two squares Firstly, what is the square root of: 4 𝑥 2 =2𝑥 ? 25 𝑦 2 =5𝑦 ? 16 𝑥 2 𝑦 2 =4𝑥𝑦 ? 𝑥 4 𝑦 4 = 𝑥 2 𝑦 2 ? 9 𝑧−6 2 =3(𝑧−6) ?

Click to Start animation TYPE 3: Difference of two squares √ √ 2𝑥 2𝑥 3 3 4 𝑥 2 −9 =( + )( − ) Click to Start animation

1− 𝑥 2 =(1+𝑥)(1−𝑥) 𝑦 2 −16=(𝑦+4)(𝑦−4) 𝑥 2 𝑦 2 −9 𝑎 2 = 𝑥𝑦+3𝑎 𝑥𝑦−3𝑎 Quickfire Examples 1− 𝑥 2 =(1+𝑥)(1−𝑥) ? ? 𝑦 2 −16=(𝑦+4)(𝑦−4) 𝑥 2 𝑦 2 −9 𝑎 2 = 𝑥𝑦+3𝑎 𝑥𝑦−3𝑎 ? ? 1− 𝑥 4 = 1+ 𝑥 2 1+𝑥 1−𝑥 4 𝑥 2 −9 𝑦 2 =(2𝑥+3𝑦)(2𝑥−3𝑦) ? 𝑥 2 −3= 𝑥+ 3 𝑥− 3 ? (Strictly speaking, this is not a valid factorisation)

𝑥 3 −𝑥=𝑥(𝑥+1)(𝑥−1) 𝑥+1 2 − 𝑥−1 2 =4𝑥 49− 1−𝑥 2 =(8−𝑥)(6+𝑥) Test Your Understanding (Working in Pairs) Tip: Sometimes you can use one type of factorisation followed by another. Perhaps common term first? 𝑥 3 −𝑥=𝑥(𝑥+1)(𝑥−1) ? ? 𝑥+1 2 − 𝑥−1 2 =4𝑥 ? 49− 1−𝑥 2 =(8−𝑥)(6+𝑥) 51 2 − 49 2 =200 ? ? 18 𝑥 2 −50 𝑦 2 =2 3𝑥+5𝑦 3𝑥−5𝑦 ? 2𝑡+1 2 −9 𝑡−6 2 = 5𝑡−17 −𝑡+19

Exercise 3 1 4 𝑝 2 −1= 2𝑝+1 2𝑝−1 4− 𝑥 2 =(2+𝑥)(2−𝑥) 144− 𝑏 2 = 12+𝑏 12−𝑏 𝑥+1 2 −25= 𝑥+6 𝑥−4 7.64 2 − 2.36 2 =52.8 2 𝑝 2 −32=2 𝑝+4 𝑝−4 3 𝑦 2 −75 𝑥 2 =3 𝑦+5𝑥 𝑦−5𝑥 4 𝑎 2 −64 𝑏 2 =4 𝑎+4𝑏 𝑎−4𝑏 9 𝑝+1 2 −4 𝑝 2 = 5𝑝+3 𝑝+3 50 2𝑥+1 2 −18 1−𝑥 2 =2 7𝑥+8 13𝑥+2 𝑥 4 −1= 𝒙 𝟐 +𝟏 𝒙+𝟏 𝒙−𝟏 32 𝑥 8 −162 =𝟐(𝟒 𝒙 𝟒 +𝟗)(𝟐 𝒙 𝟐 +𝟑)(𝟐 𝒙 𝟐 −𝟑) ? N3 [IMO] What is the highest power of 2 that is a factor of 127 2 −1? = 𝟏𝟐𝟕+𝟏 𝟏𝟐𝟕−𝟏 =𝟏𝟐𝟖×𝟏𝟐𝟔 = 𝟐 𝟕 ×𝟐×𝟔𝟑 So the highest power is 8. 2 ? ? 3 ? 4 ? 5 ? 6 ? N4 Find four prime numbers less than 100 which are factors of 3 32 − 2 32 (Hint: you can keep factorising!) 𝟑 𝟑𝟐 − 𝟐 𝟑𝟐 = 𝟑 𝟏𝟔 + 𝟐 𝟏𝟔 𝟑 𝟏𝟔 − 𝟐 𝟏𝟔 = 𝟑 𝟏𝟔 + 𝟐 𝟏𝟔 𝟑 𝟖 + 𝟐 𝟖 𝟑 𝟖 − 𝟐 𝟖 =… = 𝟑 𝟏𝟔 + 𝟐 𝟏𝟔 𝟑 𝟖 + 𝟐 𝟖 𝟑 𝟒 + 𝟐 𝟒 𝟑 𝟐 + 𝟐 𝟐 𝟑+𝟐 𝟑+𝟐 𝟑−𝟐 So clearly 5 is a factor. 𝟑𝟐+𝟐𝟐=𝟏𝟑 which is also a prime factor. 𝟑𝟒+𝟐𝟒=𝟖𝟏+𝟏𝟔=𝟗𝟕 which is prime. 𝟑𝟖+𝟐𝟖=𝟔𝟖𝟏𝟕. This fails all the divisibility tests for the primes up to 11, and dividing by 13 (by normal division) fails, but dividing by 17 (again by normal division) works, giving us our fourth prime. (Alternatively, noting that 𝟔𝟖=𝟏𝟕×𝟒, then 𝟔𝟖𝟏𝟕= 𝟏𝟕×𝟒𝟎𝟎 +𝟏𝟕=𝟒𝟎𝟏×𝟏𝟕, so 17 is a factor) 7 ? ? 8 ? 9 ? 10 ? N1 ? N2 ?

* Not official mathematical terminology. TYPE 4: 𝑎 𝑥 2 +𝑏𝑥+𝑐 2 𝑥 2 +𝑥−3 Factorise using: a. ‘Going commando’* b. Splitting the middle term Essentially ‘intelligent guessing’ of the two brackets, by considering what your guess would expand to. ⊕1 ⊗−6 2 𝑥 2 +𝑥−3 ‘Split the middle term’ Unlike before, we want two numbers which multiply to give the first times the last number. (2𝑥+3)(𝑥−1) ? ? ? ? 2 𝑥 2 +3𝑥−2𝑥−3 Factorise first and second half separately. How would we get the 2 𝑥 2 term in the expansion? =𝑥 2𝑥+3 −1(2𝑥+3) How could we get the -3? There’s a common factor of (2𝑥+1) =(2𝑥+3)(𝑥−1) * Not official mathematical terminology.

More Examples 2 𝑥 2 +11𝑥+12= 𝑥+4 2𝑥+3 6 𝑥 2 −7𝑥−3=(2𝑥−3)(3𝑥+1) ? 2 𝑥 2 +11𝑥+12= 𝑥+4 2𝑥+3 6 𝑥 2 −7𝑥−3=(2𝑥−3)(3𝑥+1) 2 𝑥 2 −5𝑥𝑦+3 𝑦 2 = 𝑥−𝑦 2𝑥−3𝑦 6 𝑥 2 −3𝑥−3=3(𝑥−1)(2𝑥+1) ? ? ?

Exercise 4 ? ? ? ? ? ? ? ? ? ? ? ? ? 2𝑥2+3𝑥+1=(2𝑥+1)(𝑥+1) 3𝑥2+8𝑥+4=(3𝑥+2)(𝑥+2) 2𝑥2−3𝑥−9=(2𝑥+3)(𝑥−3) 4𝑥2−9𝑥+2=(4𝑥−1)(𝑥−2) 2𝑥2+𝑥−15=(2𝑥−5)(𝑥+3) 2𝑥2−3𝑥−2=(2𝑥+1)(𝑥−2) 3𝑥2+4𝑥−4=(3𝑥−2)(𝑥+2) 6 𝑥 2 −13𝑥+6= 3𝑥−2 2𝑥−3 15 𝑦 2 −13𝑦−20= 5𝑦+4 3𝑦−5 12 𝑥 2 −𝑥−1= 4𝑥+1 3𝑥−1 25 𝑦 2 −20𝑦+4= 5𝑦−2 2 ? 2 ? 3 ? 4 ? 5 ? ‘Commando’ starts to become difficult from this question onwards because the coefficient of 𝑥 2 is not prime. 6 ? 7 ? 8 ? 9 ? 10 ? 11 ? N1 4 𝑥 3 +12 𝑥 2 +9𝑥=𝑥 2𝑥+3 2 𝑎 2 𝑥 2 −2𝑎𝑏𝑥+ 𝑏 2 = 𝑎𝑥−𝑏 2 ? N2 ?

RECAP :: Six different types of factorisation 1. Factoring out a single term 2. 𝒙 𝟐 +𝒃𝒙+𝒄 𝑥 2 −4𝑥=𝒙 𝒙−𝟒 ? 𝑥 2 +7𝑥−30= 𝒙+𝟏𝟎 𝒙−𝟑 ?   3. Difference of two squares 4. 𝒂 𝒙 𝟐 +𝒃𝒙+𝒄 9−16 𝑦 2 = 𝟑+𝟒𝒚 𝟑−𝟒𝒚 ? 2 𝑥 2 +𝑥−3=(2𝑥+3)(𝑥−1) Strategy: either split the middle term, or ‘go commando’.   5. Pairwise 6. Intelligent Guesswork 𝑥 3 +2 𝑥 2 −𝑥−2= 𝑥 2 𝑥+2 −1 𝑥+2 = 𝑥 2 −1 𝑥+2 =… 𝑥 2 + 𝑦 2 +2𝑥𝑦+𝑥+𝑦 = 𝒙+𝒚+𝟏 𝒙+𝒚

RECAP :: 𝑎 𝑥 2 +𝑏𝑥+𝑐 ? ? 3 𝑥 2 +10𝑥−8 =(3𝑥−2)(𝑥+4) Method A: Guessing the brackets Method B: Splitting the middle term 3 𝑥 2 +10𝑥−8 =(3𝑥−2)(𝑥+4) 3 𝑥 2 +10𝑥−8 =3 𝑥 2 +12𝑥−2𝑥−8 =3𝑥 𝑥+4 −2 𝑥+4 =(3𝑥−2)(𝑥+4) ? ? Both of these methods can be extended to more general expressions. This method of ‘intelligent guessing’ can be extended to non-quadratics. After we split the middle term, we looked at the expression in two pairs and factorised. I call more general usage of this ‘pairwise factorisation’.

𝑥 2 +𝑎𝑥+𝑏𝑥+𝑎𝑏 =( )( ) 𝑥 + 𝑎 𝑥 + 𝑏 𝑎𝑏−𝑎+𝑏−1 = 𝑎+1 𝑏−1 TYPE 5: Intelligent Guessing Just think what brackets would expand to give you expression. Look at each term one by one. 𝑥 2 +𝑎𝑥+𝑏𝑥+𝑎𝑏 It works! =( )( ) 𝑥 + 𝑎 𝑥 + 𝑏 This factorisation will become particularly important when we cover something called ‘Diophantine Equations’. 𝑎𝑏−𝑎+𝑏−1 = 𝑎+1 𝑏−1 ?

Test Your Understanding 𝑥𝑦+3𝑥−2𝑦−6 = 𝒙−𝟐 𝒚+𝟑 𝑎𝑐−𝑏𝑐−𝑏+𝑎 = 𝒂−𝒃 𝒄+𝟏 𝑎 2 𝑏−𝑎+𝑎𝑏−1 = 𝒂𝒃−𝟏 𝒂+𝟏 𝑥 2 + 𝑦 2 +2𝑥𝑦+𝑥𝑧+𝑦𝑧 = 𝒙+𝒚 𝒙+𝒚+𝒛 ? 1 ? 2 ? 3 N ? Tip: Notice that there’s an ‘algebraic symmetry’ in 𝑥 and 𝑦, as 𝑥 and 𝑦 could be swapped without changing the expression. But there’s an asymmetry in 𝑧. This gives hints about the factorisation, as the same symmetry must be seen. Tip: The 2𝑥𝑦 arose because of collecting like terms in the expansion. It might therefore be easier to first think how we get the ‘easier’ terms like the 𝑦 2 , 𝑥𝑧, 𝑦𝑧 (where the coefficient of the term is 1) when we try to fill in the brackets.

TYPE 6: Pairwise Factorisation We saw earlier with splitting the middle term that we can factorise different parts of the expression separately and hope that a common term emerges. 𝑥 2 +𝑎𝑥+𝑏𝑥+𝑎𝑏= =𝒙 𝒙+𝒂 +𝒃(𝒙+𝒂) = 𝒙+𝒂 𝒙+𝒃 ? ? 𝑥 3 −2 𝑥 2 −𝑥+2 = 𝒙 𝟐 𝒙−𝟐 −𝟏 𝒙−𝟐 = 𝒙 𝟐 −𝟏 𝒙−𝟐 =(𝒙+𝟏)(𝒙−𝟏)(𝒙−𝟐) ? ? ? 𝑥 2 − 𝑦 2 +4𝑥+4𝑦 = 𝒙+𝒚 𝒙−𝒚 +𝟒 𝒙+𝒚 = 𝒙+𝒚 𝒙−𝒚+𝟒 ? ?

Test Your Understanding 1 𝑥 2 −𝑥𝑦+2𝑥−2𝑦 =𝒙 𝒙−𝒚 +𝟐 𝒙−𝒚 = 𝒙−𝒚 𝒙+𝟐 ? 2 𝑎𝑏+𝑎+𝑏+1 =𝒂 𝒃+𝟏 +𝟏 𝒃+𝟏 =(𝒂+𝟏)(𝒃+𝟏) ? 3 𝑥 3 −3 𝑥 2 −4𝑥+12 = 𝒙 𝟐 𝒙−𝟑 −𝟒 𝒙−𝟑 = 𝒙 𝟐 −𝟒 𝒙−𝟑 = 𝒙+𝟐 𝒙−𝟐 𝒙−𝟑 ? N 𝑎 2 + 𝑏 2 +2𝑎𝑏+𝑎𝑐+𝑏𝑐 = 𝒂+𝒃 𝟐 +𝒄 𝒂+𝒃 =(𝒂+𝒃)(𝒂+𝒃+𝒄) Can you split the terms into two blocks, where in each block you can factorise? ?

Challenge Wall! ? 4 3 ? 2 ? ? 1 𝑥 2 + 𝑦 2 +2𝑥𝑦−1=(𝒙+𝒚+𝟏)(𝒙+𝒚−𝟏) Instructions: Divide your paper into four. Try and get as far up the wall as possible, then hold up your answers for me to check. Use any method of factorisation. 3 4 Warning: Pairwise factorisation doesn’t always work. You sometimes have to resort to ‘intelligent guessing’. ? 4 𝑥 2 + 𝑦 2 +2𝑥𝑦−1=(𝒙+𝒚+𝟏)(𝒙+𝒚−𝟏) 3 𝑥 3 +2 𝑥 2 −9𝑥−18=(𝒙+𝟑)(𝒙−𝟑)(𝒙+𝟐) ? 2 𝑥 𝑦 2 +3 𝑦 2 +𝑥+3=(𝒙+𝟑)( 𝒚 𝟐 +𝟏) ? 1 𝑥𝑦−𝑥−𝑦+1=(𝒙−𝟏)(𝒚−𝟏) ?

Exercise 5 Factorise the following using either ‘pairwise factorisation’ or ‘intelligent guessing’. 𝑎𝑏−2𝑎+4𝑏−8= 𝒂+𝟒 𝒃−𝟐 𝑎𝑏+𝑏𝑐−3𝑎−3𝑐= 𝒃−𝟑 𝒂+𝒄 𝑎𝑐+𝑎𝑑+𝑏𝑐+𝑏𝑑= 𝒂+𝒃 𝒄+𝒅 𝑦 2 +𝑐𝑦+𝑑𝑦+𝑐𝑑= 𝒚+𝒄 𝒚+𝒅 𝑥 2 𝑦− 𝑥 2 +2𝑦−2=( 𝒙 𝟐 +𝟐)(𝒚−𝟏) 𝑥 3 + 𝑥 2 +𝑥+1= 𝒙 𝟐 +𝟏 𝒙+𝟏 𝑎 3 − 𝑎 2 +𝑎−1= 𝒂 𝟐 +𝟏 𝒂−𝟏 𝑥 3 + 𝑥 2 −𝑥−1= 𝒙+𝟏 𝟐 𝒙−𝟏 𝑥 3 + 𝑥 2 +𝑥 𝑦 2 + 𝑦 2 = 𝒙 𝟐 + 𝒚 𝟐 𝒙+𝟏 𝑥 3 −2 𝑥 2 +3𝑥−6= 𝒙 𝟐 +𝟑 𝒙−𝟐 𝑥 2 +2𝑥𝑦+ 𝑦 2 = 𝒙+𝒚 𝟐 𝑎 2 𝑥 2 +4𝑎𝑏𝑥+4 𝑏 2 = 𝒂𝒙+𝟐𝒃 𝟐 𝑥 2 𝑦 2 − 𝑥 2 − 𝑦 2 +1= 𝒙+𝟏 𝒙−𝟏 𝒚+𝟏 𝒚−𝟏 𝑎 2 − 𝑏 2 −3𝑎+3𝑏= 𝒂−𝒃 𝒂+𝒃−𝟑 ? 1 2 ? 3 ? 4 ? 5 ? 6 ? N1 𝑥 4 + 𝑦 4 −2 𝑥 2 𝑦 2 = 𝒙+𝒚 𝟐 𝒙−𝒚 𝟐 𝑥 2 − 𝑦 2 +2𝑥+1= 𝒙+𝒚+𝟏 𝒙−𝒚+𝟏 𝑥𝑦𝑧−𝑥𝑦−𝑥𝑧+𝑦𝑧+𝑥−𝑦−𝑧+1= 𝒙+𝟏 𝒚−𝟏 𝒛−𝟏 𝑎 2 + 𝑏 2 + 𝑐 2 +2𝑎𝑏+2𝑏𝑐+2𝑐𝑎= 𝒂+𝒃+𝒄 𝟐 𝑥 4 +6 𝑥 3 +11 𝑥 2 +6𝑥+1 = 𝒙 𝟐 +𝟑𝒙+𝟏 𝟐 ? 7 ? N2 8 ? ? 9 ? N3 10 ? ? 11 ? N4 12 ? ? 13 N5 ? ? 14 ?

Summary For the following expressions, identify which of the following factorisation techniques that we use, out of: (it may be multiple!) 1 Factorising out single term: 𝑥 2 +2𝑥𝑦→𝑥(𝑥+2𝑦) 2 Simple quadratic factorisation: 𝑥 2 +3𝑥+2→ 𝑥+2 𝑥+1 3 Difference Of Two Squares: 9 𝑥 2 −4 𝑦 2 = 3𝑥+2𝑦 3𝑥−2𝑦 4 Commando/Splitting Middle Term: 2𝑥 2 −3𝑥+2→ 2𝑥+1 𝑥−2 5 Pairwise: 𝑥 4 − 𝑥 3 +𝑥−1→ 𝑥 3 +1 𝑥−1 6 Intelligent Guesswork: 𝑥𝑦+𝑥+𝑦+1→ 𝑥+1 𝑦+1 2 𝑥 2 𝑦+4𝑥 𝑦 2 (1) 1− 𝑥 2 (3) 𝑥 3 −𝑥 (1), (3) 𝑥 2 −𝑥−2 (2) 3 𝑦 2 −10𝑦−8 (4) 𝑥 4 +2 𝑥 2 +1 (2), (6) 𝑦 3 − 𝑦 2 −𝑦+1 (5) 𝑥 4 −2 𝑥 3 −8 𝑥 2 (1), (2) 𝑥 𝑦 2 −𝑥− 𝑦 2 +1 (5) or (6) 27−3 𝑥 2 (1), (3) ? ? ? ? ? ? ? ? ? ?

𝑥 𝑥+2 −3 𝑥+2 → (𝑥+2)(𝑥−3) 𝑥 𝑥+1 2 +2 𝑥+1 → 𝑥 2 +𝑥+2 𝑥+1 Factorising out an expression It’s fine to factorise out an entire expression: 𝑥 𝑥+2 −3 𝑥+2 → (𝑥+2)(𝑥−3) ? 𝑥 𝑥+1 2 +2 𝑥+1 → 𝑥 2 +𝑥+2 𝑥+1 ? 𝑎 2𝑐+1 +𝑏 2𝑐+1 → (𝑎+𝑏)(2𝑐+1) ? 2 2𝑥−3 2 +𝑥 2𝑥−3 → (5𝑥−6)(2𝑥−3) ?