Prerequisite Knowledge Check 1 1 z Simplify: 1 2 z Show all solutions Factorise: a 2𝑥 + 8 𝟐 𝒙+𝟒 −𝟒𝒙 b 5𝑦 − 10 𝟓 𝒚−𝟐 −9𝑎 − 𝑎 −𝟏𝟎𝒂 c 𝒙 𝒙−𝟗 d 5𝑥 × 2𝑥 ? 𝑥 2 − 9𝑥 𝟏𝟎𝒙𝟐 d −3𝑥 − 15 −𝟑(𝒙 + 𝟓) e −8 × 𝑦 −𝟖𝒚 −𝑏 × −11 𝟏𝟏𝒃 a 4𝑦 − 2𝑦 𝟐𝒚 b −7𝑥 + 3𝑥 c f 1 3 z ? Factorise fully: a 6𝑥 − 12 𝟔(𝒙 − 𝟐) b 2𝑥 2 + 8𝑥 𝟐𝒙(𝒙 + 𝟒) c 15𝑦 2 − 5y 𝟓𝒚(𝟑𝒚 − 𝟏) d −18𝑥 − 12𝑥 2 −𝟔𝒙(𝟑 + 𝟐𝒙) ? Recap : Factorising Quadratics We have learned how to factorise a quadratic expression such as 𝑥 2 + 5𝑥 + 6 We can think of factorising as arranging our terms into a rectangle 𝑥 1 1 1 𝑥 2 + 5𝑥 + 6 = (𝑥 + 3) × (𝑥 + 2) = (𝑥 + 3)(𝑥 + 2) 𝑥 1 1 𝑥2 𝑥 𝑥 𝑥 𝑥 𝑥 1 1 1 1 1 1 So, (𝒙 + 𝟐) and (𝒙 + 𝟑) are factors of 𝒙𝟐 + 𝟓𝒙 + 𝟔 Recap : Factorising Quadratics Factorise 𝑥 2 + 5𝑥 + 6 We can use algebra tiles and grid representations 𝑥 1 1 1 𝑥 𝑥2 𝑥 𝑥 𝑥 1 1 𝑥 𝑥 1 1 1 1 1 1 × 𝒙 +𝟑 𝒙 𝑥 2 +3𝑥 +𝟐 +2𝑥 +6 We need a pair of numbers that multiply to make +6 1 and 6 2 and 3 = (𝒙 + 𝟑 )(𝒙 + 𝟐) −1 and 6 −2 and −3 They also need to add to make +5 Quickfire Questions a c Factorise b Factorise 𝑥 2 + 7𝑥 + 10 𝑥 2 + 2𝑥 − 15 (𝒙 + 𝟓)(𝒙 ? + 𝟐) (𝒙 + 𝟓)(𝒙 ? − 𝟑) Factorise d Factorise 𝑥2 − 𝑥 − 6 𝑥 2 − 10𝑥 + 24 (𝒙 − 𝟑)(𝒙 ? + 𝟐) (𝒙 − 𝟔)(𝒙 ? − 𝟒) The Big Idea : Factorising Non-Monic Quadratics How might we factorise 2𝑥 2 + 3𝑥 + 1? 𝑥 1 2 𝑥 𝑥 1 𝑥 𝑥 𝑥2 𝑥 Can we arrange these into a rectangle? 𝑥 𝑥 2 𝑥 2 𝑥 1 𝑥 𝑥 1 2𝑥 2 + 3𝑥 + 1 = (2𝑥 + 1)(𝑥 + 1) The factors of 2𝑥 2 + 3𝑥 + 1 are (2𝑥 + 1) and (𝑥 + 1) Test Your Understanding Factorise 3𝑥 2 + 4𝑥 + 1 𝑥2 𝑥 𝑥2 2 1 𝑥 𝑥 𝑥 𝑥 𝑥 𝑥 𝑥 1 𝑥 𝑥2 𝑥2 𝑥2 𝑥 1 𝑥 𝑥 𝑥 1 3𝑥 2 + 4𝑥 + 1 ? = (3𝑥 + 1)(𝑥 + 1) There is only one way to arrange the three 𝑥 2 tiles and the one 1 tile relative to each other. The Big Idea : Factorising Non-Monic Quadratics Factorise 3𝑥 2 + 5𝑥 + 2 𝑥2 𝑥2 𝑥 2 1 𝑥 𝑥 Can we arrange these into a rectangle? How does this differ from the previous examples? 1 𝑥 𝑥 𝑥 There are two different ways we can arrange the three 𝑥 2 tiles and the two 1 tiles relative to each other Who is Correct? Factorise 3𝑥 2 + 5𝑥 + 2 Viktor and Imani are using algebra tiles to help them factorise. Who has done the correct first step? How do you know? Viktor 𝑥2 𝑥2 𝑥2 𝑥 𝑥 𝑥 𝑥 𝑥 1 1 2 𝑥 2 𝑥 2 𝑥 𝑥 𝑥 𝑥 𝑥 𝑥 𝑥 1 1 𝑥 Imani Viktor is correct as his arrangement will require 5 𝑥 tiles to form a rectangle, whereas Imani’s arrangement will require 7 𝑥 tiles. The correct factorisation is (3𝑥 + 2)(𝑥 + 1) ? Test Your Understanding Factorise 2𝑥 2 + 7𝑥 + 3 𝑥 2 𝑥 1 𝑥 𝑥2 𝑥2 𝑥 1 1 1 𝑥 𝑥 𝑥 𝑥 𝑥 𝑥 1 1 1 𝑥2 1 1 1 𝑥 𝑥 𝑥 𝑥 𝑥 𝑥 𝑥 𝑥 2𝑥 2 + 7𝑥 + 3 ? = (2𝑥 + 1)(𝑥 + 3) There are two arrangements, but only this one uses seven 𝑥 tiles. Different Representations Factorise 2𝑥 2 + 7𝑥 + 3 As well as representing quadratic factorisation with algebra tiles, we can also use the grid model 𝑥 1 1 1 𝑥 𝑥 2 2 𝑥 𝑥 𝑥 𝑥 𝑥 𝑥 𝑥 𝑥 1 𝑥 1 1 1 × 𝟐𝒙 +𝟏 𝒙 2𝑥 2 +𝑥 +𝟑 +6𝑥 +3 2𝑥 2 + 7𝑥 + 3 = ( 2𝑥 + 1 )( 𝑥 + 3 ) The Big Idea : Factorising Non-Monic Quadratics Factorise 5𝑥 2 + 8𝑥 + 3 We know that to give 5𝑥 2 we need 5𝑥 × 𝑥 × 𝟓𝒙 𝒙 5𝑥 2 +𝑥 +𝟏 +𝟑 +15𝑥 +3 We also know that to give +3 we need either + 1 × +3 or +3 × +1 We can complete the grids to see which one gives us +8𝑥 5𝑥 2 + 𝑥 + 15𝑥 + 3 = 5𝑥 2 + 16𝑥 + 3 × 𝟓𝒙 𝒙 5𝑥 2 +3𝑥 +𝟑 +𝟏 +5𝑥 +3 5𝑥 2 + 3𝑥 + 5𝑥 + 3 = 5𝑥 2 + 8𝑥 + 3 The correct factorisation is (5𝑥 + 3)(𝑥 + 1) Test Your Understanding Which grid represents the correct factorisation of 7𝑥 2 + 22𝑥 + 3? a × 𝟕𝒙 +𝟏 𝒙 7𝑥 2 +𝑥 +𝟑 +21𝑥 +3 b × 𝟕𝒙 +𝟑 𝒙 7𝑥 2 +3𝑥 +𝟏 +7𝑥 +3 a is the grid which ? gives +22𝑥 What is the correct factorisation of 7𝑥 2 + 22𝑥 + 3? (7𝑥 + 1)(𝑥 ? + 3) The Big Idea : Factorising Non-Monic Quadratics Factorise 5𝑥 2 + 14𝑥 − 3 How does this differ from previous examples? How many possible grids would there be? Can you draw them all? × 𝟓𝒙 −𝟏 × 𝟓𝒙 +𝟏 𝒙 5𝑥 2 −𝑥 𝒙 5𝑥 2 +𝑥 +𝟑 +15𝑥 −3 × 𝟓𝒙 𝒙 5𝑥 2 −𝟏 +𝟑 −𝟑 −15𝑥 −3 −𝟑 × 𝟓𝒙 +3𝑥 𝒙 5𝑥 2 −5𝑥 −3 +𝟏 +5𝑥 −3 −3𝑥 We know that to give 5𝑥 2 we need 5𝑥 × 𝑥 We also know that to give −3 there are four arrangements: −1 and +3 +3 and −1 +1 and −3 −3 and +1 Only one of these arrangements gives +14𝑥 The correct factorisation is (𝟓𝒙 − 𝟏)(𝒙 + 𝟑) Moving to an Algebraic Representation… Factorise 3𝑥 2 + 5𝑥 − 2 We know that to give 3𝑥 2 we need 3𝑥 × 𝑥 (3𝑥 )(𝑥 ) We need a pair of numbers that multiply to make −2. +1 and −2 −2 and +1 −1 and +2 +2 and −1 This method is commonly known as INSPECTION This gives four options and only one of these options gives +5𝑥 (3𝑥 + 1)(𝑥 − 2) −𝟓𝒙 (3𝑥 − 2)(𝑥 + 1) +𝒙 (3𝑥 − 1)(𝑥 + 2) +𝟓𝒙 (3𝑥 + 2)(𝑥 − 1) −𝒙 The correct factorisation is (𝟑𝒙 − 𝟏)(𝒙 + 𝟐) Test Your Understanding Factorise 2𝑥 2 + 𝑥 − 3 We know that to give 2𝑥 2 we need 2𝑥 × 𝑥 (2𝑥 )(𝑥 ) We need a pair of numbers that multiply to make −3. +1 and −3 −3 and +1 −1 and +3 +3 and −1 This gives four options and only one of these options gives +𝑥 (2𝑥 + 1)(𝑥 − 3) −𝟓𝒙 (2𝑥 − 3)(𝑥 + 1) −𝒙 (2𝑥 − 1)(𝑥 + 3) +𝟓𝒙 (2𝑥 + 3)(𝑥 − 1) +𝒙 The correct factorisation is ? − 𝟏) (𝟐𝒙 + 𝟑)(𝒙 Test Your Understanding a c Factorise b Factorise 3𝑥 2 + 16𝑥 + 5 2𝑥 2 − 23𝑥 + 11 (𝟑𝒙 + 𝟏)(𝒙 ? + 𝟓) (𝟐𝒙 − 𝟏)(𝒙 ? − 𝟏𝟏) Factorise d Factorise 5𝑥 2 − 4𝑥 − 1 2𝑥 2 + 5𝑥 − 7 (𝟓𝒙 + 𝟏)(𝒙 ? − 𝟏) (𝟐𝒙 + 𝟕)(𝒙 ? − 𝟏) The Big Idea : Factorising Non-Monic Quadratics Factorise 6𝑥 2 + 11𝑥 + 4 𝑥2 𝑥2 1 1 𝑥 𝑥2 𝑥 1 𝑥 𝑥 𝑥 2 𝑥 1 2 𝑥 𝑥 𝑥 2 𝑥 𝑥 𝑥 𝑥 𝑥 Can we arrange these into a rectangle? How does this differ from previous examples? How many possible arrangements of the six 𝑥 2 and four 1 tiles are there? The Big Idea : Factorising Non-Monic Quadratics Factorise 6𝑥 2 + 11𝑥 + 4 𝑥 2 𝑥 2 𝑥 𝑥 2 2 𝑥2 𝑥2 𝑥2 𝑥2 𝑥 2 1 1 1 1 𝑥 2 𝑥 1 1 1 1 2 𝑥 2 There are two possible arrangements of the six 𝑥 2 tiles There are also two possible arrangements of the four 1 tiles Because the number of 𝑥 2 tiles and 1 tiles are composite (non-prime), there are now multiple possible arrangements. Let me introduce you to a more efficient method… Mrs Clark The Big Idea : Splitting the Middle Term Factorise 2𝑥 2 + 7𝑥 + 3 𝑥 𝑥 𝑥 𝑥 2 𝑥 2 1 1 1 𝑥 𝑥 𝑥 𝑥 𝑥 𝑥 1 × 𝟐𝒙 +𝟏 𝑥 𝒙 2𝑥 2 +𝑥 1 1 1 +𝟑 +6𝑥 +3 When we factorised by inspection, we tried different arrangements of 𝑥 2 and 1 tiles until we found the correct number of 𝑥 tiles The 𝑥 tiles were split into two groups… 2𝑥 2 + 7𝑥 + 3 = 2𝑥 2 + 𝑥 + 6𝑥 + 3 Writing the 𝑥 term as two separate terms is called ‘splitting the middle term’ The Big Idea : Splitting the Middle Term Factorise 2𝑥 2 + 7𝑥 + 3 𝑥 𝑥 𝑥 𝑥 2 𝑥 2 1 1 1 𝑥 𝑥 𝑥 𝑥 𝑥 𝑥 We have split our expression into two pairs of terms, and we can factorise each pair separately. A common bracketed factor is revealed, leading to our fully factorised expression. 1 × 𝟐𝒙 +𝟏 𝑥 𝒙 2𝑥 2 +𝑥 1 1 1 +𝟑 +6𝑥 +3 2𝑥 2 + 𝑥 + 6𝑥 + 3 = 𝑥(2𝑥 + 1) + 3(2𝑥 + 1) = (𝟐𝒙 + 𝟏)(𝒙 + 𝟑) Example Test Your Understanding The grid shows how the middle term has been split for 3𝑥 2 + 7𝑥 + 2 Complete the factorisation. × The grid shows how the middle term has been split for 5𝑎2 + 8𝑎 + 3. Complete the factorisation. × 3𝑥 2 +6𝑥 5𝑎2 +5𝑎 +𝑥 +3𝑎 +2 +3 3𝑥 2 + 6𝑥 + 𝑥 + 2 5𝑎2 + 5𝑎?+ 3𝑎 + 3 = 3𝑥(𝑥 + 2) + 1(𝑥 + 2) = 5𝑎(𝑎 + 1) ? + 3(𝑎 + 1) = (𝒙 + 𝟐)(𝟑𝒙 + 𝟏) = (𝒂 + 𝟏)(𝟓𝒂 ? + 𝟑) The Big Idea : Splitting the Middle Term In order to know how to split the middle term, let’s spot some patterns... Consider these three grids which show correct splitting of the middle term. What do you notice about the diagonals of each of the grids? Factorise 2𝑥 2 + 7𝑥 + 3 Factorise 7𝑥 2 + 22𝑥 + 3 Factorise 5𝑥 2 + 7𝑥 + 2 × × × 2𝑥 2 +𝑥 7𝑥 2 +6𝑥 +3 +21𝑥 +3 2𝑥 2 × 3 = 6𝑥 × 𝑥 +𝑥 7𝑥 2 × 3 = 𝑥 × 21𝑥 The product of the diagonals is always equal 5𝑥 2 +5𝑥 +2𝑥 +2 5𝑥 2 × 2 = 5𝑥 × 2𝑥 The Big Idea : Splitting the Middle Term Factorise 2𝑥 2 + 11𝑥 + 5 × 𝟐𝒙 𝒙 2𝑥 2 +𝑥 +𝟏 +𝟓 +10𝑥 +5 The product of the diagonals must be equal, so the two 𝑥 terms must multiply to give +10 +1 and +10 −1 and −10 +2 and +5 −2 and −5 They must also add to give +11 2𝑥 2 + 𝑥 + 10𝑥 + 5 = 𝑥(2𝑥 + 1) + 5(2𝑥 + 1) = (𝟐𝒙 + 𝟏)(𝒙 + 𝟓) Example Test Your Understanding Factorise 5𝑥 2 + 36𝑥 + 7 × Factorise 11𝑥 2 + 14𝑥 + 3 × 5𝑥 2 +𝑥 11𝑥 2 +3𝑥 ? +35𝑥 +7 +11𝑥 ? +3 Our two 𝑥 terms must multiply to give +35 They must also add to give +36 +1 and +35 −1 and −35 +5 and +7 −5 and −7 5𝑥 2 + 𝑥 + 35𝑥 + 7 = 𝑥(5𝑥 + 1) + 7(5𝑥 + 1) = (𝟓𝒙 + 𝟏)(𝒙 + 𝟕) Our two 𝑥 terms must multiply to give +33 and must also add to give +14 +1 and +33 −1 and −33 ? +3 and +11 −3 and −11 11𝑥 2 + 3𝑥 ? + 11𝑥 + 3 = 𝑥(11𝑥 + 3)?+ 1(11𝑥 + 3) = (𝟏𝟏𝒙 +?𝟑)(𝒙 + 𝟏) Who is Correct? Basma and Nathan are both attempting to factorise by splitting the middle term. They each complete a grid, but they are different. Whose grid is correct? Factorise 4𝑥 2 + 8𝑥 + 3 × Basma × 4𝑥 2 +2𝑥 4𝑥 2 +6𝑥 +6𝑥 +2𝑥 +3 Both grids are correct. When we split the middle term, it does not matter which way around we put the 𝑥 terms. Both Basma’s and Nathan’s grids lead to a correct factorisation of (2𝑥 + 3)(2𝑥 + 1) ? +3 Nathan Example Factorise 6𝑥 2 + 11𝑥 + 4 × 𝟐𝒙 𝟑𝒙 6𝑥 2 +𝟏 +3𝑥 +𝟒 +8𝑥 +4 We need a pair of numbers that multiply to give +24 +1 and +24 +2 and +12 +3 and +8 +4 and +6 They must also add to give + 11 6𝑥 2 + 3𝑥 + 8𝑥 + 4 = 3𝑥(2𝑥 + 1) + 4(2𝑥 + 1) = (𝟐𝒙 + 𝟏)(𝟑𝒙 + 𝟒) Now we can apply our efficient method to the original problem… Mrs Clark TOP TIP! When all terms are positive, we only need consider positive factors. Test Your Understanding Factorise 5𝑥 2 + 22𝑥 + 8 × 𝟓𝒙 𝒙 5𝑥 2 +𝟐 +2𝑥 +𝟒 +20𝑥 +8 We need a pair of numbers that multiply to give +40 +1 and +40 +2 and +20 +4 and +10 +5 and +8 They must also add to give +22 5𝑥 2 + 2𝑥 + 20𝑥 + 8 = 𝑥(5𝑥 + 2) + 4(5𝑥 + 2) = (𝟓𝒙 + 𝟐)(𝒙 + 𝟒) ? Moving to an Algebraic Representation… Remember that the general form of a quadratic expression is 𝑎𝑥 2 + 𝑏𝑥 + 𝑐. In this form, we need factors of 𝑎𝑐 that sum to 𝑐. So, sometimes this is known as the ‘𝑎𝑐 method’. Factorise 6𝑥 2 + 5𝑥 − 4 We need a pair of numbers that multiply to give 𝑎 × 𝑐 = −24 −1 and +24 +1 and −24 −2 and +12 +2 and −12 −3 and +8 +3 and −8 −4 and +6 +4 and −6 They must also add to give +5 Split the middle term into − 3𝑥 and +8𝑥 6𝑥 2 − 3𝑥 + 8𝑥 − 4 = 3𝑥(2𝑥 − 1) + 4(2𝑥 − 1) = (𝟐𝒙 − 𝟏)(𝟑𝒙 + 𝟒) Example Factorise 4𝑥 2 − 9𝑥 − 9 We need a pair of numbers that multiply to give 𝑎 × 𝑐 = −36 −1 and +36 +1 and −36 −2 and +18 +2 and −18 −3 and +12 +3 and −12 −4 and +9 +4 and −9 +6 and −6 They must also add to give −9 Split the middle term into + 3𝑥 and −12𝑥 4𝑥 2 + 3𝑥 − 12𝑥 − 9 = 𝑥 4𝑥 + 3 − 3(4𝑥 + 3) = (𝟒𝒙 + 𝟑)(𝒙 − 𝟑) Test Your Understanding Factorise 8𝑥 2 − 18𝑥 − 5 We need a pair of numbers that multiply to give 𝑎 × 𝑐 = −40 −1 and +40 +1 and −40 −2 and +20 +2 and −20 −4 and +10 +4 and −10 −5 and +8 +5 and −8 They must also add to give −18 Split the middle term into + 2𝑥 and −20𝑥 8𝑥 2 + 2𝑥 − 20𝑥 − 5 = 2𝑥 4𝑥 + 1 − 5(4𝑥 + 1) = (𝟒𝒙 + 𝟏)(𝟐𝒙 − 𝟓) ? Test Your Understanding a c Factorise b Factorise 4𝑥 2 + 5𝑥 − 6 3𝑥 2 + 11𝑥 − 20 (𝟒𝒙 − 𝟑)(𝒙 ? + 𝟐) (𝟑𝒙 − 𝟒)(𝒙 ? + 𝟓) Factorise d Factorise 4𝑥 2 − 23𝑥 + 15 9𝑥 2 + 6𝑥 − 8 (𝟒𝒙 − 𝟑)(𝒙 ? − 𝟓) (𝟑𝒙 + 𝟒)(𝟑𝒙 ? − 𝟐) Spot the Mistakes Factorise 8𝑥 2 − 2𝑥 − 3 Logan has attempted to factorise by splitting the middle term. Can you spot any mistakes he has made and provide him with a correct answer? 𝑎 × 𝑐 = −3 𝑏 = −2 × to give −3 and + to give −2 :𝑎 × 𝑐 is actually − 24, so the −3 and1 middle term Split the middle term: 8𝑥 2 − 3𝑥 + 𝑥 − 3 = 8𝑥 𝑥 − 3 + 1(𝑥 − 3) Logan = 𝒙 − 𝟑 (𝟖𝒙 + 𝟏) splits into −6𝑥?+ 4𝑥. The correct factorisation is (4𝑥 − 3)(2𝑥 + 1) Spot the Mistakes Factorise 3𝑥 2 + 10𝑥 − 8 Hannah has attempted to factorise by splitting the middle term. Can you spot any mistakes they have made and provide them with a correct answer? 𝑎 × 𝑐 = −24 𝑏 = +10 × to give −24 and + to give +10: 𝟏𝟐 and −𝟐 Split the middle term: Hannah 3𝑥 2 + 12𝑥 − 2𝑥 − 8 = 3𝑥 𝑥 + 4 + 2(−𝑥 − 4) = (𝒙 + 𝟒)(𝟑𝒙 + 𝟐) When factorising each pair of terms, the bracketed factor must be the same. ? The second pair of terms factorise to give −2(𝑥 + 4), giving an answer of (𝑥 + 4)(3𝑥 − 2) Deeper Understanding Consider the general expansion of (𝑒𝑥 + 𝑓)(𝑔𝑥 + ℎ) × 𝒈𝒙 𝒆𝒙 +𝒇 𝑒𝑔𝑥 2 +𝑓𝑔𝑥 +𝒉 +𝑒ℎ𝑥 +𝑓ℎ Look at the diagonals: 𝑒𝑔𝑥 2 × 𝑓ℎ = 𝑒𝑓𝑔ℎ𝑥 2 𝑒ℎ𝑥 × 𝑓𝑔𝑥 = 𝑒𝑓𝑔ℎ𝑥 2 The product of the diagonals is always equal. We can use the product of the 𝑥 2 coefficient and the constant term to tell us what the product of the coefficients of the 𝑥 terms needs to be.. This is why, for 𝑎𝑥 2 + 𝑏𝑥 + 𝑐, we find a pair of numbers that multiply to give 𝑎 × 𝑐 and add to give 𝑏 Exercise 1 (Available as a separate worksheet) Show all solutions 1 Factorise by inspection: 2 Complete the factorisations: a 5𝑥 2 + 6𝑥 + 1 (𝟓𝒙 + 𝟏)(𝒙 + 𝟏) a 4𝑥 2 + 16𝑥 + 15 = 4𝑥 2 + 10𝑥 + 6𝑥 + 15 b 2𝑥 2 + 5𝑥 + 3 (𝟐𝒙 + 𝟑)(𝒙 + 𝟏) c 3𝑥 − 22𝑥 + 7 (𝟑𝒙 − 𝟏)(𝒙 ? − 𝟕) d 2𝑥 2 − 9𝑥 − 5 (𝟐𝒙 + 𝟏)(𝒙 − 𝟓) e 4𝑥 2 + 7𝑥 − 2 (𝟒𝒙 − 𝟏)(𝒙 + 𝟐) 2 = 2𝑥 2𝑥 + 5 + 3(2𝑥 + 5) = 2𝑥 + 5 ( 𝟐𝒙?+ 𝟑 ) b 8𝑥 2 − 10𝑥 − 3 = 8𝑥 2 + 2𝑥 − 12𝑥 − 3 = 2𝑥 4𝑥 + 1 − ? 𝟑 ( 𝟒𝒙 ?− 𝟏 ) = c 𝟒𝒙 ?+ 𝟏 ( 𝟐𝒙?− 𝟑 ) 7𝑥 2 − 23𝑥 − 20 = 7𝑥 2 − 28𝑥 + 5𝑥 − 20 = ?𝟕 3 Factorise by splitting the middle term: a 4𝑥 2 + 12𝑥 + 5 (𝟐𝒙 + 𝟓)(𝟐𝒙 + 𝟏) b 6𝑥 2 + 11𝑥 − 10 (𝟑𝒙 − 𝟐)(𝟐𝒙 + 𝟓) ? c 9𝑥 − 41𝑥 + 20 (𝟗𝒙 − 𝟓)(𝒙 − 𝟒) d 8𝑥 2 − 6𝑥 − 9 (𝟐𝒙 − 𝟑)(𝟒𝒙 + 𝟑) 2 = 𝒙? −𝟒 𝒙− ?𝟒 + 𝟓 ?( 𝒙− ?𝟒 ) ( 𝟕𝒙?+ 𝟓 ) Exercise 1 (Available as a separate worksheet) Show all solutions 4 Factorise using any method: a 2𝑥 2 − 5𝑥 + 3 (𝟐𝒙 − 𝟏)(𝒙 − 𝟑) b 3𝑥 2 + 7𝑥 + 4 (𝟑𝒙 + 𝟒)(𝒙 + 𝟏) c 2𝑥 2 − 7𝑥 − 30 (𝟐𝒙 + 𝟓)(𝒙 − 𝟔) d 3𝑥 2 − 𝑥 − 14 (𝟑𝒙 − 𝟕)(𝒙 + 𝟐) [Edexcel GCSE Jun 2004 5H Q19ai] Factorise 2𝑥 2 − 35𝑥 + 98 e 6𝑥 2 − 23𝑥 − 4 (𝟔𝒙 + 𝟏)(𝒙 − 𝟒) (𝟐𝒙 − 𝟕)(𝒙 ? − 𝟏𝟒) f 15𝑥 2 − 13𝑥 − 20 (𝟓𝒙 + 𝟒)(𝟑𝒙 − 𝟓) N Express each of these areas as a single factorised expression: a 5 (𝟔𝒙 + 𝟏)(𝒙 ? − 𝟒) 6 ? 1 − 3𝑥 10 3𝑥 − 2 b (𝟔𝒙 + 𝟏)(𝒙 ? − 𝟓) 2𝑥 + 1 2𝑥 − 3 3𝑥 + 5 [AQA GCSE Nov 2013 1H Q15b] Factorise 6𝑥 2 − 23𝑥 − 4 𝑥 𝑥+8 (𝟓𝒙 + 𝟏)(𝒙 ? − 𝟐) The Big Idea : Factorising Non-Monic Quadratics Factorise 4𝑥 2 − 1 Can we arrange these into a rectangle? 𝑥2 𝑥2 −1 How does this differ from the quadratics we have factorised so far? 𝑥2 𝑥 2 There are no 𝑥 tiles. In order to form a rectangle, we need to add pairs of +𝑥 and – 𝑥 tiles. The Big Idea : Factorising Non-Monic Quadratics Factorise 4𝑥 2 − 1 There are two possible arrangements of the 𝑥 2 and −1 tiles. Which one allows us to add equal numbers of +𝑥 and – 𝑥 tiles? 𝑥2 𝑥2 𝑥2 𝑥2 𝑥 −𝑥 −𝑥 −𝑥 −𝑥 −1 This is the correct arrangement, as we add two + 𝑥 tiles and two −𝑥 tiles to form a square. 4𝑥 2 − 1 = 2𝑥 2 − (1)2 = (2𝑥 − 1)(2𝑥 + 1) 𝑥2 𝑥2 𝑥 𝑥2 𝑥2 𝑥 −𝑥 −𝑥 −1 This is an example of the Difference of Two Squares Example Test Your Understanding Factorise 9𝑥 2 − 49 Factorise 25𝑥 2 − 121 9𝑥 2 = 3𝑥 × 3𝑥 25𝑥 2 = 5𝑥 ? × 5𝑥 49 = 7 × 7 121 = 11 ? × 11 (𝟑𝒙 + 𝟕)(𝟑𝒙 − 𝟕) ? − 𝟏𝟏) (𝟓𝒙 + 𝟏𝟏)(𝟓𝒙 Quickfire Questions a c Factorise b Factorise 9𝑥 2 − 25 36𝑦 2 − 49 (𝟑𝒙 + 𝟓)(𝟑𝒙 ? − 𝟓) (𝟔𝒚 + 𝟕)(𝟔𝒚 ? − 𝟕) Factorise d Factorise 81 − 16𝑎2 225𝑥 2 − 4 (𝟗 + 𝟒𝒂)(𝟗 ? − 𝟒𝒂) (𝟏𝟓𝒙 + 𝟐)(𝟏𝟓𝒙 − 𝟐) ? Factorising Quadratics : Progress So Far A quadratic expression has the general form 𝒂𝒙𝟐 + 𝒃𝒙 + 𝒄 where 𝑎, 𝑏 and 𝑐 are constants and 𝑎 is not zero 𝑥 2 + 7𝑥 + 10 2 𝑥 −𝑥−6 2 5𝑥 + 8𝑥 + 3 2 𝑥 − 16 2 𝑥 − 12𝑥 25 − 4𝑥 2 8 − 2𝑥 − 𝑥 2 5𝑥 2 + 20𝑥 𝑥 2 − 8𝑥 + 16 We can now factorise: Quadratics with a common factor, 𝑐 = 0 Three term quadratics where 𝑎 = 1 Difference of two squares where 𝑎 = 1 Quadratics which form perfect squares Three term quadratics where 𝑎 ≠ 1 Difference of two squares where 𝑎 ≠ 1 Quickfire Questions a c Factorise b Factorise 16𝑥 2 − 25 16𝑥 2 − 25𝑥 (𝟒𝒙 + 𝟓)(𝟒𝒙 ? − 𝟓) 𝒙(𝟏𝟔𝒙?− 𝟐𝟓) Factorise d Factorise 16𝑥 2 − 10𝑥 + 1 16𝑥 2 − 49 (𝟖𝒙 − 𝟏)(𝟐𝒙 ? − 𝟏) (𝟒𝒙 + 𝟕)(𝟒𝒙 ? − 𝟕) Find the Matching Pairs 1 9𝑥 2 − 30𝑥 + 25 a (3𝑥 − 5)(3𝑥 + 5) 2 9𝑥 2 − 25 b 𝑥(9𝑥 − 16) 3 9𝑥 2 + 16𝑥 − 25 c (3𝑥 − 5)2 4 9𝑥 2 − 16𝑥 d (9𝑥 + 25)(𝑥 − 1) 1 2 c a ? 3 4 d b Exercise 2 1 Factorise: a 4𝑥 2 − 25 b (Available as a separate worksheet) Show all solutions 2 Factorise: (𝟐𝒙 + 𝟓)(𝟐𝒙 − 𝟓) a 4𝑥 2 + 10𝑥 𝟐𝒙(𝟐𝒙 + 𝟓) 36𝑝2 − 1 (𝟔𝒑 + 𝟏)(𝟔𝒑 − 𝟏) b 4𝑥 2 + 8𝑥 + 3 (𝟐𝒙 + 𝟑)(𝟐𝒙 + 𝟏) c 121 − 9𝑦 2 (𝟏𝟏 + 𝟑𝒚)(𝟏𝟏 ? − 𝟑𝒚) c 9 − 4𝑥 2 (𝟑 + 𝟐𝒙)(𝟑 ? − 𝟐𝒙) d 9𝑥 2 − 196 (𝟑𝒙 + 𝟏𝟒)(𝟑𝒙 − 𝟏𝟒) d 4𝑥 2 − 9𝑥 𝒙(𝟒𝒙 − 𝟗) e 16𝑎2 − 49 (𝟒𝒙 + 𝟕)(𝟒𝒙 − 𝟕) e 4𝑥 2 + 11𝑥 − 20 (𝟒𝒙 − 𝟓)(𝒙 + 𝟒) 3 Factorise: 4 a 25𝑦 2 − 1 (𝟓𝒚 + 𝟏)(𝟓𝒚 − 𝟏) [WJEC GCSE June 2019 2H Q12a ] Factorise 81𝑝2 − 1 b 25𝑦 2 − 10𝑦 𝟓𝒚(𝟓𝒚 − 𝟐) c 25𝑦 2 − 10𝑦 + 1 (𝟓𝒚 − 𝟏) ?𝟐 d 9𝑦 − 25𝑦 2 𝒚(𝟗 − 𝟐𝟓𝒚) e 25𝑦 2 + 46𝑦 − 8 (𝟐𝟓𝒚 − 𝟒)(𝒚 + 𝟐) (𝟗𝒑 + 𝟏)(𝟗𝒑 ? − 𝟏) Exercise 2 5 (Available as a separate worksheet) [WJEC GCSE June 2018 2H Q10 ] Factorise 4𝑚2 − 289 (𝟐𝒎 + 𝟏𝟕)(𝟐𝒎 ? − 𝟏𝟕) 6 [CCEA GCSE June 2019 Higher M4 Q22a] Factorise 2𝑎2 + 7𝑎𝑏 − 4𝑏 2 7 Factorise: a 𝑏 2 − 𝑎2 𝑥 2 (𝒃 + 𝒂𝒙)(𝒃 − 𝒂𝒙) b 𝑎𝑥 2 + 𝑏𝑥 + 𝑎𝑏𝑥 + 𝑏 2 (𝒂𝒙 + 𝒃)(𝒙 ? + 𝒄) c 𝑐𝑥 2 − 𝑑𝑥 + 𝑐 2 𝑥 − 𝑐𝑑 (𝒄𝒙 − 𝒅)(𝒙 + 𝒄) (𝟐𝒂 − 𝒃)(𝒂 ? + 𝟒𝒃) a Write each of these areas as a single fully factorised expression: 4 (6𝑥 + 5𝑎) b 3 (𝟔𝒙 − 𝟕)(𝒙 ? − 𝟏) 𝑎 (2𝑥 − 5) (3𝑥 + 1) (2𝑥 − 𝑎) N Show all solutions 3𝑎 𝟒(𝟑𝒙 − 𝟐𝒂)(𝒙 ? + 𝒂)
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