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*Factorising harder quadratic expressions - Chatterton 2014-12-08Ψ’Β Factorising harder quadratic...*

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1 | P a g e 0 1 4 2 3 3 4 0 0 7 6 0 7 7 5 9 5 0 1 6 2 9

Factorising harder quadratic expressions

Question 1 Factorise 2π₯2 + 9π₯ + 9

Question 11 Factorise 8π₯2 + 14π₯ + 3

Question 2 Factorise 6π₯2 + 11π₯ + 3

Question 12 Factorise 5π₯2 β 17π₯ + 6

Question 3 Factorise π₯2 + 9π₯ + 14

Question 13 Factorise π₯2 β 9π₯ + 20

Question 4 Factorise 3π₯2 + 5π₯ β 2

Question 14 Factorise 6π₯2 β π₯ β 15

Question 5 Factorise 2π₯2 β 5π₯ β 3

Question 15 Factorise 2π₯2 β 13π₯ β 7

Question 6 Factorise 2π₯2 + π₯ β 10

Question 16 Factorise 8π₯2 β 2π₯ β 1

Question 7 Factorise 4π₯2 β 4π₯ β 3

Question 17 Factorise 2π₯2 β 11π₯ + 12

Question 8 Factorise 6π₯2 β 11π₯ + 4

Question 18 Factorise 3π₯2 β 11π₯ β 4

Question 9 Factorise 5π₯2 + 4π₯ β 1

Question 19 Factorise 2π₯2 β π₯ β 15

Question 10 Factorise 3π₯2 β 13π₯ + 14

Question 20 Factorise 12π₯2 β 32π₯ + 5

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WORKINGS

Question 1

Factorise 2π₯2 + 9π₯ + 9 Label π = 2, π = 9, π = 9 Multiply a by c 2 x 9 = 18 The two numbers must multiply to give +18 but add to give b (and π = +9) The two numbers will be +3 and +6 Rewrite the quadratic splitting 9π₯ into 3π₯ and 6π₯ 2π₯2 + 3π₯ + 6π₯ + 9 Factorise in pairs (partitioning) π₯(2π₯ + 3) + 3(2π₯ + 3) You should have the same in both brackets β which we do Factorise again (2π₯ + 3)(π₯ + 3)

18 1 18 2 9 3 6

Question 2

Factorise 6π₯2 + 11π₯ + 3 Label π = 6, π = 11, π = 3 Multiply a by c 6 x 3 = 18 The two numbers must multiply to give +18 but add to give b (and π = +11) The two numbers will be +2 and +9 Rewrite the quadratic splitting 9π₯ into 3π₯ and 6π₯ 6π₯2 + 2π₯ + 9π₯ + 3 Factorise in pairs (partitioning) 2π₯(3π₯ + 1) + 3(3π₯ + 1) You should have the same in both brackets β which we do Factorise again (3π₯ + 1)(2π₯ + 3)

18 1 18 2 9 3 6

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Question 3

Factorise π₯2 + 9π₯ + 14 This is actually an easier quadratic to factorise as π = 1 but we can use the same method for factorising all quadratics Label π = 1, π = 9, π = 14 Multiply a by c 1 x 14 = 14 The two numbers must multiply to give +14 but add to give b (and π = +9) The two numbers will be +2 and +7 Rewrite the quadratic splitting 9π₯ into 2π₯ and 7π₯ π₯2 + 2π₯ + 7π₯ + 14 Factorise in pairs (partitioning) π₯(π₯ + 2) + 7(π₯ + 2) You should have the same in both brackets β which we do Factorise again (π₯ + 2)(π₯ + 7)

14 1 14 2 7

Question 4

Factorise 3π₯2 + 5π₯ β 2 Label π = 3, π = 5, π = β2 Multiply a by c 3 x β 2 = β6 The two numbers must multiply to give β6 but add to give b (and π = +5) The two numbers will be β1 and +6 Rewrite the quadratic splitting 5π₯ intoβ1π₯ and 6π₯ 3π₯2 β 1π₯ + 6π₯ β 2 Factorise in pairs (partitioning) π₯(3π₯ β 1) + 2(3π₯ β 1) You should have the same in both brackets β which we do Factorise again (3π₯ β 1)(π₯ + 2)

-6 1 6 2 3

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Question 5

Factorise 2π₯2 β 5π₯ β 3 Label π = 2, π = β5, π = β3 Multiply a by c 2 x β 3 = β6 The two numbers must multiply to give β6 but add to give b (and π = β5) The two numbers will be β6 and +1 Rewrite the quadratic splitting β5π₯ intoβ6π₯ and 1π₯ 2π₯2 β 6π₯ + 1π₯ β 3 Factorise in pairs (partitioning) 2π₯(π₯ β 3) + 1(π₯ β 3) You should have the same in both brackets β which we do Factorise again (π₯ β 3)(2π₯ + 1)

-6 1 6 2 3

Question 6

Factorise 2π₯2 + π₯ β 10 Label π = 2, π = 1, π = β10 Multiply a by c 2 x β 10 = β20 The two numbers must multiply to give β20 but add to give b (and π = +1) The two numbers will be β4 and +5 Rewrite the quadratic splitting π₯ intoβ4π₯ and 5π₯

2π₯2 β 4π₯ + 5π₯ β 10 Factorise in pairs (partitioning) 2π₯(π₯ β 2) + 5(π₯ β 2) You should have the same in both brackets β which we do Factorise again (π₯ β 2)(2π₯ + 5)

-20 1 20 2 10 4 5

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Question 7

Factorise 4π₯2 β 4π₯ β 3 Label π = 4, π = β4, π = β3 Multiply a by c 4 x β 3 = β12 The two numbers must multiply to give +12 but add to give b (and π = β4) The two numbers will be β6 and +2 Rewrite the quadratic splitting π₯ intoβ6π₯ and 2π₯ 4π₯2 β 6π₯ + 2π₯ β 3 Factorise in pairs (partitioning) 2π₯(2π₯ β 3) + 1(2π₯ β 3) You should have the same in both brackets β which we do Factorise again (2π₯ β 3)(2π₯ + 1)

-12 1 12 2 6 3 4

Question 8

Factorise 6π₯2 β 11π₯ + 4 Label π = 6, π = β11, π = 4 Multiply a by c 6 x 4 = 24 The two numbers must multiply to give +24 but add to give b (and π = β11) The two numbers will be β3 and β8 Rewrite the quadratic splitting π₯ intoβ2π₯ and β8π₯

6π₯2 β 3π₯ β 8π₯ + 4 Factorise in pairs (partitioning) 3π₯(2π₯ β 1) β 4(2π₯ β 1) You should have the same in both brackets β which we do Factorise again (2π₯ β 1)(3π₯ β 4)

24 1 24 2 12 3 8 4 6

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Question 9

Factorise 5π₯2 + 4π₯ β 1 Label π = 5, π = 4, π = β1 Multiply a by c 5 x β 1 = β5 The two numbers must multiply to give β5 but add to give b (and π = +4) The two numbers will be β1 and +5 Rewrite the quadratic splitting π₯ intoβ4π₯ and 5π₯ 5π₯2 β 1π₯ + 5π₯ β 1 Factorise in pairs (partitioning) π₯(5π₯ β 1) + 1(5π₯ β 1) You should have the same in both brackets β which we do Factorise again (5π₯ β 1)(π₯ + 1)

-5 1 5

Question 10

Factorise 3π₯2 β 13π₯ + 14 Label π = 3, π = β13, π = 14 Multiply a by c 3 x 14 = 42 The two numbers must multiply to give +42 but add to give b (and π = β13) The two numbers will be β6 and β7 Rewrite the quadratic splitting π₯ intoβ6π₯ and β7π₯

3π₯2 β 6π₯ β 7π₯ + 14 Factorise in pairs (partitioning) 3π₯(π₯ β 2) β 7(π₯ β 2) You should have the same in both brackets β which we do Factorise again (π₯ β 2)(3π₯ β 7)

42 1 42 2 21 3 14 6 7

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Question 11

Factorise 8π₯2 + 14π₯ + 3 Label π = 8, π = 14, π = 3 Multiply a by c 8 x 3 = 24 The two numbers must multiply to give +24 but add to give b (and π = +14) The two numbers will be +2 and +12 Rewrite the quadratic splitting 14π₯ into+2π₯ and+12π₯ 8π₯2 + 2π₯ + 12π₯ + 3 Factorise in pairs (partitioning) 2π₯(4π₯ + 1) + 3(4π₯ + 1) You should have the same in both brackets β which we do Factorise again (4π₯ + 1)(2π₯ + 3)

24 1 24 2 12 3 8 4 6

Question 12

Factorise 5π₯2 β 17π₯ + 6 Label π = 5, π = β17, π = 6 Multiply a by c 5 x 6 = 30 The two numbers must multiply to give +30 but add to give b (and π = β17) The two numbers will be β2 and β15 Rewrite the quadratic splitting β17π₯ intoβ2π₯ and β15π₯

5π₯2 β 2π₯ β 15π₯ + 6 Factorise in pairs (partitioning) π₯(5π₯ β 2) β 3(5π₯ β 2) You should have the same in both brackets β which we do Factorise again (5π₯ β 2)(π₯ β 3)

30 1 30 2 15 3 10 5 6

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Question 13

Factorise π₯2 β 9π₯ + 20 This is actually an easier quadratic to factorise as π = 1 but we can use the same method for factorising all quadratics Label π = 1, π = β9, π = 20 Multiply a by c 1 x 20 = 20 The two numbers must multiply to give +20 but add to give b (and π = β9) The two numbers will be β4 and β5 Rewrite the quadratic splitting β9π₯ in