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

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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 βˆ’ 17π‘₯ + 6

    Question 3 Factorise π‘₯2 + 9π‘₯ + 14

    Question 13 Factorise π‘₯2 βˆ’ 9π‘₯ + 20

    Question 4 Factorise 3π‘₯2 + 5π‘₯ βˆ’ 2

    Question 14 Factorise 6π‘₯2 βˆ’ π‘₯ βˆ’ 15

    Question 5 Factorise 2π‘₯2 βˆ’ 5π‘₯ βˆ’ 3

    Question 15 Factorise 2π‘₯2 βˆ’ 13π‘₯ βˆ’ 7

    Question 6 Factorise 2π‘₯2 + π‘₯ βˆ’ 10

    Question 16 Factorise 8π‘₯2 βˆ’ 2π‘₯ βˆ’ 1

    Question 7 Factorise 4π‘₯2 βˆ’ 4π‘₯ βˆ’ 3

    Question 17 Factorise 2π‘₯2 βˆ’ 11π‘₯ + 12

    Question 8 Factorise 6π‘₯2 βˆ’ 11π‘₯ + 4

    Question 18 Factorise 3π‘₯2 βˆ’ 11π‘₯ βˆ’ 4

    Question 9 Factorise 5π‘₯2 + 4π‘₯ βˆ’ 1

    Question 19 Factorise 2π‘₯2 βˆ’ π‘₯ βˆ’ 15

    Question 10 Factorise 3π‘₯2 βˆ’ 13π‘₯ + 14

    Question 20 Factorise 12π‘₯2 βˆ’ 32π‘₯ + 5

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    2 | 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

    WORKINGS

    Question 1

    Factorise 2π‘₯2 + 9π‘₯ + 9 Label π‘Ž = 2, 𝑏 = 9, 𝑐 = 9 Multiply a by c 2 x 9 = 18 The two numbers must multiply to give +18 but add to give b (and 𝑏 = +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 – 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‘Ž = 6, 𝑏 = 11, 𝑐 = 3 Multiply a by c 6 x 3 = 18 The two numbers must multiply to give +18 but add to give b (and 𝑏 = +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 – 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‘Ž = 1 but we can use the same method for factorising all quadratics Label π‘Ž = 1, 𝑏 = 9, 𝑐 = 14 Multiply a by c 1 x 14 = 14 The two numbers must multiply to give +14 but add to give b (and 𝑏 = +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 – which we do Factorise again (π‘₯ + 2)(π‘₯ + 7)

    14 1 14 2 7

    Question 4

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

    -6 1 6 2 3

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    Question 5

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

    -6 1 6 2 3

    Question 6

    Factorise 2π‘₯2 + π‘₯ βˆ’ 10 Label π‘Ž = 2, 𝑏 = 1, 𝑐 = βˆ’10 Multiply a by c 2 x βˆ’ 10 = βˆ’20 The two numbers must multiply to give βˆ’20 but add to give b (and 𝑏 = +1) The two numbers will be βˆ’4 and +5 Rewrite the quadratic splitting π‘₯ intoβˆ’4π‘₯ and 5π‘₯

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

    -20 1 20 2 10 4 5

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    Question 7

    Factorise 4π‘₯2 βˆ’ 4π‘₯ βˆ’ 3 Label π‘Ž = 4, 𝑏 = βˆ’4, 𝑐 = βˆ’3 Multiply a by c 4 x βˆ’ 3 = βˆ’12 The two numbers must multiply to give +12 but add to give b (and 𝑏 = βˆ’4) The two numbers will be βˆ’6 and +2 Rewrite the quadratic splitting π‘₯ intoβˆ’6π‘₯ and 2π‘₯ 4π‘₯2 βˆ’ 6π‘₯ + 2π‘₯ βˆ’ 3 Factorise in pairs (partitioning) 2π‘₯(2π‘₯ βˆ’ 3) + 1(2π‘₯ βˆ’ 3) You should have the same in both brackets – which we do Factorise again (2π‘₯ βˆ’ 3)(2π‘₯ + 1)

    -12 1 12 2 6 3 4

    Question 8

    Factorise 6π‘₯2 βˆ’ 11π‘₯ + 4 Label π‘Ž = 6, 𝑏 = βˆ’11, 𝑐 = 4 Multiply a by c 6 x 4 = 24 The two numbers must multiply to give +24 but add to give b (and 𝑏 = βˆ’11) The two numbers will be βˆ’3 and βˆ’8 Rewrite the quadratic splitting π‘₯ intoβˆ’2π‘₯ and βˆ’8π‘₯

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

    24 1 24 2 12 3 8 4 6

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    6 | 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

    Question 9

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

    -5 1 5

    Question 10

    Factorise 3π‘₯2 βˆ’ 13π‘₯ + 14 Label π‘Ž = 3, 𝑏 = βˆ’13, 𝑐 = 14 Multiply a by c 3 x 14 = 42 The two numbers must multiply to give +42 but add to give b (and 𝑏 = βˆ’13) The two numbers will be βˆ’6 and βˆ’7 Rewrite the quadratic splitting π‘₯ intoβˆ’6π‘₯ and βˆ’7π‘₯

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

    42 1 42 2 21 3 14 6 7

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    Question 11

    Factorise 8π‘₯2 + 14π‘₯ + 3 Label π‘Ž = 8, 𝑏 = 14, 𝑐 = 3 Multiply a by c 8 x 3 = 24 The two numbers must multiply to give +24 but add to give b (and 𝑏 = +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 – which we do Factorise again (4π‘₯ + 1)(2π‘₯ + 3)

    24 1 24 2 12 3 8 4 6

    Question 12

    Factorise 5π‘₯2 βˆ’ 17π‘₯ + 6 Label π‘Ž = 5, 𝑏 = βˆ’17, 𝑐 = 6 Multiply a by c 5 x 6 = 30 The two numbers must multiply to give +30 but add to give b (and 𝑏 = βˆ’17) The two numbers will be βˆ’2 and βˆ’15 Rewrite the quadratic splitting βˆ’17π‘₯ intoβˆ’2π‘₯ and βˆ’15π‘₯

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

    30 1 30 2 15 3 10 5 6

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    8 | 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

    Question 13

    Factorise π‘₯2 βˆ’ 9π‘₯ + 20 This is actually an easier quadratic to factorise as π‘Ž = 1 but we can use the same method for factorising all quadratics Label π‘Ž = 1, 𝑏 = βˆ’9, 𝑐 = 20 Multiply a by c 1 x 20 = 20 The two numbers must multiply to give +20 but add to give b (and 𝑏 = βˆ’9) The two numbers will be βˆ’4 and βˆ’5 Rewrite the quadratic splitting βˆ’9π‘₯ in

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