CBSE NCERT Solutions for Class 8 Mathematics Chapter 14...CBSE NCERT Solutions for Class 8 Mathematics Chapter 14 Back of Chapter Questions Exercise 14.1 1. Find the common factors

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    CBSE NCERT Solutions for Class 8 Mathematics Chapter 14

    Back of Chapter Questions

    Exercise 14.1

    1. Find the common factors of the given terms.

    (i) 12π‘₯, 36

    (ii) 2𝑦, 22π‘₯𝑦

    (iii) 14 π‘π‘ž, 28𝑝2π‘ž2

    (iv) 2π‘₯, 3π‘₯2, 4

    (v) 6π‘Žπ‘π‘, 24π‘Žπ‘2, 12π‘Ž2𝑏

    (vi) 16π‘₯3, βˆ’4π‘₯2, 32π‘₯

    (vii) 10 π‘π‘ž, 20π‘žπ‘Ÿ, 30π‘Ÿπ‘

    (viii) 3π‘₯2𝑦3, 10π‘₯3𝑦2, 6π‘₯2𝑦2𝑧

    Solution:

    (i) 12π‘₯ = 12 Γ— π‘₯

    12π‘₯ = 12 Γ— π‘₯ = 2 Γ— 2 Γ— 3 Γ— π‘₯

    36 β‡’

    36 = 2 Γ— 2 Γ— 3 Γ— 3

    Thus,

    12π‘₯ = 2 Γ— 2 Γ— 3 Γ— π‘₯

    36 = 2 Γ— 2 Γ— 3 Γ— 3

    So, the common factors are 2, 2 and 3

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    And 2 Γ— 2 Γ— 3 = 12

    (ii) 2𝑦, 22π‘₯𝑦

    2𝑦 = 2 Γ— 𝑦

    22 π‘₯𝑦 = 22 Γ— π‘₯ Γ— 𝑦

    = 2 Γ— 11 Γ— π‘₯ Γ— 𝑦

    So, the common factors are 2 and y

    And2 Γ— 𝑦 = 2𝑦

    (iii) 14 π‘π‘ž, 28𝑝2π‘ž2

    14π‘π‘ž = 14 Γ— 𝑝 Γ— π‘ž

    = 2 Γ— 7 Γ— 𝑝 Γ— π‘ž

    28 𝑝2π‘ž2 = 28 Γ— 𝑝2 Γ— π‘ž2

    = 2 Γ— 2 Γ— 7 Γ— 𝑝2 Γ— π‘ž2

    = 2 Γ— 2 Γ— 7 Γ— 𝑝 Γ— π‘ž Γ— π‘ž Γ— π‘ž

    So,14π‘π‘ž = 2 Γ— 7 Γ— 𝑝 Γ— π‘ž

    28𝑝2π‘ž2 = 2 Γ— 2 Γ— 7 Γ— 𝑝 Γ— 𝑝 Γ— π‘ž Γ— π‘ž

    the common factor are 2, 7, 𝑝 and π‘ž

    And 2 Γ— 7 Γ— 𝑝 Γ— π‘ž = 14 Γ— π‘π‘ž

    = 14π‘π‘ž

    (iv) 2π‘₯, 3π‘₯2, 4

    2π‘₯ = 2 Γ— π‘₯

    3π‘₯2 = 3 Γ— π‘₯2

    = 3 Γ— π‘₯ Γ— π‘₯

    4 = 2 Γ— 2

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    There is no common factor visible.

    ∴ 1 is the only common factor of the given terms.

    (v) 6π‘Žπ‘π‘, 24π‘Žπ‘2, 12π‘Ž2𝑏

    6π‘Žπ‘π‘ = 6 Γ— π‘Žπ‘π‘

    = 2 Γ— 3 Γ— π‘Žπ‘π‘

    = 2 Γ— 3 Γ— π‘Ž Γ— 𝑏 Γ— 𝑐

    24π‘Žπ‘2 = 24 Γ— π‘Žπ‘2

    = 2 Γ— 2 Γ— 2 Γ— 3 Γ— π‘Žπ‘2

    = 2 Γ— 2 Γ— 2 Γ— 3 Γ— π‘Ž Γ— 𝑏 Γ— 𝑏

    12π‘Ž2𝑏 = 12 Γ— π‘Ž2𝑏

    = 2 Γ— 2 Γ— 3 Γ— π‘Ž2 Γ— 𝑏

    = 2 Γ— 2 Γ— 3 Γ— π‘Ž Γ— π‘Ž Γ— 𝑏

    So,6 π‘Žπ‘π‘ = 2 Γ— 3 Γ— π‘Ž Γ— 𝑏 Γ— 𝑐

    24π‘Žπ‘2 = 2 Γ— 2 Γ— 2 Γ— 3 Γ— π‘Ž Γ— 𝑏 Γ— 𝑐

    12 π‘Ž2𝑏 = 2 Γ— 2 Γ— 3 Γ— π‘Ž Γ— π‘Ž Γ— 𝑏

    the common factor are 2, 3, a and b

    And 2 Γ— 3 Γ— π‘Ž Γ— 𝑏 = 6 Γ— π‘Žπ‘

    = 6 π‘Žπ‘

    (vi) 16π‘₯3, βˆ’4π‘₯2, 32π‘₯

    16π‘₯3 = 16 Γ— π‘₯3

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    = 2 Γ— 2 Γ— 2 Γ— 2 Γ— π‘₯3

    = 2 Γ— 2 Γ— 2 Γ— 2 Γ— π‘₯ Γ— π‘₯ Γ— π‘₯

    βˆ’4π‘₯2 = βˆ’4 Γ— π‘₯2

    = βˆ’1 Γ— 4 Γ— π‘₯2

    = βˆ’1 Γ— 2 Γ— 2 Γ— π‘₯2

    = βˆ’1 Γ— 2 Γ— 2 Γ— π‘₯ Γ— π‘₯

    32π‘₯ = 32 Γ— π‘₯

    = 2 Γ— 2 Γ— 2 Γ— 2 Γ— 2 Γ— π‘₯

    So,16π‘₯3 = 2 Γ— 2 Γ— 2 Γ— 2 Γ— π‘₯ Γ— π‘₯ Γ— π‘₯

    βˆ’4π‘₯2 = βˆ’1 Γ— 2 Γ— 2 Γ— π‘₯ Γ— π‘₯

    32π‘₯ = 2 Γ— 2 Γ— 2 Γ— 2 Γ— π‘₯

    the common factors are 2, 2 and x

    And2 Γ— 2 Γ— π‘₯ = 4 Γ— π‘₯

    = 4π‘₯

    (vii) 10 π‘π‘ž, 20π‘žπ‘Ÿ, 30π‘Ÿπ‘

    10π‘π‘ž = 10 Γ— π‘π‘ž

    = 2 Γ— 5 Γ— π‘π‘ž

    = 2 Γ— 5 Γ— 𝑝 Γ— π‘ž

    20π‘žπ‘Ÿ = 20 Γ— π‘žπ‘Ÿ

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    = 2 Γ— 2 Γ— 5 Γ— π‘žπ‘Ÿ

    = 2 Γ— 2 Γ— 5 Γ— π‘ž Γ— π‘Ÿ

    30π‘Ÿπ‘ = 30 Γ— π‘Ÿπ‘

    = 2 Γ— 3 Γ— 5 Γ— π‘Ÿπ‘

    = 2 Γ— 2 Γ— 5 Γ— π‘Ÿ Γ— 𝑝

    So,10π‘π‘ž = 2 Γ— 5 Γ— 𝑝 Γ— π‘ž

    20π‘žπ‘Ÿ = 2 Γ— 2 Γ— 5 Γ— π‘ž Γ— π‘Ÿ

    30π‘Ÿπ‘ = 2 Γ— 2 Γ— 5 Γ— π‘Ÿ Γ— 𝑝

    the common factors are 2 and 5

    And2 Γ— 5 = 10

    (viii) 3π‘₯2𝑦3, 10π‘₯3𝑦2, 6π‘₯2𝑦2𝑧

    3π‘₯2𝑦3 = 3 Γ— π‘₯2 Γ— 𝑦2

    = 3 Γ— π‘₯ Γ— π‘₯ Γ— 𝑦 Γ— 𝑦 Γ— 𝑦

    10π‘₯3𝑦2 = 10 Γ— π‘₯3 Γ— 𝑦2

    = 2 Γ— 5 Γ— π‘₯3 Γ— 𝑦2

    = 2 Γ— 5 Γ— π‘₯ Γ— π‘₯ Γ— π‘₯ Γ— 𝑦 Γ— 𝑦

    6π‘₯2𝑦2𝑧 = 6 Γ— π‘₯2 Γ— 𝑦2 Γ— 𝑧

    = 2 Γ— 3 Γ— π‘₯2 Γ— 𝑦2 Γ— 𝑧

    = 2 Γ— 3 Γ— π‘₯ Γ— π‘₯ Γ— 𝑦 Γ— 𝑦 Γ— 𝑧

    So,3π‘₯2𝑦3 = 2 Γ— π‘₯ Γ— π‘₯ Γ— 𝑦 Γ— 𝑦 Γ— 𝑦

    10π‘₯3𝑦2 = 2 Γ— 5 Γ— π‘₯ Γ— π‘₯ Γ— π‘₯ Γ— 𝑦 Γ— 𝑦

    6π‘₯2𝑦2𝑧 = 2 Γ— 3 Γ— π‘₯ Γ— π‘₯ Γ— 𝑦 Γ— 𝑦 Γ— 𝑧

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    the common factors are x, x, y and y

    Andπ‘₯ Γ— π‘₯ Γ— 𝑦 Γ— 𝑦 = (π‘₯ Γ— π‘₯) Γ— (𝑦 Γ— 𝑦)

    = π‘₯2 Γ— 𝑦2

    = π‘₯2𝑦2

    2. Factorise the following expressions.

    (i) 7π‘₯ βˆ’ 42

    (ii) 6𝑝 βˆ’ 12π‘ž

    (iii) 7π‘Ž2 + 14π‘Ž

    (iv) βˆ’16𝑧 + 20𝑧3

    (v) 20𝑙2π‘š+ 30 π‘Ž π‘šβ„

    (vi) 5π‘₯2𝑦 βˆ’ 15π‘₯𝑦2

    (vii) 10 π‘Ž2 βˆ’ 15 𝑏2 + 20 𝑐2

    (viii) βˆ’4π‘Ž2 + 4π‘Žπ‘ βˆ’ 4 π‘π‘Ž

    (ix) π‘₯2𝑦𝑧 + π‘₯𝑦2𝑧 + π‘₯𝑦𝑧2

    (x) π‘Žπ‘₯2𝑦 + 𝑏π‘₯𝑦2 + 𝑐π‘₯𝑦𝑧

    Solution:

    (i) 7π‘₯ βˆ’ 42

    Method 1:

    7π‘₯ = 7 Γ— π‘₯

    42 = 2 Γ— 3 Γ— 7

    7 is the only common factor.

    7π‘₯ βˆ’ 42 = (7 Γ— π‘₯) βˆ’ (2 Γ— 3 Γ— 7)

    = 7(π‘₯ βˆ’ (2 Γ— 3))

    = 7(π‘₯ βˆ’ 6)

    Method 2:

    7π‘₯ βˆ’ 42

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    7π‘₯ βˆ’ 42

    = (7 Γ— π‘₯) βˆ’ (7 Γ— 6)

    = 7(π‘₯ βˆ’ 6) (taking 7 as common)

    (ii) 6𝑝 βˆ’ 12π‘ž

    Method 1:

    6𝑝 = 6 Γ— 𝑝

    = 2 Γ— 3 Γ— 𝑝

    12π‘ž = 12 Γ— π‘ž

    = 2 Γ— 2 Γ— 3 Γ— π‘ž

    So, the common factors are 2 and 3.

    6𝑝 βˆ’ 12π‘ž = (2 Γ— 3 Γ— 𝑝) βˆ’ (2 Γ— 2 Γ— 3 Γ— π‘ž)

    = 2 Γ— 3(𝑝 βˆ’ (2 Γ— π‘ž))

    = 6(𝑝 βˆ’ 2π‘ž)

    Method 2:

    6𝑝 βˆ’ 12π‘ž

    = 6𝑝 βˆ’ (6 Γ— 2)π‘ž

    = 6(𝑝 βˆ’ 2π‘ž)(taking 6 as common)

    (iii) 7π‘Ž2 + 14π‘Ž

    Method 1:

    7π‘Ž2 = 7 Γ— π‘Ž2 = 7 Γ— π‘Ž Γ— π‘Ž

    14π‘Ž = 14 Γ— π‘Ž = 7 Γ— 2 Γ— π‘Ž Γ— π‘Ž

    So, the common factors are 7 and a

    7π‘Ž2 + 14π‘Ž = (7 Γ— π‘Ž Γ— π‘Ž) + (7 Γ— 2 Γ— π‘Ž)

    = (7 Γ— π‘Ž)(π‘Ž + 2)

    = 7π‘Ž(π‘Ž + 2)

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    Method 2:

    7π‘Ž2 + 14π‘Ž

    = 7π‘Ž2 + (7 Γ— 2)π‘Ž

    = (7π‘Ž Γ— π‘Ž) + (7π‘Ž Γ— 2)

    = 7π‘Ž(π‘Ž + 2)(taking 7π‘Ž common)

    (iv) βˆ’16𝑧 + 20𝑧3

    Method 1:

    βˆ’16𝑧 = βˆ’16 Γ— 𝑧

    = βˆ’1 Γ— 2 Γ— 2 Γ— 2 Γ— 2 Γ— 𝑧

    20𝑧3 = 20 Γ— 𝑧3

    = 2 Γ— 2 Γ— 5 Γ— 𝑧 Γ— 𝑧 Γ— 𝑧

    So, the common factors are 2, 2 and 𝑧

    βˆ’16𝑧 + 20𝑧3

    = (βˆ’1 Γ— 2 Γ— 2 Γ— 2 Γ— 2 Γ— 𝑧) + (2 Γ— 2 Γ— 5 Γ— 𝑧 Γ— 𝑧 Γ— 𝑧)

    Taking 2 Γ— 2 Γ— 𝑧 common,

    = 2 Γ— 2 Γ— 𝑧((βˆ’1 Γ— 2 Γ— 2) + (5 Γ— 𝑧 Γ— 𝑧))

    = 4𝑧(βˆ’4 + 5𝑧2)

    Method 2:

    βˆ’16𝑧 + 20𝑧3

    = (4 Γ— βˆ’4)𝑧 + (4 Γ— 5)𝑧3

    = 4𝑧(βˆ’4 + 5𝑧2)(taking 4z as common)

    (v) 20𝑙2π‘š+ 30 π‘Ž π‘šβ„

    Method 1:

    20 𝑙2 π‘š = 20 Γ— 𝑙2 Γ—π‘š

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    = 2 Γ— 2 Γ— 5 Γ— l Γ— l Γ— π‘š

    30 alm = 30 Γ— π‘Ž Γ— l Γ— π‘š

    = 2 Γ— 3 Γ— 5 Γ— π‘Ž Γ— l Γ— π‘š

    So, 20𝑙2m = 2 Γ— 2 Γ— 5 Γ— l Γ— l Γ— m

    30 alm = 2 Γ— 3 Γ— 5 Γ— a Γ— l Γ— m

    So, 2, 5, l and m are the common factors.

    Now,

    20𝑙2m+ 30 alm = (2 Γ— 2 Γ— 5 Γ— l Γ— l Γ— m) + (2 Γ— 3 Γ— 5 Γ— a Γ— l Γ— m)

    Taking 2 Γ— 5 Γ— l Γ— m common

    = 2 Γ— 5 Γ— l Γ— m (2 Γ— 1) + (3 Γ— π‘Ž)

    = 10lm(2l + 3a)

    Method 2:

    20 𝑙2 π‘š + 30 alm

    = (10 Γ— 2)𝑙 Γ— 𝑙 Γ— π‘š + (10 Γ— 3) a Γ— l Γ— m

    Taking 10 Γ— l Γ— m as common,

    = 10 Γ— l Γ— m(2l + 3π‘Ž)

    = 10lm(2l + 3π‘Ž)

    (vi) 5π‘₯2𝑦 βˆ’ 15π‘₯𝑦2

    Method 1:

    5π‘₯2𝑦 = 5 Γ— π‘₯ Γ— π‘₯ Γ— 𝑦

    15 π‘₯𝑦2 = 15 Γ— π‘₯ Γ— 𝑦2

    = 3 Γ— 5 Γ— π‘₯ Γ— 𝑦 Γ— 𝑦

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    So, 5, π‘₯ and 𝑦 are the common factors.

    Now,

    5π‘₯2𝑦 βˆ’ 15π‘₯𝑦2 = (5 Γ— π‘₯ Γ— π‘₯ Γ— 𝑦) βˆ’ (3 Γ— 5 Γ— π‘₯ Γ— 𝑦 Γ— 𝑦)

    Taking 5 Γ— π‘₯ Γ— 𝑦 common

    = 5 Γ— π‘₯ Γ— 𝑦(π‘₯ βˆ’ (3 Γ— 𝑦))

    = 5π‘₯𝑦(π‘₯ βˆ’ 3𝑦)

    Method 2:

    5π‘₯2𝑦 βˆ’ 15π‘₯𝑦2 = 5π‘₯2𝑦 βˆ’ 5 Γ— 3 Γ— π‘₯𝑦2

    Taking 5 as common

    = 5(π‘₯2𝑦 βˆ’ 3π‘₯𝑦2)

    = 5((π‘₯𝑦 Γ— π‘₯) βˆ’ (π‘₯𝑦 Γ— 3𝑦))

    Taking π‘₯𝑦 as common

    = 5π‘₯𝑦(π‘₯ βˆ’ 3𝑦)

    (vii) 10 π‘Ž2 βˆ’ 15 𝑏2 + 20 𝑐2

    Method 1:

    10π‘Ž2 = 10 Γ— π‘Ž2 = 2 Γ— 5 Γ— π‘Ž2 = 2 Γ— 5 Γ— π‘Ž Γ— π‘Ž

    15𝑏2 = 15 Γ— 𝑏2 = 3 Γ— 5 Γ— 𝑏2 = 3 Γ— 5 Γ— 𝑏 Γ— 𝑏

    20𝑐2 = 20 Γ— 𝑐2 = 2 Γ— 2 Γ— 5 Γ— 𝑐2 = 2 Γ— 2 Γ— 5 Γ— 𝑐 Γ— 𝑐

    So, 5 is the common factor.

    10π‘Ž2 βˆ’ 15𝑏2 + 20𝑐2

    = (2 Γ— 5 Γ— π‘Ž Γ— π‘Ž) βˆ’ (3 Γ— 5 Γ— 𝑏 Γ— 𝑏) + (2 Γ— 2 Γ— 5 Γ— 𝑐 Γ— 𝑐)

    = 5 Γ— ((2 Γ— π‘Ž Γ— π‘Ž) βˆ’ (3 Γ— 𝑏 Γ— 𝑏) + (2 Γ— 2 Γ— 𝑐 Γ— 𝑐))

    = 5 Γ— (2π‘Ž2 βˆ’ 3𝑏2 + 4𝑐2)

    = 5(2π‘Ž2 βˆ’ 3𝑏2 + 4𝑐2)

    Method 2:

    10 π‘Ž2 βˆ’ 15 𝑏2 + 20 𝑐2

    = (5 Γ— 2)π‘Ž2 βˆ’ (5 Γ— 3)𝑏2 + (5 Γ— 4)𝑐2

    Taking 5 common,

    = 5(2π‘Ž2 βˆ’ 3𝑏2 + 4𝑐2)

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    (viii) βˆ’4π‘Ž2 + 4π‘Žπ‘ βˆ’ 4 π‘π‘Ž

    Method 1:

    βˆ’4π‘Ž2 = βˆ’4 Γ— π‘Ž2 = βˆ’1 Γ— 4 Γ— π‘Ž2

    = βˆ’1 Γ— 2 Γ— 2 Γ— π‘Ž2

    = βˆ’1 Γ— 2 Γ— 2 Γ— π‘Ž Γ— π‘Ž

    4π‘Žπ‘ = 4 Γ— π‘Ž Γ— 𝑏 = 2 Γ— 2 Γ— π‘Ž Γ— 𝑏

    4π‘π‘Ž = 4 Γ— 𝑐 Γ— π‘Ž = 2 Γ— 2 Γ— 𝑐 Γ— π‘Ž

    So, 2, 2 and π‘Ž are the common factors.

    βˆ’4π‘Ž2 + 4π‘Žπ‘ βˆ’ 4π‘π‘Ž

    = (βˆ’1 Γ— 2 Γ— 2 Γ— π‘Ž Γ— π‘Ž) + (2 Γ— 2 Γ— π‘Ž Γ— 𝑏) βˆ’ (2 Γ— 2 Γ— 𝑐 Γ— π‘Ž)

    = 2 Γ— 2 Γ— π‘Ž Γ— ((βˆ’1 Γ— π‘Ž) + 𝑏 βˆ’ 𝑐)

    = 4π‘Ž(βˆ’π‘Ž + 𝑏 βˆ’ 𝑐)

    Method 2:

    βˆ’4π‘Ž2 + 4π‘Žπ‘ βˆ’ 4 π‘π‘Ž

    Taking 4 common,

    = 4(βˆ’π‘Ž2 + π‘Žπ‘ βˆ’ π‘π‘Ž)

    = 4((βˆ’π‘Ž Γ— π‘Ž) + (π‘Ž Γ— 𝑏) βˆ’ (𝑐 Γ— π‘Ž))

    Taking a common,

    = 4π‘Ž(βˆ’π‘Ž + 𝑏 βˆ’ 𝑐)

    (ix) π‘₯2𝑦𝑧 + π‘₯𝑦2𝑧 + π‘₯𝑦𝑧2

    Method 1:

    π‘₯2𝑦𝑧 = π‘₯2 Γ— 𝑦 Γ— 𝑧 = π‘₯ Γ— π‘₯ Γ— 𝑦 Γ— 𝑧

    π‘₯2𝑦𝑧 = π‘₯ Γ— 𝑦2 Γ— 𝑧 = π‘₯ Γ— 𝑦 Γ— 𝑦 Γ— 𝑧

    π‘₯𝑦𝑧2 = π‘₯ Γ— 𝑦 Γ— 𝑧2 = π‘₯ Γ— 𝑦 Γ— 𝑧 Γ— 𝑧

    So, π‘₯, 𝑦 and 𝑧 are the common factors.

    π‘₯2𝑦𝑧 + π‘₯𝑦2𝑧 + π‘₯𝑦𝑧2

    = (π‘₯ Γ— π‘₯ Γ— 𝑦 Γ— 𝑧) + (π‘₯ Γ— 𝑦 Γ— 𝑦 Γ— 𝑧)(π‘₯ Γ— 𝑦 Γ— 𝑧 Γ— 𝑧)

    Taking π‘₯ Γ— 𝑦 Γ— 𝑧 common,

    = π‘₯ Γ— 𝑦 Γ— 𝑧(π‘₯ + 𝑦 + 𝑧)

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    = π‘₯𝑦𝑧(π‘₯ + 𝑦 + 𝑧)

    Method 2:

    π‘₯2𝑦𝑧 + π‘₯𝑦2𝑧 + π‘₯𝑦𝑧2

    = (π‘₯ Γ— π‘₯𝑦𝑧) + (𝑦 Γ— π‘₯𝑦𝑧) + (𝑧 Γ— π‘₯𝑦𝑧)

    Taking π‘₯𝑦𝑧 common,

    = π‘₯𝑦𝑧(π‘₯ + 𝑦 + 𝑧)

    (x) π‘Žπ‘₯2𝑦 + 𝑏π‘₯𝑦2 + 𝑐π‘₯𝑦𝑧

    Method 1:

    π‘Žπ‘₯2𝑦 = π‘Ž Γ— π‘₯2 Γ— 𝑦 = π‘Ž Γ— π‘₯ Γ— π‘₯ Γ— 𝑦

    𝑏π‘₯𝑦2 = 𝑏 Γ— π‘₯ Γ— 𝑦2 = 𝑏 Γ— π‘₯ Γ— 𝑦 Γ— 𝑦

    𝑐π‘₯𝑦𝑧 = 𝑐 Γ— π‘₯ Γ— 𝑦 Γ— 𝑧

    So, π‘₯ and 𝑦 are the common factors.

    π‘Žπ‘₯2𝑦 + 𝑏π‘₯𝑦2 + 𝑐π‘₯𝑦𝑧

    = (π‘Ž Γ— π‘₯ Γ— π‘₯ Γ— 𝑦) + (𝑏 Γ— π‘₯ Γ— 𝑦 Γ— 𝑦) + (𝑐 Γ— π‘₯ Γ— 𝑦 Γ— 𝑧)

    = π‘₯ Γ— 𝑦(π‘Ž Γ— π‘₯) + (𝑏 Γ— 𝑦) + (𝑐 Γ— 𝑧)

    = π‘₯𝑦(π‘Žπ‘₯ + 𝑏𝑦 + 𝑐𝑧)

    Method 2:

    π‘Žπ‘₯2𝑦 + 𝑏π‘₯𝑦2 + 𝑐π‘₯𝑦𝑧

    = (π‘₯ Γ— π‘Žπ‘₯𝑦) + (π‘₯ Γ— 𝑏𝑦2) + (π‘₯ Γ— 𝑐𝑦𝑧)

    Taking π‘₯ common,

    = π‘₯(π‘Žπ‘₯𝑦 + 𝑏𝑦2 + 𝑐𝑦𝑧)

    = π‘₯((π‘Žπ‘₯ Γ— 𝑦) + (𝑏𝑦 Γ— 𝑦) + (𝑐𝑧 Γ— 𝑦))

    Taking 𝑦 common,

    = π‘₯ Γ— 𝑦(π‘Žπ‘₯ + 𝑏𝑦 + 𝑐𝑧)

    = π‘₯𝑦(π‘Žπ‘₯ + 𝑏𝑦 + 𝑐𝑧)

    3. Factorise.

    (i) π‘₯2 + π‘₯𝑦 + 8π‘₯ + 8𝑦

    (ii) 15π‘₯𝑦 βˆ’ 6π‘₯ + 5𝑦 βˆ’ 2

    (iii) π‘Žπ‘₯ + 𝑏π‘₯ βˆ’ π‘Žπ‘¦ βˆ’ 𝑏𝑦

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    (iv) 15π‘π‘ž + 15 + 9π‘ž + 25𝑝

    (v) 𝑧 βˆ’ 7 + 7π‘₯𝑦 βˆ’ π‘₯𝑦𝑧

    Solution:

    (i) π‘₯2 + π‘₯𝑦 + 8π‘₯ + 8𝑦

    = (π‘₯2 + π‘₯𝑦)⏟ Both have x as common factor

    + (8π‘₯ + 8𝑦)⏟ Both have 8 as common factor

    = π‘₯(π‘₯ + 𝑦) + 8(π‘₯ + 𝑦)

    Taking (π‘₯ + 𝑦) common

    = (π‘₯ + 𝑦)(π‘₯ + 8)

    (ii) 15π‘₯𝑦 βˆ’ 6π‘₯ + 5𝑦 βˆ’ 2

    15π‘₯𝑦 βˆ’ 6π‘₯ + 5𝑦 βˆ’ 2

    = (15π‘₯𝑦 βˆ’ 6π‘₯)⏟ Both have 3 and π‘₯ as common factor

    + (5𝑦 βˆ’ 2)⏟ Since nothing is common,we take 1 common

    = 3π‘₯(5𝑦 βˆ’ 2) + 1(5𝑦 βˆ’ 2)

    Taking (5𝑦 βˆ’ 2) common

    = (5𝑦 βˆ’ 2)(3π‘₯ + 1)

    (iii) π‘Žπ‘₯ + 𝑏π‘₯ βˆ’ π‘Žπ‘¦ βˆ’ 𝑏𝑦

    π‘Žπ‘₯ + 𝑏π‘₯ βˆ’ π‘Žπ‘¦ βˆ’ 𝑏𝑦

    (π‘Žπ‘₯ + 𝑏π‘₯)⏟ Both have x as common factor

    βˆ’ (π‘Žπ‘¦ βˆ’ 𝑏𝑦)⏟ Both have y as common factor

    = π‘₯(π‘Ž + 𝑏) βˆ’ 𝑦(π‘Ž + 𝑏)

    Taking (π‘Ž + 𝑏) common

    = (π‘Ž + 𝑏)(π‘₯ βˆ’ 𝑦)

    (iv) 15π‘π‘ž + 15 + 9π‘ž + 25𝑝

    15π‘π‘ž + 15 + 9π‘ž + 25𝑝

    = (15π‘π‘ž + 25𝑝)⏟ Both have 5

    and p as common factor

    + (15 + 9π‘ž)⏟ Both have 3 as common factor

    = 5𝑝(3π‘ž + 5) + 3(5 + 3π‘ž)

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    = 5𝑝(3π‘ž + 5) + 3(3π‘ž + 5)

    Taking (3π‘ž + 5) Common,

    = (3π‘ž + 5)(5𝑝 + 3)

    (v) 𝑧 βˆ’ 7 + 7π‘₯𝑦 βˆ’ π‘₯𝑦𝑧

    𝑧 βˆ’ 7 + 7π‘₯𝑦 βˆ’ π‘₯𝑦𝑧

    (𝑧 βˆ’ 7) + (7π‘₯𝑦 βˆ’ π‘₯𝑦𝑧)⏟ Both have π‘₯ and 𝑦 as

    common factor

    (𝑧 βˆ’ 7) + π‘₯𝑦(7 βˆ’ 𝑧)

    = (𝑧 βˆ’ 7) + π‘₯𝑦 Γ— βˆ’(𝑧 βˆ’ 7) (As (7 βˆ’ 𝑧) = βˆ’(𝑧 βˆ’ 7))

    = (𝑧 βˆ’ 7) βˆ’ π‘₯𝑦(𝑧 βˆ’ 7)

    Taking (𝑧 βˆ’ 7) common

    = (𝑧 βˆ’ 7)(1 βˆ’ π‘₯𝑦)

    Exercise 14.2

    1. Factorise the following expressions.

    (i) π‘Ž2 + 8π‘Ž + 16

    (ii) 𝑝2 βˆ’ 10𝑝 + 25

    (iii) 25π‘š2 + 30π‘š + 9

    (iv) 49𝑦2 + 84𝑦𝑧 + 36𝑧2

    (v) 4π‘₯2 βˆ’ 8π‘₯ + 4

    (vi) 121𝑏2 βˆ’ 88𝑏𝑐 + 16𝑐2

    (vii) (𝑙 + π‘š)2 βˆ’ 4π‘™π‘š (𝐇𝐒𝐧𝐭: Expand (𝑙 + π‘š)2 first)

    (viii) π‘Ž4 + 2π‘Ž2𝑏2 + 𝑏4

    Solution:

    (i) π‘Ž2 + 8π‘Ž + 16

    = π‘Ž2 + 8π‘Ž + 42

    = π‘Ž2 + (2 Γ— π‘Ž Γ— 4) + 42

    = π‘Ž2 + 42 + (2 Γ— π‘Ž Γ— 4)

    Using (π‘₯ + 𝑦)2 = π‘₯2 + 𝑦2 + 2π‘₯𝑦

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    Here, π‘₯ = π‘Ž and 𝑦 = 4

    = (π‘Ž + 4)2

    (ii) 𝑝2 βˆ’ 10𝑝 + 25

    𝑝2 βˆ’ 10𝑝 + 25

    = 𝑝2 βˆ’ 10𝑝 + 52

    = 𝑝2 βˆ’ (2 Γ— 𝑝 Γ— 5) + 52

    = 𝑝2 + 52 βˆ’ (2 Γ— 𝑝 Γ— 5)

    Using (π‘Ž βˆ’ 𝑏)2 = π‘Ž2 + 𝑏2 βˆ’ 2π‘Žπ‘

    Here, π‘Ž = 𝑝 and 𝑏 = 5

    = (𝑝 βˆ’ 5)2

    (iii) 25π‘š2 + 30π‘š + 9

    25π‘š2 + 30π‘š + 9

    = (5π‘š)2 + 30π‘š + 32

    = (5π‘š)2 + (2 Γ— 5π‘š Γ— 3) + 32

    = (5π‘š)2 + 32 + (2 Γ— 5π‘š Γ— 3)

    Using (π‘Ž + 𝑏)2 = π‘Ž2 + 𝑏2 + 2π‘Žπ‘

    Here, π‘Ž = 5π‘š and 𝑏 = 3

    = (5π‘š + 3)2

    (iv) 49𝑦2 + 84𝑦𝑧 + 36𝑧2

    49𝑦2 + 84𝑦𝑧 + 36𝑧2

    = (7𝑦)2 + 84𝑦𝑧 + (6𝑧)2

    = (7𝑦)2 + 2 Γ— 7𝑦 Γ— 6𝑧 + (6𝑧)2

    = (7𝑦)2 + (6𝑧)2 + 2 Γ— 7𝑦 Γ— 6𝑧

    Using (π‘Ž + 𝑏)2 = π‘Ž2 + 𝑏2 + 2π‘Žπ‘

    Here, π‘Ž = 7𝑦 and 𝑏 = 6𝑧

    = (7𝑦 + 6𝑧)2

    (v) 4π‘₯2 βˆ’ 8π‘₯ + 4

    4π‘₯2 βˆ’ 8π‘₯ + 4

    = 2π‘₯2 βˆ’ (2 Γ— 2π‘₯ Γ— (βˆ’2)) + 22

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    = 2π‘₯2 + 22 βˆ’ (2 Γ— 2π‘₯ Γ— (βˆ’2))

    Using (π‘Ž βˆ’ 𝑏)2 = π‘Ž2 + 𝑏2 βˆ’ 2π‘Žπ‘

    Here, π‘Ž = 2π‘₯ and 𝑏 = 2

    = (2π‘₯ βˆ’ 2)2

    (vi) 121𝑏2 βˆ’ 88𝑏𝑐 + 16𝑐2

    121𝑏2 βˆ’ 88𝑏𝑐 + 16𝑐2 = (11𝑏)2 βˆ’ 88𝑏𝑐 + (4𝑐)2

    = (11𝑏)2 βˆ’ 2 Γ— 11𝑏 Γ— 4𝑐 + (4𝑐)2

    = (11𝑏)2 + (4𝑐)2 βˆ’ 2 Γ— 11𝑏 Γ— 4𝑐

    Using (π‘₯ βˆ’ 𝑦)2 = π‘₯2 + 𝑦2 βˆ’ 2π‘₯𝑦

    Here, π‘₯ = 11𝑏 and 𝑦 = 4𝑐

    = (11𝑏 βˆ’ 4𝑐)2

    (vii) (𝑙 + π‘š)2 βˆ’ 4π‘™π‘š(𝐇𝐒𝐧𝐭: Expand (𝑙 + π‘š)2 first)

    Using (π‘Ž + 𝑏)2 = π‘Ž2 + 𝑏2 + 2π‘Žπ‘

    Here, π‘Ž = 𝑙 and 𝑏 = π‘š

    = 𝑙2 +π‘š2 + 2π‘™π‘š βˆ’ 4π‘™π‘š

    = 𝑙2 +π‘š2 + 2π‘™π‘š(1 βˆ’ 2)

    = 𝑙2 +π‘š2 βˆ’ 2π‘™π‘š

    Using (π‘Ž βˆ’ 𝑏)2 = π‘Ž2 + 𝑏2 βˆ’ 2π‘Žπ‘

    Here, π‘Ž = 𝑙 and 𝑏 = π‘š

    = (𝑙 βˆ’ π‘š)2

    (viii) π‘Ž4 + 2π‘Ž2𝑏2 + 𝑏4

    π‘Ž4 + 2π‘Ž2𝑏2 + 𝑏4

    Using (π‘Žπ‘š)𝑛 = π‘Žπ‘šΓ—π‘›

    ∴ (π‘Ž2)2 = π‘Ž2Γ—2 = π‘Ž4

    = (π‘Ž2)2 + 2π‘Ž2𝑏2 + (𝑏2)2

    = (π‘Ž2)2 + 2(π‘Ž2 Γ— 𝑏2) + (𝑏2)2

    = (π‘Ž2)2 + (𝑏2)2 + 2(π‘Ž2 Γ— 𝑏2)

    Using (π‘₯ + 𝑦)2 = π‘₯2 + 𝑦2 + 2π‘₯𝑦

    Here, π‘₯ = π‘Ž2 and 𝑦 = 𝑏2

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    = (π‘Ž2 + 𝑏2)2

    2. Factorise

    (i) 4𝑝2 βˆ’ 9π‘ž2

    (ii) 63π‘Ž2 βˆ’ 112𝑏2

    (iii) 49π‘₯2 βˆ’ 36

    (iv) 16π‘₯5 βˆ’ 144π‘₯3

    (v) (𝑙 + π‘š)2 βˆ’ (𝑙 βˆ’ π‘š)2

    (vi) 9π‘₯2𝑦2 βˆ’ 16

    (vii) (π‘₯2 βˆ’ 2π‘₯𝑦 + 𝑦2) βˆ’ 𝑧2

    (viii) 25π‘Ž2 βˆ’ 4𝑏2 + 28𝑏𝑐 βˆ’ 49𝑐2

    Solution:

    (i) 4𝑝2 βˆ’ 9π‘ž2

    = (2𝑝)2 βˆ’ (3π‘ž)2

    Using π‘Ž2 βˆ’ 𝑏2 = (π‘Ž + 𝑏)(π‘Ž βˆ’ 𝑏)

    Here π‘Ž = 2𝑝 and 𝑏 = 3π‘ž

    = (2𝑝 + 3π‘ž)(2𝑝 βˆ’ 3π‘ž)

    (ii) 63π‘Ž2 βˆ’ 112𝑏2

    63π‘Ž2 βˆ’ 112𝑏2

    = (7 Γ— 9)π‘Ž2 βˆ’ (7 Γ— 16)𝑏2

    Taking 7 common,

    = 7(9π‘Ž2 βˆ’ 16𝑏2)

    = 7((3π‘Ž)2 βˆ’ (4𝑏)2)

    Using π‘₯2 βˆ’ 𝑦2 = (π‘₯ + 𝑦)(π‘₯ βˆ’ 𝑦)

    Here π‘₯ = 3π‘Ž and 𝑦 = 4𝑏

    = 7(3π‘Ž + 4𝑏)(3π‘Ž βˆ’ 4𝑏)

    (iii) 49π‘₯2 βˆ’ 36

    49π‘₯2 βˆ’ 36

    = (7π‘₯)2 βˆ’ (6)2

    Using π‘Ž2 βˆ’ 𝑏2 = (π‘Ž + 𝑏)(π‘Ž βˆ’ 𝑏)

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    Here π‘Ž = 7π‘₯ and 𝑏 = 6

    = (7π‘₯ + 6)(7π‘₯ βˆ’ 6)

    (iv) 16π‘₯5 βˆ’ 144π‘₯3

    16π‘₯5 βˆ’ 144π‘₯3

    = 16π‘₯2π‘₯3 βˆ’ 144π‘₯3

    Taking π‘₯3 common,

    = π‘₯3(16π‘₯2 βˆ’ 144)

    = π‘₯3((4π‘₯)2 βˆ’ (12)2)

    Using π‘Ž2 βˆ’ 𝑏2 = (π‘Ž + 𝑏)(π‘Ž βˆ’ 𝑏)

    Here π‘Ž = 4π‘₯ and 𝑏 = 12

    = π‘₯3(4π‘₯ + 12)(4π‘₯ βˆ’ 12)

    = π‘₯3 (4π‘₯ + 12)⏟ π΅π‘œπ‘‘β„Ž β„Žπ‘Žπ‘£π‘’ 4 π‘Žπ‘ 

    (4π‘₯ βˆ’ 12)⏟ π‘π‘œπ‘šπ‘šπ‘œπ‘› π‘“π‘Žπ‘π‘‘π‘œπ‘Ÿ

    = π‘₯3 Γ— 4(π‘₯ + 3) Γ— 4(π‘₯ βˆ’ 3)

    = π‘₯3 Γ— 4 Γ— 4 Γ— (π‘₯ + 3)(π‘₯ βˆ’ 3)

    = 16π‘₯3(π‘₯ + 3)(π‘₯ βˆ’ 3)

    (v) (𝑙 + π‘š)2 βˆ’ (𝑙 βˆ’ π‘š)2

    (𝑙 + π‘š)2 βˆ’ (𝑙 βˆ’ π‘š)2

    Using π‘Ž2 βˆ’ 𝑏2 = (π‘Ž + 𝑏)(π‘Ž βˆ’ 𝑏)

    Here π‘Ž = (𝑙 + π‘š) and 𝑏 = (𝑙 βˆ’ π‘š)

    = [(𝑙 + π‘š) + (𝑙 βˆ’ π‘š)][(𝑙 + π‘š) βˆ’ (𝑙 βˆ’ π‘š)]

    = [𝑙 + π‘š + 𝑙 βˆ’ π‘š][𝑙 + π‘š βˆ’ 𝑙 + π‘š]

    = (2𝑙)(2π‘š)

    = 2 Γ— 2 Γ— 𝑙 Γ— π‘š

    = 4π‘™π‘š

    (vi) 9π‘₯2𝑦2 βˆ’ 16

    9π‘₯2𝑦2 βˆ’ 16

    = (3π‘₯𝑦)2 βˆ’ (4)2

    Using π‘Ž2 βˆ’ 𝑏2 = (π‘Ž + 𝑏)(π‘Ž βˆ’ 𝑏)

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    Here π‘Ž = 3π‘₯𝑦 and 𝑏 = 4

    = (3π‘₯𝑦 + 4)(3π‘₯𝑦 βˆ’ 4)

    (vii) (π‘₯2 βˆ’ 2π‘₯𝑦 + 𝑦2) βˆ’ 𝑧2

    (π‘₯2 βˆ’ 2π‘₯𝑦 + 𝑦2) βˆ’ 𝑧2

    = (π‘₯2 + 𝑦2 βˆ’ 2π‘₯𝑦) βˆ’ 𝑧2

    Using (π‘Ž βˆ’ 𝑏)2 = π‘Ž2 + 𝑏2 βˆ’ 2π‘Žπ‘

    Here π‘Ž = π‘₯ and 𝑏 = 𝑦

    = (π‘₯ βˆ’ 𝑦)2 βˆ’ 𝑧2

    Using π‘Ž2 βˆ’ 𝑏2 = (π‘Ž + 𝑏)(π‘Ž βˆ’ 𝑏)

    Here π‘Ž = π‘₯ βˆ’ 𝑦 and 𝑏 = 𝑧

    = (π‘₯ βˆ’ 𝑦 + 𝑧)(π‘₯ βˆ’ 𝑦 βˆ’ 𝑧)

    (viii) 25π‘Ž2 βˆ’ 4𝑏2 + 28𝑏𝑐 βˆ’ 49𝑐2

    25π‘Ž2 βˆ’ 4𝑏2 + 28𝑏𝑐 βˆ’ 49𝑐2⏟ Takingβˆ’common

    = 25π‘Ž2 βˆ’ (4𝑏2 βˆ’ 28𝑏𝑐 + 49𝑐2)

    = 25π‘Ž2 βˆ’ (4𝑏2 + 49𝑐2 βˆ’ 28𝑏𝑐)

    = 25π‘Ž2 βˆ’ ((2𝑏)2 + (7𝑐)2 βˆ’ 2 Γ— 2𝑏 Γ— 7𝑐)

    Using (π‘₯ βˆ’ 𝑦)2 = π‘₯2 + 𝑦2 βˆ’ 2π‘₯𝑦

    Here π‘₯ = 2𝑏 and 𝑦 = 7𝑐

    = 25π‘Ž2 βˆ’ (2𝑏 βˆ’ 7𝑐)2

    = (5π‘Ž)2 βˆ’ (2𝑏 βˆ’ 7𝑐)2

    Using π‘₯2 βˆ’ 𝑦2 = (π‘₯ + 𝑦)(π‘₯ βˆ’ 𝑦)

    Here π‘₯ = 5π‘Ž and 𝑦 = 2𝑏 βˆ’ 7𝑐

    = (5π‘Ž + (2𝑏 βˆ’ 7𝑐))(5π‘Ž βˆ’ (2𝑏 βˆ’ 7𝑐))

    = (5π‘Ž + 2𝑏 βˆ’ 7𝑐)(5π‘Ž βˆ’ 2𝑏 + 7𝑐)

    3. Factorise the expressions:

    (i) π‘Žπ‘₯2 + 𝑏π‘₯

    (ii) 7𝑝2 + 21π‘ž2

    (iii) 2π‘₯3 + 2π‘₯𝑦2 + 2π‘₯𝑧2

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    (iv) π‘Žπ‘š2 + π‘π‘š2 + 𝑏𝑛2 + π‘Žπ‘›2

    (v) (π‘™π‘š + 𝑙) + π‘š + 1

    (vi) 𝑦(𝑦 + 𝑧) + 9(𝑦 + 𝑧)

    (vii) 5𝑦2 βˆ’ 20𝑦 βˆ’ 8𝑧 + 2𝑦𝑧

    (viii) 10π‘Žπ‘ + 4π‘Ž + 5𝑏 + 2

    (ix) 6π‘₯𝑦 βˆ’ 4𝑦 + 6 βˆ’ 9π‘₯

    Solution:

    (i) π‘Žπ‘₯2 + 𝑏π‘₯

    π‘Žπ‘₯2 = π‘Ž Γ— π‘₯ Γ— π‘₯

    𝑏π‘₯ = 𝑏 Γ— π‘₯

    So, π‘₯ is a common factor.

    Taking π‘₯ common,

    = π‘₯((π‘Ž Γ— π‘₯) + 𝑏)

    = π‘₯(π‘Žπ‘₯ + 𝑏)

    (ii) 7𝑝2 + 21π‘ž2

    7𝑝2 = 7 Γ— 𝑝2 = 7 Γ— 𝑝 Γ— 𝑝

    21π‘ž2 = 21 Γ— π‘ž2 = 3 Γ— 7 Γ— π‘ž Γ— π‘ž

    So, 7 is the only common factor.

    Taking 7 common,

    = 7 Γ— ((𝑝 Γ— 𝑝) + (3 Γ— π‘ž Γ— π‘ž))

    = 7 Γ— (𝑝2 + 3π‘ž2)

    = 7(𝑝2 + 3π‘ž2)

    Method 1:

    (iii) 2π‘₯3 + 2π‘₯𝑦2 + 2π‘₯𝑧2

    2π‘₯3 = 2 Γ— π‘₯3 = 2 Γ— π‘₯ Γ— π‘₯ Γ— π‘₯

    2π‘₯3 = 2 Γ— π‘₯ Γ— 𝑦2 = 2 Γ— π‘₯ Γ— 𝑦 Γ— 𝑦

    2π‘₯𝑧2 = 2 Γ— π‘₯ Γ— 𝑧2 = 2 Γ— π‘₯ Γ— 𝑧 Γ— 𝑧

    So, 2 and π‘₯ are the common factors.

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    2π‘₯3 + 2π‘₯𝑦2 + 2π‘₯𝑧2

    = (2 Γ— π‘₯ Γ— π‘₯ Γ— π‘₯) + (2 Γ— π‘₯ Γ— 𝑦 Γ— 𝑦) + (2 Γ— π‘₯ Γ— 𝑧 Γ—)

    Taking 2 Γ— π‘₯ common,

    = 2 Γ— π‘₯((π‘₯ Γ— π‘₯) + (𝑦 Γ— 𝑦) + (𝑧 Γ— 𝑧))

    = 2π‘₯ (π‘₯2 + 𝑦2 + 𝑧2)

    (iv) π‘Žπ‘š2 + π‘π‘š2 + 𝑏𝑛2 + π‘Žπ‘›2

    (π‘Žπ‘š2 + π‘π‘š2)⏟ π΅π‘œπ‘‘β„Ž β„Žπ‘Žπ‘£π‘’ π‘š2 π‘Žπ‘ π‘π‘œπ‘šπ‘šπ‘œπ‘› π‘“π‘Žπ‘π‘‘π‘œπ‘Ÿ

    + (𝑏𝑛2 + π‘Žπ‘›2)⏟ π΅π‘œπ‘‘β„Ž β„Žπ‘Žπ‘£π‘’ 𝑛2 π‘Žπ‘ π‘π‘œπ‘šπ‘šπ‘œπ‘› π‘“π‘Žπ‘π‘‘π‘œπ‘Ÿ

    = π‘š2(π‘Ž + 𝑏) + 𝑛2(π‘Ž + 𝑏)

    Taking (π‘Ž + 𝑏) common,

    = (π‘Ž + 𝑏)(π‘š2 + 𝑛2)

    (v) (π‘™π‘š + 𝑙) + π‘š + 1

    (π‘™π‘š + 𝑙) + π‘š + 1

    Taking 𝑙 common,

    = 𝑙(π‘š + 1) + 1(π‘š + 1)

    Taking (π‘š + 1) common,

    = (π‘š + 1)(𝑙 + 1)

    (vi) 𝑦(𝑦 + 𝑧) + 9(𝑦 + 𝑧)

    𝑦(𝑦 + 𝑧) + 9(𝑦 + 𝑧)

    Taking (𝑦 + 𝑧) common,

    = (𝑦 + 𝑧)(𝑦 + 9)

    (vii) 5𝑦2 βˆ’ 20𝑦 βˆ’ 8𝑧 + 2𝑦𝑧

    5𝑦2 βˆ’ 20𝑦 βˆ’ 8𝑧 + 2𝑦𝑧

    = (5𝑦2 βˆ’ 20𝑦)⏟ Both have 5 only 𝑦

    as common factors

    + (βˆ’8𝑧 + 2𝑦𝑧)⏟ Both have 2 only 𝑧

    as common factors

    = 5𝑦(𝑦 βˆ’ 4) + 2𝑧(βˆ’4 + 𝑦)

    = 5𝑦(𝑦 βˆ’ 4) + 2𝑧(𝑦 βˆ’ 4)

    Taking (𝑦 βˆ’ 4) as common,

    = (𝑦 βˆ’ 4)(5𝑦 + 2𝑧)

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    (viii) 10π‘Žπ‘ + 4π‘Ž + 5𝑏 + 2

    10π‘Žπ‘ + 4π‘Ž + 5𝑏 + 2

    (10π‘Žπ‘ + 4π‘Ž)⏟ Both have 2 andas common factors

    + (5𝑏 + 2)⏟ Since nothing is common,we

    take 1 common

    = 2π‘Ž(5𝑏 + 2) + 1(5𝑏 + 2)

    Taking (5𝑏 + 2) as common,

    = (5𝑏 + 2)(2π‘Ž + 1)

    (ix) 6π‘₯𝑦 βˆ’ 4𝑦 + 6 βˆ’ 9π‘₯

    (6π‘₯𝑦 βˆ’ 4𝑦)⏟ Both have 2 and 𝑦as common factors

    + (6 βˆ’ 9π‘₯)⏟ Both have 3

    as common factors

    = 2𝑦(3π‘₯ βˆ’ 2) + 3(2 βˆ’ 3π‘₯)

    = 2𝑦(3π‘₯ βˆ’ 2) + 3 Γ— βˆ’1(3π‘₯ βˆ’ 2)(𝐴𝑠 (2 βˆ’ 3π‘₯) = βˆ’1 Γ— (3π‘₯ βˆ’ 2))

    = 2𝑦 (3π‘₯ βˆ’ 2) βˆ’ 3(3π‘₯ βˆ’ 2)

    Taking (3π‘₯ βˆ’ 2) as common,

    = (3π‘₯ βˆ’ 2)(2𝑦 βˆ’ 3)

    4. Factorise:

    (i) π‘Ž4 βˆ’ 𝑏4

    (ii) 𝑝4 βˆ’ 81

    (iii) π‘₯4 βˆ’ (𝑦 + 𝑧)4

    (iv) π‘₯4 βˆ’ (π‘₯ βˆ’ 𝑧)4

    (v) π‘Ž4 βˆ’ 2π‘Ž2𝑏2 + 𝑏4

    Solution:

    (i) π‘Ž4 βˆ’ 𝑏4

    = (π‘Ž2)2 βˆ’ (𝑏2)2

    Using π‘₯2 βˆ’ 𝑦2 = (π‘₯ + 𝑦)(π‘₯ βˆ’ 𝑦)

    Here π‘₯ = π‘Ž2 and 𝑦 = 𝑏2

    = (π‘Ž2 + 𝑏2)(π‘Ž2 βˆ’ 𝑏2)

    Using π‘₯2 βˆ’ 𝑦2 = (π‘₯ + 𝑦)(π‘₯ βˆ’ 𝑦)

    Here π‘₯ = π‘Ž and 𝑦 = 𝑏

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    = (π‘Ž2 + 𝑏2)(π‘Ž + 𝑏)(π‘Ž βˆ’ 𝑏)

    = (π‘Ž βˆ’ 𝑏)(π‘Ž + 𝑏)(π‘Ž2 + 𝑏2)

    (ii) 𝑝4 βˆ’ 81

    = (𝑝2)2 βˆ’ (9)2

    Using π‘Ž2 βˆ’ 𝑏2 = (π‘Ž + 𝑏)(π‘Ž βˆ’ 𝑏)

    Here π‘Ž = 𝑝2 and 𝑏 = 9

    = (𝑝2 + 9)(𝑝2 βˆ’ 9)

    = (𝑝2 + 9)(𝑝2 βˆ’ 32)

    Again Using π‘Ž2 βˆ’ 𝑏2 = (π‘Ž + 𝑏)(π‘Ž βˆ’ 𝑏)

    Here π‘Ž = 𝑝 and 𝑏 = 3

    = (𝑝2 + 9)(𝑝 + 3)(𝑝 βˆ’ 3)

    = (𝑝 βˆ’ 3)(𝑝 + 3)(𝑝2 + 9)

    (iii) π‘₯4 βˆ’ (𝑦 + 𝑧)4

    = (π‘₯2)2 βˆ’ ((𝑦 + 𝑧)2)2

    Using π‘Ž2 βˆ’ 𝑏2 = (π‘Ž + 𝑏)(π‘Ž βˆ’ 𝑏)

    Here π‘Ž = π‘₯2 and 𝑏 = (𝑦 + 𝑧)2

    = [π‘₯2 + (𝑦 + 𝑧)2] [π‘₯2 βˆ’ (𝑦 + 𝑧)2]

    Again Using π‘Ž2 βˆ’ 𝑏2 =(π‘Ž + 𝑏)(π‘Ž βˆ’ 𝑏)

    Here π‘Ž = π‘₯ and 𝑏 = (𝑦 + 𝑧)

    = [π‘₯2 + (𝑦 + 𝑧)2](π‘₯ βˆ’ (𝑦 + 𝑧))(π‘₯ + (𝑦 + 𝑧))

    = [π‘₯2 + (𝑦 + 𝑧)2](π‘₯ βˆ’ 𝑦 βˆ’ 𝑧)(π‘₯ + 𝑦 + 𝑧)

    (iv) π‘₯4 βˆ’ (π‘₯ βˆ’ 𝑧)4

    = (π‘₯ 2)2 βˆ’ [(π‘₯ βˆ’ 𝑧)2]2

    Using π‘Ž2 βˆ’ 𝑏2 = (π‘Ž + 𝑏)(π‘Ž βˆ’ 𝑏)

    Here π‘Ž = π‘₯2 and 𝑏 = (π‘₯ βˆ’ 𝑧)2

    = [π‘₯2 + (π‘₯ βˆ’ 𝑧)2] [π‘₯2 βˆ’ (π‘₯ βˆ’ 𝑧)2]

    Again Using π‘Ž2 βˆ’ 𝑏2 =(π‘Ž + 𝑏)(π‘Ž βˆ’ 𝑏)

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    Here π‘Ž = π‘₯ and 𝑏 = (π‘₯ βˆ’ 𝑧)

    = [π‘₯2 + (π‘₯ βˆ’ 𝑧)2][π‘₯ + (π‘₯ βˆ’ 𝑧)][π‘₯ βˆ’ (π‘₯ βˆ’ 𝑧)]

    = [π‘₯2 + (π‘₯ βˆ’ 𝑧)2][π‘₯ + π‘₯ βˆ’ 𝑧][π‘₯ βˆ’ π‘₯ + 𝑧]

    = [π‘₯2 + (π‘₯ βˆ’ 𝑧)2][2π‘₯ βˆ’ 𝑧][𝑧]

    Using (π‘Ž βˆ’ 𝑏)2 = π‘Ž2 + 𝑏2 βˆ’ 2π‘Žπ‘

    Here π‘Ž = π‘₯ and 𝑏 = 𝑧

    = [π‘₯2 + (π‘₯2 + 𝑧2 βˆ’ 2π‘₯𝑧)][2π‘₯ βˆ’ 𝑧][𝑧]

    = [π‘₯2 + π‘₯2 + 𝑧2 βˆ’ 2π‘₯𝑧][2π‘₯ βˆ’ 𝑧][𝑧]

    = [2π‘₯2 + 𝑧2 βˆ’ 2π‘₯𝑧][2π‘₯ βˆ’ 𝑧][𝑧]

    = 𝑧(2π‘₯ βˆ’ 𝑧)(2π‘₯2 + 𝑧2 βˆ’ 2π‘₯𝑧)

    (v) π‘Ž4 βˆ’ 2π‘Ž2𝑏2 + 𝑏4

    = (π‘Ž2)2 βˆ’ 2π‘Ž2𝑏2 + (𝑏2)2

    = (π‘Ž2)2 + (𝑏2)2 βˆ’ 2(π‘Ž2 Γ— 𝑏2)

    Using (π‘₯ βˆ’ 𝑦)2 = π‘₯2 + 𝑦2 βˆ’ 2π‘₯𝑦

    Here π‘₯ = π‘Ž2 and 𝑦 = 𝑏2

    = (π‘Ž2 βˆ’ 𝑏2)2

    Using π‘₯2 βˆ’ 𝑦2 = (π‘₯ + 𝑦)(π‘₯ βˆ’ 𝑦)

    Here π‘₯ = π‘Ž and 𝑦 = 𝑏

    = [(π‘Ž + 𝑏)(π‘Ž βˆ’ 𝑏)]2

    = (π‘Ž + 𝑏)2(π‘Ž βˆ’ 𝑏)2( 𝑆𝑖𝑛𝑐𝑒 (π‘Žπ‘)π‘š = π‘Žπ‘š Γ— π‘π‘š)

    5. Factorise the following expressions:

    (i) 𝑝2 + 6𝑝 + 8

    (ii) π‘ž2 βˆ’ 10π‘ž + 21

    (iii) 𝑝2 + 6𝑝 βˆ’ 16

    Solution:

    (I) 𝑝2 + 6𝑝 + 8

    = 𝑝2 + 2𝑝 + 4𝑝 + 8(here 6p can be written as 2p + 4p)

    = (𝑝2 + 2𝑝) + (4𝑝 + 8)

    = 𝑝(𝑝 + 2) + 4(𝑝 + 2)

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    Taking (𝑝 + 2) common,

    = (𝑝 + 2)(𝑝 + 4)

    (ii) π‘ž2 βˆ’ 10π‘ž + 21

    = π‘ž2 βˆ’ 3π‘ž βˆ’ 7π‘ž + 21 (here -10q can be written as- 3q – 7q)

    = (π‘ž2 βˆ’ 3π‘ž) βˆ’ (7π‘ž βˆ’ 21)

    = π‘ž(π‘ž βˆ’ 3) βˆ’ 7(π‘ž βˆ’ 3)

    Taking (π‘ž βˆ’ 3) common,

    = (π‘ž βˆ’ 3)(π‘ž βˆ’ 7)

    (iii) 𝑝2 + 6𝑝 βˆ’ 16

    = 𝑝2 βˆ’ 2𝑝 + 8𝑝 βˆ’ 16(here, 6p can be written as -2p + 8p)

    = (𝑝2 βˆ’ 2𝑝) + (8𝑝 βˆ’ 16)

    = 𝑝(𝑝 βˆ’ 2) + 8(𝑝 βˆ’ 2)

    Taking (𝑝 βˆ’ 2) common

    = (𝑝 βˆ’ 2)(𝑝 + 8)

    Exercise: 14.3

    1. Carryout the following divisions.

    (i) 28π‘₯4 Γ· 56π‘₯

    (ii) βˆ’36𝑦3 Γ· 9𝑦2

    (iii) 66π‘π‘ž2π‘Ÿ3 Γ· 11π‘žπ‘Ÿ2

    (iv) 34π‘₯3𝑦3𝑧3 Γ· 51π‘₯𝑦2𝑧3

    (v) 12π‘Ž8𝑏8 Γ· (βˆ’6π‘Ž6𝑏4)

    Solution:

    (i) 28π‘₯4 Γ· 56π‘₯

    =28 π‘₯4

    56 π‘₯

    =28

    56Γ—π‘₯4

    π‘₯

    =1

    2Γ— π‘₯4βˆ’1 (

    π‘Žπ‘š

    π‘Žπ‘›= π‘Žπ‘šβˆ’π‘›)

    =1

    2Γ— π‘₯3

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    =1

    2π‘₯3

    (ii) βˆ’36𝑦3 Γ· 9𝑦2

    =βˆ’36𝑦3

    9𝑦2

    =βˆ’36

    9×𝑦3

    𝑦2

    = βˆ’4 Γ— 𝑦3βˆ’2 (π‘Žπ‘š

    π‘Žπ‘›= π‘Žπ‘šβˆ’π‘›)

    = βˆ’4𝑦

    (iii) 66π‘π‘ž2π‘Ÿ3 Γ· 11π‘žπ‘Ÿ2

    =66π‘π‘ž2π‘Ÿ3

    11π‘žπ‘Ÿ2

    =66

    11Γ— 𝑝 Γ—

    π‘ž2

    π‘žΓ—π‘Ÿ3

    π‘Ÿ2

    = 6 Γ— 𝑝 Γ— π‘ž2βˆ’1 Γ— π‘Ÿ3βˆ’2 (π‘Žπ‘š

    π‘Žπ‘›= π‘Žπ‘šβˆ’π‘›)

    = 6 Γ— 𝑝 Γ— π‘ž Γ— π‘Ÿ

    = 6π‘π‘žπ‘Ÿ

    (iv) 34π‘₯3𝑦3𝑧3 Γ· 51π‘₯𝑦2𝑧3

    =34π‘₯3𝑦3𝑧3

    51π‘₯𝑦2𝑧3

    =34

    51Γ—π‘₯3

    π‘₯×𝑦3

    𝑦2×𝑧3

    𝑧3

    =2

    3Γ— π‘₯3βˆ’1 Γ— 𝑦3βˆ’2 Γ— 𝑧3βˆ’3 (

    π‘Žπ‘š

    π‘Žπ‘›= π‘Žπ‘šβˆ’π‘›)

    =2

    3Γ— π‘₯2 Γ— 𝑦 Γ— 𝑧0

    =2

    3Γ— π‘₯2 Γ— 𝑦 Γ— 1

    =2

    3π‘₯2𝑦

    (v) 12π‘Ž8𝑏8 Γ· (βˆ’6π‘Ž6𝑏4)

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    =12π‘Ž8𝑏8

    βˆ’6π‘Ž6𝑏4

    =12

    βˆ’6Γ—π‘Ž8

    π‘Ž6×𝑏8

    𝑏4

    = βˆ’2 Γ— π‘Ž8βˆ’6 Γ— 𝑏8βˆ’4 (π‘Žπ‘š

    π‘Žπ‘›= π‘Žπ‘šβˆ’π‘›)

    = βˆ’2 Γ— π‘Ž2 Γ— 𝑏4

    = βˆ’2π‘Ž2𝑏4

    2. (Method 1:) Separating each term

    Divide the given polynomial by the given monomial.

    (i) (5π‘₯2 βˆ’ 6π‘₯) Γ· 3π‘₯

    Solution:

    5π‘₯2 βˆ’ 6π‘₯

    Taking π‘₯ common,

    = π‘₯(5π‘₯ βˆ’ 6)

    5π‘₯2βˆ’6π‘₯

    3π‘₯=π‘₯(5π‘₯βˆ’6)

    3π‘₯

    =π‘₯

    π‘₯Γ—5π‘₯ βˆ’ 6

    3

    =5π‘₯ βˆ’ 6

    3

    (Method 2:) Cancelling the terms

    Divide the given polynomial by the given monomial.

    (i) (5π‘₯2 βˆ’ 6π‘₯) Γ· 3π‘₯

    Solution:

    5π‘₯2 βˆ’ 6π‘₯

    3π‘₯

    =5π‘₯2

    3π‘₯ βˆ’

    6π‘₯

    3π‘₯

    = (5

    3Γ—π‘₯2

    π‘₯) βˆ’ (

    6

    3Γ—π‘₯

    π‘₯)

    = (5

    3Γ— π‘₯) βˆ’ 2

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    =5

    3π‘₯ βˆ’ 2

    =5π‘₯ βˆ’ (2 Γ— 3)

    3=5π‘₯ βˆ’ 6

    3

    (ii) (Method 1:)

    Divide the given polynomial by the given monomial.

    (ii) (3𝑦8 βˆ’ 4𝑦6 + 5𝑦4) Γ· 𝑦4

    Solution:

    3𝑦8 βˆ’ 4𝑦6 + 5𝑦4

    = (3𝑦4 Γ— 𝑦4) βˆ’ (4𝑦2 Γ— 𝑦4) + (5 Γ— 𝑦4)

    Taking 𝑦4 common

    = 𝑦4(3𝑦4 βˆ’ 4𝑦2 + 5)

    3𝑦8βˆ’4𝑦6+5𝑦4

    𝑦4

    =𝑦4(3𝑦4 βˆ’ 4𝑦4 + 5)

    𝑦4

    = 3𝑦4 βˆ’ 4𝑦2 + 5

    (Method 2:)

    Divide the given polynomial by the given monomial.

    (ii) (3𝑦8 βˆ’ 4𝑦6 + 5𝑦4) Γ· 𝑦4

    Solution:

    3𝑦8 βˆ’ 4𝑦6 + 5𝑦4

    𝑦4

    =3𝑦8

    𝑦4βˆ’4𝑦6

    𝑦4+5𝑦4

    𝑦4

    = 3 Γ— 𝑦8βˆ’4 βˆ’ 4 Γ— 𝑦6βˆ’4 + 5 Γ— 𝑦4βˆ’4 (π‘Žπ‘š

    π‘Žπ‘›= π‘Žπ‘šβˆ’π‘›)

    = 3 Γ— 𝑦4 βˆ’ 4 Γ— 𝑦2 + 5𝑦0

    = 3𝑦4 βˆ’ 4𝑦2 + 5(π‘Ž0 = 1)

    (iii) (Method 1:)

    Divide the given polynomial by the given monomial.

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    8(π‘₯3𝑦2𝑧2 + π‘₯2𝑦3𝑧2 + π‘₯2𝑦2𝑧3) Γ· 4π‘₯2𝑦2𝑧2

    Solution:

    8(π‘₯3𝑦2𝑧2 + π‘₯2𝑦3𝑧2 + π‘₯2𝑦2𝑧3)

    = 8(π‘₯ Γ— π‘₯2𝑦2𝑧2) + (𝑦 Γ— π‘₯2𝑦2𝑧2) + (𝑧 Γ— π‘₯2𝑦2𝑧2)

    Taking π‘₯2𝑦2𝑧2 common

    = 8π‘₯2𝑦2𝑧2(π‘₯ + 𝑦 + 𝑧)

    8(π‘₯3𝑦2𝑧2+π‘₯2𝑦3𝑧2+π‘₯2𝑦2𝑧3)

    4π‘₯2𝑦2𝑧2

    =8π‘₯2𝑦2𝑧2(π‘₯ + 𝑦 + 𝑧)

    4π‘₯2𝑦2𝑧2

    =8

    4Γ—π‘₯2𝑦2𝑧2

    π‘₯2𝑦2𝑧2Γ— (π‘₯ + 𝑦 + 𝑧)

    = 2 Γ— (π‘₯ + 𝑦 + 𝑧)

    = 2(π‘₯ + 𝑦 + 𝑧)

    (Method 2:)

    Divide the given polynomial by the given monomial.

    (iii) 8(π‘₯3𝑦2𝑧2 + π‘₯2𝑦3𝑧2 + π‘₯2𝑦2𝑧3) Γ· 4π‘₯2𝑦2𝑧2

    Solution:

    =8(π‘₯3𝑦2𝑧2 + π‘₯2𝑦3𝑧2 + π‘₯2𝑦2𝑧3)

    4π‘₯2𝑦2𝑧2

    =8π‘₯3𝑦2𝑧2

    4π‘₯2𝑦2𝑧2+8π‘₯2𝑦3𝑧2

    4π‘₯2𝑦2𝑧2+8π‘₯2𝑦2𝑧3

    4π‘₯2𝑦2𝑧2

    = 2π‘₯ + 2𝑦 + 2𝑧

    Taking 2 common

    = 2(π‘₯ + 𝑦 + 𝑧)

    (iv) (Method 1:)

    Divide the given polynomial by the given monomial.

    (iv) (π‘₯3 + 2π‘₯2 + 3π‘₯) Γ· 2π‘₯

    Solution:

    π‘₯3 + 2π‘₯2 + 3π‘₯ = (π‘₯2 Γ— π‘₯) + (2π‘₯ Γ— π‘₯) + (3 Γ— π‘₯)

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    Taking π‘₯ common,

    = π‘₯(π‘₯2 + 2π‘₯ + 3)

    β‡’π‘₯3 + 2π‘₯2 + 3π‘₯

    2π‘₯

    =π‘₯(π‘₯2 + 2π‘₯ + 3)

    2π‘₯

    =π‘₯

    π‘₯Γ—π‘₯2 + 2π‘₯ + 3

    2

    =π‘₯2 + 2π‘₯ + 3

    2

    =1

    2(π‘₯2 + 2π‘₯ + 3)

    (Method 2:)

    Divide the given polynomial by the given monomial.

    (iv) (π‘₯3 + 2π‘₯2 + 3π‘₯) Γ· 2π‘₯

    Solution:

    π‘₯2 + 2π‘₯ + 3

    2π‘₯

    =π‘₯3

    2π‘₯+2π‘₯2

    2π‘₯+3π‘₯

    2π‘₯

    = (1

    2Γ—π‘₯3

    π‘₯) + (

    2

    2Γ—π‘₯2

    π‘₯) + (

    3

    2Γ—π‘₯

    π‘₯)

    = (1

    2Γ— π‘₯2) + (1 Γ— π‘₯) + (

    3

    2Γ— 1)

    =1

    2π‘₯2 + π‘₯ +

    3

    2

    =π‘₯2 + 2π‘₯ + 3

    2

    =1

    2(π‘₯2 + 2π‘₯ + 3)

    (v) (Method 1:)

    Divide the given polynomial by the given monomial.

    (𝑝3π‘ž6 βˆ’ 𝑝6π‘ž3) Γ· 𝑝3π‘ž3

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    Solution:

    𝑝3π‘ž6 βˆ’ 𝑝6π‘ž3

    = (𝑝3π‘ž3 Γ— π‘ž3) βˆ’ (𝑝3π‘ž3 Γ— 𝑝3)

    Taking 𝑝3π‘ž3 common,

    = 𝑝3π‘ž3(π‘ž3 βˆ’ 𝑝3)

    ⇒𝑝3π‘ž6 βˆ’ 𝑝6π‘ž3

    𝑝3π‘ž3

    =𝑝3π‘ž3(π‘ž3 βˆ’ 𝑝3)

    𝑝3π‘ž3

    = π‘ž3 βˆ’ 𝑝3

    (Method 2:)

    Divide the given polynomial by the given monomial.

    (v) (𝑝3π‘ž6 βˆ’ 𝑝6π‘ž3) Γ· 𝑝3π‘ž3

    Solution:

    𝑝3π‘ž6 βˆ’ 𝑝6π‘ž3

    𝑝3π‘ž3

    =𝑝3π‘ž6

    𝑝3π‘ž3βˆ’π‘6π‘ž3

    𝑝3π‘ž3

    =π‘ž6

    π‘ž3βˆ’π‘6

    𝑝3

    = π‘ž6βˆ’3 βˆ’ 𝑝6βˆ’3 (π‘Žπ‘š

    π‘Žπ‘›= π‘Žπ‘šβˆ’π‘›)

    = π‘ž3 βˆ’ 𝑝3

    3. Work out the following divisions.

    (i) (10π‘₯ βˆ’ 25) Γ· 5

    Solution:

    10π‘₯ βˆ’ 25

    = (5 Γ— 2)π‘₯ βˆ’ (5 Γ— 5)

    Taking 5 common,

    = 5(2π‘₯ βˆ’ 5)

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    Dividing,10π‘₯βˆ’25

    5

    =5(2π‘₯ βˆ’ 5)

    5

    = (2π‘₯ βˆ’ 5)

    (ii) (10π‘₯ βˆ’ 25) Γ· (2π‘₯ βˆ’ 5)

    Solution:

    10π‘₯ βˆ’ 25

    = (5 Γ— 2)π‘₯ βˆ’ (5 Γ— 5)

    Taking 5 common,

    = 5(2π‘₯ βˆ’ 5)

    Dividing,(10π‘₯βˆ’25)

    (2π‘₯βˆ’5)

    =5(2π‘₯ βˆ’ 5)

    (2π‘₯ βˆ’ 5)

    = 5

    (iii) 10𝑦(6𝑦 + 21) Γ· 5(2𝑦 + 7)

    Solution:

    10𝑦(6𝑦 + 21)

    = 10𝑦[(3 Γ— 2)𝑦 + (3 Γ— 7)]

    Taking 3 common,

    = 10𝑦 Γ— 3(2𝑦 + 7)

    Dividing, 10𝑦(6𝑦+21)

    5(2𝑦+7)

    =10𝑦 Γ— 3(2𝑦 + 7)

    5 Γ— (2𝑦 + 7)

    = 3 Γ—10

    5Γ— 𝑦 Γ—

    (2𝑦 + 7)

    (2𝑦 + 7)

    = 3 Γ— 2 Γ— 𝑦 Γ— 1

    = 6𝑦

    (iv) 9π‘₯2𝑦2(3𝑧 βˆ’ 24) Γ· 27π‘₯𝑦(𝑧 βˆ’ 8)

    Solution:

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    9π‘₯2𝑦2(3𝑧 βˆ’ 24)

    = 9π‘₯2𝑦2 Γ— [3𝑧 βˆ’ (3 Γ— 8)]

    Taking 3 common,

    = 9π‘₯2𝑦2 Γ— 3(𝑧 βˆ’ 8)

    = 27π‘₯2𝑦2(𝑧 βˆ’ 8)

    Dividing,9π‘₯2𝑦2(3π‘§βˆ’24)

    27π‘₯𝑦(π‘§βˆ’8)

    =27π‘₯2𝑦2(𝑧 βˆ’ 8)

    27π‘₯𝑦(𝑧 βˆ’ 8)

    =27

    27Γ—π‘₯2

    π‘₯×𝑦2

    𝑦×(𝑧 βˆ’ 8)

    (𝑧 βˆ’ 8)

    = 1 Γ— π‘₯ Γ— 𝑦 Γ— 1 (π‘Žπ‘š

    π‘Žπ‘›= π‘Žπ‘šβˆ’π‘›)

    = π‘₯𝑦

    (v) 96 π‘Žπ‘π‘ (3π‘Ž βˆ’ 12)(5𝑏 βˆ’ 30) Γ· 144 (π‘Ž βˆ’ 4)(𝑏 βˆ’ 6)

    Solution:

    96 π‘Žπ‘π‘ (3π‘Ž βˆ’ 12)(5𝑏 βˆ’ 30)

    = 96 π‘Žπ‘π‘ (3π‘Ž βˆ’ (3 Γ— 4))(5𝑏 βˆ’ 30)

    Taking 3 common,

    = 96 π‘Žπ‘π‘ Γ— 3(π‘Ž βˆ’ 4)(5𝑏 βˆ’ 30)

    = 288 π‘Žπ‘π‘ (π‘Ž βˆ’ 4)(5𝑏 βˆ’ 30)

    = 288 π‘Žπ‘π‘ (π‘Ž βˆ’ 4)(5𝑏 βˆ’ 5 Γ— 6)

    Taking 5 common,

    = 288 π‘Žπ‘π‘(π‘Ž βˆ’ 4) Γ— 5(𝑏 βˆ’ 6)

    = 288 Γ— 5 π‘Žπ‘π‘(π‘Ž βˆ’ 4)(𝑏 βˆ’ 6)

    = 1440 π‘Žπ‘π‘ (π‘Ž βˆ’ 4)(𝑏 βˆ’ 6)

    Dividing,

    96 π‘Žπ‘π‘ (3π‘Ž βˆ’ 12)(5𝑏 βˆ’ 30)

    144(π‘Ž βˆ’ 4)(𝑏 βˆ’ 6)

    =1440 π‘Žπ‘π‘ (π‘Ž βˆ’ 4)(𝑏 βˆ’ 6)

    144 (π‘Ž βˆ’ 4)(𝑏 βˆ’ 6)

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    =1440

    144Γ— π‘Žπ‘π‘ Γ—

    (π‘Ž βˆ’ 4)

    (π‘Ž βˆ’ 4)Γ—(𝑏 βˆ’ 6)

    (𝑏 βˆ’ 6)

    = 10 Γ— π‘Žπ‘π‘ Γ— 1 Γ— 1

    = 10 π‘Žπ‘π‘

    4. Divide as directed

    (i) 5(2π‘₯ + 1)(3π‘₯ + 5) Γ· (2π‘₯ + 1)

    Solution:

    5(2π‘₯ + 1)(3π‘₯ + 5) Γ· (2π‘₯ + 1)

    5(2π‘₯ + 1)(3π‘₯ + 5)

    (2π‘₯ + 1)

    = 5(3π‘₯ + 5)

    (ii) 26π‘₯𝑦(π‘₯ + 5)(𝑦 βˆ’ 4) Γ· 13π‘₯(𝑦 βˆ’ 4)

    Solution:

    26π‘₯𝑦(π‘₯ + 5)(𝑦 βˆ’ 4) Γ· 13π‘₯(𝑦 βˆ’ 4)

    =26π‘₯𝑦(π‘₯ + 5)(𝑦 βˆ’ 4)

    13π‘₯(𝑦 βˆ’ 4)

    =26𝑦(π‘₯ + 5)

    13

    =26

    13Γ— 𝑦(π‘₯ + 5)

    = 2 Γ— 𝑦(π‘₯ + 5)

    = 2𝑦(π‘₯ + 5)

    (iii) 52π‘π‘žπ‘Ÿ(𝑝 + π‘ž)(π‘ž + π‘Ÿ)(π‘Ÿ + 𝑝) Γ· 104π‘π‘ž(π‘ž + π‘Ÿ)(π‘Ÿ + 𝑝)

    Solution:

    52π‘π‘žπ‘Ÿ(𝑝 + π‘ž)(π‘ž + π‘Ÿ)(π‘Ÿ + 𝑝) Γ· 104π‘π‘ž(π‘ž + π‘Ÿ)(π‘Ÿ + 𝑝)

    =52π‘π‘žπ‘Ÿ(𝑝 + π‘ž)(π‘ž + π‘Ÿ)(π‘Ÿ + 𝑝)

    104π‘π‘ž(π‘ž + π‘Ÿ)(π‘Ÿ + 𝑝)

    =52

    104Γ—π‘π‘žπ‘Ÿ

    π‘π‘žΓ— (𝑝 + π‘ž) Γ—

    (π‘ž + π‘Ÿ)

    (π‘ž + π‘Ÿ)Γ—(π‘Ÿ + 𝑝)

    (π‘Ÿ + 𝑝)

    =1

    2Γ— π‘Ÿ Γ— (𝑝 + π‘ž) Γ— 1 Γ— 1

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    =1

    2π‘Ÿ(𝑝 + π‘ž)

    (iv) 20(𝑦 + 4)(𝑦2 + 5𝑦 + 3) Γ· 5(𝑦 + 4)

    Solution:

    20(𝑦 + 4)(𝑦2 + 5𝑦 + 3) Γ· 5(𝑦 + 4)

    =20(𝑦 + 4)(𝑦2 + 5𝑦 + 3)

    5(𝑦 + 4)

    20

    5Γ—(𝑦 + 4)

    (𝑦 + 4)Γ— (𝑦2 + 5𝑦 + 3)

    = 4 Γ— 1 Γ— (𝑦2 + 5𝑦 + 3)

    = 4(𝑦2 + 5𝑦 + 3)

    (v) π‘₯(π‘₯ + 1)(π‘₯ + 2)(π‘₯ + 3) Γ· π‘₯(π‘₯ + 1)

    Solution:

    π‘₯(π‘₯ + 1)(π‘₯ + 2)(π‘₯ + 3) Γ· π‘₯(π‘₯ + 1)

    =π‘₯(π‘₯ + 1)(π‘₯ + 2)(π‘₯ + 3)

    π‘₯(π‘₯ + 1)

    =π‘₯

    π‘₯Γ— (π‘₯ + 1

    π‘₯ + 1) Γ— (π‘₯ + 2)(π‘₯ + 3)

    = 1 Γ— 1 Γ— (π‘₯ + 2)(π‘₯ + 3)

    = (π‘₯ + 2)(π‘₯ + 3)

    5. Factorise the expressions and divide them as directed.

    (i) (𝑦2 + 7𝑦 + 10) Γ· (𝑦 + 5)

    Solution:

    𝑦2 + 7𝑦 + 10

    = 𝑦2 + 2𝑦 + 5𝑦 + 10(here, the middle term can be split as7y = 2y + 5y)

    = (𝑦2 + 2𝑦) + (5𝑦 + 10)

    = 𝑦(𝑦 + 2) + 5(𝑦 + 2)

    Taking (𝑦 + 2) common,

    = (𝑦 + 2)(𝑦 + 5)

    Now, dividing

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    (𝑦2 + 7𝑦 + 10) Γ· (𝑦 + 5)

    =𝑦2 + 7𝑦 + 10

    (𝑦 + 5)

    =(𝑦 + 2)(𝑦 + 5)

    (𝑦 + 5)

    = (𝑦 + 2) Γ—(𝑦 + 5)

    (𝑦 + 5)

    = (𝑦 + 2)

    Hint: To split the middle term

    We need to find two numbers whose

    Sum = 7

    Product = 10

    Sum Product

    1 and 10 11 10

    2 and 5 7 10

    So, we write 7𝑦 = 2𝑦 + 5𝑦

    (ii) (π‘š2 βˆ’ 14π‘š βˆ’ 32) Γ· (π‘š + 2)

    Solution:

    π‘š2 βˆ’ 14π‘š βˆ’ 32

    = π‘š2 + 2π‘š βˆ’ 16π‘š βˆ’ 32(here, the middle term can be split as βˆ’14π‘š = 2π‘š βˆ’16π‘š)

    = (π‘š2 + 2π‘š) βˆ’ (16π‘š + 32)

    = π‘š(π‘š + 2) βˆ’ 16(π‘š + 2)

    Taking (π‘š + 2) common,

    = (π‘š + 2)(π‘š βˆ’ 16)

    Now, dividing

    (π‘š2 βˆ’ 14π‘š βˆ’ 32) Γ· (π‘š + 2)

    =π‘š2 βˆ’ 14π‘š βˆ’ 32

    (π‘š + 2)

    =(π‘š + 2)(π‘š βˆ’ 16)

    (π‘š + 2)

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    =(π‘š + 2)

    (π‘š + 2)Γ— (π‘š βˆ’ 16)

    = (π‘š βˆ’ 16)

    Hint: To split the middle term

    We need to find two numbers whose

    Sum = βˆ’14

    Product = βˆ’32

    Sum Product

    1 and βˆ’32 βˆ’31 βˆ’32

    2 and βˆ’16 βˆ’14 βˆ’32

    So, we write βˆ’14π‘š = 2π‘š βˆ’ 16π‘š

    (iii) (5𝑝2 βˆ’ 25𝑝 + 20) Γ· (𝑝 βˆ’ 1)

    Solution:

    5𝑝2 βˆ’ 25𝑝 + 20

    Taking 5 common,

    = 5(𝑝2 βˆ’ 5𝑝 + 4)

    = 5(𝑝2 βˆ’ 𝑝 βˆ’ 4𝑝 + 4)(here, the middle term can be split as- 5p = - p – 4p)

    = 5[(𝑝2 βˆ’ 𝑝) βˆ’ (4𝑝 βˆ’ 4)]

    5[𝑝(𝑝 βˆ’ 1) βˆ’ 4(𝑝 βˆ’ 1)]

    Taking (𝑝 βˆ’ 1) common,

    = 5(𝑝 βˆ’ 1)(𝑝 βˆ’ 4)

    Now, dividing

    (5𝑝2 βˆ’ 25𝑝 + 20) Γ· (𝑝 + 1)

    =5𝑝2 βˆ’ 25𝑝 + 20

    (𝑝 βˆ’ 1)

    =5(𝑝 βˆ’ 1)(𝑝 βˆ’ 4)

    (𝑝 βˆ’ 1)

    = 5 Γ—(𝑝 βˆ’ 1)

    (𝑝 βˆ’ 1)Γ— (𝑝 βˆ’ 4)

    = 5(𝑝 βˆ’ 4)

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    Hint: To split the middle term

    We need to find two numbers whose

    Sum = βˆ’5

    Product = 4

    Sum Product

    βˆ’1 and βˆ’4 βˆ’5 4

    So, we write βˆ’5𝑝 = βˆ’π‘ βˆ’ 4𝑝

    (iv) 4𝑦𝑧(𝑧2 + 6𝑧 βˆ’ 16) Γ· 2𝑦(𝑧 + 8)

    Solution:

    4𝑦𝑧(𝑧2 + 6𝑧 βˆ’ 16)

    = 4𝑦𝑧(𝑧2 βˆ’ 2𝑧 + 8𝑧 βˆ’ 16)(here, the middle term can be split as6z = - 2z + 8z)

    = 4𝑦𝑧[(𝑧2 βˆ’ 2𝑧) + (8𝑧 βˆ’ 16)]

    = 4𝑦𝑧[𝑧(𝑧 βˆ’ 2) + 8(𝑧 βˆ’ 2)]

    Taking (𝑧 βˆ’ 2) common,

    = 4𝑦𝑧(𝑧 βˆ’ 2)(𝑧 + 8)

    Now, dividing

    4𝑦𝑧(𝑧2 + 6𝑧 βˆ’ 16) Γ· 2𝑦(𝑧 + 8)

    =4𝑦𝑧(𝑧 βˆ’ 2)(𝑧 + 8)

    2𝑦(𝑧 + 8)

    =4

    2×𝑦

    𝑦× 𝑧 Γ— (𝑧 βˆ’ 2) Γ—

    (𝑧 + 8)

    (𝑧 + 8)

    = 2 Γ— 𝑧 Γ— (𝑧 βˆ’ 2)

    = 2𝑧(𝑧 βˆ’ 2)

    Hint: To split the middle term

    We need to find two numbers whose

    Sum = 6

    Product = βˆ’16

    Sum Product

    βˆ’1 and 16 15 βˆ’16

    βˆ’2 and 8 βˆ’15 βˆ’16

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    βˆ’2 and 8 6 βˆ’16

    So, we write 6𝑧 = βˆ’2𝑧 + 8𝑧

    (v) 5π‘π‘ž(𝑝2 βˆ’ π‘ž2) Γ· 2𝑝(𝑝 + π‘ž)

    Solution:

    5π‘π‘ž(𝑝2 βˆ’ π‘ž2)

    Using a2 βˆ’ b2 = (a + b)(a βˆ’ b)

    Here a = p and b = q

    = 5π‘π‘ž(𝑝 + π‘ž)(𝑝 βˆ’ π‘ž)

    Now, dividing

    5π‘π‘ž(𝑝2 βˆ’ π‘ž2) Γ· 2𝑝(𝑝 + π‘ž)

    =5π‘π‘ž(𝑝2 βˆ’ π‘ž2)

    2𝑝(𝑝 + π‘ž)

    =5π‘π‘ž(𝑝 + π‘ž)(𝑝 βˆ’ π‘ž)

    2𝑝(𝑝 + π‘ž)

    =5

    2×𝑝

    𝑝× π‘ž Γ—

    (𝑝 + π‘ž)

    (𝑝 + π‘ž)Γ— (𝑝 βˆ’ π‘ž)

    =5

    2Γ— π‘ž Γ— (𝑝 βˆ’ π‘ž)

    =5

    2π‘ž(𝑝 βˆ’ π‘ž)

    (vi) 12π‘₯𝑦(9π‘₯2 βˆ’ 16𝑦2) Γ· 4π‘₯𝑦(3π‘₯ + 4𝑦)

    Solution:

    12π‘₯𝑦(9π‘₯2 βˆ’ 16𝑦2)

    = 12π‘₯𝑦[(3π‘₯)2 βˆ’ (4𝑦)2]

    Using a2 βˆ’ b2 = (a + b)(a βˆ’ b)

    Here a = 3x and b = 4y

    = 12π‘₯𝑦(3π‘₯ + 4𝑦)(3π‘₯ βˆ’ 4𝑦)

    Now, dividing

    12π‘₯𝑦(9π‘₯2 βˆ’ 16𝑦2) Γ· 4π‘₯𝑦(3π‘₯ + 4𝑦)

    =12π‘₯𝑦(9π‘₯2 βˆ’ 16𝑦2)

    4π‘₯𝑦(3π‘₯ + 4𝑦)

    http://www.embibe.com/

  • Class- VIII-CBSE-Mathematics Factorisation

    Practice more on Factorisation Page - 40 www.embibe.com

    =12π‘₯𝑦(3π‘₯ + 4𝑦)(3π‘₯ βˆ’ 4𝑦)

    4π‘₯𝑦(3π‘₯ + 4𝑦)

    =12

    4Γ—π‘₯𝑦

    π‘₯𝑦×(3π‘₯ + 4𝑦)

    (3π‘₯ + 4𝑦)Γ— (3π‘₯ βˆ’ 4𝑦)

    = 3(3π‘₯ βˆ’ 4𝑦)

    (vii) 39𝑦3(50𝑦2 βˆ’ 98) Γ· 26𝑦2(5𝑦 + 7)

    Solution:

    39𝑦3(50𝑦2 βˆ’ 98)

    = 39𝑦3(2 Γ— 25𝑦2 βˆ’ 2 Γ— 49)

    Taking 2 common,

    = 39𝑦3 Γ— 2(25𝑦2 βˆ’ 49)

    = 78𝑦3(25𝑦2 βˆ’ 49)

    = 78𝑦3[(5𝑦)2 βˆ’ (7)2]

    Using a2 βˆ’ b2 = (a + b)(a βˆ’ b)

    Here a = 5y and b = 7

    = 78𝑦3(5𝑦 βˆ’ 7)(5𝑦 + 7)

    Now, dividing

    39𝑦3(50𝑦2 βˆ’ 98) Γ· 26𝑦2(5𝑦 + 7)

    =39𝑦3(50𝑦2 βˆ’ 98)

    26𝑦2(5𝑦 + 7)

    =78𝑦3(5𝑦 + 7)(5𝑦 βˆ’ 7)

    26𝑦2(5𝑦 + 7)

    =78

    26×𝑦3

    𝑦2Γ—(5𝑦 + 7)

    (5𝑦 + 7)Γ— (5𝑦 βˆ’ 7)

    = 3 Γ— 𝑦 Γ— (5𝑦 βˆ’ 7)

    = 3𝑦(5𝑦 βˆ’ 7)

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