## ML Aggarwal Class 8 Solutions for ICSE Maths Chapter 11 Factorisation Ex 11.3

Question 1.

Factorise the following expressions using algebraic identities:

(i) x^{2} – 12x + 36

(ii) 36p^{2} – 60pq + 25q^{2}

(iii) 9y^{2} + 66xy + 121y^{2}

(iv) a^{4} + 6a^{2}b^{2} + 9b^{4}

(v) x^{2} + \(\frac{1}{x^{2}}\) + 2

(vi) x^{2} + x + \(\frac{1}{4}\)

Solution:

Using (a + b)^{2} = a^{2} + 2ab +b^{2} and (a – b)^{2} = a^{2} – 2ab + b^{2}

(i) y^{2} – 12x + 36

= (x)^{2} – 2 × x × 6 + (6)2^{2}

= (x – 6)^{2}

(ii) 36p^{2} – 60pq + 25q^{2}

= (6p)^{2} – 2 × 6p × 5q + (5q)^{2}

= (6p – 5q)^{2}

(iii) 9x^{2} + 66xy + 121 y^{2}

= (3x)^{2} + 2 × 3x × 11y + (11y)^{2}

= (3x + 11 y)^{2}

(iv) a^{4} + 6a^{2}b^{2} + 9b^{4}

= (a^{2})^{2} + 2 × 2a^{2} × 3b^{2} + (3b^{2})^{2}

= (a^{2} + 3b^{2})^{2}

(v) x^{2} + \(\frac{1}{x^{2}}\) + 2

(vi) x^{2} + x + \(\frac{1}{4}\)

**Factorise the following (2 to 13) expressions:
**Question 2.

(i) 4p

^{2}– 9

(ii) 4x

^{2}– 169y

^{2}

Solution:

(i) 4p

^{2}– 9

= (2p)

^{2}– (3)

^{2}

= (2p + 3) (2p – 3)

(ii) 4x^{2} – 169y^{2}

= (2x)^{2} – (13y)^{2}

= (2x + 13y) (2x – 13y)

Question 3.

(i) 9x^{2}y^{2} – 25

(ii) 16x^{2} – \(\frac{1}{144}\)

Solution:

(i) 9x^{2}y^{2} – 25

= (3xy)^{2} – (5)^{2}

= (3xy + 5) (3xy – 5)

Question 4.

(i) 20x^{2} – 45y^{2}

(ii) \(\frac{9}{16}\) – 25a^{2}b^{2}

Solution:

(i) 20x^{2} – 45y^{2}

= 5 (4x^{2} – 9y^{2})

= 5[(2x)^{2} – (3y)^{2}]

= 5 (2x + 3y) (2x – 3y)

Question 5.

(i) (2a + 3b)^{2} – 16c^{2}

(ii) 1 – (b – c)^{2}

Solution:

(i) (2a + 3b)^{2} – 16c^{2}

= (2a + 3b)^{2} – (4c)^{2}

= (2a + 3b + 4c) (2a + 3b – 4c)

(ii) 1 – (b – c)^{2}

= (1)^{2} – (b – c)^{2}

= [1 + b – c)] [1 – (b – c)]

= (1 +b – c)(1 – b + c)

Question 6.

(i) 9 (x + y)^{2} – x^{2}

(ii) (2m + 3n)^{2} – (3m + 2n)^{2}

Solution:

(i) 9 (x + x)^{2} – x^{2}

= [3 (x + y)]^{2} – [x]^{2}

= [3 (x + y) + x] [3 (x + y) – x]

= (3x + 3y + x) (3x + 3y – x)

= (4x + 3y) (2x + 3x)

(ii) (2m + 3n)^{2} – (3m + 2n)^{2}

= (4m^{2} + 9n^{2} + 12mn) – (9m^{2} + 4n^{2} + 12mn)

= 4m^{2} + 9n^{2} + 12mn – 9m^{2} – 4m^{2} – 12mn

= 4m^{2} + 9n^{2} – 9m^{2} – 4n^{2}

= – 5m^{2} + 5n^{2} = 5 (n^{2} – m^{2})

= 5 (m + n) (n – m)

Question 7.

(i) 25 (a + b)^{2} – 16 (a – b)^{2}

(ii) 9 (3x + 2)^{2} – 4 (2x – 1)^{2}

Solution:

(i) 25 (a + b)^{2} – 16 (a – b)^{2}

= [5 (a + b)]^{2} – [4 (a – b)]^{2}

= (5a + 5b)^{2} – (4a – 4b)^{2}

= [(5a + 5b)^{2} + (4a – 4b)] [(5a + 5b) – (4a – 4b)]

= (5a + 5b + 4a – 4b) (5a + 5b – 4a + 4b)

= (9a + ft) (a + 9ft)

(ii) 9 (3x + 2)^{2} – 4 (2x – 1)^{2}

= [3 (3x + 2)]^{2} – [2 (2x – 1)]^{2}

= (9x + 6)^{2} – (4x – 2)^{2}

= [(9x + 6) + (4x – 2)] [(9x + 6) – (4x – 2)]

= (9x + 6 + 4x – 2) (9x + 6 – 4x + 2)

= (13x + 4) (5x + 8)

Question 8.

(i) x^{3} – 25x

(ii) 63p^{2}q^{2} – 7

Solution:

(i) x^{3} – 25x

= x (x^{2} – 25) = x [(x)^{2} – (5)^{2}]

= x (x + 5) (x – 5)

(ii) 63p^{2}q^{2} – 7

= 7 (9p^{2}q^{2} – 1)

= 7 [(3pq)^{2} – (1)^{2}]

= 7 (3pq + 1) (3pq – 1)

Question 9.

(i) 32a^{2}b – 72b^{3}

(ii) 9 (a + b)^{3} – 25 (a + b)

Solution:

(i) 32 a^{2}b – 72b^{3}

= 8b (4a^{2} – 9b^{2}) ⇒ 8b [(2a)^{2} – (3b)^{2}]

= 8b (2a + 3b) (2a – 3b)

(ii) 9 (a + b)^{3} – 25 (a + b)

= (a + b) [9 (a + b)^{2} – 25]

= (a + b) [{3 (a + b)}^{2} – (5)^{2}]

= (a + 6) [(3a + 3b)^{2} – (5)^{2}]

= (a + b) [(3a + 3b + 5) (3a + 36 – 5)]

= (a + b) (3a + 3b + 5) (3a + 3b – 5)

Question 10.

(i) x^{2} – y^{2} – 2y – 1

(ii) p^{2}– 4pq + 4q^{2} – r^{2}

Solution:

(i) x^{2} – y^{2} – 2y – 1

= x^{2} – (y^{2} + 2y + 1)

= (x)^{2} – (y + 1)^{2}

= [x + (y + 1)] [x – (y + 1)]

= (x + y + 1)(x – y – 1)

(ii) p^{2} – 4pq + 4q^{2} – r^{2}

= (p)^{2} – 2 × p × 2q + (2q)^{2} – r^{2}

{∵ (a – b)^{2} = a2 – 2ab + b^{2}

a^{2} – b^{2} = (a + b)(a – b)}

= (p – 2q)^{2} – (r)^{2}

= (p – 2q + r)(p – 2q – r)

Question 11.

(i) 9x^{2} – y^{2} + 4y – 4

(ii) 4a^{2} – 4b^{2} + 4a + 1

Solution:

(i) 9x^{2} – y^{2} + 4y – 4

= 9x^{2} – (y^{2} – 4y + 4)

= 9x^{2} – (y – 2)^{2}

= (3x)^{2} (y – 2)^{2}

= {3x + (y – 2)} {3x – (y – 2)}

= (3x + y – 2) (3x – y + 2)

(ii) 4a^{2} – 4b^{2} + 4a + 1

= (4a^{2} + 4a + 1) – 4b^{2}

= (2a + 1)^{2} – (2b)^{2}

= (2a + 2b + 1) (2a – 2b + 1)

Question 12.

(i) 625 – p^{4}

(ii) 5y^{5} – 405y

Solution:

(i) 625 – p^{4}

= (25)^{2} – (p^{2})^{2}

= (25 + p^{2}) (25 – p^{2})

= (25 + p^{2}) [(5)^{2} – (p)^{2}]

= (25 +p^{2}) (5 + p) (5 – p)

(ii) 5y^{5} – 405y

= 5y(y^{4} – 81)

= 5y [(y^{2})^{2} – (9)^{2}]

= 5y (y^{2} + 9) (y^{2} – 9)

= 5y (y^{2} + 9) [(y)^{2} – (3)^{2}

= 5y (y^{2} + 9) (y + 3) (y – 3)

Question 13.

(i) x^{4} – y^{4} + x^{2} – y^{2}

(ii) 64a^{2} – 9b^{2} + 42bc – 49c^{2}

Solution:

(i) x^{4} – y^{4} + x^{2} – y^{2}

= [(x^{2})^{2} – (y^{2})^{2}] + (x^{2} – y^{2})

{a^{2} – b^{2} = (a + b) (a – b)}

= (x^{2} + y^{2}) (x^{2} – y^{2}) + 1(x^{2} – y^{2})

= (x^{2} – y^{2}) (x^{2} + y^{2} + 1)

= (x + y(x – y)(x^{2} + y^{2} + 1)

(ii) 64a^{2} – 9b^{2} + 42bc – 49c^{2}

= 64a^{2} – [9b^{2} – 42bc + 49c^{2}]

= (8a)^{2} – [(3b)^{2} – 2 × 3b × 7c + (7c)^{2}]

{∵ a^{2} + b^{2} – 2ab = (a – b)^{2}

a^{2} – b^{2} = (a + b)(a – ^{2})}

= (8a)^{2} – (3b – 7c)^{2}

= (8a + 3b – 7c) (8a – 3b + 7c)