# Factorization EXERCISE 3.2.3 – Class 9

1. Factorize

(i) x2 + 9x + 18

Solution:

= x2 + 6x + 3x + 18

= x (x + 6) + 3 (x + 6)

= (x + 6) (x + 3)

(ii) y2 + 5y – 24

Solution:

= y2 + 8y – 3y – 24

= y (y + 8) – 3 (y + 8)

= (y + 8) (y – 3)

(iii) 7y2 + 49y +84

Solution:

= 7 (y2 + 7y + 12)

= 7 (y2 + 4y + 3y + 12)

=7 2017

=7 (y + 4) (y + 3)

(iv) 40 + 3x – x2

Solution:

= – [x2 – 3x – 40]

= 3 – [x2 – 8x + 5x – 40]

= – [x (x – 8) + 5 (x – 8)]

= [(x – 8) (x + 5)]

= (8 – x) (x + 5)

(v) m2 + 17mn – 84n2

Solution:

= m2 + 21mn – 4mn – 84n2

= m (m + 21n) – 4n (m + 21n)

= (m + 21m) (m – 4n)

(vi) 117p2 + 2pq – 24q2

Solution:

= 117p2 + 54pq – 52pq – 24q2

= 9p (13p + 6q) – 3q (13p + 6q)

= (9p – 3q) (13p + 6q)

(vii) 15x2 – r – 28

Solution:

= 15x2 – 21x + 20x – 28

= 3x (5x – 7) + 4 (5x – 7)

= (3x + 4) (5x – 7)

(viii) 2x2 – x – 21

Solution:

= 2x2 – 7x + 6x – 21

= x (2x – 7) + 3 (2x – 7)

= (2x – 7) (x + 3)

(ix) 8k2 – 22k – 21

Solution:

= 8k2 – 28k + 6k – 21

= 4k (2k – 7) + 3(2k – 7)

= (2k – 7) (4k + 3)

(x) 𝟏/𝟑 x2 – 2x – 9

Solution:

= (x2+ 6x−27)/3

= 1/3 [x2 – 6x – 27]

= 1/3 [x2 – 9x + 3x – 27]

= 1/3 [x (x – 9) + 3 (x – 9)]

=1/3 (x – 9) (x + 3)

1. Factorize

(i) √𝟓 x2 + 2x – 3√𝟓

Solution:

= √5 x2 + 5x – 3x – 3√5

= √5 x (x + √5 ) – 3 ((x +√ 5 )

= (x + √5 ) (√5 x – 3)

(ii) √𝟑 a2 + 2a – 5√𝟑

Solution:

= √3a2 + 5a – 3a – 5√3 – 5√3 – √3

= a (√3 a + 5) – √3 (√3 a + 5)

= (√3 a + 5) (a – √3 )

(iii) 7√2 y2 – 10y – 4√2

Solution:

= 7√2 y2 – 14y + 4y – 4√2

= 7√2 y (y – √2) + 4 (y – 2)

= (y – √2) (7√2 y + 4)

(iv) 6√𝟑 z2 – 47z – 5√𝟑

Solution:

= 6√3 z2 – 45z – 2z – 5√3

= 3√3 z (2z –5√3) – (2z – 5√3)

= (2z – 5√3) (3√3z –1)

(v) 4√𝟑x2 + 5x – 2√𝟑

Solution:

= 4√3 x2 + 8x – 3x – 2√3

= 4x (√3x + 2) – √3(√3x + 2)

= (√3x + 2) (4x –√3)

1. Factorize

(i) 2 (x + y)2 – 9 (x + y) – 5

Solution:

Let x + y = p

= 2p2 – 9p – 5

= 2p2 – 10p + p – 5

= 2p(p – 5) + 1 (p – 5)

= (p – 5) (2p + 1)

= (x + y – 5) [2(x + y) + 1]

= (x + y – 5) (2x + 2y + 1)

(ii) 2(a – 2b)2 – 25(a – 2b) + 12

Solution:

Put a – 2b = x

= 2x2 – 25x + 12

= 2x2 – 24x – x + 12

= 2x (x – 12) – 1 (x – 12)

= (x – 12) (2x – 1)

= (a – 2b – 12) [2(a – 2b) – 1]

= (a – 2b – 12) [2a – 4b – 1]

(iii) 12(z + 1)2 – 25(z + 1) (x + 2) + 12 (x + 2)2

Solution:

Let z + 1 = a x + 2 = b

= 12a2 – 25ab + 12b2

= 12a2 – 16ab – 9ab + 12b2

= 4a (3a – 4b) – 3b (3a – 4b)

= (3a – 4b) (4a – 3b)

= [3 (z + 1) – 4 (x + 2)] [4 (z + 1) – 3 (x + 2)]

= (3z + 3 – 4x – 8) (4z + 4 3x – 2)

= (3z – 4x – 5) (4z – 3x – 2)

(iv) 9(2x – y) – 4(2x – y) – 13

Solution:

9(2x – y) – 4(2x – y) – 13

Put 2x – y = a

= 9a + 9a – 13a – 13

= 9a (a + 1) – 13 (a +1)

= (a + 1) (9a – 13)

= (2x – y + 1) [9 (2x – y) – 13]

= (2x – y + 1) (18x – 9y – 13)

1. Factorize

(i) x4 – 3x2 + 2

Solution:

Put x2 = a

a2 – 3a + 2

= a2 – 2a – a + 2

= a (a – 2) – 1 (a – 2)

= (a – 2) (a – 1)

= (x2 – 2) (x2 – 1)

= (x – 2 ) (x + 2 ) (x – 1) (x + 1) [∵a2 – b2 = (a + b) (a – b)]

(ii) 4x4 + 7x2 – 2

Solution:

Put x2 = a

= 4a2 + 7a – 2

= 4a2 + 8a – a – 2

= 4a (a + 2) – 1 (a + 2)

= (a + 2) (4a – 1) [∵a2 – b2 = (a + b) (a – b)]

= (x2 + 2) (4x2 – 1)

= (x2 + 2) (2x + 1) (2x – 1)

(iii) 3x3 – x2 – 10x

Solution:

= x (3x2 – x – 10)

= x (3x2 – 6x + 5x – 10)

= x [3x (x – 2) + 5 (x – 2)]

= x (x – 2) (3x + 5)

(iv) 8x3 + 2x2y – 15xy2

Solution:

= x (8x2 – 2xy – 15y2)

= x (8x2 – 12xy + 10xy – 15y2)

= x [4x (2x – 3y) + 5y (2x – 3y)]

= x (2x – 3y) (4x + 5y)

(v) x6 – 7x3 – 8

Solution:

Put x3 = a

a3 – 7a – 8 [∵a3 + b3 = (a + b) (a2 – ab + b2)]

= a3 – 8a + a – 8 [∵a3 – b3 = (a – b) (a2 – ab + b2)]

= a (a – 8) + 1 (a – 8)

= (a – 8) (a + 1)

= (x3 – 3) (x3 + 1)

= (x3 – 3) (x3 + 1)

= (x + 1) (x2 – x + 1) (x – 2) (x2 + 2x + 4)