OP Malhotra Class 9 Maths Solutions Chapter 4 Factorisation Ex 4(F)

Practicing OP Malhotra Class 9 Solutions Chapter 4 Factorisation Ex 4(F) is the ultimate need for students who intend to score good marks in examinations.

S Chand Class 9 ICSE Maths Solutions Chapter 4 Factorisation Ex 4(F)

Factorise :

Question 1.
a² + 5a + 6
Solution:
a² + 5a + 6
= a² + 2a + 3a + 6

= a (a + 2) + 3 (a + 2)
= (a + 2) (a + 3)

Question 2.
a² + 6a + 8
Solution:
a² + 6a + 8
= a² + 2 a + 4a + 8

= a (a + 2) + 4 (a + 2) = (a + 2) (a + 4)

Question 3.
p² + 10p + 16
Solution:
p² + 10p + 16
= p² + 2p + 8p + 16

= p(p + 2) + 8(p + 2)
= (p + 2)(p + 8)

Question 4.
a² + 13a + 42
Solution:
a² + 13a + 42
= a² + 7a + 6a + 42

= a (a + 7) + 6 (a + 7)
= (a + 7) (a + 6)

Question 5.
a² + 25a – 54
Solution:
a² + 25a – 54

= a (a + 27) – 2 (a + 27)
= (a + 27) (a – 2)

Question 6.
x² + 5x – 176
Solution:
x² + 5x – 176
= x² + 16a – 11a – 176

= x (x + 16) – 11 (a + 16)
= (x + 16) (x – 11)

Question 7.
y² – 18y + 65
Solution:
y² – 18y + 65
= y² – 13y – 5y + 65

= y (y – 13) – 5 (y – 13)
= (y – 13) (y – 5)

Question 8.
m² – 29m + 204
Solution:
m² – 29m + 204
= m² – 17a – 12m + 204

= m (m – 17) – 12 (m – 17)
= (m – 17) (m – 12)

Question 9.
b² – 2b – 48
Solution:
b² – 2b – 48
= b² – 8b + 6b – 48

= b (b – 8) + 6 (b – 8)
= (b – 8) (b + 6)

Question 10.
x² – 11x – 102
Solution:
x² – 11x – 102
= x² – 17x + 6x – 102

= x (x – 17) + 6 (x – 17)
= (x – 17) (x + 6)

Question 11.
3 – 4t + t²
Solution:
3 – 4t + t²
= 3 – 3t – t + t²

= 3(1 – t) – t(1 – t)
= (1 – t) (3 – t)

Question 12.
51 – 20k + k²
Solution:
51 – 20k + k²
= 51 – 17k – 3k + k²

= 17 (3 – k) – k (3 – k)
= (3 – k) (17 – k)

Question 13.
2x² – 10x + 12
Solution:
2x² – 10x + 12
= 2 (x² – 5x + 6)
= 2 [x² – 2x – 3x + 6]

= 2 [x (x – 2) – 3 (x – 2)]
= 2 (x – 2) (x – 3)

Question 14.
3x³ – 33x² + 84x
Solution:
3x³ – 33x² + 84x = 3x [x² – 11x + 28]
= 3x [x² – 7x – 4x + 28]

= 3x [x (x – 7) – 4 (x – 7)]
= 3x (x – 7) (x – 4)

Question 15.
5y² – 45y – 110
Solution:
5y² – 45y – 110 = 5 (y² – 9y – 22)
= 5 [y² – 11y + 2y – 22]

= 5 [y (y – 11) + 2(y – 11)]
= 5 (y – 11) (y + 2)

Question 16.
x⁴ – 13x² + 36
Solution:
x⁴ – 13x² + 36
= x⁴ – 9x² – 4x² + 36

= x² (x² – 9) – 4 (x² – 9)
= (x² – 9) (x² – 4)
= [(A’)² – (3)²] [(x)² – (2)²]
= (x + 3) (x – 3) (x + 2) (x – 2)

Question 17.
x² + 3xy – 88y²
Solution:
x² + 3xy – 88y²
= x² + 11 xy – 8xy – 88y²

= x (x + 11y) – 8y (x + 11y)
= (x + 11 y) (x – 8y)

Question 18.
x⁴ – x²y² – 72y⁴
Solution:
x⁴ – x²y² – 72y⁴
= x⁴ – 9x²y² + 8x² y² – 72y⁴
= x² (x² – 9y²) + 8y² (x² – 9y²)

= (x² – 9y²) (x² + 8y²)
= (x² + 8y²) [(x)² – (3y)²]
= (x² + 8y) (x + 3y) (x – 3y)
= (x + 3y) (x – 3y) (x² + 8y²)

Question 19.
a³b³ – 9a²b² + 20ab
Solution:
a³b³ – 9a²b² + 20ab
= ab [a²b² – 9ab + 20]
= ab [a²b² – 4ab – 5ab + 20]

= ab [ab (ab – 4) – 5 (ab – 4)]
= ab (ab – 4) (ab – 5)

Question 20.
(x² + x)² + 4 (x² + x) – 21
Solution:
(x² + x)² + 4 (x² + x) – 21
Let x² + x = a, then
a² + 4a – 21 = a² + 7a – 3a – 21

= a (a + 7) – 3 (a – 7)
= (a + 7) (a – 3)
= (x² + x + 7) (x² + x – 3)