Parents can use OP Malhotra Class 9 Solutions Chapter 4 Factorisation Ex 4(C) to provide additional support to their children.

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

Factorise :

Question 1.

x² + 4x + 4

Solution:

x² + 4x + 4

= (x)² + 2 × x × 2 + (2)²

= (x + 2)²

Question 2.

x² + 6x + 9

Solution:

x² + 6x + 9

= (x)² + 2 x x x 3 + (3)² = (x + 3)²

Question 3.

x² – 10x + 25

Solution:

x² – 10x + 25

= (x)² – 2 × x × 5 + (5)²

= (x – 5)²

Question 4.

4x² – 4x + 1

Solution:

4x² – 4x + 1

= (2x)² – 2 × 2x × 1 + (1)²

= (2x – 1)²

Question 5.

1 – 8x + 16x²

Solution:

1 – 8x + 16x²

= (1)² – 2 x 1 x 4x + (4x)²

= (1 – 4x)²

Question 6.

49x^{4} + 168x²y² + 144y^{4}

Solution:

49x^{4} + 168x²y² + 144y^{4}

= (7x²)² + 2 x 7x² x 12y² + (12y)²

= (7x² + 12y²)²

Question 7.

x² + x + \(\frac { 1 }{ 4 }\)

Solution:

x² + x + \(\frac { 1 }{ 4 }\)

= (x)² + 2 × x × \(\frac { 1 }{ 2 }\) + (\(\frac { 1 }{ 2 }\))²

= (x + \(\frac { 1 }{ 2 }\))²

Question 8.

25p² + \(\frac { 5p }{ 2q }\) + \(\frac { 1 }{ 16q² }\)

Solution:

25p² + \(\frac { 5p }{ 2q }\) + \(\frac { 1 }{ 16q² }\)

= (5p)² + 2 x 5p x \(\frac { 1 }{ 4q }\) x (\(\frac { 1 }{ 4q }\))²

= (5p + \(\frac { 1 }{ 4q }\))²