NCERT Solutions for Class 8 Maths Chapter 14 Exercise 14.3 (Ex. 14.3) Factorisation updated for 2022-2023 CBSE and state boards like MP board, UP board and other also. All the contents are prepared by subject experts using easier methods. Step by step solutions are given in PDF as well as videos solutions. Contact us if you are not getting solutions in videos or text, we here to help you.

## Class 8 Maths Chapter 14 Exercise 14.3 Solution

Class: 8 | Mathematics |

Chapter: 14 | Factorisation |

Exercise: 14.3 | NCERT Book Solutions |

### CBSE NCERT Class 8 Maths Chapter 14 Exercise 14.3 Solution in Hindi and English Medium

### Class 8 Maths Chapter 14 Exercise 14.3 Solution in Videos

##### Factorisation When Given Expression is Difference of Two Squares

FORMULA: (a – b)² = (a + b) (a – b)

### Factorise: (i) 49x² – 16y² (ii) 64 – x² (iii) y² – 121

We have:

(i) 49x² – 16y²

= (7x)² – (4y)²

= (7x + 4y) (7x – 4y) {Applying formula (a – b)² = (a + b) (a – b)}

(ii) 64 – x²

= (8)² – (x)²

= (8 + x) (8 – x) {Applying formula (a – b)² = (a + b) (a – b)}

(iii) y² – 121

= (y)² – (11)²

= (y + 11) (y – 11) {Applying formula (a – b)² = (a + b) (a – b)}

### Factorise: 25 (x + y)² – 36 (x – 2y)²

We have:

25 (x + y)² – 36 (x – 2y)²

= {5(x + y)}² – {6(x – 2y)}²

= {5 (x + y) + 6(x – 2y)} {5(x + y) – 6(x – 2y)} {Applying formula (a – b)² = (a + b) (a – b)}

= (11x – 7y) (7y – x)

##### Factorisation When Expression is a Perfect Square (Using Identilies)

FORMULAE:

(i) a² + b² + 2ab = (a + b)²

(ii) a² + b² – 2ab = (a – b)²

### Factorise: (i) x² + 10x + 25 (ii) x² – 20x + 100

We have:

(i) x² + 10x + 25

= X2 + (5)² + 2 X x X 5 = (x + 5)² {By applying formula: a² + b² + 2ab = (a + b)²}

(ii) x² – 20x + 100

= x² + 10² – 2 X x X 10 = (x – 10)² {By applying formula: a² + b² – 2ab = (a – b)²}

### Factorise: 4x² + 9y² + 12xy

We have:

4x² + 9y² + 12xy

= (2x)² + (3y)² + 2 X 2x X 3y

= (2x + 3y)²

### Factorise: x² + 8x + 16

We have:

x² + 8x + 16

we can write this expression in the form of a² + b2 + 2ab

= x² + 4² + 2 X 4 X x

= (x + 2)²

### Factorise: p² – 10p + 25

We have:

p² – 10p + 25 we can write this expression in the form of a² + b² – 2ab

= p² + 5² – 2 x p x 5

= (p – 5)² {By applying formula: a² + b² – 2ab = (a – b)²}

### Factorise: a⁴ + 25b⁴ – 10a² b²

we can write this expression in the form of a² + b² – 2ab

= (a²)² + (5b²)² – 2 x a² x 5b²

= (a² – 5b²)² {By applying formula: a² + b² – 2ab = (a – b)²}