NCERT Solutions for Class 8 Maths Chapter 14 Exercise 14.3 in Hindi and English Medium updated for CBSE session 2022-2023.

## 8th Maths Exercise 14.3 Solution in Hindi and English Medium

### Class 8 Maths Chapter 14 Exercise 14.3 Solution

Class VIII Mathematics NCERT textbook Ex. 14.3 Factorisation updated for 2022-23 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 | Mathematics |

Chapter: 14 | Exercise: 14.3 |

Topic: | Factorisation |

Content: | NCERT Book Solutions |

Medium: | Hindi and English Medium |

#### 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:

- a² + b² + 2ab = (a + b)²
- 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}