# NCERT Solutions for Class 8 Maths Chapter 14 Exercise 14.3

NCERT Solutions for Class 8 Maths Chapter 14 Exercise 14.3 (Ex. 14.3) Factorisation updated for 2021-2022 CBSE and state boards like MP board, UP board and other also. All the contents are prepared by subject experts using easier methods.

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## 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}