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: 8Mathematics
Chapter: 14Factorisation
Exercise: 14.3NCERT Book Solutions

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Class 8 Maths Chapter 14 Exercise 14.3 Solution in Videos

Class 8 Maths Chapter 14 Exercise 14.3 Solution
Class 8 Maths Chapter 14 Exercise 14.3 Explanation
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}

NCERT Solutions for Class 8 Maths Chapter 14 Exercise 14.3
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8 Maths Exercise 14.3 solutions
8 Maths Exercise 14.3 solutions in pdf form




8 Maths Exercise 14.3 solutions in english medium
8 Maths Exercise 14.3 solutions download