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

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: 8Mathematics
Chapter: 14Exercise: 14.3
Topic:Factorisation
Content:NCERT Book Solutions
Medium:Hindi and English Medium

NCERT Solution App Class 8

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:

    1. a² + b² + 2ab = (a + b)²
    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}

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