NCERT Solutions for Class 8 Maths Exercise 14.3

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.

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)

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}