Question

1 Lesson no. 5 Expansion formulae Expand using using the formulae 1) (a + 18) (a-20) ​

Answer 1

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Maharashtra Board Class 8 Maths Solutions Chapter 5 Expansion Formulae Practice Set 5.1

August 13, 2019 by Laxmi

Maharashtra State Board Class 8 Maths Solutions Chapter 5 Expansion Formulae Practice Set 5.1

Question 1.

Expand :

i. (a + 2)(a – 1)

ii. (m – 4)(m + 6)

iii. (p + 8) (p – 3)

iv. (13 + x)(13 – x)

v. (3x + 4y) (3x + 5y)

vi. (9x – 5t) (9x + 3t)

vii. (m+23)(m−73)

viii. (x+1x)(x−1x)

ix. (1y+4)(1y−9)

Solution:

i. (a + 2)(a – 1)

= a² + (2 – 1) a + 2 × (-1)

..[∵ (x + A) (x + B) = x² + (A + B)x + AB]

= a² + a – 2

ii. (m – 4)(m + 6)

= m² + (- 4 + 6) m + (-4) × 6

…[∵ (x + a) (x + b) = x² + (a + b)x + ab]

= m² + 2m – 24

iii. (p + 8) (p – 3)

= p² + (8 – 3) p + 8 x (-3)

…[∵ (x + a) (x + b) = x² + (a + b)x + ab]

= p² + 5p – 24

iv. (13 + x) (13 – x)

= (13)² + (x – x) 13 + x × (-x)

…[∵ (x + a) (x + b) = x² + (a + b)x + ab]

= 169 + 0 × 13 – x²

= 169 – x²

v. (3x + 4y) (3x + 5y)

= (3x)² + (4y + 5y) 3x + 4y × 5y

…[∵ (x + a) (x + b) = x² + (a + b)x + ab]

= 9x² + 9y × 3x + 20y²

= 9x² + 27xy + 20y²

vi. (9x – 5t) (9x + 3t)

= (9x)² + [(-5t) + 3t] 9x + (-5t) × 3t

…[∵ (x + a) (x + b) = x² + (a + b)x + ab]

= 81x² + (-2t) × 9x – 15t²

= 81x² – 18xt – 15t²

vii. (m+23)(m−73)

Maharashtra Board Class 8 Maths Solutions Chapter 5 Expansion Formulae Practice Set 5.1 1

viii. (x+1x)(x−1x)

Maharashtra Board Class 8 Maths Solutions Chapter 5 Expansion Formulae Practice Set 5.1 2