Balbharti Maharashtra State Board Class 8 Maths Solutions covers the Practice Set 5.1 8th Std Maths Answers Solutions Chapter 5 Expansion Formulae.
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. \(\left(m+\frac{2}{3}\right)\left(m-\frac{7}{3}\right)\)
viii. \(\left(x+\frac{1}{x}\right)\left(x-\frac{1}{x}\right)\)
ix. \(\left(\frac{1}{y}+4\right)\left(\frac{1}{y}-9\right)\)
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. \(\left(m+\frac{2}{3}\right)\left(m-\frac{7}{3}\right)\)
viii. \(\left(x+\frac{1}{x}\right)\left(x-\frac{1}{x}\right)\)
ix. \(\left(\frac{1}{y}+4\right)\left(\frac{1}{y}-9\right)\)
Maharashtra Board Class 8 Maths Chapter 5 Expansion Formulae Practice Set 5.1 Intext Questions and Activities
Question 1.
Use the above formulae to fill proper terms in the following boxes. (Textbook pg. no. 23)
Solution:
Question 2.
Expand (x + a) (x + b) using formulae for areas of a square and a rectangle. (Textbook pg. no. 23)
(x + a) (x + b) = x² + ax + bx + ab
(x + a) (x + b) = x² + (a + b) x + ab
Solution:
∴ (x + a) (x + b) = x² + ax + bx + ab
∴ (x + a) (x + b) = x² + (a + b) x + ab