## Maharashtra State Board Class 7 Maths Solutions Chapter 14 Algebraic Formulae – Expansion of Squares Practice Set 53

Question 1.

Factorize the following expressions:

i. p² – q²

ii. 4x² – 25y²

iii. y² – 4

iv. \(\mathrm{p}^{2}-\frac{1}{25}\)

v. \(9 x^{2}-\frac{1}{16} y^{2}\)

vi. \(x^{2}-\frac{1}{x^{2}}\)

vii. a²b – ab

viii. 4x²y – 6x²

ix. \(\frac{1}{2} y^{2}-8 z^{2}\)

x. 2x² – 8y²

Solution:

i. p² – q²

Here, a = p, b = q

∴ p² – q² = (p + q)(p – q)

….[(a² – b²) = (a + b)(a – b)]

ii. 4x² – 25y²

= (2x)² – (5y)²

Here, a = 2x, b = 5y

∴ (2x)² – (5y)² = (2x + 5y)(2x – 5y)

….[(a² – b²) = (a + b)(a – b)]

iii. y² – 4

= y² – 2²

Here, a = y, b = 2

∴ y² – 2² = (y + 2)(y – 2)

….[(a² – b²) = (a + b)(a – b)]

iv. \(\mathrm{p}^{2}-\frac{1}{25}\)

Here a = \(\frac { 1 }{ 25 }\), b = \(\frac { 1 }{ 5 }\)

\(p^{2}-\left(\frac{1}{5}\right)^{2}=\left(p+\frac{1}{5}\right)\left(p-\frac{1}{5}\right)\)

….[(a² – b²) = (a + b)(a – b)]

v. \(9 x^{2}-\frac{1}{16} y^{2}\)

Here a = 3x, b = \(\frac { 1 }{ 4 }y\)

∴\((3 x)^{2}-\left(\frac{1}{4} y\right)^{2}=\left(3 x+\frac{1}{4} y\right)\left(3 x-\frac{1}{4} y\right)\)

….[(a² – b²) = (a + b)(a – b)]

vi. \(x^{2}-\frac{1}{x^{2}}\)

Here a = x, b = \(\frac { 1 }{ x }\)

\(x^{2}-\left(\frac{1}{x}\right)^{2}=\left(x+\frac{1}{x}\right)\left(x-\frac{1}{x}\right)\)

….[(a² – b²) = (a + b)(a – b)]

vii. a²b – ab

= a (ab – b)

= ab (a – 1)

viii. 4x²y – 6x²

= 2 (2x²y – 3x²)

= 2x² (2y – 3)

ix. \(\frac{1}{2} y^{2}-8 z^{2}\)

x. 2x² – 8y²

= 2 (x² – 4y²)

= 2 [x² – (2y)²]

= 2(x + 2y)(x – 2y)

….[(a² – b²) = (a + b)(a – b)]