Chapter 7: Algebraic Expressions and Identities & Factorisation

Solution:-

(d) x² +3x + 9

The value of (3x³ +9x² + 27x) ÷ 3x = (3x³ + 9x² + 27x)/3x

Takeout 3x as common,

= 3x (x² + 3x + 9)/3x

= x² + 3x + 9

Solution:-

(c) 2a² + 2b²

(a + b)² + (a – b)² = (a² + b² + 2ab) + (a² + b² – 2ab)

= (a² + a²) + (b² + b²) + (2ab – 2ab)

= 2a² + 2b²

Solution:-

(a) 4ab

The value of (a + b)² – (a – b)² = (a² + b² + 2ab) – (a² + b² – 2ab)

= a² – a² + b² – b² + 2ab + 2ab

= 4ab

In questions 34 to 58, fill in the blanks to make the statements true:

Solution:-

The product of two terms with like signs is a positive term.

Let us assume two like terms are, 3p and 2q

= 3p × 2q

= 6pq

Solution:-

The product of two terms with unlike signs is a negative term.

Let us assume two unlike terms are, – 3p and 2q

= -3p × 2q

= – 6pq

Solution:-

a (b + c) = a × b + a × c. … [by using left distributive law]

= ab + ac

Solution:-

(a – b) (a – b) = (a – b)²= a² – 2ab + b²

(a – b) (a – b)= a × (a – b) – b × (a – b)

= a² – ab – ba + b²

= a² – 2ab + b²

Solution:-

a² – b² = (a + b) (a – b) … [from the standard identities]

Solution:-

(a – b)² + (2ab – 2b²) = a² – b²

= (a – b)² + (2ab – 2b²)

= a² + b² – 2ab + 2ab – 2b²

= a² – b²

Solution:-

(a + b)² – 2ab = a² + b²

= (a + b)² – 2ab

= a² + 2ab + b² – 2ab

= a² + b²