Question

All the formula of polynomials (like a - b whole squre , a+ b whole squre, a squre - b square

Answer 1

How this identity of (a + b)² = a² + b² + 2ab is obtained:

Taking LHS of the identity:
(a + b)²

This can also be written as:
= (a + b) (a + b)

Multiply as we do multiplication of two binomials and we get:
= a(a + b) + b(a + b)
= a² + ab + ab + b²

Add like terms and we get:
= a² + 2ab + b²

Rearrange the terms and we get:
= a² + b² + 2ab

Hence, in this way we obtain the identity i.e. (a + b)² = a² + b² + 2ab
How this identity of (a - b)² = a² + b² - 2ab is obtained:

Taking LHS of the identity:
(a - b)²

This can also be written as:
= (a - b) (a - b)

Multiply as we do multiplication of two binomials and we get:
= a(a - b) - b(a - b)
= a² - ab - ab + b²

Add like terms and we get:
= a² - 2ab + b²

Rearrange the terms and we get:
= a² + b² - 2ab

Hence, in this way we obtain the identity i.e. (a - b)² = a² + b² - 2ab
How this identity of a² - b² = (a + b) (a - b) is obtained:

Taking RHS of the identity:
(a + b) (a - b)

Multiply as we do multiplication of two binomials and we get:
= a(a - b) + b(a - b)
= a² - ab + ab - b²

Solve like terms and we get:
= a² - b²

Hence, in this way we obtain the identity i.e. a² - b² = (a + b) (a - b)

• (a + b)3 = a3 + 3a2b + 3ab2 + b3
• (a - b)3 = a3 - 3a2b + 3ab2 - b3

• a2 + b2 = (a – b)2 + 2ab
• (a + b + c)2 = a2 + b2 + c2 + 2ab + 2ac + 2bc
• (a – b – c)2 = a2 + b2 + c2 – 2ab – 2ac + 2bc

hope you understand