Welcome back, bright minds! Today, we are diving deep into Chapter 2: Polynomials (Exercise 2.5) of the Class 9 MP Board / NCERT syllabus. This exercise is one of the most crucial parts of algebra, focusing on algebraic identities. Master these step-by-step solutions to ace your school exams and build a solid foundation for competitive exams like NTSE and Olympiads!

Key Algebraic Identities to Remember

• Identity I: (x + y)² = x² + 2xy + y²
• Identity II: (x - y)² = x² - 2xy + y²
• Identity III: x² - y² = (x + y)(x - y)
• Identity IV: (x + a)(x + b) = x² + (a + b)x + ab
• Identity V: (x + y + z)² = x² + y² + z² + 2xy + 2yz + 2zx
• Identity VI: (x + y)³ = x³ + y³ + 3xy(x + y)
• Identity VII: (x - y)³ = x³ - y³ - 3xy(x - y)
• Identity VIII: x³ + y³ + z³ - 3xyz = (x + y + z)(x² + y² + z² - xy - yz - zx)

Step-by-Step Solutions

Question 1: Use suitable identities to find the following products:

(i) (x + 4)(x + 10)

Solution:
Using Identity IV: (x + a)(x + b) = x² + (a + b)x + ab
Here, a = 4 and b = 10.
Therefore, (x + 4)(x + 10) = x² + (4 + 10)x + (4 × 10)
= x² + 14x + 40

Question 2: Evaluate the following products without multiplying directly:

(i) 103 × 107

Solution:
We can write 103 × 107 as (100 + 3)(100 + 7).
Using Identity IV: (x + a)(x + b) = x² + (a + b)x + ab, where x = 100, a = 3, and b = 7.
= (100)² + (3 + 7)(100) + (3 × 7)
= 10000 + (10)(100) + 21
= 10000 + 1000 + 21
= 11021

Question 3: Factorise the following using appropriate identities:

(i) 9x² + 6xy + y²

Solution:
We can rewrite the expression as:
9x² + 6xy + y² = (3x)² + 2(3x)(y) + (y)²
Using Identity I: a² + 2ab + b² = (a + b)², where a = 3x and b = y.
= (3x + y)²
= (3x + y)(3x + y)

Question 4: Expand using suitable identities:

(i) (x + 2y + 4z)²

Solution:
Using Identity V: (a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca
Here, a = x, b = 2y, and c = 4z.
= (x)² + (2y)² + (4z)² + 2(x)(2y) + 2(2y)(4z) + 2(4z)(x)
= x² + 4y² + 16z² + 4xy + 16yz + 8zx

Question 5: Factorise:

(i) 4x² + 9y² + 16z² + 12xy - 24yz - 16xz

Solution:
Notice that the negative signs are associated with terms containing 'z' (i.e., -24yz and -16xz). This indicates that the 'z' term is negative.
We rewrite the expression as:
(2x)² + (3y)² + (-4z)² + 2(2x)(3y) + 2(3y)(-4z) + 2(-4z)(2x)
Using Identity V in reverse: a² + b² + c² + 2ab + 2bc + 2ca = (a + b + c)²
= (2x + 3y - 4z)²
= (2x + 3y - 4z)(2x + 3y - 4z)

Question 6: Evaluate using suitable identities:

(i) (99)³

Solution:
We can express 99 as (100 - 1).
So, (99)³ = (100 - 1)³
Using Identity VII: (a - b)³ = a³ - b³ - 3ab(a - b), where a = 100 and b = 1.
= (100)³ - (1)³ - 3(100)(1)(100 - 1)
= 1000000 - 1 - 300(99)
= 1000000 - 1 - 29700
= 1000000 - 29701
= 970299

Great job on working through these identity-based problems! Practice is the key to mastering algebra. Keep these identities handy in your formula notebook. If you have any doubts, feel free to drop them in the comments section or ask during our next class session. Happy learning!