Class 9 Mathematics
Chapter 2: Polynomials
Exercise 2.4 – Stepwise Solutions
Important Identities Used
(a + b)² = a² + 2ab + b²
(a − b)² = a² − 2ab + b²
(a + b)(a − b) = a² − b²
(a + b)³ = a³ + 3a²b + 3ab² + b³
a³ + b³ = (a + b)(a² − ab + b²)
a³ − b³ = (a − b)(a² + ab + b²)
Q1. Use identities to find products
(i) (x + 4)(x + 10)
= x² + 10x + 4x + 40
= x² + 14x + 40
(ii) (x + 8)(x − 10)
= x² − 10x + 8x − 80
= x² − 2x − 80
(iii) (3x + 4)(3x − 5)
= 9x² − 15x + 12x − 20
= 9x² − 3x − 20
(iv) (y² + 3/2)(y² − 3/2)
= y⁴ − 9/4
(v) (3 − 2x)(3 + 2x)
= 9 − 4x²
Q2. Evaluate without multiplying directly
(i) 103 × 107
= (105 − 2)(105 + 2)
= 105² − 4
= 11025 − 4 = 11021
(ii) 95 × 96
= 95(100 − 4)
= 9500 − 380 = 9120
(iii) 104 × 96
= (100 + 4)(100 − 4)
= 10000 − 16 = 9984
Q3. Factorise
(i) 9x² + 6xy + y²
= (3x + y)²
(ii) 4y² − 4y + 1
= (2y − 1)²
(iii) x² − y²/100
= (x − y/10)(x + y/10)
Q4. Expand
(i) (x + 2y + 4z)²
= x² + 4y² + 16z² + 4xy + 8xz + 16yz
(ii) (2x − y + z)²
= 4x² + y² + z² − 4xy + 4xz − 2yz
(iii) (−2x + 3y + 2z)²
= 4x² + 9y² + 4z² − 12xy − 8xz + 12yz
(iv) (3a − 7b − c)²
= 9a² + 49b² + c² − 42ab − 6ac + 14bc
(v) (−2x + 5y − 3z)²
= 4x² + 25y² + 9z² − 20xy + 12xz − 30yz
(vi) (1/4 a − 1/2 b + 1)²
= a²/16 + b²/4 + 1 − ab/4 + a/2 − b
Q5. Factorise
(i) = (2x + 3y + 4z)²
(ii) = (√2x − y − 2z)²
Q6. Expand cubes
(i) (2x + 1)³
= 8x³ + 12x² + 6x + 1
(ii) (2a − 3b)³
= 8a³ − 36a²b + 54ab² − 27b³
(iii) (3/2 x + 1)³
= 27/8 x³ + 27/4 x² + 9/2 x + 1
(iv) (x − 2/3 y)³
= x³ − 2x²y + 4/3 xy² − 8/27 y³
Q7. Evaluate
(i) (99)³
= 970299
(ii) (102)³
= 1061208
(iii) (998)³
= 994011992
Q8. Factorise
(i) = (2a + b)³
(ii) = (2a − b)³
(iii) = (3 − 5a)³
(iv) = (4a − 3b)³
Q9.
Identities verified
Q10.
(i) = (3y + 5z)(9y² − 15yz + 25z²)
(ii) = (4m − 7n)(16m² + 28mn + 49n²)
Q11.
= (3x + y + z)(9x² + y² + z² − 3xy − 3yz − 3zx)
Q12.
Identity verified
Q13.
x³ + y³ + z³ = 3xyz
Q14.
(i) = 0
(ii) = 0
Q15.
(i) (5a − 3)(5a − 4)
(ii) (5y − 3)(7y + 4)