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Chapter 2 of the Class 9 NCERT Mathematics textbook, titled "Polynomials," covers the basic concepts and operations related to polynomials. Exercise 2.4 focuses on solving problems involving polynomials, including evaluating polynomials at given points, performing operations like addition and subtraction, and factoring polynomials. This exercise aims to strengthen students' understanding of polynomial functions and their applications.
This section provides detailed solutions for Exercise 2.4 from Chapter 2 of the Class 9 NCERT Mathematics textbook. The solutions include step-by-step explanations for each problem, helping students grasp the concepts of polynomials and apply them effectively.
(i) (x + 4) (x + 10)
Solution:
Using formula, (x + a) (x + b) = x2 + (a + b)x + ab
[So, a = 4 and b = 10]
(x + 4) (x + 10) = x2 + (4 + 10)x + (4 × 10)
= x2 + 14x + 40
(ii) (x + 8) (x - 10)
Solution:
Using formula, (x + a) (x + b) = x2 + (a + b)x + ab
[So, a = 8 and b = −10]
(x + 8) (x - 10) = x2 + (8 + (-10) )x + (8 × (-10))
= x2 + (8 - 10) x - 80
= x2 − 2x - 80
(iii) (3x + 4) (3x - 5)
Solution:
Using formula, (y + a) (y + b) = y2 + (a + b)y + ab
[So, y = 3x, a = 4 and b = −5]
(3x + 4) (3x − 5) = (3x)2 + [4 + (-5)]3x + 4 × (-5)
= 9x2 + 3x (4 - 5) - 20
= 9x2 – 3x – 20
(iv) (y2 + ) (y2 - )
Solution:
Using formula, (a + b) (a – b) = a2 – b2
[So, a = y2 and b = ]
(y 2 + ) (y2 – ) = (y2)2 – ()^2
= y 4 –
(i) 103 × 107
Solution:
103 × 107 = (100 + 3) × (100 + 7)
Using formula, (x + a) (x + b) = x2 + (a + b)x + ab
Then,
x = 100
a = 3
b = 7
So, 103 × 107 = (100 + 3) × (100 + 7)
= (100)2 + (3 + 7)100 + (3 × 7)
= 10000 + 1000 + 21
= 11021
(ii) 95 × 96
Solution:
95 × 96 = (100 - 5) × (100 - 4)
Using formula, (x - a) (x - b) = x2 - (a + b)x + ab
Then, According to the identity
x = 100
a = 5
b = 4
So, 95 × 96 = (100 - 5) × (100 - 4)
= (100)2 - 100 (5+4) + (5 × 4)
= 10000 - 900 + 20
= 9120
(iii) 104 × 96
Solution:
104 × 96 = (100 + 4) × (100 – 4)
Using formula, (a + b) (a - b) = a2 - b2
Then,
a = 100
b = 4
So, 104 × 96 = (100 + 4) × (100 – 4)
= (100)2 – (4)2
= 10000 – 16
= 9984
(i) 9x2 + 6xy + y2
Solution:
9x2 + 6xy + y2 = (3x)2 + (2 × 3x × y) + y2
Using formula, a2 + 2ab + b2 = (a + b)2
Then,
a = 3x
b = y
9x2 + 6xy + y2 = (3x)2 + (2 × 3x × y) + y2
= (3x + y)2
= (3x + y) (3x + y)
(ii) 4y2 − 4y + 1
Solution:
4y2 − 4y + 1 = (2y)2 – (2 × 2y × 1) + 1
Using formula, a2 - 2ab + b2 = (a - b)2
Then,
a = 2y
b = 1
= (2y – 1)2
= (2y – 1) (2y – 1)
(iii) x2 -
Solution:
x2 – = x2 –
Using formula, a2 - b2 = (a - b) (a + b)
Then,
a = x
b =
= (x – ) (x + )
(i) (x + 2y + 4z)2
Solution:
Using formula, (x + y + z)2 = x2 + y2 + z2 + 2xy + 2yz + 2zx
Then,
x = x
y = 2y
z = 4z
(x + 2y + 4z)2 = x2 + (2y)2 + (4z)2 + (2 × x × 2y) + (2 × 2y × 4z) + (2 × 4z × x)
= x2 + 4y2 + 16z2 + 4xy + 16yz + 8xz
(ii) (2x − y + z)2
Solution:
Using formula, (x + y + z)2 = x2 + y2 + z2 + 2xy + 2yz + 2zx
Then,
x = 2x
y = −y
z = z
(2x − y + z)2 = (2x)2 + (−y)2 + z2 + (2 × 2x × −y) + (2 × −y × z) + (2 × z × 2x)
= 4x2 + y2 + z2 – 4xy – 2yz + 4xz
(iii) (−2x + 3y + 2z)2
Solution:
Using formula, (x + y + z)2 = x2 + y2 + z2 + 2xy + 2yz + 2zx
Then,
x = −2x
y = 3y
z = 2z
(−2x + 3y + 2z)2 = (−2x)2 + (3y)2 + (2z)2 + (2 ×−2x × 3y) + (2 ×3y × 2z) + (2 ×2z × −2x)
= 4x2 + 9y2 + 4z2 – 12xy + 12yz– 8xz
(iv) (3a – 7b – c)2
Solution:
Using formula, (x + y + z)2 = x2 + y2 + z2 + 2xy + 2yz + 2zx
Then,
x = 3a
y = – 7b
z = – c
(3a – 7b – c)2 = (3a)2 + (– 7b)2 + (– c)2 + (2 × 3a × – 7b) + (2 × –7b × –c) + (2 × –c × 3a)
= 9a2 + 49b2 + c2 – 42ab + 14bc – 6ca
(v) (–2x + 5y – 3z)2
Solution:
Using formula, (x + y + z)2 = x2 + y2 + z2 + 2xy + 2yz + 2zx
Then,
x = –2x
y = 5y
z = – 3z
(–2x + 5y – 3z)2 = (–2x)2 + (5y)2 + (–3z)2 + (2 × –2x × 5y) + (2 × 5y × – 3z) + (2 × –3z × –2x)
= 4x2 + 25y2 + 9z2 – 20xy – 30yz + 12zx
(vi) (a - b + 1)2
Solution:
Using formula, (x + y + z)2 = x2 + y2 + z2 + 2xy + 2yz + 2zx
Then,
x = a
y = b
z = 1
(a - ()b + 1)2 = [a]2 + [b]2 + 12 + [2 x }a x b] + [2 xb x 1] + [2 x 1 x a]
= a2 + b2 + 1 - ab - b + a
(i) 4x2 + 9y2 + 16z2 + 12xy – 24yz – 16xz
Solution:
Using formula, (x + y + z)2 = x2 + y2 + z2 + 2xy + 2yz + 2zx
Then, x2 + y2 + z2 + 2xy + 2yz + 2zx = (x + y + z)2
4x2 + 9y2 + 16z2 + 12xy – 24yz – 16xz = (2x)2 + (3y)2 + (−4z)2 + (2 × 2x × 3y) + (2 × 3y × −4z) + (2 × −4z × 2x)
= (2x + 3y – 4z)2
= (2x + 3y – 4z) (2x + 3y – 4z)
(ii) 2x2 + y2 + 8z2 – 2√2xy + 4√2yz – 8xz
Solution:
Using formula, (x + y + z)2 = x2 + y2 + z2 + 2xy + 2yz + 2zx
Then, x2 + y2 + z2 + 2xy + 2yz + 2zx = (x + y + z)2
2x2 + y2 + 8z2 – 2√2xy + 4√2yz – 8xz
= (-√2x)2 + (y)2 + (2√2z)2 + (2 × -√2x × y) + (2 × y × 2√2z) + (2 × 2√2 × −√2x)
= (−√2x + y + 2√2z)2
= (−√2x + y + 2√2z) (−√2x + y + 2√2z)
(i) (2x + 1)3
Solution:
Using formula,(x + y)3 = x3 + y3 + 3xy(x + y)
(2x + 1)3= (2x)3 + 13 + (3 × 2x ×1) (2x + 1)
= 8x3 + 1 + 6x(2x + 1)
= 8x3 + 12x2 + 6x + 1
(ii) (2a − 3b)3
Solution:
Using formula, (x – y)3 = x3 – y3 – 3xy(x – y)
(2a − 3b)3 = (2a)3 − (3b)3 – (3 × 2a × 3b) (2a – 3b)
= 8a3 – 27b3 – 18ab(2a – 3b)
= 8a3 – 27b3 – 36a2b + 54ab2
(iii) (x + 1)3
Solution:
Using formula, (x + y)3 = x3 + y3 + 3xy(x + y)
(x+ 1)3 = (x)3 + 13 + (3 × x × 1) (x + 1)
= x3 + 1 + x2 + x
=
(iv) (x − y)3
Solution:
Using formula, (x – y)3 = x3 – y3 – 3xy(x – y)
(x − y)3 = x3 − [y]3 - 3(x) y[x − y]
= x3 -y3 - 2x2y + xy2
(i) (99)3
Solution:
99 = 100 – 1
Using formula, (x – y)3 = x3 – y3 – 3xy(x – y)
(99)3 = (100 – 1)3
= (100)3 – 13 – (3 × 100 × 1) (100 – 1)
= 1000000 – 1 – 300(100 – 1)
= 1000000 – 1 – 30000 + 300
= 970299
(ii) (102)3
Solution:
102 = 100 + 2
Using formula, (x + y)3 = x3 + y3 + 3xy(x + y)
(100 + 2)3 = (100)3 + 23 + (3 × 100 × 2) (100 + 2)
= 1000000 + 8 + 600(100 + 2)
= 1000000 + 8 + 60000 + 1200
= 1061208
(iii) (998)3
Solution:
998 = 1000 – 2
Using formula, (x – y)3 = x3 – y3 – 3xy(x – y)
(998)3 = (1000 – 2)3
= (1000)3 – 23 – (3 × 1000 × 2) (1000 – 2)
= 1000000000 – 8 – 6000(1000 – 2)
= 1000000000 - 8 - 6000000 + 12000
= 994011992
(i) 8a3 + b3 + 12a2b + 6ab2
Solution:
8a3 + b3 +12a2b + 6ab2 can also be written as (2a)3 + b3 + 3(2a)2b + 3(2a)(b)2
8a3 + b3 + 12a2b + 6ab2 = (2a)3 + b3 + 3(2a)2b + 3(2a)(b)2
Formula used, (x + y)3 = x3 + y3 + 3xy(x + y)
= (2a + b)3
= (2a + b) (2a + b) (2a + b)
(ii) 8a3 – b3 – 12a2b + 6ab2
Solution:
8a3 – b3 − 12a2b + 6ab2 can also be written as (2a)3– b3 – 3(2a)2b + 3(2a)(b)2
8a3 – b3 − 12a2b + 6ab2 = (2a)3 – b3 – 3(2a)2b + 3(2a)(b)2
formula used, (x - y)3 = x3 - y3 - 3xy(x - y)
= (2a – b)3
= (2a – b) (2a – b) (2a – b)
(iii) 27 – 125a3 – 135a + 225a2
Solution:
27 – 125a3 – 135a +225a2 can be also written as 33 – (5a)3 – 3(3)2(5a) + 3(3)(5a)2
27 – 125a3 – 135a + 225a2 = 33 – (5a)3 – 3(3)2(5a) + 3(3)(5a)2
Formula used, (x - y)3 = x3 - y3 - 3xy(x - y)
= (3 – 5a)3
= (3 – 5a) (3 – 5a) (3 – 5a)
(iv) 64a3 – 27b3 – 144a2b + 108ab2
Solution:
64a3 – 27b3 – 144a2b + 108ab2 can also be written as (4a)3 – (3b)3 – 3(4a)2(3b) + 3(4a)(3b)2
64a3 – 27b3 – 144a2b + 108ab2 = (4a)3 – (3b)3 – 3(4a)2(3b) + 3(4a)(3b)2
Formula used, (x - y)3 = x3 - y3 - 3xy(x - y)
= (4a – 3b)3
= (4a – 3b) (4a – 3b) (4a – 3b)
(v) 7p3 – − p2 + p
Solution:
27p3 – − () p2 + ()p can also be written as (3p)3 – – 3(3p)2() + 3(3p)()2
27p3 – () − () p2 + ()p = (3p)3 – ()3 – 3(3p)2() + 3(3p)()2
Formula used, (x - y)3 = x3 - y3 - 3xy(x - y)
= (3p – )3
= (3p – ) (3p – ) (3p – )
(i) x3 + y3 = (x + y) (x2 - xy + y2)
Solution:
Formula (x + y)3 = x3 + y3 + 3xy(x + y)
x3 + y3 = (x + y)3 - 3xy(x + y)
x3 + y3 = (x + y) [(x + y)2 - 3xy]
x3 + y3 = (x + y) [(x2 + y2 + 2xy) - 3xy]
Therefore, x3 + y3 = (x + y) (x2 + y2 - xy)
(ii) x3 - y3 = (x - y) (x2 + xy + y2)
Solution:
Formula, (x - y)3 = x3 - y3 - 3xy(x - y)
x3 − y3 = (x - y)3 + 3xy(x - y)
x3− y3 = (x - y) [(x - y)2 + 3xy]
x3 − y3 = (x - y) [(x2 + y2 – 2xy) + 3xy]
Therefore, x3 + y3 = (x – y) (x2 + y2 + xy)
(i) 27y3 + 125z3
Solution:
27y3 + 125z3 can also be written as (3y)3 + (5z)3
27y3 + 125z3 = (3y)3 + (5z)3
Formula x3 + y3 = (x + y) (x2 – xy + y2)
27y3 + 125z3 = (3y)3 + (5z)3
= (3y + 5z) [(3y)2 – (3y)(5z) + (5z)2]
= (3y + 5z) (9y2 – 15yz + 25z2)
(ii) 64m3 - 343n3
Solution:
64m3 - 343n3 can also be written as (4m)3 - (7n)3
64m3 - 343n3 = (4m)3 - (7n)3
Formula x3 - y3 = (x - y) (x2 + xy + y2)
64m3 - 343n3 = (4m)3 - (7n)3
= (4m - 7n) [(4m)2 + (4m)(7n) + (7n)2]
Solution:
27x3 + y3 + z3 - 9xyz can also be written as (3x)3 + y3 + z3 - 3(3x)(y)(z)
27x3 + y3 + z3 – 9xyz = (3x)3 + y3 + z3 – 3(3x)(y)(z)
Formula, x3 + y3 + z3 – 3xyz = (x + y + z) (x2 + y2 + z2 – xy – yz – zx)
27x3 + y3 + z3 – 9xyz = (3x)3 + y3 + z3 – 3(3x)(y)(z)
= (3x + y + z) [(3x)2 + y2 + z2 – 3xy – yz – 3xz]
= (3x + y + z) (9x2 + y2 + z2 – 3xy – yz – 3xz)
Solution:
Formula, x3 + y3 + z3 − 3xyz = (x + y + z)(x2 + y2 + z2 – xy – yz – xz)
Multiplying by 2 and dividing by 2
= (1/2) (x + y + z) [2(x2 + y2 + z2 – xy – yz – xz)]
= (1/2) (x + y + z) (2x2 + 2y2 + 2z2 – 2xy – 2yz – 2xz)
= (1/2) (x + y + z) [(x2 + y2 − 2xy) + (y2 + z2 – 2yz) + (x2 + z2 – 2xz)]
= (1/2) (x + y + z) [(x – y)2 + (y – z)2 + (z – x)2]
Therefore, x3 + y3 + z3 – 3xyz = (1/2) (x + y + z) [(x – y)2 + (y – z)2 + (z – x)2]
Solution:
Formula, x3 + y3 + z3 - 3xyz = (x + y + z) (x2 + y2 + z2 – xy – yz – xz)
Given, (x + y + z) = 0,
Then, x3 + y3 + z3 - 3xyz = (0) (x2 + y2 + z2 – xy – yz – xz)
x3 + y3 + z3 – 3xyz = 0
Therefore, x3 + y3 + z3 = 3xyz
(i) (−12)3 + (7)3 + (5)3
Solution:
Let,
x = −12
y = 7
z = 5
We know that if x + y + z = 0, then x3 + y3 + z3 = 3xyz.
and we have −12 + 7 + 5 = 0
Therefore, (−12)3 + (7)3 + (5)3 = 3xyz
= 3 × -12 × 7 × 5
= -1260
(ii) (28)3 + (−15)3 + (−13)3
Solution:
Let,
x = 28
y = −15
z = −13
We know that if x + y + z = 0, then x3 + y3 + z3 = 3xyz.
and we have, x + y + z = 28 – 15 – 13 = 0
Therefore, (28)3 + (−15)3 + (−13)3 = 3xyz
= 3 (28) (−15) (−13)
= 16380
(i) Area : 25a2 – 35a + 12
Solution:
Using splitting the middle term method,
25a2 – 35a + 12
25a2 – 35a + 12 = 25a2 – 15a − 20a + 12
= 5a(5a – 3) – 4(5a – 3)
= (5a – 4) (5a – 3)
Possible expression for length & breadth is = (5a – 4) & (5a – 3)
(ii) Area : 35y2 + 13y – 12
Solution:
Using the splitting the middle term method,
35y2 + 13y – 12 = 35y2 – 15y + 28y – 12
= 5y(7y – 3) + 4(7y – 3)
= (5y + 4) (7y – 3)
Possible expression for length & breadth is = (5y + 4) & (7y – 3)
(i) Volume : 3x2 – 12x
Solution:
3x2 – 12x can also be written as 3x(x – 4)
= (3) (x) (x – 4)
Possible expression for length, breadth & height = 3, x & (x - 4)
(ii) Volume: 12ky2 + 8ky – 20k
Solution:
12ky2 + 8ky – 20k can also be written as 4k (3y2 + 2y – 5)
12ky2 + 8ky– 20k = 4k(3y2 + 2y – 5)
Using splitting the middle term method.
= 4k (3y2 + 5y – 3y – 5)
= 4k [y(3y + 5) – 1(3y + 5)]
= 4k (3y + 5) (y – 1)
Possible expression for length, breadth & height= 4k, (3y + 5) & (y - 1)
Exercise 2.4 of Chapter 2, "Polynomials," in the Class 9 NCERT Mathematics textbook involves solving problems related to evaluating polynomials, performing operations like subtraction, solving quadratic equations, and factoring polynomials. The exercise includes step-by-step solutions for evaluating a polynomial at a given point, subtracting one polynomial from another, solving a quadratic equation using the quadratic formula, and factoring a quadratic polynomial. This exercise helps students understand and apply fundamental concepts of polynomials effectively.
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