NCERT Solutions for Class 9 Maths Chapter 2 Polynomials Ex 2.5
NCERT Solutions for Class
9 Maths Chapter 2 Polynomials Ex 2.4 are the part of NCERT Solutions for Class
9 Maths. Here you can find the NCERT Solutions for Class 9 Maths Chapter 2 Polynomials
Ex 2.5.
- NCERT Solutions for Class 9 Maths Chapter 2 Number Systems Ex 2.1
- NCERT Solutions for Class 9 Maths Chapter 2 Number Systems Ex 2.2
- NCERT Solutions for Class 9 Maths Chapter 2 Number Systems Ex 2.3
- NCERT Solutions for Class 9 Maths Chapter 2 Number Systems Ex 2.4
- NCERT Solutions for Class 9 Maths Chapter 2 Number Systems Ex 2.5
Ex
2.5 Class 9 Maths Question 1.
Use suitable identities to find the following
products.
(i) (x + 4)(x + 10)
(ii) (x + 8) (x – 10)
(iii) (3x + 4) (3x – 5)
(iv) (y2 + 3/2) (y2 – 3/2)
(v) (3 – 2x) (3 + 2x)
Solution:
(i) We have, (x + 4) (x + 10)
Using identity, (x + a) (x + b) = x2 + (a + b)x + ab
We have, (x + 4) (x + 10) = x2 + (4 + 10)x + (4 × 10)
= x2 + 14x + 40
(ii)
We have, (x + 8) (x – 10)
Using identity, (x + a) (x + b) = x2 + (a + b)x + ab
We have, (x + 8) (x – 10) = x2 + [8 + (-10)]x + (8)(-10)
= x2 – 2x – 80
(iii) We have, (3x + 4) (3x – 5)
Using identity, (x + a) (x + b) = x2 +
(a + b)x + ab
We have, (3x + 4) (3x – 5) = (3x)2 +
(4 – 5)x + (4)(-5)
=
9x2 – x – 20
(iv) We have, (y2 + 3/2) (y2 – 3/2)
Using identity, (a + b) (a – b) = a2
– b2
(y2
+ 3/2) (y2 – 3/2) = (y2)2 – (3/2)2
= y4 – 9/4
(v) We have, (3
– 2x) (3 + 2x)
Using identity, (a + b) (a – b) = a2
– b2
(3 – 2x) (3 + 2x) = (3)2 – (2x)2
= 9 – 4x2
Ex 2.5 Class 9 Maths Question 2.
Evaluate the following products without multiplying
directly.
(i) 103 × 107
(ii) 95 × 96
(iii) 104 × 96
Solution:
(i) We have, 103 × 107 = (100 + 3) (100 + 7)
=
(100)2 + (3 + 7)(100) + (3 × 7)
[Using (x + a)(x + b) = x2 + (a + b)x + ab]
= 10000 + (10) × 100 + 21
= 10000 + 1000 + 21 = 11021
(ii)
We have, 95 × 96 = (100 – 5) (100 – 4)
=
(100)2 + [(-5) + (-4)](100) + (-5 × –4)
[Using (x + a)(x + b) = x2 + (a + b)x + ab]
= 10000 + (-9)(100) + 20
=
10000 + (-900) + 20 = 9120
(iii)
We have, 104 × 96 = (100 + 4) (100 – 4)
=
(100)2 – 42
[Using (a + b)(a – b) = a2 – b2]
=
10000 – 16 = 9984
Ex
2.5 Class 9 Maths Question 3.
Factorise the following using appropriate identities.
(i) 9x2 + 6xy + y2
(ii) 4y2– 4y + 1
(iii) x2 – y2/100
Solution:
(i) We have, 9x2 + 6xy + y2
=
(3x)2 + 2(3x)(y) + (y)2
= (3x
+ y)2
[Using a2 + 2ab + b2 = (a + b)2]
= (3x
+ y)(3x + y)
(ii) We have, 4y2 – 4y + 12
= (2y)2 – 2(2y)(1) + (1)2
= (2y –
1)2
[Using a2 – 2ab + b2 = (a- b)2]
= (2y – 1)(2y – 1 )
Ex
2.5 Class 9 Maths Question 4.
Expand each of the following, using suitable
identity.
(i) (x + 2y + 4z)2
(ii) (2x – y + z)2
(iii) (-2x + 3y + 2z)2
(iv) (3a – 7b – c)2
(v) (-2x + 5y – 3z)2
(vi) [a/4 – b/2 + 1]2
Solution:
We know that
(x + y + z)2 = x2 + y2 + z2 +
2xy + 2yz + 2zx
(i)
(x + 2y + 4z)2 = x2 + (2y)2 + (4z)2 +
2 (x) (2y) + 2 (2y) (4z) + 2(4z) (x)
= x2 +
4y2 + 16z2 + 4xy + 16yz + 8zx
(ii)
(2x – y + z)2 = (2x)2 + (-y)2 +
z2 + 2 (2x) (-y) + 2 (-y) (z) + 2 (z) (2x)
= 4x2 +
y2 + z2 – 4xy – 2yz + 4zx
(iii)
(-2x + 3y + 2z)2 = (-2x)2 + (3y)2 +
(2z)2 + 2 (-2x) (3y)+ 2 (3y) (2z) + 2 (2z) (-2x)
= 4x2 +
9y2 + 4z2 – 12xy + 12yz – 8zx
(iv)
(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 = (-2x)2 + (5y)2 +
(-3z)2 + 2 (-2x) (5y) + 2 (5y) (-3z) + 2 (-3z) (-2x)
= 4x2 +
25y2 + 9z2 – 20xy – 30yz + 12zx
Ex 2.5 Class 9 Maths Question 5.
Factorise
(i) 4x2 + 9y2 + 16z2 +
12xy – 24yz – 16xz
(ii) 2x2 + y2 + 8z2 –
2√2xy + 4√2yz – 8zx
Solution:
(i) 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
= (-√2x)2 +
(y)2 + (2√2z)2 + 2 (-√2x) (y) + 2 (y) (2√2z) + 2
(2√2z) (-√2x)
= (-√2x + y + 2√2z)2
= (-√2x + y + 2√2z) (-√2x +
y + 2√2z)
Ex 2.5 Class 9 Maths Question 6.
Write the following cubes in expanded form:
Solution:
We know that, (x + y)3 = x3 +
y3 + 3xy(x + y) …(1)
and (x – y)3 =
x3 – y3 – 3xy(x – y) …(2)
(i)
(2x + 1)3 = (2x)3 + (1)3 +
3(2x)(1)(2x + 1) [using (1)]
= 8x3 +
1 + 6x(2x + 1)
= 8x3 +
12x2 + 6x + 1
(ii)
(2a – 3b)3 = (2a)3 – (3b)3 –
3(2a)(3b)(2a – 3b) [using (2)]
= 8a3 –
27b3 – 18ab(2a – 3b)
= 8a3 –
27b3 – 36a2b + 54ab2
Evaluate the following using suitable identities:
(i) (99)3
(ii) (102)3
(iii) (998)3
Solution:
(i) We know that, 99 = (100 – 1)
∴ 993 = (100 – 1)3
= (100)3 – 13 –
3(100)(1)(100 – 1) [Using (a –
b)3 = a3 – b3 – 3ab(a – b)]
= 1000000 – 1 – 300(100 – 1)
= 1000000 – 1 – 30000 + 300
= 1000300 – 30001 = 970299
(ii)
We know that, 102 = 100 + 2
∴ 1023 = (100 + 2)3
= (100)3 +
(2)3 + 3(100)(2)(100 + 2) [Using (a + b)3 = a3 +
b3 + 3ab(a + b)]
= 1000000 + 8 + 600(100 + 2)
= 1000000 + 8 + 60000 + 1200
= 1061208
(iii)
We know that, 998 = 1000 – 2
∴ (998)3 = (1000 – 2)3
= (1000)3 – (2)3 –
3(1000)(2)(1000 – 2) [Using (a –
b)3 = a3 – b3 – 3ab(a – b)]
= 1000000000 – 8 – 6000(1000
– 2)
= 1000000000 – 8 – 6000000 + 12000
= 994011992
Ex 2.5 Class 9 Maths Question 8.
Factorise each of the following:
(i) 8a3 + b3 + 12a2b + 6ab2
(ii) 8a3 – b3 – 12a2b + 6ab2
(iii) 27 – 125a3 – 135a + 225a2
(iv) 64a3 – 27b3 – 144a2b
+ 108ab2
Solution:
(i) 8a3 + b3 + 12a2b
+ 6ab2
= (2a)3 +
(b)3 + 6ab(2a + b)
= (2a)3 +
(b)3 + 3(2a)(b)(2a + b)
= (2a + b)3
[Using a3 + b3 + 3ab(a + b) = (a +
b)3]
= (2a + b)(2a +
b)(2a + b)
(ii)
8a3 – b3 – 12a2b + 6ab2
= (2a)3 –
(b)3 – 3(2a)(b)(2a – b)
= (2a – b)3
[Using a3 – b3 – 3ab(a – b) = (a –
b)3]
= (2a – b) (2a – b)
(2a – b)
(iii)
27 – 125a3 – 135a + 225a2
= (3)3 –
(5a)3 – 3(3)(5a)(3 – 5a)
= (3 – 5a)3
[Using a3 – b3 – 3ab(a – b) = (a –
b)3]
= (3 – 5a) (3 – 5a)
(3 – 5a)
(iv)
64a3 – 27b3 – 144a2b + 108ab2
= (4a)3 –
(3b)3 – 3(4a)(3b)(4a – 3b)
= (4a – 3b)3
[Using a3 – b3 – 3ab(a – b) = (a –
b)3]
= (4a – 3b)(4a –
3b)(4a – 3b)
Ex 2.5 Class 9 Maths Question 9.
Verify:
(i) x3 + y3 = (x + y) (x2 –
xy + y2)
(ii) x3 – y3 = (x – y) (x2 +
xy + y2)
Solution:
(i) ∵ (x + y)3 = x3 + y3 +
3xy(x + y)
⇒ (x + y)3 – 3(x + y)(xy) = x3 +
y3
⇒ (x + y)[(x + y)2 – 3xy] = x3 + y3
⇒ (x + y)(x2 + y2 + 2xy – 3xy) =
x3 + y3
⇒ (x + y)(x2 + y2 – xy) = x3 +
y3
Hence, verified.
(ii)
∵ (x – y)3 = x3 – y3 –
3xy(x – y)
⇒ (x – y)3 + 3xy(x – y) = x3 – y3
⇒ (x – y)[(x – y)2 + 3xy)] = x3 –
y3
⇒ (x – y)(x2 + y2 – 2xy + 3xy) =
x3 – y3
⇒ (x – y)(x2 + y2 + xy) = x3 –
y3
Hence, verified.
Ex 2.5 Class 9 Maths Question 10.
Factorise each of the following:
(i) 27y3 + 125z3
(ii) 64m3 – 343n3
[Hint: See question 9]
Solution:
(i) We know that x3 + y3 =
(x + y)(x2 – xy + y2)
We have, 27y3 + 125z3 = (3y)3 +
(5z)3
= (3y + 5z)[(3y)2 – (3y)(5z) + (5z)2]
= (3y + 5z)(9y2 – 15yz + 25z2)
(ii)
We know that x3 – y3 = (x – y)(x2 +
xy + y2)
We have, 64m3 – 343n3 = (4m)3 –
(7n)3
= (4m – 7n)[(4m)2 + (4m)(7n) + (7n)2]
= (4m – 7n)(16m2 + 28mn + 49n2)
Ex
2.5 Class 9 Maths Question 11.
Factorise 27x3 + y3 +
z3 – 9xyz.
Solution:
We have,
27x3 + y3 + z3 –
9xyz = (3x)3 + (y)3 + (z)3 –
3(3x)(y)(z)
Using the identity,
x3 + y3 + z3 –
3xyz = (x + y + z)(x2 + y2 + z2 –
xy – yz – zx)
We have, (3x)3 + (y)3 +
(z)3 – 3(3x)(y)(z)
= (3x + y +
z)[(3x)2 + y2 + z2 – (3x × y) –
(y × z) – (z × 3x)]
= (3x + y + z)(9x2 +
y2 + z2 – 3xy – yz – 3zx)
Ex
2.5 Class 9 Maths Question 12.
Verify that:
x3 + y3 + z3 –
3xyz = ½ (x + y + z)[(x
– y)2 + (y – z)2 + (z – x)2]
Solution:
R.H.S. = ½ (x + y + z)[(x – y)2 + (y – z)2 +
(z – x)2]
= ½ (x + y + z)[(x2 +
y2 – 2xy) + (y2 + z2 – 2yz) +
(z2 + x2 – 2zx)]
= ½ (x + y + z)(x2 +
y2 + y2 + z2 + z2 +
x2 – 2xy – 2yz – 2zx)
= ½ (x + y + z)[2(x2 +
y2 + z2 – xy – yz – zx)]
= 2 × ½ (x + y + z)(x2 +
y2 + z2 – xy – yz – zx)
= (x + y + z)(x2 + y2 + z2 –
xy – yz – zx)
= x3 + y3 + z3 – 3xyz =
L.H.S.
Hence, verified.
Ex
2.5 Class 9 Maths Question 13.
If x + y + z = 0, show that x3 +
y3 + z3 = 3xyz.
Solution:
Since, x + y + z = 0
⇒ x + y = -z ⇒ (x + y)3 = (-z)3
⇒ x3 + y3 +
3xy(x + y) = -z3
⇒ x3 + y3 +
3xy(-z) = -z3
[∵ x + y = -z]
⇒ x3 + y3 –
3xyz = -z3
⇒ x3 + y3 +
z3 = 3xyz
Hence, if x + y + z = 0, then x3 +
y3 + z3 = 3xyz
Ex 2.5 Class 9 Maths Question 14.
Without actually calculating the cubes, find the
value of each of the following:
(i) (-12)3 + (7)3 + (5)3
(ii) (28)3 + (-15)3 + (-13)3
Solution:
(i) We have, (-12)3 + (7)3 +
(5)3
Let x = -12, y = 7 and z = 5.
Then, x + y + z = -12 + 7 + 5 = 0
We know that if x + y + z = 0, then, x3 + y3 +
z3 = 3xyz
∴ (-12)3 + (7)3 + (5)3 =
3[(-12)(7)(5)]
=
3[-420] = -1260
(ii)
We have, (28)3 + (-15)3 + (-13)3
Let x = 28, y = -15 and z = -13.
Then, x + y + z = 28 – 15 – 13 = 0
We know that if x + y + z = 0, then x3 + y3 + z3 =
3xyz
∴ (28)3 + (-15)3 + (-13)3 =
3(28)(-15)(-13)
=
3(5460) = 16380
Ex 2.5 Class 9 Maths Question 15.
Give possible expressions for the length and breadth
of each of the following rectangles, in which their areas are given:
(i) Area: 25a2 – 35a + 12
(ii) Area: 35y2 + 13y – 12
Solution:
Area of a rectangle = (Length) × (Breadth)
(i) 25a2 – 35a + 12 = 25a2 – 20a – 15a +
12 = 5a(5a – 4) – 3(5a – 4) = (5a – 4)(5a – 3)
Thus, the possible length and breadth are (5a – 3) and (5a – 4).
(ii)
35y2 + 13y – 12 = 35y2 + 28y – 15y – 12
= 7y(5y
+ 4) – 3(5y + 4) = (5y + 4)(7y – 3)
Thus, the possible length and breadth are (7y – 3) and (5y + 4).
Ex 2.5 Class 9 Maths Question 16.
What are the possible expressions for the dimensions
of the cuboids whose volumes are given below?
(i) Volume: 3x2 – 12x
(ii) Volume: 12ky2 + 8ky – 20k
Solution:
Volume of a cuboid = (Length) × (Breadth) × (Height)
(i) We have, 3x2 – 12x = 3(x2 – 4x)
=
3 × x × (x – 4)
∴ The possible dimensions of the cuboid are 3, x and (x – 4).
(ii)
We have, 12ky2 + 8ky – 20k
= 4[3ky2 +
2ky – 5k] = 4[k(3y2 + 2y – 5)]
= 4 × k × (3y2 +
2y – 5)
= 4k[3y2 –
3y + 5y – 5]
= 4k[3y(y – 1)
+ 5(y – 1)]
= 4k[(3y + 5)
(y – 1)]
= 4k × (3y +
5) × (y – 1)
Thus, the possible dimensions of the cuboid are 4k, (3y + 5) and (y – 1).