NCERT Solutions Class 9 – Chapter 2 Polynomials – Exercise 2.4
Question 1. Use suitable identities to find the following products:
(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 –
Question 2. Evaluate the following products without multiplying directly:
(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
Question 3. Factorize the following using appropriate identities:
(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 + )
Question 4. Expand each of the following, using suitable identities:
(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
Question 5. Factorize:
(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)
Question 6. Write the following cubes in expanded form:
(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
Question 7. Evaluate the following using suitable identities:
(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
Question 8. Factorise each of the following:
(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 – )
Question 9. Verify:
(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)
Question 10. Factorize each of the following:
(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]
Question 11. Factorise: 27x3 + y3 + z3 – 9xyz
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)
Question 12. Verify that: 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)
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]
Question 13. If x + y + z = 0, show that x3 + y3 + z3 = 3xyz.
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
Question 14. Without actually calculating the cubes, find the value of each of the following:
(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
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
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)
Question 16. What are the possible expressions for the dimensions of the cuboids whose volumes are given below?
(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)
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