NCERT Solutions Class 9th Maths Chapter – 2 Polynomials Exercise – 2.4

Question 1. Use suitable identities to find the following products:
(i) (x + 4) (x + 10)

Solution – 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 x 10)
= x² + 14x + 40

(ii) (x+8) (x -10)

Solution – 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) (3x + 4) (3x – 5)

Solution – 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) (y²  3/2) (y²– 3/2)

Solution – We have, (y²+ 3/2) (y²– 3/2)
Using the identity, (x+y)(x–y) = x²–y ²
We get, (y² + 3/2) (y² – 3/2)
= (y²)² – (3/2)²
= y⁴ – 9/4

(v) (3 – 2x) (3 + 2x)

Solution – We have, (3 – 2x) (3 + 2x)
Using identity, (x + y) (x – y) = $x^2$ – $y^2$
(3 – 2x) (3 + 2x)
= ${\left(3\right)}^2$ – ${\left(2x\right)}^2$
= 9 – ${4x}^2$

Question 2. Evaluate the following products without multiplying directly:
(i) 103 x 107

Solution – We have, 103 x 107
[Using (x + a)(x + b) = x² + (a + b)x + ab]
= (100 + 3) (100 + 7)
= ( 100)² + (3 + 7) (100) + (3 x 7)
= 10000 + (10) x 100 + 21
= 10000 + 1000 + 21
= 11021

(ii) 95 x 96

Solution – We have, 95 x 96
[Using (x + a)(x + b) = x² + (a + b)x + ab]
= (100 – 5) (100 – 4)
= ( 100)² + (- 5) + (- 4) 100 + (- 5) (- 4)
= 10000 – 500 – 400 + 20
= 10020 – 900
= 9120

(iii) 104 x 96

Solution – We have 104 x 96
[Using (a + b)(a -b) = a²– b²]
= (100 + 4) (100 – 4)
= (100)² – 4²
= 10000 – 16
= 9984

Question 3. Factorise the following using appropriate identities:
(i) 9x² + 6xy + y²

Solution – We have, 9x² + 6xy + y²
[Using a² + 2ab + b² = (a + b)²]
= (3x)² + 2(3x) (y) + (y)²
= (3x + y)²
= (3x + y) (3x + y)

(ii) 4y² – 4y + 1

Solution – We have, 4y² – 4y + 1²
[Using a² – 2ab + b² = (a- b)²]
= (2y)² + 2(2y)(1) + (1)²
= (2y -1)²
= (2y – 1) (2y – 1)

(iii) x² – y² / 100

Solution – We have, x² – y²/100
[Using identity, x² – y² = (x – y) (x + y)]
= x²–y²/100
= x²–(y/10)²
= (x–y/10) (x+y/10)

Question 4. Expand each of the following, using suitable identity
(i) (x+2y+ 4z)²

Solution – We know that,
[Using identity, (x + y + z)² = x² + y² + z² + 2xy + 2yz + 2zx]
(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 – (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 – (- 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)^z

Solution – (3a -7b- c)²
= (3a)² + (- 7b)² + (- c)² + 2(3a)(- 7b) + 2(- 7b)(- c) + 2(- c)(3a)
= 9a² + 49b² + c² – 42ab + 14bc – 6ac

(v) (- 2x + 5y – 3z)²

Solution – (- 2x + 5y- 3z)²
= (- 2x)² + (5y)² + (- 3z)² + 2(- 2x)(5y) + 2(5y)(- 3z) + 2(- 3z)(- 2x)
= 4x² + 25y² + 9z² – 20xy – 30yz + 12xz

(vi) [1/4a –1/4b + 1] ²

Solution

Question 5. Factories:
(i) 4 x² + 9y² + 16z² + 12xy – 24yz – 16xz

Solution – 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 – 2x² + y² + 8z² – 2√2xy + 4√2yz – 8xz
= (- √2x)² + (y)² + (2 √2z)²y + 2(- √2x)(y) + 2(y) (2√2z) + 2(2√2z)(- √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 identity,(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 identity,(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) ((3/2)x+1)³

Solution – Using identity,(x+y)³ = x³+y³+3xy(x+y)
((3/2)x+1)³
= ((3/2)x)³ + 1³+(3×(3/2)x×1) ((3/2)x +1)
= ((27/8)x)³ + 1 + (9/2)x (3/2)x + 1)
=  ((27/8)x)³ + 1 + (27/8)x)² + (9/2)x
= ((27/8)x)³ + (27/8)x)² + (9/2)x + 1

(iv) (x−(2/3)y)³

Solution – Using identity, (x –y)³ = x³–y³–3xy(x–y)
(x−(2/3)y)³
= (x)³ – (2/3)y)³ – (3×x×2/3y) (x-2/3y)
= (x)³ – (8/27)y)³ – 2xy (x-2/3y)
= (x)³ – (8/27)y)³ – 2x²y + 4/3xy²

Question 7. Evaluate the following using suitable identities:
(i) (99)³

Solution – We have, 99 = (100 -1)
= 99³ = (100 – 1)³ [Using (a – b)³ = a³ – b³ – 3ab (a – b)]
= (100)³ – 1³ – 3(100)(1)(100 -1)
= 1000000 – 1 – 300(100 – 1)
= 1000000 -1 – 30000 + 300
= 1000300 – 30001
= 970299

(ii) (102)³

Solution – We have, 102 =100 + 2
= 102³ = (100 + 2)³ [Using (a + b)³ = a³ + b³ + 3ab (a + b)]
= (100)³ + (2)³ + 3(100)(2)(100 + 2)
= 1000000 + 8 + 600(100 + 2)
= 1000000 + 8 + 60000 + 1200
= 1061208

(iii) (998)³

Solution – We have, 998 = 1000 – 2
= (998)³ = (1000-2)³ [Using (a – b)³ = a³ – b³ – 3ab (a – b)]
= (1000)³– (2)³ – 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
= (2a)³ + (b)³ + 6ab(2a + b)
= (2a)³ + (b)³ + 3(2a)(b)(2a + b)
= (2 a + b)³
[Using a³ + b³ + 3 ab(a + b) = (a + b)³]
= (2a + b)(2a + b)(2a + b)

(ii) 8a³ -b³-12a²b+6ab²

Solution – 8a³ – b³ – 12o²b + 6ab²
= (2a)³ – (b)³ – 3(2a)(b)(2a – b)
= (2a – b)³
[Using a³ + b³ + 3 ab(a + b) = (a + b)³]
= (2a – b) (2a – b) (2a – b)

(iii) 27-125a³ -135a+225a²

Solution – 27 – 125a³ – 135a + 225a²
= (3)³ – (5a)³ – 3(3)(5a)(3 – 5a)
= (3 – 5a)³
[Using a³ + b³ + 3 ab(a + b) = (a + b)³]
= (3 – 5a) (3 – 5a) (3 – 5a)

(iv) 64a³ -27b³ -144a²b + 108ab²

Solution – 64a³ -27b³ -144a²b + 108ab²
= (4a)³ – (3b)³ – 3(4a)(3b)(4a – 3b)
= (4a – 3b)³
[Using a³ – b³ – 3 ab(a – b) = (a – b)³]
= (4a – 3b)(4a – 3b)(4a – 3b)

(v) 27p³–(1/216)−(9/2) p²+(1/4)p

Solution – The expression, 27p³–(1/216)−(9/2) p²+(1/4)p can be written as
(3p)³–(1/6)³−(9/2) p²+(1/4)p = (3p)³–(1/6)³−3(3p)(1/6)(3p – 1/6)
Using (x – y)³ = x³ – y³ – 3xy (x – y)
27p³–(1/216)−(9/2) p²+(1/4)p = (3p)³–(1/6)³−3(3p)(1/6)(3p – 1/6)
Taking x = 3p and y = 1/6
= (3p–1/6)³
= (3p–1/6)(3p–1/6)(3p–1/6)

Question 9. Verify
(i) x³ + y³ = (x + y)-(x² – xy + y²)

Solution – (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 – xy) = x³ + y³
Hence, verified.

(ii) x³ – y³ = (x – y) (x² + xy + y²)

Solution – (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 + xy) = x³ – y³
Hence, verified.

Question 10. Factorise each of the following:
(i) 27y³ + 125z³
[Hint: See question 9.]

Solution – 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) 64m³ – 343n³

Solution – 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)

Question 11. Factories: 27x³ +y³ +z³ -9xyz

Solution – We know that,
Using the identity, x³ + y³ + z³ – 3xyz = (x + y + z)(x² + y² + z² – xy – yz – zx)
We have, 27x³ + y³ + z³ – 9xyz
= (3x)³ + (y)³ + (z)³ – 3(3x)(y)(z)
= (3x)³ + (y)³ + (z)³ – 3(3x)(y)(z)
= (3x + y + z)[(3x)³ + y³ + z³ – (3x × y) – (y × 2) – (z × 3x)]
= (3x + y + z)(9x² + y² + z² – 3xy – yz – 3zx)

Question 12. Verify that x³ +y³ +z³ – 3xyz = 1/2 (x + y+z)[(x-y)² + (y – z)² +(z – x)²]

Solution – R.H.S
= 1/2(x + y + z)[(x – y)²+(y – z)²+(z – x)²]
= 1/2 (x + y + 2)[(x² + y² – 2xy) + (y² + z² – 2yz) + (z² + x² – 2zx)]
= 1/2 (x + y + 2)(x² + y² + y² + z² + z² + x² – 2xy – 2yz – 2zx)
= 1/2 (x + y + z)[2(x² + y² + z² – xy – yz – zx)]
= 2 x 1/2 x (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.

Question 13. If x + y + z = 0, show that x³ + y³ + z³ = 3 xyz.

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

Question 14. Without actually calculating the cubes, find the value of each of the following:
(i) (- 12)³ + (7)³ + (5)³

Solution – We have, (-12)³ + (7)³ + (5)³
Let x = -12, y = 7 and z = 5.
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) (28)³ + (- 15)³ + (- 13)³

Solution – We have, (28)³ + (-15)³ + (-13)³
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

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 – Area of a rectangle = (Length) x (Breadth)
We have, 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) Area: 35y² + 13y – 12

Solution – We have, 35y²+ 13y -12
= 35y² + 28y – 15y -12
= 7y(5y + 4) – 3(5y + 4)
= (5 y + 4)(7y – 3)
Thus, the possible length and breadth are (7y – 3) and (5y + 4).

Question 16. What are the possible expressions for the dimensions of the cuboids whose volumes are given below?
(i) Volume 3x² – 12x

Solution – Volume of a cuboid = (Length) × (Breadth) × (Height)
We have, 3x² – 12x = 3(x² – 4x)
= 3 x x x (x – 4)
∴ The possible dimensions of the cuboid are 3, x and (x – 4).

(ii) Volume 12ky² + 8ky – 20k

Solution – We have, 12ky² + 8ky – 20k
= 4[3ky² + 2ky – 5k] = 4[k(3y² + 2y – 5)]
= 4 x k x (3y² + 2y – 5)
= 4k[3y² – 3y + 5y – 5]
= 4k[3y(y – 1) + 5(y – 1)]
= 4k[(3y + 5) x (y – 1)]
= 4k x (3y + 5) x (y – 1)
Thus, the possible dimensions of the cuboid are 4k, (3y + 5) and (y -1).