NCERT Solutions Class 9 Chapter 2 Polynomials Ex 2.4

Polynomials Class 9 Exercise 2.4 introduces the Remainder Theorem. It is one of the most important concepts in Class 9 Chapter 2 Polynomials. This theorem helps you find the remainder when a polynomial is divided by a linear polynomial without performing the full division process. 

The NCERT solutions provided here explain each question in a step-by-step format. This helps students understand how to substitute the value of x and compute the remainder correctly.

Class 9 Chapter 2 Polynomials Exercise 2.4 Questions

NCERT Class 9 Chapter 2 Polynomials Exercise 2.4 Questions revolve around applying the Remainder Theorem to find remainders quickly and accurately. These questions help students understand how polynomial functions behave when specific values are substituted and how linear polynomial divisors relate to the remainder. Below are the  NCERT solutions for Polynomials Class 9 Exercise 2.4: 

1. Use suitable identities to find the following products:

(i) (x+4)(x +10)

Solution: Using the identity, (x+a)(x+b) = x 2 +(a+b)x+ab [Here, a = 4 and b = 10]
We get, (x+4)(x+10)
= x 2 +(4+10)x+(4×10)
= x 2 +14x+40

(ii) (x+8)(x –10)

Solution: Using the identity, (x+a)(x+b)
= x 2 +(a+b)x+ab [Here, a = 8 and b = −10]
We get, (x+8)(x−10) = x 2 +(8+(−10))x+(8×(−10))
= x 2 +(8−10)x–80
= x 2 −2x−80

(iii) (3x+4)(3x–5)

Solution: Using the identity, (x+a)(x+b)
= x 2 +(a+b)x+ab [Here, x = 3x, a = 4 and b = −5]
We get, (3x+4)(3x−5) = (3x) 2 +[4+(−5)]3x+4×(−5)
= 9x 2 +3x(4–5)–20
= 9x 2 –3x–20

(iv) (y 2 +3/2)(y 2 -3/2)

Solution: Using the identity, (x+y)(x–y)
= x 2 –y 2 [Here, x = y 2 and y = 3/2] We get, (y 2 +3/2)(y 2 –3/2)
= (y 2 ) 2 –(3/2) 2
= y 4 –9/4

(v) (3 – 2x) (3 + 2x)  

Solution: Using the identity, (a − b)(a + b) = a² − b² [Here, a = 3 and b = 2x]

We get, (3 − 2x)(3 + 2x)
= 3² − (2x)² = 9 − 4x²
= 9 − 4x²

2. Evaluate the following products without multiplying directly:

(i) 103×107

Solution: 103×107= (100+3)×(100+7)
Using identity, [(x+a)(x+b) = x 2 +(a+b)x+ab
Here, x = 100 a = 3 b = 7
We get, 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 identity, [(x-a)(x-b) = x 2 -(a+b)x+ab
Here, x = 100 a = -5 b = -4
We get, 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 identity, [(a+b)(a-b)= a 2 -b 2 ]
Here, a = 100 b = 4 We get, 104×96
= (100+4)×(100–4)
= (100) 2 –(4) 2
= 10000–16 = 9984

3. Factorise the following using appropriate identities:

(i) 9x 2 +6xy+y 2

Solution: 9x 2 +6xy+y 2 = (3x) 2 +(2×3x×y)+y 2 Using identity, x 2 +2xy+y 2 = (x+y) 2 Here, x = 3x y = y 9x 2 +6xy+y 2 = (3x) 2 +(2×3x×y)+y 2 = (3x+y) 2 = (3x+y)(3x+y)

(ii) 4y 2 −4y+1

Solution: 4y 2 −4y+1 = (2y) 2 –(2×2y×1)+1 Using identity, x 2 – 2xy + y 2 = (x – y) 2 Here, x = 2y y = 1 4y 2 −4y+1 = (2y) 2 –(2×2y×1)+1 2 = (2y–1) 2 = (2y–1)(2y–1)

(iii)  x 2 –y 2 /100

Solution: x 2 –y 2 /100 = x 2 –(y/10) 2 Using identity, x 2 -y 2 = (x-y)(x+y) Here, x = x y = y/10 x 2 –y 2 /100 = x 2 –(y/10) 2 = (x–y/10)(x+y/10)

4. Expand each of the following using suitable identities:

(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) ((1/4)a-(1/2)b +1) 2

Solution:

(i) (x+2y+4z) 2

Using identity, (x+y+z) 2 = x 2 +y 2 +z 2 +2xy+2yz+2zx Here, x = x y = 2y z = 4z (x+2y+4z) 2 = x 2 +(2y) 2 +(4z) 2 +(2×x×2y)+(2×2y×4z)+(2×4z×x) = x 2 +4y 2 +16z 2 +4xy+16yz+8xz

(ii) (2x−y+z) 2

Using identity, (x+y+z) 2 = x 2 +y 2 +z 2 +2xy+2yz+2zx Here, x = 2x y = −y z = z (2x−y+z) 2 = (2x) 2 +(−y) 2 +z 2 +(2×2x×−y)+(2×−y×z)+(2×z×2x) = 4x 2 +y 2 +z 2 –4xy–2yz+4xz

(iii) (−2x+3y+2z) 2

Solution: Using identity, (x+y+z) 2 = x 2 +y 2 +z 2 +2xy+2yz+2zx Here, 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) = 4x 2 +9y 2 +4z 2 –12xy+12yz–8xz

(iv) (3a –7b–c) 2

Solution: Using identity (x+y+z) 2 = x 2 +y 2 +z 2 +2xy+2yz+2zx Here, 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) = 9a 2 + 49b 2 + c 2 – 42ab+14bc–6ca

(v) (–2x+5y–3z) 2

Solution: Using identity, (x+y+z) 2 = x 2 +y 2 +z 2 +2xy+2yz+2zx Here, 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) = 4x 2 +25y 2 +9z 2 – 20xy–30yz+12zx

(vi) ((1/4)a-(1/2)b+1) 2

Solution: Using identity, (x+y+z) 2 = x 2 +y 2 +z 2 +2xy+2yz+2zx Here, x = (1/4)a y = (-1/2)b z = 1

5. Factorise:

(i) 4x 2 +9y 2 +16z 2 +12xy–24yz–16xz

(ii) 2x 2 +y 2 +8z 2 –2√2xy+4√2yz–8xz

Solution:

(i) 4x 2 +9y 2 +16z 2 +12xy–24yz–16xz

Using identity, (x+y+z) 2 = x 2 +y 2 +z 2 +2xy+2yz+2zx We can say that, x 2 +y 2 +z 2 +2xy+2yz+2zx = (x+y+z) 2 4x 2 +9y 2 +16z 2 +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) 2x 2 +y 2 +8z 2 –2√2xy+4√2yz–8xz

Using identity, (x +y+z) 2 = x 2 +y 2 +z 2 +2xy+2yz+2zx We can say that, x 2 +y 2 +z 2 +2xy+2yz+2zx = (x+y+z) 2 2x 2 +y 2 +8z 2 –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)

6. Write the following cubes in expanded form:

(i) (2x+1) 3

(ii) (2a−3b) 3

(iii) ((3/2)x+1) 3

(iv) (x−(2/3)y) 3

Solution:

(i) (2x+1) 3

Using identity,(x+y) 3 = x 3 +y 3 +3xy(x+y) (2x+1) 3 = (2x) 3 +1 3 +(3×2x×1)(2x+1) = 8x 3 +1+6x(2x+1) = 8x 3 +12x 2 +6x+1

(ii) (2a−3b) 3

Using identity,(x–y) 3 = x 3 –y 3 –3xy(x–y) (2a−3b) 3 = (2a) 3 −(3b) 3 –(3×2a×3b)(2a–3b) = 8a 3 –27b 3 –18ab(2a–3b) = 8a 3 –27b 3 –36a 2 b+54ab 2

(iii) ((3/2)x+1) 3

Using identity,(x+y) 3 = x 3 +y 3 +3xy(x+y) ((3/2)x+1) 3 =((3/2)x) 3 +1 3 +(3×(3/2)x×1)((3/2)x +1)

(iv)  (x−(2/3)y) 3

Using identity, (x –y) 3 = x 3 –y 3 –3xy(x–y)

7. Evaluate the following using suitable identities:

(i) (99) 3

(ii) (102) 3

(iii) (998) 3

Solutions:

(i) (99) 3

Solution: We can write 99 as 100–1 Using identity, (x –y) 3 = x 3 –y 3 –3xy(x–y) (99) 3 = (100–1) 3 = (100) 3 –1 3 –(3×100×1)(100–1) = 1000000 –1–300(100 – 1) = 1000000–1–30000+300 = 970299

(ii) (102) 3

Solution: We can write 102 as 100+2 Using identity,(x+y) 3 = x 3 +y 3 +3xy(x+y) (100+2) 3 =(100) 3 +2 3 +(3×100×2)(100+2) = 1000000 + 8 + 600(100 + 2) = 1000000 + 8 + 60000 + 1200 = 1061208

(iii) (998) 3

Solution: We can write 99 as 1000–2 Using identity,(x–y) 3 = x 3 –y 3 –3xy(x–y) (998) 3 =(1000–2) 3 =(1000) 3 –2 3 –(3×1000×2)(1000–2) = 1000000000–8–6000(1000– 2) = 1000000000–8- 6000000+12000 = 994011992

8. Factorise each of the following:

(i) 8a 3 +b 3 +12a 2 b+6ab 2

(ii) 8a 3 –b 3 –12a 2 b+6ab 2

(iii) 27–125a 3 –135a +225a 2

(iv) 64a 3 –27b 3 –144a 2 b+108ab 2

(v) 27p 3 –(1/216)−(9/2) p 2 +(1/4)p

Solutions:

(i) 8a 3 +b 3 +12a 2 b+6ab 2

Solution: The expression, 8a 3 +b 3 +12a 2 b+6ab 2 can be written as (2a) 3 +b 3 +3(2a) 2 b+3(2a)(b) 2 8a 3 +b 3 +12a 2 b+6ab 2 = (2a) 3 +b 3 +3(2a) 2 b+3(2a)(b) 2 = (2a+b) 3 = (2a+b)(2a+b)(2a+b) Here, the identity, (x +y) 3 = x 3 +y 3 +3xy(x+y) is used.

(ii) 8a 3 –b 3 –12a 2 b+6ab 2

Solution: The expression, 8a 3 –b 3 −12a 2 b+6ab 2 can be written as (2a) 3 –b 3 –3(2a) 2 b+3(2a)(b) 2 8a 3 –b 3 −12a 2 b+6ab 2 = (2a) 3 –b 3 –3(2a) 2 b+3(2a)(b) 2 = (2a–b) 3 = (2a–b)(2a–b)(2a–b) Here, the identity,(x–y) 3 = x 3 –y 3 –3xy(x–y) is used.

(iii) 27–125a 3 –135a+225a 2

Solution: The expression, 27–125a 3 –135a +225a 2 can be written as 3 3 –(5a) 3 –3(3) 2 (5a)+3(3)(5a) 2 27–125a 3 –135a+225a 2 = 3 3 –(5a) 3 –3(3) 2 (5a)+3(3)(5a) 2 = (3–5a) 3 = (3–5a)(3–5a)(3–5a) Here, the identity, (x–y) 3 = x 3 –y 3 -3xy(x–y) is used.

(iv) 64a 3 –27b 3 –144a 2 b+108ab 2

Solution: The expression, 64a 3 –27b 3 –144a 2 b+108ab 2 can be written as (4a) 3 –(3b) 3 –3(4a) 2 (3b)+3(4a)(3b) 2 64a 3 –27b 3 –144a 2 b+108ab 2 = (4a) 3 –(3b) 3 –3(4a) 2 (3b)+3(4a)(3b) 2 =(4a–3b) 3 =(4a–3b)(4a–3b)(4a–3b) Here, the identity, (x – y) 3 = x 3 – y 3 – 3xy(x – y) is used.

(v) 27p 3 – (1/216)−(9/2) p 2 +(1/4)p

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

9. Verify:

(i) x 3 +y 3 = (x+y)(x 2 –xy+y 2 )

(ii) x 3 –y 3 = (x–y)(x 2 +xy+y 2 )

Solutions:

(i) x 3 +y 3 = (x+y)(x 2 –xy+y 2 )

We know that, (x+y) 3 = x 3 +y 3 +3xy(x+y) ⇒ x 3 +y 3 = (x+y) 3 –3xy(x+y) ⇒ x 3 +y 3 = (x+y)[(x+y) 2 –3xy] Taking (x+y) common ⇒ x 3 +y 3 = (x+y)[(x 2 +y 2 +2xy)–3xy] ⇒ x 3 +y 3 = (x+y)(x 2 +y 2 –xy)

(ii) x 3 –y 3 = (x–y)(x 2 +xy+y 2 )

We know that, (x–y) 3 = x 3 –y 3 –3xy(x–y) ⇒ x 3 −y 3 = (x–y) 3 +3xy(x–y) ⇒ x 3 −y 3 = (x–y)[(x–y) 2 +3xy] Taking (x+y) common ⇒ x 3 −y 3 = (x–y)[(x 2 +y 2 –2xy)+3xy] ⇒ x 3 +y 3 = (x–y)(x 2 +y 2 +xy)

10. Factorise each of the following:

(i) 27y 3 +125z 3

(ii) 64m 3 –343n 3

Solutions:

(i) 27y 3 +125z 3

The expression, 27y 3 +125z 3 can be written as (3y) 3 +(5z) 3 27y 3 +125z 3 = (3y) 3 +(5z) 3 We know that, x 3 +y 3 = (x+y)(x 2 –xy+y 2 ) 27y 3 +125z 3 = (3y) 3 +(5z) 3 = (3y+5z)[(3y) 2 –(3y)(5z)+(5z) 2 ] = (3y+5z)(9y 2 –15yz+25z 2 )

(ii) 64m 3 –343n 3

The expression, 64m 3 –343n 3 can be written as (4m) 3 –(7n) 3 64m 3 –343n 3 = (4m) 3 –(7n) 3 We know that, x 3 –y 3 = (x–y)(x 2 +xy+y 2 ) 64m 3 –343n 3 = (4m) 3 –(7n) 3 = (4m-7n)[(4m) 2 +(4m)(7n)+(7n) 2 ] = (4m-7n)(16m 2 +28mn+49n 2 )

11. Factorise: 27x 3 +y 3 +z 3 –9xyz.

Solution: The expression 27x 3 +y 3 +z 3 –9xyz can be written as (3x) 3 +y 3 +z 3 –3(3x)(y)(z) 27x 3 +y 3 +z 3 –9xyz  = (3x) 3 +y 3 +z 3 –3(3x)(y)(z) We know that, x 3 +y 3 +z 3 –3xyz = (x+y+z)(x 2 +y 2 +z 2 –xy –yz–zx) 27x 3 +y 3 +z 3 –9xyz  = (3x) 3 +y 3 +z 3 –3(3x)(y)(z) = (3x+y+z)[(3x) 2 +y 2 +z 2 –3xy–yz–3xz] = (3x+y+z)(9x 2 +y 2 +z 2 –3xy–yz–3xz)

12. Verify that:

x 3 +y 3 +z 3 –3xyz = (1/2) (x+y+z)[(x–y) 2 +(y–z) 2 +(z–x) 2 ]

Solution: We know that, x 3 +y 3 +z 3 −3xyz = (x+y+z)(x 2 +y 2 +z 2 –xy–yz–xz) ⇒ x 3 +y 3 +z 3 –3xyz = (1/2)(x+y+z)[2(x 2 +y 2 +z 2 –xy–yz–xz)] = (1/2)(x+y+z)(2x 2 +2y 2 +2z 2 –2xy–2yz–2xz) = (1/2)(x+y+z)[(x 2 +y 2 −2xy)+(y 2 +z 2 –2yz)+(x 2 +z 2 –2xz)] = (1/2)(x+y+z)[(x–y) 2 +(y–z) 2 +(z–x) 2 ]

13. If  x+y+z = 0, show that x 3 +y 3 +z 3 = 3xyz.

Solution: We know that, x 3 +y 3 +z 3 -3xyz = (x +y+z)(x 2 +y 2 +z 2 –xy–yz–xz) Now, according to the question, let (x+y+z) = 0, Then, x 3 +y 3 +z 3 -3xyz = (0)(x 2 +y 2 +z 2 –xy–yz–xz) ⇒ x 3 +y 3 +z 3 –3xyz = 0 ⇒ x 3 +y 3 +z 3 = 3xyz Hence Proved

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) (−12) 3 +(7) 3 +(5) 3

Let a = −12 b = 7 c = 5 We know that if x+y+z = 0, then x 3 +y 3 +z 3 =3xyz. Here, −12+7+5=0 (−12) 3 +(7) 3 +(5) 3 = 3xyz = 3×-12×7×5 = -1260

(ii) (28) 3 +(−15) 3 +(−13) 3

Solution: (28) 3 +(−15) 3 +(−13) 3 Let a = 28 b = −15 c = −13 We know that if x+y+z = 0, then x 3 +y 3 +z 3 = 3xyz. Here, x+y+z = 28–15–13 = 0 (28) 3 +(−15) 3 +(−13) 3 = 3xyz = 0+3(28)(−15)(−13) = 16380

15. Give possible expressions for the length and breadth of each of the following rectangles, in which their areas are given:

(i) Area: 25a 2 –35a+12

(ii) Area: 35y 2 +13y–12

Solution: (i) Area: 25a 2 –35a+12 Using the splitting the middle term method, We have to find a number whose sum = -35 and product =25×12 = 300 We get -15 and -20 as the numbers [-15+-20=-35 and -15×-20 = 300] 25a 2 –35a+12 = 25a 2 –15a−20a+12 = 5a(5a–3)–4(5a–3) = (5a–4)(5a–3) Possible expression for length  = 5a–4 Possible expression for breadth  = 5a –3

(ii) Area: 35y 2 +13y–12 Using the splitting the middle term method, We have to find a number whose sum = 13 and product = 35×-12 = 420 We get -15 and 28 as the numbers [-15+28 = 13 and -15×28=420] 35y 2 +13y–12 = 35y 2 –15y+28y–12 = 5y(7y–3)+4(7y–3) = (5y+4)(7y–3) Possible expression for length  = (5y+4) Possible expression for breadth  = (7y–3)

16. What are the possible expressions for the dimensions of the cuboids whose volumes are given below?

(i) Volume: 3x 2 –12x

(ii) Volume: 12ky 2 +8ky–20k

Solution: (i) Volume: 3x 2 –12x 3x 2 –12x can be written as 3x(x–4) by taking 3x out of both the terms. Possible expression for length = 3 Possible expression for breadth = x Possible expression for height = (x–4)

(ii) Volume: 12ky 2 +8ky–20k 12ky 2 +8ky–20k can be written as 4k(3y 2 +2y–5) by taking 4k out of both the terms. 12ky 2 +8ky–20k = 4k(3y 2 +2y–5) [Here, 3y 2 +2y–5 can be written as 3y 2 +5y–3y–5 using splitting the middle term method.] = 4k(3y 2 +5y–3y–5) = 4k[y(3y+5)–1(3y+5)] = 4k(3y+5)(y–1) Possible expression for length = 4k Possible expression for breadth = (3y +5) Possible expression for height = (y -1)

NCERT Maths Class 9 Polynomials Exercise 2.4 PDF

NCERT Maths Class 9 Polynomials Exercise 2.4 PDF includes neatly solved examples demonstrating the correct method to apply the Remainder Theorem. Each solution shows how to substitute the value of x and simplify it step by step to reach the correct result. NCERT Polynomials Class 9 Exercise 2.4 PDF includes simple explanations and is aligned with the Class 9 syllabus. 

NCERT Solutions Maths Class 9 PDF

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