NCERT Solutions for Class 9 Maths Chapter 2 - Polynomials Exercise 2.5

NCERT Solutions Class 9 Maths Chapter 2 – Polynomials Exercise 2.5 are given here. These NCERT Maths solutions are created by our subject experts, who make it easy for students to learn. The students use NCERT Class 9 Maths Solutions for reference while solving the exercise problems. The fifth exercise in Polynomials- Exercise 2.5 discusses the Algebraic Identities.

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NCERT Solutions for Class 9 Maths Chapter 2- Polynomials Exercise 2.5

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Exercise 2.1 Solutions 5 Questions
Exercise 2.2 Solutions 4 Questions
Exercise 2.3 Solutions 3 Questions
Exercise 2.4 Solutions 5 Questions

Access Answers to NCERT Class 9 Maths Chapter 2 – Polynomials Exercise 2.5

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

We get,

(x+4)(x+10) = x²+(4+10)x+(4×10)

= x²+14x+40

(ii) (x+8)(x –10)     

Solution:

Using the identity, (x+a)(x+b) = x ²+(a+b)x+ab

We get,

(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 the identity, (x+a)(x+b) = x ²+(a+b)x+ab

We get,

(3x+4)(3x−5) = (3x)²+[4+(−5)]3x+4×(−5)

= 9x²+3x(4–5)–20

= 9x²–3x–20

(iv) (y²+3/2)(y²-3/2)

Solution:

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

2. Evaluate the following products without multiplying directly:

(i) 103×107

Solution:

103×107= (100+3)×(100+7)

Using identity, [(x+a)(x+b) = x²+(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) = x²-(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)= a²-b²]

Here, a = 100

b = 4

We get, 104×96 = (100+4)×(100–4)

= (100)²–(4)²

= 10000–16

= 9984

3. Factorize the following using appropriate identities:

(i) 9x²+6xy+y²

Solution:

9x²+6xy+y² = (3x)²+(2×3x×y)+y²

Using identity, x²+2xy+y² = (x+y)²

Here, x = 3x

y = 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 identity, x² – 2xy + y² = (x – y)²

Here, x = 2y

y = 1

4y²−4y+1 = (2y)²–(2×2y×1)+1²

= (2y–1)²

= (2y–1)(2y–1)

(iii)  x²–y²/100

Solution:

x²–y²/100 = x²–(y/10)²

Using identity, x²-y² = (x-y)(x+y)

Here, x = x

y = y/10

x²–y²/100 = x²–(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 = x²+y²+z²+2xy+2yz+2zx

Here, 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 

Using identity, (x+y+z)² = x²+y²+z²+2xy+2yz+2zx

Here, 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 identity, (x+y+z)² = x²+y²+z²+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)

= 4x²+9y²+4z²–12xy+12yz–8xz

(iv) (3a –7b–c)²

Solution:

Using identity (x+y+z)² = x²+y²+z²+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)

= 9a² + 49b² + c²– 42ab+14bc–6ca

(v) (–2x+5y–3z)²

Solution:

Using identity, (x+y+z)² = x²+y²+z²+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)

= 4x²+25y² +9z²– 20xy–30yz+12zx

(vi) ((1/4)a-(1/2)b+1)²

Solution:

Using identity, (x+y+z)² = x²+y²+z²+2xy+2yz+2zx

Here, x = (1/4)a

y = (-1/2)b

z = 1

5. Factorize:

(i) 4x²+9y²+16z²+12xy–24yz–16xz

(ii ) 2x²+y²+8z²–2√2xy+4√2yz–8xz

Solution:

(i) 4x²+9y²+16z²+12xy–24yz–16xz

Using identity, (x+y+z)² = x²+y²+z²+2xy+2yz+2zx

We can say that, 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

Using identity, (x +y+z)² = x²+y²+z²+2xy+2yz+2zx

We can say that, 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)

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 = 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

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

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)

(iv)  (x−(2/3)y)³

Using identity, (x –y)³ = x³–y³–3xy(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 = 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:

We can write 102 as 100+2

Using identity,(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:

We can write 99 as 1000–2

Using identity,(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

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

(v) 27p³–(1/216)−(9/2) p²+(1/4)p

Solutions:

(i) 8a³+b³+12a²b+6ab²

Solution:

The expression, 8a³+b³+12a²b+6ab² can 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

= (2a+b)³

= (2a+b)(2a+b)(2a+b)

Here, the identity, (x +y)³ = x³+y³+3xy(x+y) is used.

(ii) 8a³–b³–12a²b+6ab²

Solution:

The expression, 8a³–b³−12a²b+6ab² can 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

= (2a–b)³

= (2a–b)(2a–b)(2a–b)

Here, the identity,(x–y)³ = x³–y³–3xy(x–y) is used.

(iii) 27–125a³–135a+225a² 

Solution:

The expression, 27–125a³–135a +225a² can be 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

= (3–5a)³

= (3–5a)(3–5a)(3–5a)

Here, the identity, (x–y)³ = x³–y³-3xy(x–y) is used.

(iv) 64a³–27b³–144a²b+108ab²

Solution:

The expression, 64a³–27b³–144a²b+108ab²can 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

=(4a–3b)³

=(4a–3b)(4a–3b)(4a–3b)

Here, the identity, (x – y)³ = x³ – y³ – 3xy(x – y) is used.

(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)

9. Verify:

(i) x³+y³ = (x+y)(x²–xy+y²)

(ii) x³–y³ = (x–y)(x²+xy+y²)

Solutions:

(i) x³+y³ = (x+y)(x²–xy+y²)

We know that, (x+y)³ = x³+y³+3xy(x+y)

⇒ x³+y³ = (x+y)³–3xy(x+y)

⇒ x³+y³ = (x+y)[(x+y)²–3xy]

Taking (x+y) common ⇒ x³+y³ = (x+y)[(x²+y²+2xy)–3xy]

⇒ x³+y³ = (x+y)(x²+y²–xy)

(ii) x³–y³ = (x–y)(x²+xy+y²) 

We know that,(x–y)³ = x³–y³–3xy(x–y)

⇒ x³−y³ = (x–y)³+3xy(x–y)

⇒ x³−y³ = (x–y)[(x–y)²+3xy]

Taking (x+y) common ⇒ x³−y³ = (x–y)[(x²+y²–2xy)+3xy]

⇒ x³+y³ = (x–y)(x²+y²+xy)

10. Factorize each of the following:

(i) 27y³+125z³

(ii) 64m³–343n³

Solutions:

(i) 27y³+125z³

The expression, 27y³+125z³ can be written as (3y)³+(5z)³

27y³+125z³ = (3y)³+(5z)³

We know that, 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

The expression, 64m³–343n³can be written as (4m)³–(7n)³

64m³–343n³ =
(4m)³–(7n)³

We know that, x³–y³ = (x–y)(x²+xy+y²)

64m³–343n³ = (4m)³–(7n)³

= (4m–7n)[(4m)²+(4m)(7n)+(7n)²]

= (4m–7n)(16m²+28mn+49n²)

11. Factorise: 27x³+y³+z³–9xyz 

Solution:

The expression 27x³+y³+z³–9xyz can be written as (3x)³+y³+z³–3(3x)(y)(z)

27x³+y³+z³–9xyz  = (3x)³+y³+z³–3(3x)(y)(z)

We know that, 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)

12. Verify that:

x³+y³+z³–3xyz = (1/2) (x+y+z)[(x–y)²+(y–z)²+(z–x)²]

Solution:

We know that,

x³+y³+z³−3xyz = (x+y+z)(x²+y²+z²–xy–yz–xz)

⇒ x³+y³+z³–3xyz = (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]

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

Solution:

We know that,

x³+y³+z³-3xyz = (x +y+z)(x²+y²+z²–xy–yz–xz)

Now, according to the question, let (x+y+z) = 0,

then, x³+y³+z³ -3xyz = (0)(x²+y²+z²–xy–yz–xz)

⇒ x³+y³+z³–3xyz = 0

⇒ x³+y³+z³ = 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 x³+y³+z³=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 x³+y³+z³ = 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 : 25a²–35a+12

(ii) Area : 35y²+13y–12

Solution:

(i) Area : 25a²–35a+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]

25a²–35a+12 = 25a²–15a−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 : 35y²+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]

35y²+13y–12 = 35y²–15y+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 : 3x²–12x

(ii) Volume : 12ky²+8ky–20k

Solution:

(i) Volume : 3x²–12x

3x²–12x 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:
12ky²+8ky–20k

12ky²+8ky–20k can be written as 4k(3y²+2y–5) by taking 4k out of both the terms.

12ky²+8ky–20k = 4k(3y²+2y–5)

= 4k(3y²+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)