__Exercise 2.5 Page: 48__

__Exercise 2.5 Page: 48__

**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

**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. Factorize 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)+12

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}**

**((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. Factorize:**

**(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) 64a3–27b3–144a ^{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) 7p ^{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}–3(3p)^{2}(1/6)+3(3p)(1/6)^{2}

27p^{3}–(1/216)−(9/2) p^{2}+(1/4)p =

(3p)^{3}–(1/6)^{3}–3(3p)^{2}(1/6)+3(3p)(1/6)^{2}

= (3p–16)^{3}

= (3p–16)(3p–16)(3p–16)

**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. Factorize 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 expression27x^{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)

**Class 9th Maths Polynomial**