# NCERT Solutions for class 9 Maths Chapter 2- Polynomials Exercise 2.5

NCERT Solutions Class 9 Maths Chapter 2 â€“ Polynomials Exercise 2.5 are given here. These NCERT Maths solutions are created by our subject experts which makes it easy for students to learn. The students use it for reference while solving the exercise problems. The fifth exercise in Polynomials- Exercise 2.5 discusses the Algebraic Identities. They provide a detailed and stepwise explanation of each answer to the questions given in the exercises in the NCERT textbook for Class 9. The NCERT solutions are always prepared by following NCERT guidelines so that it should cover the whole syllabus accordingly. These are very helpful in scoring well in first and second term examinations.

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1. Use suitable identities to find the following products:

(i) (x+4)(xÂ +10)Â

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) = x2+(4+10)x+(4Ã—10)

= x2+14x+40

(ii) (x+8)(xÂ â€“10)Â Â Â Â Â

Solution:

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

[Here, a = 8 and b = âˆ’10]

We get,

(x+8)(xâˆ’10) = x2+(8+(âˆ’10))x+(8Ã—(âˆ’10))

= x2+(8âˆ’10)xâ€“80

= x2âˆ’2xâˆ’80

(iii) (3x+4)(3xâ€“5)

Solution:

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

[Here, x = 3x, a = 4 and b = âˆ’5]

We get,

(3x+4)(3xâˆ’5) = (3x)2+[4+(âˆ’5)]3x+4Ã—(âˆ’5)

= 9x2+3x(4â€“5)â€“20

= 9x2â€“3xâ€“20

(iv) (y2+3/2)(y2-3/2)

Solution:

Using the identity, (x+y)(xâ€“y) = x2â€“y 2

[Here, x = y2and y = 3/2]

We get,

(y2+3/2)(y2â€“3/2) = (y2)2â€“(3/2)2

= y4â€“9/4

2. Evaluate the following products without multiplying directly:

(i) 103Ã—107

Solution:

103Ã—107= (100+3)Ã—(100+7)

Using identity, [(x+a)(x+b) = x2+(a+b)x+ab

Here, x = 100

a = 3

b = 7

We get, 103Ã—107 = (100+3)Ã—(100+7)

= (100)2+(3+7)100+(3Ã—7)

= 10000+1000+21

= 11021

(ii) 95Ã—96 Â

Solution:

95Ã—96 = (100-5)Ã—(100-4)

Using identity, [(x-a)(x-b) = x2-(a+b)x+ab

Here, x = 100

a = -5

b = -4

We get, 95Ã—96 = (100-5)Ã—(100-4)

= (100)2+100(-5+(-4))+(-5Ã—-4)

= 10000-900+20

= 9120

(iii) 104Ã—96

Solution:

104Ã—96 = (100+4)Ã—(100â€“4)

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

Here, a = 100

b = 4

We get, 104Ã—96 = (100+4)Ã—(100â€“4)

= (100)2â€“(4)2

= 10000â€“16

= 9984

3. Factorize the following using appropriate identities:

(i) 9x2+6xy+y2

Solution:

9x2+6xy+y2 = (3x)2+(2Ã—3xÃ—y)+y2

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

Here, x = 3x

y = y

9x2+6xy+y2 = (3x)2+(2Ã—3xÃ—y)+y2

= (3x+y)2

= (3x+y)(3x+y)

(ii) 4y2âˆ’4y+1

Solution:

4y2âˆ’4y+1 = (2y)2â€“(2Ã—2yÃ—1)+1

Using identity, x2 â€“ 2xy + y2 = (x â€“ y)2

Here, x = 2y

y = 1

4y2âˆ’4y+1 = (2y)2â€“(2Ã—2yÃ—1)+12

= (2yâ€“1)2

= (2yâ€“1)(2yâ€“1)

(iii) Â x2â€“y2/100

Solution:

x2â€“y2/100 = x2â€“(y/10)2

Using identity, x2-y2 = (x-y)(x+y)

Here, x = x

y = y/10

x2â€“y2/100 = x2â€“(y/10)2

= (xâ€“y/10)(x+y/10)

4. Expand each of the following, using suitable identities:

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

(ii) (2xâˆ’y+z)2

(iii) (âˆ’2x+3y+2z)2

(iv) (3a â€“7bâ€“c)2

(v) (â€“2x+5yâ€“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+(2Ã—xÃ—2y)+(2Ã—2yÃ—4z)+(2Ã—4zÃ—x)

= x2+4y2+16z2+4xy+16yz+8xz

(ii) (2xâˆ’y+z)2Â

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

Here, x = 2x

y = âˆ’y

z = z

(2xâˆ’y+z)2 = (2x)2+(âˆ’y)2+z2+(2Ã—2xÃ—âˆ’y)+(2Ã—âˆ’yÃ—z)+(2Ã—zÃ—2x)

= 4x2+y2+z2â€“4xyâ€“2yz+4xz

(iii) (âˆ’2x+3y+2z)2

Solution:

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

Here, x = âˆ’2x

y = 3y

z = 2z

(âˆ’2x+3y+2z)2 = (âˆ’2x)2+(3y)2+(2z)2+(2Ã—âˆ’2xÃ—3y)+(2Ã—3yÃ—2z)+(2Ã—2zÃ—âˆ’2x)

= 4x2+9y2+4z2â€“12xy+12yzâ€“8xz

(iv) (3a â€“7bâ€“c)2

Solution:

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

Here, x = 3a

y = â€“ 7b

z = â€“ c

(3a â€“7bâ€“ c)2 = (3a)2+(â€“ 7b)2+(â€“ c)2+(2Ã—3a Ã—â€“ 7b)+(2Ã—â€“ 7b Ã—â€“ c)+(2Ã—â€“ c Ã—3a)

= 9a2 + 49b2 + c2â€“ 42ab+14bcâ€“6ca

(v) (â€“2x+5yâ€“3z)2

Solution:

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

Here, x = â€“2x

y = 5y

z = â€“ 3z

(â€“2x+5yâ€“3z)2 = (â€“2x)2+(5y)2+(â€“3z)2+(2Ã—â€“2x Ã— 5y)+(2Ã— 5yÃ—â€“ 3z)+(2Ã—â€“3z Ã—â€“2x)

= 4x2+25y2 +9z2â€“ 20xyâ€“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+12xyâ€“24yzâ€“16xz

(ii ) 2x2+y2+8z2â€“2âˆš2xy+4âˆš2yzâ€“8xz

Solution:

(i) 4x2+9y2+16z2+12xyâ€“24yzâ€“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+12xyâ€“24yzâ€“16xzÂ = (2x)2+(3y)2+(âˆ’4z)2+(2Ã—2xÃ—3y)+(2Ã—3yÃ—âˆ’4z)+(2Ã—âˆ’4zÃ—2x)

= (2x+3yâ€“4z)2

= (2x+3yâ€“4z)(2x+3yâ€“4z)

(ii) 2x2+y2+8z2â€“2âˆš2xy+4âˆš2yzâ€“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+8z2â€“2âˆš2xy+4âˆš2yzâ€“8xz

= (-âˆš2x)2+(y)2+(2âˆš2z)2+(2Ã—-âˆš2xÃ—y)+(2Ã—yÃ—2âˆš2z)+(2Ã—2âˆš2Ã—âˆ’âˆš2x)

= (âˆ’âˆš2x+y+2âˆš2z)2

= (âˆ’âˆš2x+y+2âˆš2z)(âˆ’âˆš2x+y+2âˆš2z)

6. Write the following cubes in expanded form:

(i) (2x+1)3

(ii) (2aâˆ’3b)3

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

(iv) (xâˆ’(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+(3Ã—2xÃ—1)(2x+1)

= 8x3+1+6x(2x+1)

= 8x3+12x2+6x+1

(ii) (2aâˆ’3b)3

Using identity,(xâ€“y)3 = x3â€“y3â€“3xy(xâ€“y)

(2aâˆ’3b)3 = (2a)3âˆ’(3b)3â€“(3Ã—2aÃ—3b)(2aâ€“3b)

= 8a3â€“27b3â€“18ab(2aâ€“3b)

= 8a3â€“27b3â€“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+(3Ã—(3/2)xÃ—1)((3/2)x +1)

(iv) Â (xâˆ’(2/3)y)3

Using identity, (x â€“y)3 = x3â€“y3â€“3xy(xâ€“y)

7. Evaluate the following using suitable identities:Â

(i) (99)3

(ii) (102)3

(iii) (998)3

Solutions:

(i) (99)3

Solution:

We can write 99 as 100â€“1

Using identity, (x â€“y)3 = x3â€“y3â€“3xy(xâ€“y)

(99)3 = (100â€“1)3

= (100)3â€“13â€“(3Ã—100Ã—1)(100â€“1)

= 1000000 â€“1â€“300(100 â€“ 1)

= 1000000â€“1â€“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+(3Ã—100Ã—2)(100+2)

= 1000000 + 8 + 600(100 + 2)

= 1000000 + 8 + 60000 + 1200

= 1061208

(iii) (998)3

Solution:

We can write 99 as 1000â€“2

Using identity,(xâ€“y)3 = x3â€“y3â€“3xy(xâ€“y)

(998)3 =(1000â€“2)3

=(1000)3â€“23â€“(3Ã—1000Ã—2)(1000â€“2)

= 1000000000â€“8â€“6000(1000â€“ 2)

= 1000000000â€“8- 6000000+12000

= 994011992

8. Factorise each of the following:

(i) 8a3+b3+12a2b+6ab2

(ii) 8a3â€“b3â€“12a2b+6ab2

(iii) 27â€“125a3â€“135a +225a2Â  Â

(iv) 64a3â€“27b3â€“144a2b+108ab2

(v) 27p3â€“(1/216)âˆ’(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.

Â

(ii) 8a3â€“b3â€“12a2b+6ab2

Solution:

The expression, 8a3â€“b3âˆ’12a2b+6ab2 can be written as (2a)3â€“b3â€“3(2a)2b+3(2a)(b)2

8a3â€“b3âˆ’12a2b+6ab2 = (2a)3â€“b3â€“3(2a)2b+3(2a)(b)2

= (2aâ€“b)3

= (2aâ€“b)(2aâ€“b)(2aâ€“b)

Here, the identity,(xâ€“y)3 = x3â€“y3â€“3xy(xâ€“y) is used.

Â

(iii) 27â€“125a3â€“135a+225a2Â

Solution:

The expression, 27â€“125a3â€“135a +225a2 can be written as 33â€“(5a)3â€“3(3)2(5a)+3(3)(5a)2

27â€“125a3â€“135a+225a2 =
33â€“(5a)3â€“3(3)2(5a)+3(3)(5a)2

= (3â€“5a)3

= (3â€“5a)(3â€“5a)(3â€“5a)

Here, the identity, (xâ€“y)3 = x3â€“y3-3xy(xâ€“y) is used.

(iv) 64a3â€“27b3â€“144a2b+108ab2

Solution:

The expression, 64a3â€“27b3â€“144a2b+108ab2can be written as (4a)3â€“(3b)3â€“3(4a)2(3b)+3(4a)(3b)2

64a3â€“27b3â€“144a2b+108ab2=
(4a)3â€“(3b)3â€“3(4a)2(3b)+3(4a)(3b)2

=(4aâ€“3b)3

=(4aâ€“3b)(4aâ€“3b)(4aâ€“3b)

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

(v) 27p3â€“ (1/216)âˆ’(9/2) p2+(1/4)p

Solution:

The expression, 27p3â€“(1/216)âˆ’(9/2) p2+(1/4)p can be written as

(3p)3â€“(1/6)3âˆ’(9/2) p2+(1/4)p =Â (3p)3â€“(1/6)3âˆ’3(3p)(1/6)(3p â€“ 1/6)

Using (x â€“ y)3 = x3 â€“ y3 â€“ 3xy (x â€“ y)

27p3â€“(1/216)âˆ’(9/2) p2+(1/4)p =Â (3p)3â€“(1/6)3âˆ’3(3p)(1/6)(3p â€“ 1/6)

Taking x = 3p and y = 1/6

= (3pâ€“1/6)3

= (3pâ€“1/6)(3pâ€“1/6)(3pâ€“1/6)

9. Verify:

(i) x3+y3 = (x+y)(x2â€“xy+y2)

(ii) x3â€“y3 = (xâ€“y)(x2+xy+y2)

Solutions:

(i) x3+y3 = (x+y)(x2â€“xy+y2)

We know that, (x+y)3 = x3+y3+3xy(x+y)

â‡’ x3+y3 = (x+y)3â€“3xy(x+y)

â‡’ x3+y3 = (x+y)[(x+y)2â€“3xy]

Taking (x+y) common â‡’ x3+y3 = (x+y)[(x2+y2+2xy)â€“3xy]

â‡’ x3+y3 = (x+y)(x2+y2â€“xy)

(ii) x3â€“y3 = (xâ€“y)(x2+xy+y2)Â

We know that,(xâ€“y)3 = x3â€“y3â€“3xy(xâ€“y)

â‡’ x3âˆ’y3 = (xâ€“y)3+3xy(xâ€“y)

â‡’ x3âˆ’y3 = (xâ€“y)[(xâ€“y)2+3xy]

Taking (x+y) common â‡’ x3âˆ’y3 = (xâ€“y)[(x2+y2â€“2xy)+3xy]

â‡’ x3+y3 = (xâ€“y)(x2+y2+xy)

10. Factorize each of the following:

(i) 27y3+125z3

(ii) 64m3â€“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)(x2â€“xy+y2)

27y3+125z3 = (3y)3+(5z)3

= (3y+5z)[(3y)2â€“(3y)(5z)+(5z)2]

= (3y+5z)(9y2â€“15yz+25z2)

(ii) 64m3â€“343n3

The expression, 64m3â€“343n3can be written as (4m)3â€“(7n)3

64m3â€“343n3 =
(4m)3â€“(7n)3

We know that, x3â€“y3 = (xâ€“y)(x2+xy+y2)

64m3â€“343n3 = (4m)3â€“(7n)3

= (4mâ€“7n)[(4m)2+(4m)(7n)+(7n)2]

= (4mâ€“7n)(16m2+28mn+49n2)

11. Factorise:Â 27x3+y3+z3â€“9xyzÂ

Solution:

The expression 27x3+y3+z3â€“9xyzÂ can be written as (3x)3+y3+z3â€“3(3x)(y)(z)

27x3+y3+z3â€“9xyzÂ  = (3x)3+y3+z3â€“3(3x)(y)(z)

We know that, x3+y3+z3â€“3xyz = (x+y+z)(x2+y2+z2â€“xy â€“yzâ€“zx)

27x3+y3+z3â€“9xyzÂ  = (3x)3+y3+z3â€“3(3x)(y)(z)

= (3x+y+z)[(3x)2+y2+z2â€“3xyâ€“yzâ€“3xz]

= (3x+y+z)(9x2+y2+z2â€“3xyâ€“yzâ€“3xz)

12. Verify that:

x3+y3+z3â€“3xyz = (1/2) (x+y+z)[(xâ€“y)2+(yâ€“z)2+(zâ€“x)2]

Solution:

We know that,

x3+y3+z3âˆ’3xyz = (x+y+z)(x2+y2+z2â€“xyâ€“yzâ€“xz)

â‡’ x3+y3+z3â€“3xyz = (1/2)(x+y+z)[2(x2+y2+z2â€“xyâ€“yzâ€“xz)]

= (1/2)(x+y+z)(2x2+2y2+2z2â€“2xyâ€“2yzâ€“2xz)

= (1/2)(x+y+z)[(x2+y2âˆ’2xy)+(y2+z2â€“2yz)+(x2+z2â€“2xz)]

= (1/2)(x+y+z)[(xâ€“y)2+(yâ€“z)2+(zâ€“x)2]

13. IfÂ  x+y+z =Â 0, show thatÂ x3+y3+z3 = 3xyz.

Solution:

We know that,

x3+y3+z3-3xyz = (x +y+z)(x2+y2+z2â€“xyâ€“yzâ€“xz)

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

then, x3+y3+z3 -3xyz = (0)(x2+y2+z2â€“xyâ€“yzâ€“xz)

â‡’ x3+y3+z3â€“3xyz = 0

â‡’ x3+y3+z3 = 3xyz

Hence Proved

14. Without actually calculating the cubes, find the value of each of the following:

(i) (âˆ’12)3+(7)3+(5)3

(ii) (28)3+(âˆ’15)3+(âˆ’13)3

Solution:

(i) (âˆ’12)3+(7)3+(5)3

Let a = âˆ’12

b = 7

c = 5

We know that ifÂ x+y+z =Â 0, then x3+y3+z3=3xyz.

Here, âˆ’12+7+5=0

(âˆ’12)3+(7)3+(5)3 = 3xyz

= 3Ã—-12Ã—7Ã—5

= -1260

(ii) (28)3+(âˆ’15)3+(âˆ’13)3

Solution:

(28)3+(âˆ’15)3+(âˆ’13)3

Let a = 28

b = âˆ’15

c = âˆ’13

We know that ifÂ x+y+z =Â 0, then x3+y3+z3 = 3xyz.

Here, x+y+zÂ = 28â€“15â€“13 = 0

(28)3+(âˆ’15)3+(âˆ’13)3 = 3xyz

= 0+3(28)(âˆ’15)(âˆ’13)

= 16380

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

(i) Area :Â 25a2â€“35a+12

(ii) Area :Â 35y2+13yâ€“12

Solution:

(i) Area :Â 25a2â€“35a+12

Using the splitting the middle term method,

We have to find a number whose sum = -35 and product =25Ã—12=300

We get -15 and -20 as the numbers [-15+-20=-35 and -15Ã—-20=300]

25a2â€“35a+12 = 25a2â€“15aâˆ’20a+12

= 5a(5aâ€“3)â€“4(5aâ€“3)

= (5aâ€“4)(5aâ€“3)

Possible expression for lengthÂ  = 5aâ€“4

Possible expression for breadthÂ  = 5a â€“3

(ii) Area :Â 35y2+13yâ€“12

Using the splitting the middle term method,

We have to find a number whose sum = 13 and product = 35Ã—-12 = 420

We get -15 and 28 as the numbers [-15+28 = 13 and -15Ã—28=420]

35y2+13yâ€“12 = 35y2â€“15y+28yâ€“12

= 5y(7yâ€“3)+4(7yâ€“3)

= (5y+4)(7yâ€“3)

Possible expression for lengthÂ  =Â (5y+4)

Possible expression for breadthÂ  =Â (7yâ€“3)

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

(i) Volume :Â 3x2â€“12x

(ii) Volume :Â 12ky2+8kyâ€“20k

Solution:

(i) Volume :Â 3x2â€“12x

3x2â€“12xÂ can be written as 3x(xâ€“4) by taking 3x out of both the terms.

Possible expression for lengthÂ =Â 3

Possible expression for breadthÂ =Â x

Possible expression for heightÂ =Â (xâ€“4)

(ii) Volume:
12ky2+8kyâ€“20k

12ky2+8kyâ€“20kÂ can be written as 4k(3y2+2yâ€“5) by taking 4k out of both the terms.

12ky2+8kyâ€“20k = 4k(3y2+2yâ€“5)

[Here, 3y2+2yâ€“5 can be written as 3y2+5yâ€“3yâ€“5 using splitting the middle term method.]

= 4k(3y2+5yâ€“3yâ€“5)

= 4k[y(3y+5)â€“1(3y+5)]

= 4k(3y+5)(yâ€“1)

Possible expression for lengthÂ = 4k

Possible expression for breadthÂ =Â (3y +5)

Possible expression for heightÂ =Â (y -1)

1. Thank you!

1. Nabed khan

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5. Sanchita