Class IX Math NCERT Solutions For Polynomials Exercise– 2.5

Exercise– 2.5

1.  Use suitable identities to find the following products:

         (i)   (x + 4) (x + 10)

         (ii)   (x + 8) (x – 10)

         (iii)   (3x + 4)(3x – 5)

         (iv)   

         (v)   (3 – 2x) (3 + 2x)

Ans. (i)   (x + 4) (x + 10):

                   Using the identity

                (x + a)(x + b) = x2 + (a + b)x + ab, we have:

                (x + 4)(x + 10) = x2 + (14 + 10)x + (4 × 10)

                = x2 + 14x + 40

         (ii)   (x + 8) (x – 10):

                Here, a = 8 and b = (–10)

                ∴ Using

                (x + a)(x + b) = x2 + (a + b) + ab,

                we have:

                (x + 8)(x – 10) = x2 + [8 + (–10)]x + [8 × (–10)]

                = x2 + [–2]x + [–80]

                = x2 + 2x – 80

         (iii)   (3x + 4)(3x – 5):

                ∴ Using the identity

                (x + a)(x + b) = x2 + (a + b)x + ab,

                we have:

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

                = 9x2 + [–1]3x + [–20]

                = 9x2 – 3x – 20

         (iv)   

                Using the identity

                (a + b)(a – b) = a2 – b2, we have:

         (v)   (3 – 2x)(3 + 2x):

                Using the identity

                (a + b)(a – b) = a2 – b2, we have:

                (3 – 2x)(3 + 2x) = (3)2 – (2x)2

                = 9 – 4x2

2.  Evaluate the following products without multiplying directly:

         (i)   103 × 107

         (ii)   95 × 96

         (iii)   104 × 96

Ans. (i)   We have 103 × 107 = (100 + 3)(100 + 7)

                = (100)2 + (3 + 7) × 100 + (3 × 7)

                [Using (x + a)(x + b) = x2 + (a + b)x +ab]

                = 10000 + (10) × 100 + 21

                = 10000 + 1000 + 21

                = 11021

         (ii)   We have 95 × 96 = (100 – 5)(100 – 4)

                = (100)2 + [(–5) + (–4)] × 100 + [(–5) × (–4)]

                [Using (x + a)(x + b) = x2 + (a + b)x +ab]

                = 10000 + [–9] × 100 + 21

                = 10000 + (–900) + 20

                = 9120

         (iii)   We have 104 × 96 = (100 + 4)(100 – 4)

                = (100)2 – (4)2

                [Using (a + b)(a – b) = a2 – b2]

                = 10000 – 16

                = 9984

3.  Factorise the following using appropriate identities:

         (i)   9x2 + 6xy + y2

         (ii)   4y2 – 4y + 1

         (iii)   

Ans. (i)   We have 9x2 + 6xy + y2

                = (3x)2 + 2(3x)(y) + (y)2

                = (3x + y)2

                [Using a2 + 2ab + b2 = (a + b)2]

                = (3x + y)(3x + y)

         (ii)   We have 4y2 + 4y + 1

                = (2y)2 – 2(2y)(1) + (1)2

                = (2y – 1)2

                [∴ a2 – 2ab + b2 = (a – b)2]

                = (2y – 1)(2y – 1)

          

                [Using a2 – b2 = (a + b)(a – b)]

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

       

Ans. (i)   (x + 2y + 4z)2

                We have

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

                ∴ (x + 2y + 4z)2 = (x)2 + (2y)2 + (4y)2 + 2(x)(2y) + 2(2y)(4z) + 2(4z)(x)

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

         (ii)   (xx – y + z)2

                Using

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

                we have

                (2x – y + z)2 = (2x)2 + (–y)2 + (z)2 + 2(2x)(–y) + 2(–y)(z) + 2(z)(2x)

                = 4x2 + y2 + z2 – 4xy – 2yz + 4zx

         (iii)   (–2x + 3y + 2z)2

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

                (–2x + 3y + 2z)2 = (–2x)2 + (3y)2 + (2z)2 + 2(–2x)(3y) + 2(3y)(2z) + 2(2z)(–2x)

                = 4x2 + 9y2 + 4z2 – 12xy + 12yz – 8zx

         (iv)   (3a – 7b – c)2

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

                (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

                = 9a2 + 49b2 + c2 – 42ab + 14bc – 6ca

         (v)   (–2x + 5y – 3z)2

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

                ∴ (–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]

                = 4x2 + 25y2 + 9z2 – 20xy – 30yz + 12zx

         

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

                

5.  Factorise:

        

Ans. (i)   4x2 + 9y2 + 16z2 + 12xy – 24yz – 16xz

                = (2x)2 + (3y)2 + (–4z)2 + 2(2x)(3y)

                + 2(3y)(–4z) + 2(–4z)(2x)

                          [Using Identity V]

                = (2x + 3y – 4z)2

                = (2x + 3y – 4z)(2x + 3y – 4z)

         

6.  Write the following cubes in expanded form:

         (i)   (2x + 1)3

         (ii)   2a – 3b)3

         (iii)   

         (iv)   

Ans. Using Identity VI and Identity VII, we have

                (x + y)3 = x3 + y3 + 3xy (x + y), and

                (x + y)3 = x3 + y3 + 3xy (x – y).

         (i)   (2x + 1)3 = (2x)3 + (1)3 + 3(2x)(1)

                [(2x) + (1)]

                       = 8x3 + 1 + 6x[2x + 1]

                [Using Identity VI]

                       = 8x3 + 1 + 12x2 + 6x

                       = 8x3 + 12x2 + 6x + 1

         (ii)   (2a – 3b)3 = (2a)3 – (3b)3 – 3(2a)(3b)

                [(2a) + (3b)]

                 = 8a3 – 27b3 – 18ab[2a – 3b]

                [Using Identity VII]

                 = 8a3 – 27b3 – [36a2b – 54ab2]

                 = 8a3 – 27b3 – 36a2b + 54ab2

         

                 

         

7.  Evaluate the following using suitable identities:

         (i)   (99)3

         (ii)   (102)3

         (iii)   (998)3

Ans. (i)   (99)3

                We have 99 = 100 – 1

                ∴ 993 = (100 – 1)3

                = 1003 – 13 – 3(100)(1)(100 – 1)

                = 1000000 – 1 – 30000 + 300

                = 100300 – 30001

                = 970299

         (ii)   (102)3

                We have 102 = 100 + 2

                (102)3 = (100 + 2)3

                = (100)3 + (2)3 + 3(100)(2)[100 2 1]

                = 1000000 + 8 + 600[100 + 2]

                = 1000000 + 8 + 60000 + 1200

                = 1061208

         (ii)   (998)3

                 We have 998 = 100 – 2

                ∴ (999)3 = (1000 – 2)3

                = (1000)3 – (2)3 – 3(1000)(2)[1000 – 2]

                = 10000000 – 8 – 6000[1000 – 2]

                = 10000000 – 8 – 600000 – 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

Ans. (i)   8a3 + b3 + 12a2b + 6ab2

                = (2a)3 + (b)3 + 6ab(2a + b)

                = (2a)3 + (b)3 + 3(2a)(b)(2a + b)

                = (2a + b)3              [Using Identify VI]

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

         (ii)   8a3 – b3 – 12a2b – 6ab2

                = (2a)3 – (b)3 – 3(2a)(b)(2a – b)

                = (2a – b)3              [Using Identify VII]

                = (2a – b)(2a – b)(2a – b)

         (iii)   27 – 125a3 – 135a + 225a2

                = (3)3 – (5a)3 – 3(3)(5a)[3 – 5a]

                = (3 – 5a)3              [Using Identify VII]

                = (3 – 5a)(3 – 5a)(3 – 5a)

         (iv)   64a3 – 27b3 – 144a2b + 108ab2

                = (4a)3 – (3b)3 – 3(4a)(3b)[4a – 3b]

                = (4a – 3b)3              [Using Identify VII]

                = (4a – 3b)(4a – 3b)(4a – 3b)

         

9.  Verify:

         (i)   x3 + y3 = (x + y)(x2 – xy + y2)

         (ii)   x3 – y3 = (x – y)(x2 + xy + y2)

Ans. (i)   R.H.S. = (x + y)(x2 – xy + y2)

                = x(x2 – xy + y2) + y(x2 – xy + y2)

                = x3 – x2y + xy2 + x2y – xy2 + y3

                = x3 + y3

                = L.H.S

         (ii)   R.H.S. = (x – y)(x2 + xy + y2)

                = x(x2 + xy + y2) – y(x2 + xy + y2)

                = x3 + x2y + xy2 – x2y – xy2 – y3

                = x3 – y3

                = L.H.S

10.  Factorise each of the following:

         (i)   27y3 + 125z3

         (ii)   64m3 – 343n3

Remember

         I.   x3 + y3 = (x + y)(x2 + y2 – xy)

         II.   x3 – y3 = (x – y)(x2 + y2 + xy)

Ans. (i)   Using the identity

                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)   Using the identity

                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 + 19n2)

11.  Factorise 27x3 + y3 + z3 – 9xyz.

Ans. Remember

                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)

                ∴ Using the identity

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

                – xy – yz – zx), we have

                (3x)3 + y3 +z3 – 3(3x)(y)(z)

                = (3x + y +z)[(3x)2 + y2 + z2 – (3x × y)

                                                    – (y × z) – (z × 3x)]

                = (3x + y +z)(9x2 + y2 + z2 – 3xy – yz – 3zx)

12.  Verify that

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

Ans. Ans. Since x + y + z = 0

                ∴ x + y = –z

                or (x + y)3 = (–z)3

                or x3 + y3 + 3xy(x + y) = –z3

                or x3 + y3 + 3xy(–z) = –z3

                [ x + y = (–z)]

                or x3 + y3 – 3xy = –z3

                or (x3 + y3 + z3) – 3xy = 0

                or (x3 + y3 + z3) = 3xy

                Hence,

                         If x + y + z = 0, then (x3 + y3 + z3 = 3xy.

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

Ans. (i)   (–12)3 + (7)3 + (5)3

                Let x = –12, y = 7 and z = 5

                Then x + y + z = –12 + 7 + 5 = 0

                We know that if

                x + y + z =0, then x3 + y3 + z3 = 3xyz.

                ∴ (–12)3 + (7)3 + (5)3 = 3[(–12)(7)(5)]

                [∴ (–12) + 7 + 5 = 0]

                = 3[–420]

                = –1260

                Thus, (–12)3 + (7)3 + (5)3 = –1260

         (ii)   (28)3 + (–15)3 + (–13)3

                Let x = 28, y = –15 and z = –13

                ∴ x + y + z =0, then x3 + y3 + z3 = 3xyz.

                We know that if

                x + y + z =0, then x3 + y3 + z3 = 3xyz.

                (28)3 + (–15)3 + (–13)3 = 3(28)(–15)(–13) [28 + (–15) + (–13) = 0]

                = 3(5460)

                = 16380

                Thus, (28)3 + (–15)3 + (–13)3 = 1638

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

Area: 25a – 35a 2 + 12 Area: 35y2 + 13y – 12

                        (i)                                        (ii)

Remember

                Area of a rectangle = (Length) × (Breadth)

Ans. (i)   Area = 25a2 – 35a + 12

                We have to factorise the polynomial:

                25a2 – 35a + 12

                Splitting the co-efficient of a, we have

                –35 = (–20) + (–15)

                [∴ 25 × 12 = 300 and (–20) × (–15) = 300]

                = 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

                 We have to factorise the polynomial:

                 35y2 + 13y – 12

                 Splitting the middle term, we get

                 13y = 28y – 15y

                [∴ 28 × (–15) = –420 and –12 × 35 = –420]

                ∴ 35y2 + 13y – 12 = 35y2 + 28y – 15y – 12

                 = 7y(5y + 4) – 3(5y + 4)

                 = (5y + 4)(7y – 3)

                 Thus, the possible length and breadth are                 (7y – 3) and (5y + 4).

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

         (i)   Volume: 3x – 122 x

         (ii)   Volume: 12ky2 + 18ky – 12k

Remember

          Volume of a cuboid = (Length) × (Breadth) × (Height)

Ans. (i)   Volume = 3x2 – 12x

                 On factorising 3x2 – 12x, we have

                 3x2 – 12x = 3[x2 – 4x]

                 = 3 × [x(x – 4)]

                 = 3 × x × (x – 4)

                 ∴The possible dimensions of the cuboid are: 3, x and (x – 4) units.

         (ii)   Volume = 12ky2 + 8ky – 20k

                 We have

                 12ky2 + 8ky – 20k = 4[3ky2 + 2ky – 5k]

                 = 4[2(3y2 + 2y – 5]

                 (Splitting the middle term)

                 = 4k[3y(y – 1) + 5(y – 1)]

                 = 4k[(3y + 5)(y – 1)]

                 = 4k × (3y + 5) × (y – 1)

                 Thus, the possible dimensions are:

                 4k, (3y + 5) and (y – 1) units.