# Polynomial – Exercise 2.5 – Class IX

Previous exercise – Polynomials – Exercise 2.4 – Class IX

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) (y2+ 3/2 ) (y23/2 )

(v) (3 – 2x) (3 + 2x)

Solution:

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

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

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

= x2 + 14x + 40

(ii) (x + 8) (x – 10)

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

(x + 8) (x – 10) = x2 + (8 – 10)x + 8x(-10)

= x2 – 2x – 80x

(iii) (3x + 4) (3x – 5)

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

(3x + 4) (3x – 5) = (3x)2 + (4 – 5)3x + 4x(-5)

= 9x2 – 3x – 20

(iv) (y2+ 3/2 ) (y23/2 )

Using identity (x + a)(x – a) = x2 – a2

(y2+ 3/2 ) (y23/2 ) = y2  – (3/2)2

= y2  – 9/4

(v) (3 – 2x) (3 + 2x)

Using identity (x + a)(x – a) = x2 – a2

(3 – 2x) (3 + 2x)= 32 – (2x)2 = 92 – 4x2

1. Evaluate the following products without multiplying directly:

(i) 103 × 107

(ii) 95 × 96

(iii) 104 × 96

Solution:

(i) 103 × 107 = (100 + 3)(100 + 7)

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

(100 + 3)(100 + 7) = 1002 + (3 + 7)100 + 3×7

= 10000 + 10(100) + 21

= 10000 + 1000 + 21

= 11021

(ii) 95 × 96

= (100 – 5)(100 – 4)

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

(100 – 5)(100 – 4) = 1002 + (-5 – 4)*100 + (-4)(-5)

= 10000 – 900 + 20

= 9120

(iii) 104 × 96

=(100 + 4)(100 – 4)

Using the identity (x + a)(x – a) = x2 – a2

(100 + 4)(100 – 4) = 1002 – 42 = 10000 – 16 = 9984

1. Factorise the following using appropriate identities:

(i) 9x2+ 6xy + y2

(ii) 4y2– 4y + 1

(iii) x21/100 .y2

Solution:

(i) 9x2+ 6xy + y2

=(3x)2 + 2(3x)(y) + y2

Using the identity a2 + 2ab + b2 = (a + b)2

= (3x + y)2

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

(ii) 4y2– 4y + 1

= (2y)2 – 2(2y)(1) + 12

Using the identity a2 + 2ab + b2 = (a + b)2

= (2y – 1)2

= (2y – 1)(2y – 1)

(iii) x21/100 .y2

= x2 – 2.x.1/10y + (1/10y)2

Using the identity a2 + 2ab + b2 = (a + b)2

= (x – 1/10y)2

=(x – 1/10y) (x – 1/10y)

1. 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

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

here, x = x ; y = 2y ; z = 4z

= x2 + (2y)2 + (4z)2 + 2(x)(2y) + 2(2y)(4z) + 2(4z)(x)

= x2 + 4y2 + 4z2 + 4xy + 16yz + 8xz

(ii) (2x – y + z)2

Identity V : (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 + 2z(2x)

= 4x2 – y2 + z2 – 4xy – 2yz + 4xz

(iii) (-2x + 3y +  2z)2

Identity V : (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 – 8zx

(iv) (3a – 7b – c)2

Identity V : (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

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

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

Here x =1/4 a ; y =  – 1/2 b ; z = 1

[1/4 a – 1/2 b + 1]2 = (1/4 a)2 + (– 1/2 b)2 + 12 + 2(1/4 a)( – 1/2 b) + 2(– 1/2 b)(1) + 2(1)( 1/4 a)

= 1/16 a2 + 1/4b2 + 1 – 1/4 ab – b + 1/2 a

1. Factorise:

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

(ii) 2x2 + y2+ 8z2– 2√2 xy + 4√2 yz – 8xz

Solution:

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

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

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

= (2x + 3y + z)2

(ii) 2x2 + y2+ 8z2– 2√2 xy + 4√2 yz – 8xz

= (√2x)2 + y2 + (2√2z)2 + 2(√2x)y + 2(2√2z)y + 2(2√2z)(√2x)

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

= (√2x + y + 2√2z)2

1. 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

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 + 1+ 12x2 + 6x

= 8x3 + 12x2 + 6x + 1

(ii) (2a – 3b)3

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

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)

= 27/8 x3 + 1 + 9/2 x(3/2 x + 1)

= 27/8 x3 + 1 + 27/4 x + 9/2 x

(iv) [ x – 2/3 y]3

Identity : (x – y)3 = x3 – y3 – 3xy(x ­- y)

[ x – 2/3 y]3 = x3 – (2/3y)3 – 3(x)(2/3y)(x – 2/3y)

= x38/27 y3 – 2x2y + 4/3xy2

1. Evaluate the following using suitable identities:

(i) (99)3

(ii) (102)3

(iii) (998)3

Solution:

(i) (99)3 = (100 – 1)3

Identity : (x – y)3 = x3 – y3 – 3xy(x – y)

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

= 10000 – 1 – 300(100 – 1)

= 10000 – 1 – 30000 – 300

= 970299

(ii) (102)3 = (100 + 2)3

Identity : (x + y)3 = x3 + y3 + 3xy(x + y)

(100 + 2)3 = 1003 + 23 + 3*100*2(100 + 2)

= 10000 + 8 + 600(100 + 2)

= 10000 + 8 + 60000 + 1200

= 1061208

(iii) (998)3

Identity : (x – y)3 = x3 – y3 – 3xy(x – y)

(1000 – 2)3 = 10003 – 23 – 3*1000*2(1000 – 2)

= 1000000 – 8 – 6000(1000 – 2)

=1000000 – 8 – 6000000 + 12000

= 994011992

1. 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) 27p31/2169/2p2 + 1/4 p

Solution:

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

Identity : (x + y)3 = x3 + y3 + 3xy(x + y)

= (2a)3 + b3 + 3(2a)(b) (2a + b)

= (2a + b)3

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

(ii) 8a3 – b3 – 12a2b + 6ab2

Identity : (x – y)3 = x3 – y3 – 3xy(x – y)

= (2a)3 – (b)3 – 3(2a)(b)(ab – b)

= (2a – b)3

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

(iii) 27 – 125a3 – 135a + 225a2

Identity : (x – y)3 = x3 – y3 – 3xy(x – y)

=(3)3– (5a)3 – 3(3)(5a)(3 – 5a)

= (3 – 5a)3

(iv) 64a3 – 27b3 – 144a2b + 108ab2

Identity : (x – y)3 = x3 – y3 – 3xy(x – y)

=(4a)3 – (3b)3 – 3(4a)(3b)(4a – 3b)

=(4a – 3b)3

=(4a – 3b) (4a – 3b) (4a – 3b)

(v) 27p31/2169/2p2 + 1/4 p

Identity : (x – y)3 = x3 – y3 – 3xy(x – y)

=(3p)3 – (1/6)3 – 3(3p)(1/6)(3p – 1/6)

= (3p – 1/6)3

=(3p – 1/6) (3p – 1/6) (3p – 1/6)

1. Verfiy:

(i) x3 + y3 = (x + y)(x2 – xy + y2)

(ii) x3 – y3 = (x – y)(x2 + xy + y2)

Solution:

(i) x3 + y3 = (x + y)(x2 – xy + y2)

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) x3 – y3 = (x – y)(x2 + xy + y2)

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

1. Factorise each of the following:

(i) 27y3 + 125z3

(ii) 64m3 – 343n3

Solution:

(i) 27y3 + 125z3

Using the identity:  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

sing the identity: 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)

1. Factorise 27x3 + y3 + z3 – 9xyz

Solution:

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

Using identity: a3 + b3 + c3 – 3abc = (a + b + c)(a2 + b2 + c2 – ab – bc – ca)

a = 3x ; b = y ; c = z

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

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

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

Solution:

R.H.S. = 1/2(x + y + z)[(x – y)2 + (y – z)2 + (z – x)2]

= 1/2(x + y + z)[x2 – 2xy + y2 + y2 – 2yz + z2 + z2 – 2zx + x2]

= 1/2(x + y + z)(2x2 + 2y2 + 2z2 – 2xy – 2yz – 2zx)

= 1/2 (x + y + z)2(x2 + y2 + z2 – xy – yz – zx)

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

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

= x3 + xy2 + xz2 – x2 y – xyz – zx2 + x2y + y3 + z2y – xy2 – y2z – xyz + x2z + zy2 + z3 – xyz – yz2 – xz2

= x3 + y3 + z3 – 3xyz

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

Solution:

We know the identity x3 + y3 + z3 – 3xyz = (x + y + z)(x2 + y2 + z2 – xy – yz – zx) ………….(*)

It is given that x + y + z = 0 ……………(1)

Substitute eq(1) in (*)

x3 + y3 + z3 – 3xyz = (0)(x2 + y2 + z2 – xy – yz – zx)

x3 + y3 + z3 – 3xyz = 0

Therefore, x3 + y3 + z3 = 3xyz

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

(i) (-12)3 + 73 + 53

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

Solution:

(i) (-12)3 + 73 + 53

We know that x3 + y3 + z3 = 3xyz if x + y + z = 0

Here, -12 + 7 + 5 = 0

(-12)3 + 73 + 53 = 3(-12)(7)(5) = – 1260

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

We know that x3 + y3 + z3 = 3xyz if x + y + z = 0

Here, 28 – 15 – 13 = 0

(28)3 + (-15)3 + (-13)3 = 3(28)(-15)(-13) = 16380

1. 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

= 25a2 – 20a – 15a + 12

= 5a(5a – 4) – 3(5a – 4)

=(5a – 4)(5a – 3)

So, one possible answer is length = (5a – 4), breadth = (5a – 3)

(ii) area = 35y2 + 13y – 12

= 35y2 + 28y – 15y – 12

= 7y(5y + 4) – 3(5y + 4)

=(5y + 4)(7y – 3)

So, (5y + 4) can be taken as breadth and (7y – 3) as length.

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

(i) Volume: 3x2 – 12x

(ii) Volume 12ky2 + 8ky – 20 k

Solution:

(i) Volume: 3x2 – 12x

abc = 3x2 – 12x = 3x(x – 4)

Therefore, 3, x and (x – 4) are the three factors so they can be three dimensions

(ii) Volume 12ky2 + 8ky – 20 k

abc = 12ky2 + 8ky – 20k

= 4k(3y2 + 2y – 5)

= 4k (3y2 – 3y + 5y – 5)

= 4k (3y(y – 1)+5( y – 1))

= 4k(3y + 5)(y – 1)

Therefore, 4k , (y – 1), (3y + 5) are the three factors of three dimensions.