## Sunday, June 24, 2012

### CBSE Class 9 Maths Polynomials Exercise 2.5

Algebraic Identities

1.  (x + y)2 = x2 + 2xy + y2

2.  (x – y)2  = x2 – 2xy + y2

3.  x2  – y2  = (x + y) (x – y)

4.  (x + a) (x + b) =  x2 + (a+b)x + ab

5.  (x - a)(x + b) = x2 + (b-a)x - ab

6.  (x + a) (x - b) =  x2 + (a-b)x - ab

7. (x - a)(x - b) =  x2 - (a+b)x + ab

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

9.  (x + y)3 = x3 + y3 + 3xy(x + y) = x3 + 3x2y + 3xy2 + y3

10. (x - y)3 = x3 - y3 - 3xy(x - y)  = x3 - 3x2y + 3xy2 - y3

11.  x3 + y3 + z3- 3xyz = (x + y + z)(x2 + y2 + z2- xy - yz -zx)

12.   x2 + y2 = ½[ (x + y)2 +  (x – y)2 ]

13.  xy = ¼[ (x + y)2 -  (x – y)2 ]

14.  x2 + y2  =  (x + y)2- 2xy

15.  (x – y)2 = (x + y)2- 4xy

16.  x2 + y2 =  (x – y)2 + 2xy

17.  (x + y)2 = (x – y)2 + 4xy

18. (x + a)(x + b)(x + c) = x3  + (a + b + c)x2 + (ab + bc + ca)x + abc

19.  x3 + y3 = (x + y) (x2- xy + y2)

20.  x3 - y3 = (x - y) (x2+ xy + y2)

21.  x2 + y2 + z2 -  xy - yz -zx = ½[ (x - y)2 +  (y – z)2 + (z – x)2]

Exercise 2.5

Q1: 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)(y2 – 3/2)

(v) (3 – 2x) (3 + 2x)

(i) 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)  Using identity, (x + a) (x + b) =  x2 + (a+b)x + ab
(x + 8) (x – 10) = x2 + (8+(-10))x + (8)(-10)
=  x2 +(8-10)x - 80
=  x2 -2x - 80
(Note, you may use identity, (x + a) (x - b) =  x2 + (a-b)x - ab directly here)

(iii) Using identity, (x + a) (x + b) =  x2 + (a+b)x + ab
Here x = 3x, a = 4 and b = -5
(3x + 4) (3x – 5) = (3x)2 + (4+(-5))(3x) + (4)(-5)
= 9x2 + (4-5)(3x) + (-20)
= 9x2 -3x -20

(iv) Using identity,  x2  – y2  = (x + y) (x – y)

(v)  Using identity, (x + y) (x – y) =  x2  – y2
(3 - 2x)(3 + 2x) = (3)2  – (2x)2  = 9 - 4x2

Q2: Evaluate the following products without multiplying directly:
(i) 103 × 107

(ii) 95 × 96
(iii) 104 × 96

(i) 103  ✕ 107 = (100 + 3)(100 + 7)
Using identity, (x + a) (x + b) =  x2 + (a+b)x + ab
= (100)2 + (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)2 - (5 + 4)(100) + (5)(4)
= 10000 - 900 + 20
= 9120

(iii)  104 ✕ 96 = (100 + 4)(100 - 4)
Using identity, (x + y) (x – y) =  x2  – y2

Here x = 100, y = 4
=  (100)2  – (4)2 = 10000 - 16 = 9984

Q3: Factorise the  following using appropriate identities:
(i) 9x2 + 6xy + y2
(ii) 4y2 – 4y + 1
(iii) x2 - y2/100

(i)  9x2 + 6xy + y2
=  (3x)2 + 2(3x)(y) + (y)2
(a + b)2 = a2 + 2ab + b2
∴ = (3x + y)2
∴ = (3x + y)(3x + y)

(ii)  4y2 – 4y + 1
= (2y)2 – 2(2y)(1) + 12
(x – y)2  = x2 – 2xy + y2
= (2y – 1)2 = (2y - 1)(2y - 1)

Q4: 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) [¼a - ½b + 1]2

Answer: Using identity (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca here,

(i)  (x + 2y + 4z)2
Here a = x, b = 2y and x = 4z
=  x2 + (2y)2 + (4z)2 + 2x(2y) + 2(2y)(4z) + 2(4z)x
=  x2 + 4y2 + 16z2 + 4xy + 16yz + 8zx

(ii) (2x – y + z)2
Here a = 2x, b = -y and c = z
=  (2x)2 + (-y)2 + (z)2 + 2(2x)(-y) + 2(-y)(z) + 2(z)(2x)
=  4x2 + y2 + z2 - 4xy -2yz + 4xz

(iii) (–2x + 3y + 2z)2
Here a = -2x, b= 3y and c = 2z
= (-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
Here  a= 3a, b = -7b and c = -c
= (3a)2 + (-7b)2 + (-c)2 + 2(3a)(-7b) + 2(-7b)(-c) + 2(-c)(3a)
= 9a2 + 49b2 + c2- 42ab + 14bc - 6ac

(v) (–2x + 5y – 3z)2
Here a = -2x, b = 5y and c = -3z
= (-2x)2 + (5y)2 + (-3z)2 + 2(-2x)(5y) + 2(5y)(-3z) + 2(-3z)(-2x)
= 4x2 + 25y2 + 9z2 - 20xy -30yz + 12zx

(vi)

Q5: Factorise:
(i) 4x2 + 9y2 + 16z2 + 12xy – 24yz – 16xz
(ii) 2x2 + y2  + 8z2  – 2√2 xy + 4√2 yz – 8xz

(i) 4x2 + 9y2 + 16z2 + 12xy – 24yz – 16xz
= (2x)2 + (3y)2 + (4z)2 + 2(2x)(3y) – 2(3y)(4z) – 2(2x)(4z)
= (2x)2 + (3y)2 + (-4z)2 + 2(2x)(3y) + 2(3y)(-4z) + 2(2x)(-4z)
Using identity, (x + y + z)2 = x2 + y2 + z2 + 2xy + 2yz + 2zx
= (2x + 3y -4z)2
=  (2x + 3y -4z)(2x + 3y -4z)

(ii) 2x2 + y2  + 8z2  – 2√2 xy + 4√2 yz – 8xz
= (√2x)2 + y2  + (2√2z)2  – 2(√2 x)(y) + 2(y)(2√2z) – 2(√2x)(2√2z)
= (-√2x)2 + y2  + (2√2z)2  + 2(-√2 x)(y) + 2(y)(2√2z) + 2(-√2x)(2√2z)
Using identity, (x + y + z)2 = x2 + y2 + z2 + 2xy + 2yz + 2zx
= (-√2x + y + 2√2z )2
= (-√2x + y + 2√2z )(-√2x + y + 2√2z )

Q6: Write the following cubes in expanded form:
(i)  (2x + 1)3
(ii) (2a – 3b)3
(iii) (3x/2 + 1)3
(iv) (x - 2y/3)3

(i)  (2x + 1)3
Using identity (x + y)3 = x3 + y3 + 3xy(x + y)
=  (2x)3 + (1)3 + 3(2x)(1)(2x + 1)
=  8x3 + 1 + 6x(2x + 1)
=  8x3 + 1 + 12x2 + 6x

(ii) (2a – 3b)3
Using identity (x - y)3 = x3 - y3 - 3xy(x - y)
= (2a)3 - (3b)3 - 3(2a)(3b)(2a - 3b)
= 8a3 - 27b3 - 18ab(2a - 3b)
= 8a3 - 27b3 -36a2b + 54ab2

(iii)  (3x/2 + 1)3
Using identity (x + y)3 = x3 + y3 + 3xy(x + y)
=  (3x/2)3 + (1)3 + 3(3x/2)(1)(3x/2 + 1)
= 27x3/8 + 1 + (9x/2)(3x/2 + 1)
= 27x3/8 + 27x2/4 - 9x/2 + 1

(iv) (x - 2y/3)3
Using identity (x - y)3 = x3 - y3 - 3xy(x - y)
= x3 - (2y/3)3 - 3x(2y/3)(x - 2y/3)
= x3 - 8y3/27 - 2xy/3(x - 2y/3)
=  x3 - 8y3/27 - 2x2y + 4xy2/3

Q7: Evaluate the following using suitable identities:
(i) (99)3
(ii) (102)3
(iii) (998)3

(i) (99)3
= (100-1)3
Using identity (x - y)3 = x3 - y3 - 3xy(x - y)
= (100)3 - (1)3 - 3(100)(1)(100 - 1)
= 1000000 - 1 -300(99)
= 1000000 − 1 − 29700
= 970299

(ii) (102)3
= (100 + 2)3
Using identity (x + y)3 = x3 + y3 + 3xy(x + y)
= (100)3 + (2)3 + 3(100)(2)(100+2)
= 1000000 + 8 + 600(102)
= 1000000 + 8 + 61200
= 1061208

(iii) (998)3
= (1000 - 2)3
Using identity (x - y)3 = x3 - y3 - 3xy(x - y)
= (1000)3 + (2)3 + 3(100)(2)(1000-2)
= 1000000000 − 8 − 6000(998)
= 1000000000 − 8 − 5988000
= 1000000000 − 5988008
= 994011992

Q8: 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

(i) 8a3 + b3 + 12a2b + 6ab2
= (2a)3 + b3 + 3(2a)2b + 3(2a)b2

(x + y)3 =  x3 + 3x2y + 3xy2 + y3
= (2a + b)3

(ii) 8a3 – b3 – 12a2b + 6ab2
=  (2a)3 - b3 - 3(2a)2b + 3(2a)b2
∵  (x - y)3 = x3 - y3 - 3xy(x - y)  = x3 - 3x2y + 3xy2 - y3

=  (2a - b)3

(iii) 27 – 125a3  – 135a + 225a2
= 33 – (5a)3 – 3(3)2(5a) + 3(3)(5a)2
∵  (x - y)3 = x3 - y3 - 3xy(x - y)  = x3 - 3x2y + 3xy2 - y3
= (3 - 5a)3

(iv) 64a3 – 27b3 – 144a2b + 108ab2
= (4a)3 - (3b)3 –3(4a)2(3b) + 3(4a)(3b)2
= (4a - 3b)3                            [∵  (x - y)3 = x3 - 3x2y + 3xy2 - y3]

Q9. Verify :
(i) x3 + y3 = (x + y) (x2  – xy + y2)
(ii) x3 –  y3 = (x – y) (x2 + xy +  y2)

(i) x3 + y3 = (x + y) (x2  – xy + y2)
∵  (x + y)3 = x3 + y3 + 3xy(x + y)
⇒  x3 + y3 = (x + y)3 - 3xy(x + y)
⇒  x3 + y3 = (x+y) [(x + y)2 - 3xy]
⇒  x3 + y3 = (x+y) [x2 + y2 + 2xy - 3xy]
⇒  x3 + y3 = (x+y) (x2 + y2 -xy)                      ... (answer)

(ii) ∵ (x - y)3 = x3 - y3 - 3xy(x - y)
⇒  x3 - y3 = (x - y)3 + 3xy(x - y)
⇒  x3 - y3 = (x - y)[(x - y)2 + 3xy]
⇒  x3 - y3 = (x - y)(x2 + y2 - 2xy+ 3xy)
⇒  x3 - y3 = (x - y)(x2 + y2 + xy)                     ... (answer)

Q10: Factorise each of the following
(i) 27y3 + 125z3
(ii) 64m3 - 343n3

(i) 27y3 + 125z3
=  (3y)3 + (5z)3
Using identity x3 + y3 = (x+y) (x2 + y2 -xy)
= (3y + 5z)((3y)2 + (5z)2 -(3y)(5z))
= (3y +5z)(9y2 + 25z2 -15yz)

(ii) 64m3 - 343n3
=  (4m)3 - (7n)3
Using identity x3 - y3 = (x - y)(x2 + y2 + xy)
= (4m - 7n)((4m)2 + (7n)2 + (4m)(7n))
= (4m - 7n)(16m2 + 49n2 + 28mn)

Q11: Factorise 27x3 + y3 + z3 - 9xyz

Answer: ∵ x3 + y3 + z3- 3xyz = (x + y + z)(x2 + y2 + z2- xy - yz -zx)
∴ = (3x)3 + y3 + z3 - 3(3x)yz
= (3x + y + z)((3x)2 + y2 + z2- 3xy - yz -3zx)
= (3x + y + z)(9x2 + y2 + z2- 3xy - yz -3zx)

Q12: Verify that
x3 + y3 + z3- 3xyz = ½((x + y + z)[(x-y)2 + (y-z)2 + (z - x)2]

Answer: ∵ x3 + y3 + z3- 3xyz = (x + y + z)(x2 + y2 + z2- xy - yz -zx)
∴ RHS =  ½(x + y + z)(2x2 + 2y2 + 2z2- 2xy -2yz -2zx)
=  ½(x + y + z)(x2 + x2 +y2 + y2 +z2 + z2 - 2xy -2yz -2zx)
=  ½(x + y + z)[x2 + y2 - 2xy + y2 +z2 -2yz + z2 + x2 -2zx]
= ½(x + y + z)[(x2 + y2 - 2xy) + (y2 +z2 -2yz) + (z2 + x2 -2zx)]
= ½(x + y + z)[(x - y)2 + (y - z)2 + (z - x)2 ]    ... (answer)

Q13: If x + y + z = 0, show that x3 + y3 + z3 = 3xyz.

Answer: ∵ x3 + y3 + z3- 3xyz = (x + y + z)(x2 + y2 + z2- xy - yz -zx)
Also it is given  x + y + z = 0
⇒  x3 + y3 + z3- 3xyz = (0)(x2 + y2 + z2- xy - yz -zx)
⇒  x3 + y3 + z3- 3xyz = 0
⇒  x3 + y3 + z3= 3xyz

Q14. Without actually calculating the cubes, find the value of each of the following:
(i) (–12)3 + (7)3 + (5)3
(ii) (28) x3 + (–15) x3 + (–13) x3

(i) (–12)3 + (7)3 + (5)3

∵ (-12) + (7) + (5) = 0
Using identity,  if x + y + z = 0, then x3 + y3 + z3 = 3xyz.

(ii) (28) x3 + (–15) x3 + (–13) x3
∵  (28) + (-15) + (-13) = 0
Using identity,  if x + y + z = 0, then x3 + y3 + z3 = 3xyz.

Q15: 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

Answer: Since area of rectangle = length ✕ breadth. Let us factorize the following equations into two terms.
(i)  Area =  25a2 – 35a + 12
=  25a2 – 15a  - 20a + 12             (Using splitting method)
=  5a(5a -3) - 4(5a - 3)
=  (5a - 3)(5a -4)
∴ Possible length = (5a - 3)
and Possible width =  (5a -4)

(ii) Area : 35y2 + 13y –12
= 35y2 + 28y -15y –12
= 7y(5y + 4)-3(5y + 4)
= (5y +4)(7y - 3)
∴ Possible length = (5y +4)
and Possible width = (7y - 3)

Q16: 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

Since, volume of cuboid = length ✕ breadth ✕ height
Let us factorise the equations into three terms.
(i) Volume : 3x2 – 12x
= 3x(x-4)
⇒ Possible length = 3, width = x and height = (x- 4)
or Possible length = 1, width = 3x and height = (x- 4)

(ii) Volume : 12ky2  + 8ky – 20k
= 4k(3y2  + 2y – 5)
= 4k( 3y2 - 3y + 5y - 5)
= 4k( 3y(y - 1) + 5(y -1))
= 4k(y-1)(3y + 5)
⇒ Possible length = 4k, width = (y-1) and height = (3y + 5)

✪ Extra Problems ✪

Q17: If x + x-1 = 5, evaluate x3 – x-3

Answer: x + x-1 = 5
Cubing both sides, we get
(x + x-1)3 = 53
Using identity,  (x - y)3 = x3 - y3 - 3xy(x - y)
x3 - x-3 -3(x)(x-1)(x - x-1) = 125

⇒ x3 - x-3 -3(x - x-1) = 125
⇒ x3 - x-3 -3(5) = 125
⇒ x3 - x-3  = 125 + 15
⇒ x3 - x-3  =140                ...(answer)

Q18: if x2 + x-2 = 102, evaluate  x + x-1

Answer: x2 + x-2 = 102
⇒  x2 + x-2 - 2 = 102 - 2
⇒  x2 + x-2 - 2(x)(x-1) = 100
⇒  (x - x-1)2 = 100
⇒  x - x-1 = 10

RD Sharma Exercise 4.1
Q19: Evaluate (a - 0.1)(a + 0.1)

Answer: Using identity, x2  – y2  = (x + y) (x – y)
(a - 0.1)(a + 0.1) = a2  – (0.1)2 = a2  – 0.01

Q20: Evaluate
(i) (399)2
(ii) (0.98)2
(iii) 991 ✕ 1009
(iv) 117 ✕ 83

(i) (399)2 = (400 -1)
Using identity, (x – y)2  = x2 – 2xy + y2
= (400)2 - 2(400)(1) + (1)2
= 160000 - 800 + 1
= 159201

(ii) (0.98)2 = (1 - 0.02)2
Using identity,  (x – y)2  = x2 – 2xy + y2
= (1)2 – 2(1)(0.02) + (0.02)2
= 1 -0.04 + 0.0004
= 0.9604

(iii) 991 ✕ 1009 = (1000 - 9)(1000 + 9)
Using identity, x2  – y2  = (x + y) (x – y)
= (1000)2  – (9)2
= 106 - 81
= 999919

(iv) 117 ✕ 83 = (100 + 17)(100 - 17)
Using identity, x2  – y2  = (x + y) (x – y)
= (100)2  – (17)2 = 10000 - 289 = 9711

Q21: Simplify 0.76 ✕ 0.76 + 2 ✕ 0.76 ✕ 0.24 + 0.24 ✕ 0.24

Answer: Using identity (x + y)2  = x2 + 2xy + y2
= (0.76 + 0.24)2 = (1)2 = 1

Q22: If x + x-1 = 11, evaluate x2 + x-2

Answer: x + x-1 = 11
(x + x-1)2 = (x)2 +(x-1)2 + 2(x)(x-1) = 112
⇒  x2 + x-2+ 2 = 121
⇒  x2 + x-2 = 121 - 2 = 119

Q23: Prove that a2 + b2 + c2  –ab -bc -ca is always non-negative for all values of a, b and c.

Answer: To prove a2 + b2 + c2  –ab -bc -ca ≥ 0.
We know that square of any number (+ve or -ve) is always +ve.
a2 + b2 + c2- ab - bc -ca
= ½[2a2 + 2b2 + 2c2 -2ab -2bc -2ca]
⇒ = ½[a2 + b2 -2ab + b2 + c2 -2bc + a2 + c2 -2ca]
⇒ = ½[(a2 + b2 -2ab) + (b2 + c2 -2bc) + (a2 + c2 -2ca)]
⇒ = ½[(a - b)2 + (b - c)2 + (c - a)2 ]
Here, all the terms are always be positive,
⇒ a2 + b2 + c2  –ab -bc -ca ≥ 0.

Q24: If x + x-1 = √5, evaluate x2 + x-2 and x4 + x-4

Answer:  x + x-1 = √5
(x + x-1 )2 = (√5)2
⇒  x2 + x-2 + 2 = 5
⇒  x2 + x-2 = 5 -3 = 3
(x2 + x-2)2 = (3)2
⇒  x4 + x-4 + 2 = 9
⇒  x4 + x-4 = 9 -2 = 7

Q25: If 9x2+ 25y2 = 181 and xy = -6. Find the value of 3x + 5y

(3x)2 + (5y)2 = 181
⇒ (3x)2 + (5y)2 + 30xy - 30xy = 181
⇒ (3x)2 + (5y)2 + 2(5x)(6y) = 181 + 30xy
⇒  (3x + y)2 = 181 + 30(-6) = 181 - 180
⇒ (3x + y)2 =1
⇒ 3x + y = ∓1

Q26: Simplify (x3- 3x2 - x)(x2- 3x + 1)

Answer:  (x3- 3x2 - x)(x2- 3x + 1)
⇒ = x3(x2- 3x + 1) - 3x2(x2- 3x + 1) -x(x2- 3x + 1)
=  x5 - 3x4 + x3- 3x4+ 9x3- 3x2- x3 + 3x2- x
= x5 - 6x4 + 9x3 - x

miscellaneous problems

Q27:  Factorise (x - y)3 + (y - z)3 + (z - x)3

Answer: Let (x - y) = a, (y - z) = b and (z - x) = c
⇒ = a3 + b3 + c3
∵  a + b + c = (x - y) + (y -z) + (z -x) = 0
Using identity, p3 + q3 + r3 = 3xyz (if p + q + r = 0)
⇒= 3abc
= 3(x -y)(y - z)(z - x)

Q28: Factorise: abx + aby - bcx - bcy

Answer: abx + aby - bcx - bcy
= ab(x + y) -bc(x + y)
= (x + y)(ab -ac)
= a(b - c)(x + y)

See here the video tutorial by Khanacademy.org on Factorization of sum of cubes:

Q29(CBSE 2011):  Show that (xa-b)a+b + (xb-c)b+c + (xc-a)c+a = 1

Answer: (xa-b)a+b + (xb-c)b+c + (xc-a)c+a
Using identity  (x + y) (x – y) = x2  – y2

⇒ = xa2-b2 + xb2-c2 + xc2-a2
Using identity ap.ap = ap+q
⇒ = xa2- b2+ b2- c2 + c2-a2
⇒ = x0 = 1

Q30: Evaluate 5252- 4752
(a) 100
(b) 10000
(c) 50000
(d) 100000

x2  – y2  = (x + y) (x – y)
⇒ (525 + 475)(525 - 475) = (1000)(50) = 50000

Q31: If a+ b + c = 0, then (a3 + b3 + c3 ) = ?

(a) abc
(b) 2abc
(c) 3abc
(d) 4abc

Answer: (c) 3abc [Hint: See Q 13 above]

1. thanks for the video

2. very very thanks and now im good in maths thanks very very much

3. there is a mistake in exercise 2.5 Q16 (i) it is 3x(x-4) not 3x(x-12). pls correct

1. fixed. thanks for pointing it.

4. you forgot for both the heights in the same question it should also be (x-4)

5. helped me to clear all my confusions

6. thanks for list of identities

7. can give me answer of any another question not of ncert

8. in question 24 ,it should be 5-2=3 not 5-3=3.

9. iam well in maths

10. THANK FOR A WONDER FULL HELP

11. Thnks it helped to clear all my confusions