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) 

Answer:  
(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

Answer:
(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

Answer:
(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


Answer:
(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

Answer:
(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


Answer:
(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

Answer:
(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)

Answer:
(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

Answer:
(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



Answer:
(i) (–12)³ + (7)³ + (5)³

∵ (-12) + (7) + (5) = 0
Using identity,  if x + y + z = 0, then x³ + y³ + z³ = 3xyz.
 = 3(-12)(7)(5) = -1260      ...(answer)

(ii) (28) x³ + (–15) x³ + (–13) x³
∵  (28) + (-15) + (-13) = 0
Using identity,  if x + y + z = 0, then x³ + y³ + z³ = 3xyz.
 = 3(28)(-15)(-13) = 16380     ...(answer)

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

Answer:
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

Answer:
(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

Answer:  9x²+ 25y² = 181
       (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 (x^a-b)^a+b + (x^b-c)^b+c + (x^c-a)^c+a = 1

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

⇒ = x^a2-b2 + x^b2-c2 + x^c2-a2 
 Using identity a^p.a^p = a^p+q
⇒ = x^{a2- b2+ b2- c2 + c2-a2}
⇒ = x⁰ = 1

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

Answer: (c) 50000
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]