Exercise 2.4
1. Use suitable identities to find the following products.
Match the following
(x+4)(x+10)
(x+8)(x-10)
(3x+4)(3x-5)
(y^2) + (3/2)(y^2 -3/2)
(3-2x)(3+2x)
y^4-9/4
x^2-2x-80
9-4x^2
x^2+14x+40
9x^2-3x-20
2. Evaluate the following products without multiplying directly.
(i)103 x 107
Use the identity (x + a)(x + b)=x 2
= 103 x 107 = (100 + 3)(100 + 7)
Substitute x=100, a=3, b=7 in the above identity.
= 100 2
= 10000 + 1000 + 21 =
(ii)95 x 96
Use the identity (x + a)(x + b)=x 2
= 95 x 96 = (100 - 5)(100 - 4)
Substitute x=100, a=-5, b=-4 in the above identity.
= 100 2
= 10000 - 900 + 20 =
(iii)104 x 96
Use the identity (a+b)(a-b) = a 2 b 2
= 104 x 96 = (100+4)(100-4)
Substitute a=100,b=4 in the above identity.
= 100 2 4 2
3. Factorise the following using appropriate identities.
(i)9 x 2 6 xy +y 2
Using Algebraic Identity a + b 2 a 2 + 2 ab + b 2
Simplifying : 3 x 2 + 2 3 x y + y 2
Thus, 9 x 2 6 xy +y 2 3 x + y 2
(ii)4 y 2 4 y + 1
Using Algebraic Identity a − b 2 a 2 − 2 ab + b 2
Simplifying : 2 2 − 2 2 y 1 + 1 2
Thus, 4 y 2 4 y + 1 = 2 y − 1 2
(iii)x 2 y 2 100
Using Algebraic Identity a + b a − b a 2 b 2
Simplifying : x 2 y 10 2
Thus, x 2 − y 2 100 x + y 10 x − y 10
4. Expand each of the following, using suitable identities.
Hint: Use the Identity for all the given Quetions
(i)x + 2 y + 4 z 2
(ii)2 x − y + z 2
(iii)− 2 x + 3 y + 2 z 2
(iv)3 a − 7 b − c 2
(v)− 2 x + 5 y − 3 z 2
(vi)1 4 a − 1 2 b + 1 2
5. Factorise the given equation.
Hint:
(i)4 x 2 9 y 2 16 z 2
Comparing with the identity we get: a =
Simplifying: 2 x 2 3 y 2 − 4 z 2 2 2 x 3 y 2 3 y − 4 z 2 − 4 z 2 x
Hence, 4 x 2 9 y 2 16 z 2 2 x + 3 y − 4 z 2
(ii)2 x 2 y 2 8 z 2 2 2
We know that, 2 = 8
Simplifying: 2 x 2 y 2 8 z 2 2 2 2 x 2 y 2 2 2 z 2 − 2 x 2 z ) + (22 z )(-2 x ) =
Hence 2 x 2 y 2 8 z 2 2 2 − 2 x + y + 2 2 z 2
6. Write the following cubes in expanded form.
Hint: Use
(i)2 x + 1 3
Expanding we get: 2 x + 1 3 2 x 3 3 2 x 2 3 2 x 1 2 1 3
(ii)2 a − 3 b 3
Expanding we get: 2 a − 3 b 3 2 a 3 3 2 a 2 3 2 a − 3 b 2 − 3 b 3
(iii)3 2 x + 1 3
Expanding we get: 3 2 x + 1 3 3 2 x 3 3 2 x 2 3 2 x 1 2 1 3
(iv)x − 2 3 y 3
Expanding we get: x − 2 3 y 3 x 3 3 x 2 − 2 3 3 x − 2 3 y 2 − 2 3 y 3
- Evaluate the following using suitable identities:
(i) 99 3
(ii)102 3
(iii) 998 3
We have found all the answers.
- Factorise each of the following:
(i) 8 a 3 + b 3 + 12 a 2 b + 6 ab 2
Using x + y 3 x 3 + 3 xy x + y + y 3
(ii) 8 a 3 − b 3 − 12 a 2 b + 6 ab 2
Using x − y 3 x 3 − 3 xy x − y − y 3
(iii) 27 − 125 a 3 − 135 a + 225 a 2
Using x − y 3 x 3 − 3 xy x − y − y 3
(iv) 64 a 3 − 27 b 3 − 144 a 2 b + 108 ab 2
Using x − y 3 x 3 − 3 xy x − y − y 3
(v) 27 p 3 − 1 216 − 9 2 p 2 + 1 4 p
Using x − y 3 x 3 − 3 xy x − y − y 3
- Verify:
(i) x 3 + y 3 x + y x 2 − xy + y 2
Taking x + y x 2 − xy + y 2 x x 2 − xy + y 2 y x 2 − xy + y 2
= x 3 + y 3
(ii) x 3 − y 3 = x − y x 2 + xy + y 2
Taking x − y x 2 + xy + y 2 x x 2 + xy + y 2 y x 2 + xy + y 2
= x 3 − y 3
- Factorise each of the following:
(i) 27 y 3 + 125 z 3
Using the formula for the sum of cubes: a 3 + b 3
= 3 y + 5 z
(ii) 64 m 3 − 343 n 3
Using the formula for the difference of cubes: a 3 − b 3
= 4 m − 7 n
- Factorise :
27 x 3 + y 3 + z 3 − 9 xyz
Using a 3 + b 3 + c 3 − 3 abc = a + b + c a 2 + b 2 + c 2 − ab − bc − ca 27 x 3
Thus, the expression can be rewritten as: 27 x 3 + y 3 + z 3 − 9 xyz = y 3 z 3
= 3 x + y + z
= 3 x + y + z
- Verify that
x 3 + y 3 + z 3 − 3 xyz =1 2 x + y + z x − y 2 + y − z 2 + z − x 2
We know: x 3 + y 3 + z 3 − 3 xyz = x + y + z x 2 + y 2 + z 2 − xy − yz − zx
= 1 2 x + y + z 2 x 2 + 2 y 2 + 2 z 2 − 2 xy − 2 yz − 2 zx 1 2 x + y + z
Thus, we get: x 3 + y 3 + z 3 − 3 xyz = 1 2 x + y + z x − y 2 + y − z 2 + z − x 2
- If x + y + z = 0, show that
x 3 + y 3 + z 3 = 3 xyz
Given: x + y + z = 0.
Substituting x + y + z = 0 into the identity: x 3 + y 3 + z 3 x 2 + y 2 + z 2 − xy − yz − zx + 3 xyz =
Thus, x 3 + y 3 + z 3 3 xyz
- Without actually calculating the cubes, find the value of each of the following:
We know that: To find the value of the given expressions without actually calculating the cubes, we can use the identity for the sum of cubes: a 3 + b 3 + c 3 a + b + c a 2 + b 2 + c 2 − ab − bc − ca + 3 abc . If a + b + c = 0 , the identity simplifies to: a 3 + b 3 + c 3
(i) − 12 3 + 7 3 + 5 3
First, check if the sum of the numbers is zero: − 12 5 =
Since the sum is zero, we can use the simplified identity: − 12 3 + 7 3 + 5 3 3 × − 12 × 7 × 5 =
Therefore, the value of − 12 3 + 7 3 + 5 3 − 1260 .
(ii) 28 3 + − 15 3 + − 13 3
First, check if the sum of the numbers is zero: 28 + (−15) + (−13) =
Since the sum is zero, we can use the simplified identity: 28 3 + − 15 3 + − 13 3 3 × 28 × − 15 × − 13
Therefore, the value of 28 3 + − 15 3 + − 13 3 16380 .
- Give possible expressions for the length and breadth of each of the following rectangles, in which their areas are given:
(i) Area : 25 a 2 − 35 a + 12
Using the split middle term method: 25 a 2 − 35 a + 12 = 25 a 2 12
This give us: 25 a 2 − 35 a + 12 =
Therefore, the possible expressions for the length and breadth of the rectangle are: Length: 5 a − 4 and Breadth: 5 a − 3
(ii) Area : 35 y 2 + 13 y − 12
We get 35 y 2 + 13 y − 12 = 35 y 2
Thus, 35 y 2 + 13 y − 12 =
Therefore, the possible expressions for the length and breadth of the rectangle are: Length: 7 y − 3 , Breadth: 5 y + 4
- What are the possible expressions for the dimensions of the cuboids whose volumes are given below?
(i) Volume : 3 x 2 − 12 x
To factor 3 x 2 − 12 x , we first look for common factors: 3 x 2 − 12 x =
The possible expressions for the dimensions of the cuboid are
Since the volume of a cuboid is the product of its length, width, and height, we can express these dimensions as: Length: 3, Width: x−4, Height: x (any permutation of these factors)
(ii) Volume: Volume : 12 ky 2 + 8 ky − 20 k
To factor 12 ky 2 + 8 ky − 20 k , we first factor out the common factor: 12 ky 2 + 8 ky − 20 k = 3 y 2 + 2 y − 5
Factorising the quadratic expression 3 y 2 + 2 y − 5 :
Thus, the factored form of the volume is: 12 ky 2 + 8 ky − 20 k =
The possible expressions for the dimensions of the cuboid are: Length: 4 k , Width: 3 y + 5 , Height: y − 1 (any permutation of these factors)