Chapter 2 Polynomials Ex 2.5 eng

Chapter 2 Polynomials Ex 2.5

Solution:
(i) (x+ 4) (x + 10)
Using identity (iv), i.e., (x+ a) (x+ b) = x² + (a + b) x+ ab.
We have, (x+4) (x + 10) = x²+(4 + 10) x + (4x 10) (∵ a = 4, b = 10)
= x² + 14x+40

(ii) (x+ 8) (x -10)
Using identity (iv), i.e., (x + a) (x + b) = x² + (a + b) x + ab
We have, (x + 8) (x – 10) = x² + [8 + (-10)] x + (8) (- 10)(∵a = 8, b = -10)
= x² – 2x – 80

(iii) (3x + 4) (3x – 5)
Using identity Eq. (iv), i.e.,
(x + a) (x + b) = x² + (a + b) x + ab
We have, (3x + 4) (3x – 5) = (3x)² + (4 – 5) x + (4) (- 5) (∵a = 4, b = -5)
= 9x² – x – 20

Question 2.
Evaluate the following products without multiplying directly
(i) 103 x 107
(ii) 95 x 96
(iii) 104 x 96
Solution:
(i) 103 x 107 = (100 + 3) (100 + 7)
= 100 x 100+ (3 + 7) (100)+ (3 x 7) [Using identity (iv)]
= 10000+ 1000+21 = 11021
(ii) 95 x 96 = (100-5) (100-4)
= 100 x 100 + [(- 5) + (- 4)] 100 + (- 5 x – 4) [Using identity (iv)]
= 10000 – 900 + 20 = 9120
(iii) 104 x 96 = (100 + 4) (100 – 4)
= (100)²-4² [Using identity (iii)]
= 10000-16 = 9984

Solution:

Solution:
(i) (x + 2y + 4z)² = x² + (2y)² + (4z)² + 2 (x) (2y) + 2 (2y) (4z) + 2(4z) (x) [Using identity (v)]
= x² + 4y² + 16z² + 4xy + 16yz + 8 zx
(ii) (2x – y + z)² = (2x)² + (- y)² + z² + 2 (2x) (- y)+ 2 (- y) (z) + 2 (z) (2x) [Using identity (v)]
= 4x² + y² + z² – 4xy – 2yz + 4zx
(iii) (- 2x + 3y + 2z)² = (- 2x)² + (3y)² + (2z)² + 2 (- 2x) (3y)+ 2 (3y) (2z) + 2 (2z) (- 2x) [Using identity (v)]
= 4x² + 9y² + 4z² – 12xy + 12yz – 8zx
(iv) (3a -7b- c)² = (3a)² + (- 7b)² + (- c)² + 2 (3a) (- 7b) + 2 (- 7b) (- c) + 2 (- c) (3a) [Using identity (v)]
= 9a² + 49b² + c² – 42ab + 14bc – 6ac
(v) (- 2x + 5y- 3z)² = (- 2x)² + (5y)² + (- 3z)² + 2 (- 2x) (5y) + 2 (5y) (- 3z) + 2 (- 3z) (- 2x) [Using identity (v)]
= 4x² + 25y² + 9z² – 20xy – 30yz + 12zx

Question 5.
Factorise
(i) 4 x² + 9y² + 16z² + 12xy – 24yz – 16xz
(ii) 2x² + y² + 8z² – 2√2xy + 4√2yz – 8xz
Solution:
(i) 4x² + 9y² + 16z² +12xy-24yz-16xz
= (2x)² + (3y)² + (- 4z)² + 2 (2x) (3y) + 2 (3y) (- 4z) + 2 (- 4z) (2x)
= (2x + 3y – 4z)²
(ii) 2x² + y² + 8z² – 2√2xy + 4√2yz – 8xz
= (- √2x)² + (y)² + (2 √2z)² + 2(- √2x) (y)+ 2 (y) (2√2z) + 2 (2√2z) (- √2x)
= (- √2x + y + 2 √2z)²

Question 6.
Write the following cubes in expanded form

Solution:
(i) (2x + 1)³ = (2x)³ + 1³ + 3 (2x) (1) (2x + 1) [Using identity (x + y)³ = x³ + y³ + 3xy (x + y)]
= 8x³ + 1 + 6x (2x + 1)
= 8x³ + 1 + 12x² + 6x= 8x³ + 12x² + 6x + 1
(ii) (2a – 3b)³ = (2a)³ – (3b)³ -3(2a)(3b)(2a-3b) [Using identity (x-y)³=x³-y³ -3xy(x-y)]
= 8a³-27b³-18ab(2a-3b)
= 8a³ – 27 b³ – 36a²b + 54ab²
=8a³ – 36a²b + 54ab² – 27 b³

Question 7.
Evaluate the following using suitable identities
(i) (99)³
(ii) (102)³
(iii) (998)³
Solution:
(i) (99)³ =(100-1)³ = 100³ -(1)³ – 3x 100x 1(100-1) [Using identity (x-y)³ =x³-y³-3xy (x-y)]
=1000000-1-300(100-1)
=1000000-1-30000+300
=970299
(ii) (102)³ = (100+ 2)³ = 100³ + 2³ + 3x 100x 2 (100+ 2 [Using identity (x + y)³ = x³ + y³ + 3xy (x + y)]
=1000000 + 8 + 600(100+ 2)
=1000000 + 8 + 60000+ 1200=1061208
(iii) (998)³ = (1000-2)³ =1000³ – 2³ – 3x 1000x 2(1000-2) [Using identity (x-y)³=x³-y³-3xy (x-y)]
=1000000000-8-6000(1000-2)
= 1000000000- 8 – 6000000+12000
=994011992

Question 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

Solution:
(i) 8a³ +b³ +12a²b+6ab² = (2a)³ + b³ + 3x 2axb (2a+ b)
=(2a+b)³ [Using identity (vi)]
(ii) 8a³ -b³ -12a²b+ 6ab² = (2a)³ + (-b)³ + 3x2ax (-b)[(2a)+ (-b)]
= (2a)³ – (b)³ – 3 x 2ax b (2a-b) [Using identity (vii)]
=(2 a-b)³
(iii) 27-125a³-135a + 225a² =(3)³ + (-5a)³ + 3x3x(-5a)[(3)+(-5a)]
= (3)³ -(5a)³– 3x3x5a(3-5a) [Using identity (vii)]
= (3=5 a)³
(iv) 64a³ – 27b³ -144a²b +108ab² =(4a)³ + (-3b)³ + 3x4ax(-3b)[4a+ (-3b)]
=(4a)³-(3b)³-3x4ax3b(4a-3b)
=(4a-3b)³ [Using identity (vii)]

Question 9.
Verify
(i) x³ + y³ = (x + y)-(x² – xy + y²)
(ii) x³ – y³ = (x – y) (x² + xy + y²)
Solution:
(i) We know that,
(x + y)³ = x³ + y³ + 3xy (x + y)
⇒ x3 + y3 = (x + y)3 – 3xy (x + y)
= (x + y) [(x + y)² – 3xy]
= (x + y) [x² + y² + 2xy – 3xy]
= (x + y) [x² + y² – xy]
= RHS
Hence proved,
(ii) We .know that,
(x – y)³ = x³ – y³ – 3xy (x – y) x³ – y³
= (x – y)³ + 3xy (x – y)
= (x – y) [(x – y)² + 3xy]
= (x – y) [x² + y² + 2xy + 3xy]
= (x – y) [x² + y² + xy]
= RHS
Hence proved.

Question 10.
Factorise each of the following
(i) 27y³ + 125z
(ii) 64rh³ – 343n [Hint See question 9]
Solution:
(i) 27y³ + 125z³ = (3y)³ + (5z)³
= (3y + 5z) [(3y)² – (3y) (5z) + (5z)²]
= (3y + 5z) (9y² – 15yz + 25z²)
(ii) 64 m³ – 343 n³ = (4m)³ – (7n)³
= (4m – 7n) [(4m)² + (4m) (7n) + (7n)²]
= (4m – 7 n) [16m² + 28mn + 49n²]

Question 11.
Factorise 27x³ +y³ +z³ -9xyz.
Solution:
27x³ + y³ + z³-9xyz=(3x)³ + y³ + z³ -3x 3xx yx z
=(3x+y+z)[(3x)² + y² + z²– 3xy – yz -z(3x)][Using identity (viii)]
=(3x+ y + z) (9x² + y² + z² -3xy -yz -3zx)

Question 13.
If x + y + z = 0, show that x³ + y³ + z³ = 3 xyz.
Solution:
We know that,
x³ + y³ + z³ – 3 xyz = (x + y + z) (x² + y² + z² – xy – yz – zx [Using identity (viii)]
= 0(x² + y² + z² – xy – yz – zx) (∵ x + y + z = 0 given)
= 0
⇒ x³ + y³ + z³ = 3 xyz
Hence proved.

Question 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
Solution:
We know that, x³ + y³ + z³ -3xyz
=(x+y+z)(x² + y² + z² – xy – yz – zx)
If x+y+z=0, then x³ + y³ + z³-3xyz=0
or x³ + y³ + z³ = 3xyz
(i) We have to find the value of (-12)³ + (7)³ + (5)³
Here, -12+7+5=0
So, (-12)³ + (7)³ + (5)³= 3x(-12)(7)(5)
=-1260
(ii) We have to find the value of (28)³ + (-15)³ + (-13)³.
Here, 28+(-15)+(-13)=28-15-13 =0
So, (28)³ + (-15)³ + (-13)³
= 3x (28) (-15) (-13)
= 16380

Question 15.
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) We have, area of rectangle
= 25a² – 35a+12
= 25a² – 20a – 15a+12
= 5a(5a – 4)-3(5a – 4)=(a-4)(5a – 3)
Possible expression for length =5a-3
and breadth =5a-4
(ii) We have, Area of rectangle = 35y²+13y – 12=35y² -15y + 28y – 12
=5y(7y – 3)+4(7y – 3) = (7y – 3)(5y+4)
Possible expression on for length =7y-3
and breadth =5y+4

Question 16.
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
Solution:
(i) We have, volume of cuboid =3x²-12x=3x(x – 4)
One possible expressions for the dimensions of the cuboid is 3, x and x-4.
(ii) We have, volume of cuboid =12ky² + 8ky-20k
=12ky² + 20ky-12ky-20k
=4ky (3y+5) – 4k(3y+5)
=(3y+5)(4ky – 4k)
=(3y+5)4k(y – 1)
One possible expressions for the dimensions of the cuboid is 4k, 3y+5 and y – 1.