Telangana SCERT Class 9 Math Solution Chapter 2 Polynomials and Factorisation Exercise 2.5

Telangana SCERT Solution Class IX (9) Math Chapter 2 Polynomials and Factorisation Exercise 2.5

(1) (i) (x + 5) (x+2)

Using identity-iv we get

X2 + (5+2)x + 5×2

= x2 + 7x + 10

(ii) (x – 5) (x – 5)

= (x – 5)2

= using identity-ii we get

X2 – 2 X x X 5 + 52

= x2 – 10x + 25

(iii) (3x + 2) (3x – 2)

= on comparing it with (a+b) (a-b) we get.

a = 3x, b = 2

∴ (3x)2 – 22

= 9x2 – 4

(iv) (x2 + 1/x2) (x2 – 1/x2)

On comparing it with (a + b) (a –b) we get

a = x2 , b = 1/x2

∴ using identity iii we get

(x2)2 – (1/x2)2

= x4 – 1/x4

(v) (1+x) (1+x)

= (1+x)2

∴ using identity I we get

12 + 2 X 1 X x + x2

= 1 + 2x + x2

(2) (i) 101 x 99

= (100 + 1) (100 – 1)

∴on comparing it with identity III we get

1002 – 12

= 10000 – 1

= 9999

(ii) 999 x 999

= (1000 – 1) (1000 – 1)

= (1000 – 1)2

∴ using identity II we get

10002 – 2 x 1000 x 1 + 12

= 1000000 – 2000 + 1

= 998000 + 1

= 998001

(iv) 501 x 501

= (500 + 1) x (500+1)

= (500 + 1)2

∴ using identity  I we get

5002 + 2 x 500 x 1 + 12

= 250000 + 1000 + 1

= 251001

(v) 30.5 x 29.7

= (30 + 0.5) (30 – 0.5)

∴ using identity III we get

302 – 0.52

= 900 – 0.25

= 899.75

(3) (i) 16x2 + 24xy + 9y2

On comparing with identity I

(a + b)2 = a2 + 2ab + b2 we get,

16x2 + 24xy + 9y2

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

∴ a = 4x, b = 3y

∴ (4x + 3y)2 = 16x2 + 24xy + 9y2

Hence 4x + 3y is the factor of the above equation.

(ii) 4y2 – 4y + 1

= (2y)2 – 2 X 2y X 1 + 12

On comparing it with (a – b)2 = a2 – 2ab + b2 we get

a = 2y, b = 1

∴ (2y + 1)2 = 4y2 – 4y + 1

Hence 2y + 1 is the factor of the above sum.

(iii) 4x2 – y2/25

= (2x)2 – (y/5)2

On comparing with a2 – b2 = (a+b) (a-b) we get

(2x + y/5) (2x – y/5)

(iv) 18a2 – 50

= 2(9a2 – 25)

= 2{(3a)2 – 52}

On comparing with a2 – b2 = (a+b) (a-b)  we get

2(3a + 5) (3a – 5)

∴ 2, (3a + 5) (3a – 5) are the factor of 18a2 – 50

(v) x2 + 5x + 6

= x2 + (2+3)x + 2 X 3

On comparing it with identity IV (x+a)(x+b) = x2 +(a+b)x + ab

We get,

(x + 2) (x + 3)

∴ (x + 2), (x + 3) are the factors of x2 + 5x + 6

(4) (i) (x + 2y+4z)2

Using identity V we get

(x + 2y + 4z)2

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

= x2 + 4y2 + 16z2 + 4xy + 16yz + 8xz

(ii) (2a – 3b)2

On comparing with identity VII we get

a = (2a), b = (3b)

∴ using identity VII we get

(2a)2 – (3b)3 – 3 x 2a x 3b(2a – 3b)

= 8a3 – 27b3 – 18ab(2a – 3b)

= 8a3 – 27b3 – 36a2b + 54ab2

(iii) (-2a + 5b – 3c)2

= On comparing it with identity V we get

a = (-2a), b = 5b, c = (-3c)

Using identity V we get,

(-2a)2 + (5b)2 + (-3c)2 + 2 X -2a X 5b + 2 X 5b X -3c + 2 X -2a X -3c

= 4a2 + 25b2 + 9c2 – 20ab – 30bc + 12ac

(5) 25x2 + 16y2 + 4z2 – 40xy + 16yz – 20xz

= (-5x)2 + (4y)2 + (27)2 + 2 X -5x X 4y + 2 X 4y X 27 + 2 X -5x X 27

On comparing it with identity V where,

(a+b+c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ac

We get,

a = -5x, b = 4y, c = 27

∴ (-5x + 4y + 27)2 = 25x2 + 16y2 + 4z2 – 40xy + 16yz – 20az

∴ (-5x + 4y + 2z) is a factor of the above given sum.

(ii) 9a2 + 4b2 + 16c2 + 12ab – 16bc – 24ca

= (3a)2 + (2b)2 + (-4c)2 + 2 X 3a X 2b + 2 X 2b X -4c + 2 X 3a X -4c

On comparing it with identity V where

(x + y + z)2 = (x2 + y2 + z2 + 2xy + 2yz + 2xz)

We get,

x = 3a, y = 2b, z = -4c

Hence,

(3a + 2b + 4c)2 = 9a2 + 4b2 + 16c2 + 12ab – 16bc – 24ac

∴ (3a + 2b – 4c) is a factor of the above given sum.

(6.) If a + b + c = 9 and ab + bc + ca = 26, then find a2 + b2 + c2

Given, a + b + c = 9, ab + bc + ac = 26.

a + b + c = 9

Or, (a + b + c)2 = 92

Or, a2 + b2 + c2 + 2ab + 2bc + 2ac = 81

Or, a2 + b2 + c2 + 2(ab + bc + ac) = 81

Putting the value of ab+ bc + ac = 26 we get

a2 + b2 + c2 + 2 x 26 = 81

Or, a2 + b2 + c2 = 81 – 52

Or, a2 + b2 + c2 = 29.

(7) (i) (99)3

= (100 – 1)3

Using identity VII we get

1003 – 13 – 3 x 100 x 1 (100 – 1)

= 1000000 – 1 – 300 x 99

= 1000000 – 1 – 29700

= 1000000 – 29701

= 970299

(ii) (102)3

= (100 + 2)3

Using identity VI we get

(100 + 2)3

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

= 1000000 + 8 + 600 x 102

= 1000008 + 61200

= 1061208

(iii) (998)3

= (1000 – 2)3

Using identity VII we get

10003 – 23 – 3 x 1000 x 2(1000 -2)

= 1000000000 – 8 – 6000 x 998

= 1000000000 – 8 – 5988000

= 1000000008 – 5988000

= 994011992

(iv) (1001)3

= (1000 + 1)3

Using identity VI we get

(1000 + 1)3

= 10003 + 13 + 3 x 1000 x 1 (1000 + 1)

= 1000000000 + 1 + 3000 x 1001

= 1000000001 + 3003000

= 1003003001

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

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

On comparing it with identity VI we get

x = 2a, y = b

∴ (2a + b)3 = 8a3 + b3 + 12a2b + 6ab2

∴ (2a + b)3 is a factor of the above given sum.

(ii) 8a3 – b3 – 12a2b + 6ab2

= (2a)3 – b3 – 3 x 2a x b(2a – b)

On comparing it with identity VII we get

(x – y)3 = x3 – y3 – 3 X x X y (x – y) we get,

x = 2a, y = b

∴ (2a – b)3 = 8a3 – b3 – 12a2b + 6ab2

∴ (2a – b) is a factor of the above given sum.

(iii) 1 – 64a3 – 12a + 48a2

= 13 – (4a)3 – 3 x 1 x 4a (1 – 4a)

On comparing it with identity VI where

(x – y)3 = x3 – y3 – 3 X x X y (x – y) we get,

x = 1, y = 4a.

∴ (1 – 4a)3 = 1 – 64a3 – 12a + 48a2

∴ (1 – 4a) is a factor of the above given sum.

(9.) Verify (i) x3 + y3 = (x + y) (x2 − xy + y2 ) (ii) x3 – y3 = (x − y) (x2 + xy + y2 ) using some non-zero positive integers and check by actual multiplication. Can you call these as identites ?

(i) To prove x3 + y3 = (x + y) (x2 – xy + y2)

Now, using identity VI we can say

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

Or, (x+y)3 – 3xy (x + y) = x3 + y3

Or, x3 + y3 = (x + y) {(x + y)2 – 3xy}

[Taking (x+y) common]

Or, x3 + y3 = (x+y) (x2 + 2xy  + y2 – 3xy)

Or, x3 + y3 = (x + y) (x2 – xy + y2) (Hence proved)

(ii) To prove x3 – y3 = (x – y) (x2 + xy + y2)

Now, using identity VII we can say that,

(x – y)3 = x3 – y3 – 3xy(x – y)

Or, (x – y)3 + 3xy(x – y) = x3 – y3

Or, x3 – y3 = (x – y) {(x – y)2 + 3xy}

[Taking (x –y) common]

Or, x3 – y3 = (x – y) (x2 – 2xy + y2 + 3xy)

Or, x3 – y3 = (x – y) (x2 – 2xy + y2 + 3xy)

Or, x3 – y3 = (x – y) (x2 + xy + y2) Hence proved

Yes, we can call both these generalized form of the equations as identities.

(10) (i) 27a3 + 64b3

= (3a)3 + (4b)3

Using the above identity in equation 9.1 we get

= (3a)3 + (4b)3

= (3a + 4b) {(3a)2 – 3a x 4b +(4b)2}

= (3a + 4b) (9a2 – 12ab + 16b2)

∴ (3a + 4b) and (9a2 – 12ab + 16b2) are the factors of the above given sum.

(ii) 343y3 – 1000

= (7y)3 – (10)3

Using the identity in equation 9.2 we get

(7y)3 – (10)3

= (7y – 10) {(7y)2 + 7y x 10 + 102}

= (7y – 10) (49y2 + 70y + 10)

∴ (7y – 10) and (49y2 + 70y + 10) are the factors of the above given sum.

(11.) Factorise 27x3 + y3 + z3 − 9xyz using identity

= (3x)3 + y3 + z3 – 3 X 3x X y X z

∴ a = 3x, b = y, c = z

On comparing it with identity VIII

(a + b+ c) (a2 + b2 + c2 – ab – bc – ac)

= a3 + b3 + c3 – 3abc we get,

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

= 27x3 + y3 + z3– 9xyz

∴ (3x + y + z) and (9x2 + y2 + z – 3xy – yz – 3xz)

Are the factors of the above given sum.

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

(13.) (a) If x + y + z = 0, show that x3 + y3 + z3 = 3xyz.

(b) Show that (a – b)3 + (b – c)3 + (c – a)3 = 3 (a – b) (b – c) (c – a)

Or, x3 + y3 + z3 = 3xyz …(iv)

Now, lets put the value of x, y, z in equation (iv)

We get,

(a –b)3 + (b – c)3 + (c – a)3 = 3(a – b) (b – c) (c – a) Hence proved.

(14) (i) (-10)3 + 73 + 33

We can write the above sum as

-(7 + 3)3 + 73 + 33

= -[73 + 33 + 3 x 7 x 3 (7 + 3)] + 73 + 33

= -73 – 33 – 63 x 10 + 73 + 33

= – 630

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

Now,

We can write this expression as,

(15 + 13)3 – 153 – 133 [∵ the cube of any negative integer is always negative]

= 153 + 133 + 3 x 15 x 13 (15 + 3) – 153 – 133

= 3 x 15 x 13 x 18

= 45 x 13 x 18

= 10530

(16.) What are the possible polynomial expressions for the dimensions of the cuboids whose volumes are given below?

(i) 3x3 − 12x (ii) 12y2 + 8y − 20.

(i) let, l = length, b = breadth, h = height

Volume of cuboid = 3x3 – 12x

Or, l x b x h = 3x(x2 – 4)

= 3x(x + 4) (x – 4) [from identity III]

∴ l = 3x, b = x+4, h = x -4

(ii) let, l = length, b = breadth, h = height

∴ Volume of cuboid =12y2 + 8y – 20

Or, l x b x h = 12y2 + 8y -20

= 4(3y2 + 2y – 5)

=4[3y2 – 3y + 5y – 5]

= 4[3y(y – 1) + 5(y – 1)]

= 4[3y(y-1) + 5(y – 1)]

= 4(y-1) (3y + 5)

∴ l = 4, b = y – 1, h = 3y + 5

(17.) If 2(a2 +b2 ) = (a+b)2 , then show that a = b

Given, 2(a2 + b2 ) = (a + b)2 to prove a = b

Or, 2(a2 + b2) = a2 + b2 + 2ab [from identity I]

Or, 2a2 + 2b2 = a2 + b2 + 2ab

Or, 2a2 – a2 + 2b2 – b2 – 2ab = 0

Or, a2 + b2 – 2ab = 0

Or, (a – b)2 = 0 [from identity II]

Or, a – b = 0

Or, a = b Hence proved.


Updated: September 18, 2021 — 4:45 pm

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