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.
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