NCERT Solution: Polynomials
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
In (x + 4) (x + 10), a = 4 and b = 10
Now,
(x + 4) (x + 10) = x2 + (4 + 10)x + (4 × 10)
= x2 + 14x + 40
(ii) (x + 8) (x – 10)
Using identity, (x + a) (x + b) = x2 + (a + b) x + ab
Here, a = 8 and b = –10
(x + 8) (x – 10) = x2 + {8 +(– 10)}x + {8×(– 10)}
= x2 + (8 – 10)x – 80
= x2 – 2x – 80
(iii) (3x + 4) (3x – 5)
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 + 3x(4 - 5) - 20
= 9x2 - 3x - 20
(iv) (y2 + 3/2) (y2 - 3/2)
Using identity, (x + y) (x -y) = x2 - y2
Here, x = y2 and y = 3/2
(y2 + 3/2) (y2 - 3/2) = (y2)2 - (3/2)2
= y4 - 9/4
(v) (3 - 2x) (3 + 2x)
Using identity, (x + y) (x -y) = x2 - y2
Here, x = 3 and y = 2x
(3 - 2x) (3 + 2x) = 32 - (2x)2
= 9 - 4x2
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
Here, x = 100, a = 3 and b = 7
103 × 107 = (100 + 3) (100 + 7) = (100)2 + (3 + 7)10 + (3 × 7)
= 10000 + 100 + 21
= 10121
(ii) 95 × 96 = (90 + 5) (90 + 4)
Using identity, (x + a) (x + b) = x2 + (a + b) x + ab
Here, x = 90, a = 5 and b = 4
95 × 96 = (90 + 5) (90 + 4) = 902 + 90(5 + 6) + (5 × 6)
= 8100 + (11 × 90) + 30
= 8100 + 990 + 30 = 9120
(iii) 104 × 96 = (100 + 4) (100 - 4)
Using identity, (x + y) (x -y) = x2 - y2
Here, x = 100 and y = 4
104 × 96 = (100 + 4) (100 - 4) = (100)2 - (4)2 = 10000 - 16 = 9984
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) + y2
Using identity, (a + b)2 = a2 + 2ab + b2
Here, a = 3x and b = y
9x2 + 6xy + y2 = (3x) 2 + (2×3x×y) + y2 = (3x + y)2 = (3x + y) (3x + y)
(ii) 4y2 - 4y + 1 = (2y)2 - (2×2y×1) + 12
Using identity, (a - b)2 = a2 - 2ab + b2
Here, a = 2y and b = 1
4y2 - 4y + 1 = (2y)2 - (2×2y×1) + 12 = (2y - 1)2 = (2y - 1) (2y - 1)
(iii) x2 - y2/100 = x2 - (y/10)2
Using identity, a2 - b2 = (a + b) (a - b)
Here, a = x and b = (y/10)
x2 - y2/100 = x2 - (y/10)2 = (x - y/10) (x + y/10)
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) [1/4 a - 1/2 b + 1]2
Answer
(i) (x + 2y + 4z)2
Using identity, (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca
Here, a = x, b = 2y and c = 4z
(x + 2y + 4z)2 = x2 + (2y)2 + (4z)2 + (2×x×2y) + (2×2y×4z) + (2×4z×x)
= x2 + 4y2 + 16z2 + 4xy + 16yz + 8xz
(ii) (2x – y + z)2
Using identity, (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca
Here, a = 2x, b = -y and c = z
(2x – y + z)2 = (2x)2 + (-y)2 + z2 + (2×2x×-y) + (2×-y×z) + (2×z×2x)
= 4x2 + y2 + z2 - 4xy - 2yz + 4xz
(iii) (–2x + 3y + 2z)2
Using identity, (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca
Here, a = -2x, b = 3y and c = 2z
(–2x + 3y + 2z)2 = (-2x)2 + (3y)2 + (2z)2 + (2×-2x×3y) + (2×3y×2z) + (2×2z×-2x)
= 4x2 + 9y2 + 4z2 - 12xy + 12yz - 8xz
(iv) (3a – 7b – c)2
Using identity, (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca
Here, a = 3a, b = -7b and c = -c
(3a – 7b – c)2 = (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
Using identity, (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca
Here, a = -2x, b = 5y and c = -3z
(–2x + 5y – 3z)2 = (-2x)2 + (5y)2 + (-3z)2 + (2×-2x×5y) + (2×5y×-3z) + (2×-3z×-2x)
= 4x2 + 25y2 + 9z2 - 20xy - 30yz + 12xz
(vi) [1/4 a - 1/2 b + 1]2
Using identity, (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca
Here, a = 1/4 a, b = -1/2 b and c = 1
[1/4 a - 1/2 b + 1]2 = (1/4 a)2 + (-1/2 b)2 + 12 + (2×1/4 a×-1/2 b) + (2×-1/2 b×1) + (2×1×1/4 a)
= 1/16 a2 + 1/4 b2 + 1 - 1/4 ab - b + 1/2 a
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
Using identity, (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca
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
= (2x + 3y - 4z) (2x + 3y - 4z)
(ii) 2x2 + y2 + 8z2 - 2√2 xy + 4√2 yz - 8xz
Using identity, (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca
2x2 + y2 + 8z2 - 2√2 xy + 4√2 yz - 8xz
= (-√2x)2 + (y)2 + (2√2z)2 + (2×-√2x×y) + (2×y×2√2z) + (2×2√2z×-√2x)
= (-√2x + y + 2√2z)2
= (-√2x + y + 2√2z) (-√2x + y + 2√2z)
Write the following cubes in expanded form:
(i) (2x + 1)3 (ii) (2a – 3b)3 (iii) [3/2 x + 1]3 (iv) [x - 2/3 y]3
Answer
(i) (2x + 1)3
Using identity, (a + b)3 = a3 + b3 + 3ab(a + b)
(2x + 1)3 = (2x)3 + 13 + (3×2x×1)(2x + 1)
= 8x3 + 1 + 6x(2x + 1)
= 8x3 + 12x2 + 6x + 1
(ii) (2a – 3b)3
Using identity, (a - b)3 = a3 - b3 - 3ab(a - b)
(2a – 3b)3 = (2a)3 - (3b)3 - (3×2a×3b)(2a - 3b)
= 8a3 - 27b3 - 18ab(2a - 3b)
= 8a3 - 27b3 - 36a2b + 54ab2
(iii) [3/2 x + 1]3
Using identity, (a + b)3 = a3 + b3 + 3ab(a + b)
[3/2 x + 1]3 = (3/2 x)3 + 13 + (3×3/2 x×1)(3/2 x + 1)
= 27/8 x3 + 1 + 9/2 x(3/2 x + 1)
= 27/8 x3 + 1 + 27/4 x2 + 9/2 x
= 27/8 x3 + 27/4 x2 + 9/2 x + 1
(iv) [x - 2/3 y]3
Using identity, (a - b)3 = a3 - b3 - 3ab(a - b)
[x - 2/3 y]3 = (x)3 - (2/3 y)3 - (3×x×2/3 y)(x - 2/3 y)
= x3 - 8/27y3 - 2xy(x - 2/3 y)
= x3 - 8/27y3 - 2x2y + 4/3xy2
Evaluate the following using suitable identities:
(i) (99)3 (ii) (102)3 (iii) (998)3
Answer
(i) (99)3 = (100 - 1)3
Using identity, (a - b)3 = a3 - b3 - 3ab(a - b)
(100 - 1)3 = (100)3 - 13 - (3×100×1)(100 - 1)
= 1000000 - 1 - 300(100 - 1)
= 1000000 - 1 - 30000 + 300
= 970299
(ii) (102)3 = (100 + 2)3
Using identity, (a + b)3 = a3 + b3 + 3ab(a + b)
(100 + 2)3 = (100)3 + 23 + (3×100×2)(100 + 2)
= 1000000 + 8 + 600(100 + 2)
= 1000000 + 8 + 60000 + 1200
= 1061208
(iii) (998)3
Using identity, (a - b)3 = a3 - b3 - 3ab(a - b)
(1000 - 2)3 = (1000)3 - 23 - (3×1000×2)(1000 - 2)
= 100000000 - 8 - 6000(1000 - 2)
= 100000000 - 8- 600000 + 12000
= 994011992
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