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### Chapter 2: Polynomials NCERT Solution

#### TEXTBOOK’S EXERCISE – 2.5

1. 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 )

Sol. (i) Using identity (x + a) (x + b)

= x2 + (a + b) x + ab (x + 4) (x + 10)

= x2 + (4 + 10)x + 4 × 10 = x2 + 14x + 40

(ii) Using the same identity as in (i) above (x + 8) (x – 10)

= x2 + (8 – 10)x + 8 × (–10)

= x2 – 2x – 80

(iii) Using the same identity

(3x + 4) (3x – 5)

= 3x × 3x + (–1) (3x) – 20 = 9x2 – 3x – 20.

(iv) Using (x + y) (x y) = x2y2

2. Evaluate the following products without multiplying directly :

(i) 103 × 107 (ii) 95 × 96 (iii) 104 × 96

Sol. (i) 103 × 107 = (100 + 3) (100 + 7)

= (100)2 + (3 + 7) × 100 + 3 × 7

= 10000 + 1000 + 21 = 11021

(ii) 95 × 96 = (100 – 5) (100 – 4)

= (100)2 – (5 + 4) × 100 + 5 × 4

= 10000 – 900 + 20 = 9120

(iii) 104 × 96 = (100 + 4) (100 – 4)

= 1002 – 42

= 10000 – 16 = 9984

3. Factories the following using appropriate identities:

(i) 9x2 + 6xy + y2 (ii) 4y2 – 4y + 1

Sol. (i) 9x2 + 6xy + y2

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

= (3x + y)2

[Using a2 + 2ab + b2 = (a + b)2]

= (3x + y) (3x + y)

(ii) 4y2 – 4y + 1

= (2y)2 – 2(2y) (1) + (1)2

= (2y – 1)2 = (2y – 1) (2y – 1)

[Using a2 – 2ab + b2 = (a – b)2]

[Using a2 – b2 = (a + b) (a – b)]

4. 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

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

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

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

(ii) (2x – y + z)2

= (2x)2 + (–y)2 + (z)2 + 2 × (2x) (–y) + 2 (–y) (z) + 2 (z) × 2x

= 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)

= 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)

= 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)

= 4x2 + 25y2 + 9z2 – 20xy – 30yz + 12zx

5. Factorise :

(i) 4x2 + 9y2 + 16z2 + 12xy – 24yz –16xz

(ii) 2x2 + y2 + 8z2 – √2 xy + 4 √2 yz – 8xz

Sol. (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

= (2x + 3y – 4z) (2x + 3y – 4z)

(ii) 2x2 + y2 + 8z2 – 2 √2 xy + 4√2 yz – 8xz

= (–√2 x)2 + y2 + (2 √2 z)2 + 2 (– 2 x) y + 2y (2 √2 z) + 2 (– 2 x) (2 √2 z)

= (– 2 x + y + 2 √2 z)2

= (– 2 x + y + 2 √2 z) (– 2 x + y + 2 √2 z)

6. Write the following cubes in expanded form:

(i) (2x + 1)3 (ii) (2a – 3b)3

Sol. (i) (2x + 1)3

= (2x)3 + 13 + 3(2x) (1) (2x + 1)

= 8x3 + 1 + 6x (2x + 1)

= 8x3 + 12x2 + 6x + 1

(ii) (2a – 3b)3

= (2a)3 – (3b)3 – 3 × 2a × 3b (2a – 3b)

= 8a3 – 27b3 – 18ab (2a – 3b)

= 8a3 – 27b3 – 36a2b + 54ab2

7. Evaluate the following using suitable identities :

(i) (99)3 (ii) (102)3 (iii) (998)3

Sol. (i) (99)3 = (100 – 1)3

= (100)3 + (–1)3 + 3(100) (–1) (100 – 1)

= 1000000 – 1 – 300 (100 – 1)

= 1000000 – 1 – 30000 + 300

= 970299

(ii) (102)3 = (100 + 2)3

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

= 1000000 + 8 + 600 (100 + 2)

= 1000000 + 8 + 60000 + 1200

= 1061208

(iii) (998)3 = (1000 – 2)3

= (1000)3 + (–2)3 + 3(1000) (–2) (998)

= (1000)3 – 8 – 6000 (998)

= 1000000000 – 8 – 5988000

= 994011992

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

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

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

(ii) 8a3 – b3 – 12a2b + 6ab2

= (2a)3 + (–b)3 + 3 (2a) (–b) (2a – b)

= (2a – b)3 = (2a – b) (2a – b) (2a – b)

(iii) 27 – 125a3 – 135a + 225a2

= 33 + (–5a)3 + 3 × (3) (–5a) (3 – 5a)

= (3 – 5a)3 = (3 – 5a) (3 – 5a) (3 – 5a)

(iv) 64a3 – 27b3 – 144a2b + 108ab2

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

= (4a – 3b)3 = (4a – 3b) (4a – 3b) (4a – 3b)

9. Verify :

(i) x3 + y3 = (x + y) (x2 – xy + y2)

(ii) x3 – y3 = (x – y) (x2 + xy + y2)

Sol. (i) x3 + y3 = (x + y) (x2 – xy + y2)

R.H.S. = x (x2 – xy + y2) + y (x2 – xy + y2)

= x3 – x2y + xy2 + yx2 – xy2 + y3

= x3 + y3 = L.H.S.

(ii) x3 – y3 = (x – y) (x2 + xy + y2)

R.H.S. = x (x2 + xy + y2) – y (x2 + xy + y2)

= x3 + x2y + xy2 – yx2 – xy2 – y3

= x3 – y3 = L.H.S.

10. Factorise each of the following :

(i) 27y3 + 125z3 (ii) 64m3 – 343n3

Sol. (i) 27y3 + 125z3 = (3y)3 + (5z)3

= (3y + 5z) [(3y)2 – (3y) (5z) + (5z)2]

= (3y + 5z) (9y2 – 15yz + 25z2)

(ii) 64m3 – 343n3 = (4m)3 – (7n)3

= (4m – 7n) [(4m)2 + (4m) (7n) + (7n)2

= (4m – 7n) (16m2 + 28mn + 49n2)

11. Factorise : 27x3 + y3 + z3 – 9xyz

Sol. 27x3 + y3 + z3 – 9xyz

= (3x)3 + y3 + z3 – 3 (3x) (y) (z)

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

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

Sol. To verify : x3 + y3 + z3 – 3xyz

= 1/2 (x + y + z) [(x – y)2 + (y – z)2 + (z – x)2]

R.H.S. = 1/2 (x + y + z) [x2 + y2 – 2xy + y2 + z2 – 2yz + z2 + x2 – 2zx]

= 1/2 (x + y + z) [2x2 + 2y2 + 2z2 – 2xy – 2yz – 2zx]

= (x + y + z) [x2 + y2 + z2 – xy – yz – zx]

= x (x2 + y2 + z2 – xy – yz – zx) + y (x2 + y2 + z2 – xy – yz – zx) + z (x2 + y2 + z2 – xy – yz – zx)

= x3 + xy2 + xz2 – x2y – xyz – zx2 + yx2 + y3 + yz2 – xy2 – y2z – zxy + zx2 + zy2 + z3 – zxy – yz2 – z2x

= x3 + y3 + z3 – 3xyz = L.H.S. Hence verified.

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

Sol. x + y + z = 0

⇒ (x + y + z)3 = x3 + y3 + z3 – 3xyz = 0

⇒ x3 + y3 + z3 = 3xyz.

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

Sol. From the above question, we have x3 + y3 + z3 = 3xyz, if x + y + z = 0

(i) Here –12 + 7 + 5 = 0

∴ (–12)3 + (7)3 + (5)3 = 3 (–12) (7) (5) = – 1260

(ii) Here 28 + (–15) + (–13) = 0

So, (28)3 + (–15)3 + (–13)3

= 3 × 28 (–15) (–13) = 16380

Therefore, p (3/5) gives zero value and (5a – 3) is a factor.

Second factor (5a – 4), length = (5a – 3); breadth = (5a – 4).

(ii) Area = 35y2 + 13y – 12

= 35y2 + 28y – 15y – 12 = 7y (5y + 4) – 3 (5y + 4)

= (5y + 4) (7y – 3) So, (5y + 4) may be taken as breadth and (7y – 3) as length.

Sol. (i) abc = 3x2 – 12x = 3x (x – 4)

3, x (x – 4) are the three factors so they can be three dimensions.

(ii) abc = 12ky2 + 8ky – 20k

= 4k (3y2 + 2y – 5)

= 4k {3y (y – 1) + 5 (y – 1)}

= 4k (y – 1) (3y + 5)

4k, (y – 1) and (3y + 5) are the three factors, so they can be three dimensions.
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