Polynomials (Exercise 2.5)

EX 2.5 QUESTION 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) (y2– )
(v) (3 – 2x) (3 + 2x)
Solution:

(i) (x+ 4) (x + 10)
Using identity, (x+ a) (x+ b) = x2 + (a + b) x+ ab.
(x + 4) (x + 10) = x2+(4 + 10) x + (4 x 10)
= x2 + 14x+40

(ii) (x+ 8) (x -10)
Using identity, (x + a) (x + b) = x2 + (a + b) x + ab
(x + 8) (x – 10) = x2 + [8 + (-10)] x + (8) (- 10)
= x2 – 2x – 80

(iii) (3x + 4) (3x – 5)
Using identity,(x + a) (x + b) = x2 + (a + b) x + ab
(3x + 4) (3x – 5) = (3x)2 + (4 – 5) x + (4) (- 5)
= 9x2 – x – 20

(iv) (y2+3/2)(y2-3/2)

Using identity, (x+y)(x–y) = x2–y 2

(y2+3/2)(y2–3/2) = (y2)2–(3/2)2

= y4–9/4

(v) (3 – 2x) (3 + 2x)

Using identity, (a+b) (a-b) = a² – b²

(3 – 2x) (3 + 2x) = (3)² – (2x)² = 9 – 4x²

EX 2.5 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)2 + (3 + 7) (100)+ (3 x 7)
Using –  (x + a)(x + b) = x2 + (a + b)x + ab
= 10000 + (10) x 100 + 21
= 10000 + 1000 + 21=11021

(ii) 95 x 96 = (100 – 5) (100 – 4)
= ( 100)2 + [(- 5) + (- 4)] 100 + (- 5 x – 4)
Using – (x + a)(x + b) = x2 + (a + b)x + ab
= 10000 + (-9) + 20 = 9120
= 10000 + (-900) + 20 = 9120

(iii)104 x 96 = (100 + 4) (100 – 4)
= (100)2-42
Using – (a + b)(a -b) = a2– b2
= 10000 – 16 = 9984

EX 2.5 QUESTION 3.

Factorise the following using appropriate identities
(i) 9x+ 6xy + y2
(ii) 4y2-4y + 1
(iii) x2 – y2/100
Solution:
(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 + 12
= (2y)2 + 2(2y)(1) + (1)2
= (2y -1)2
Using  – a2 – 2ab + b2 = (a- b)2
= (2y – 1)(2y – 1 )

(iii) x2 – y2/100

= x2–(y/10)2

Using –  x2-y= (x-y)(x+y)

Here, x = x

y = y/10

x2–y2/100 = x2–(y/10)2

= (x–y/10)(x+y/10)

EX 2.5 QUESTION 4.

Expand each of the following, using suitable identity
(i) (x+2y+ 4z)2
(ii) (2x – y + z)2
(iii) (- 2x + 3y + 2z)2
(iv) (3a -7b – c)z
(v) (- 2x + 5y – 3z)2
(vi) [(1/4)a-(1/2)b +1]2
Solution:

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

Using identity, (x+y+z)2 = x2+y2+z2+2xy+2yz+2zx
= x2 + (2y)2 + (4z)2 + 2 (x) (2y) + 2 (2y) (4z) + 2(4z) (x)
= x2 + 4y2 + 16z2 + 4xy + 16yz + 8 zx

(ii) (2x – y + z)2

Using identity, (x+y+z)2 = x2+y2+z2+2xy+2yz+2zx

= (2x)2 + (- y)2 + z2 + 2 (2x) (- y)+ 2 (- y) (z) + 2 (z) (2x)
= 4x2 + y2 + z2 – 4xy – 2yz + 4zx

(iii) (- 2x + 3y + 2z)2

Using identity, (x+y+z)2 = x2+y2+z2+2xy+2yz+2zx

= (- 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)

Using identity, (x+y+z)2 = x2+y2+z2+2xy+2yz+2zx

=(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, (x+y+z)2 = x2+y2+z2+2xy+2yz+2zx

= (- 2x)2 + (5y)2 + (- 3z)2 + 2 (- 2x) (5y) + 2 (5y) (- 3z) + 2 (- 3z) (- 2x)
= 4x2 + 25y2 + 9z2 – 20xy – 30yz + 12zx

(vi) [(1/4)a-(1/2)b +1]2

Using identity, (x+y+z)2 = x2+y2+z2+2xy+2yz+2zx

EX 2.5 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 = (2x + 3y + 4z) (2x + 3y – 4z)

(ii) 2x2 + y2 + 8z2 – 2√2xy + 4√2yz – 8xz
= (- √2x)2 + (y)2 + (2 √2z)2y + 2(- √2x) (y)+ 2 (y) (2√2z) + 2 (2√2z) (- √2x)
= (- √2x + y + 2 √2z)2
= (- √2x + y + 2 √2z) (- √2x + y + 2 √2z)

EX 2.5 QUESTION 6.

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

Solution:

(i) (2x+1)3

Using identity,(x+y)3 = x3+y3+3xy(x+y)

(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,(x–y)3 = x3–y3–3xy(x–y)

(2a−3b)= (2a)3−(3b)3–(3×2a×3b)(2a–3b)

= 8a3–27b3–18ab(2a–3b)

= 8a3–27b3–36a2b+54ab2

EX 2.5 QUESTION 7.

Evaluate the following using suitable identities
(i) (99)3
(ii) (102)3
(iii) (998)3
Solution:
(i) 99 = (100 -1)
∴ 993 = (100 – 1)3
= (100)3 – 13 – 3(100)(1)(100 -1)
Using (a – b)3 = a3 – b3 – 3ab (a – b)
= 1000000 – 1 – 300(100 – 1)
= 1000000 -1 – 30000 + 300
= 1000300 – 30001 = 970299

(ii) 102 =100 + 2
∴ 1023 = (100 + 2)3
= (100)3 + (2)3 + 3(100)(2)(100 + 2)
Using (a + b)3 = a3 + b3 + 3ab (a + b)
= 1000000 + 8 + 600(100 + 2)
= 1000000 + 8 + 60000 + 1200 = 1061208

(iii)  998 = 1000 – 2
∴ (998)3 = (1000-2)3
= (1000)3– (2)3 – 3(1000)(2)(1000 – 2)
Using (a – b)3 = a3 – b3 – 3ab (a – b)
= 1000000000 – 8 – 6000(1000 – 2)
= 1000000000 – 8 – 6000000 +12000
= 994011992

EX 2.5 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

(v) 27p3–(1/216)−(9/2) p2+(1/4)p

Solution:

(i) 8a3 +b3 +12a2b+6ab2
= (2a)3 + (b)3 + 6ab(2a + b)
= (2a)3 + (b)3 + 3(2a)(b)(2a + b)
= (2 a + b)3
Using a3 + b3 + 3 ab(a + b) = (a + b)3
= (2a + b)(2a + b)(2a + b)

(ii) 8a3 – b3 – 12o2b + 6ab2
= (2a)3 – (b)3 – 3(2a)(b)(2a – b)
= (2a – b)3
Using a3 + b3 + 3 ab(a + b) = (a + b)3
= (2a – b) (2a – b) (2a – b)

(iii) 27 – 125a3 – 135a + 225a2
= (3)3 – (5a)3 – 3(3)(5a)(3 – 5a)
= (3 – 5a)3
Using a3 + b3 + 3 ab(a + b) = (a + b)3
= (3 – 5a) (3 – 5a) (3 – 5a)

(iv) 64a3 -27b3 -144a2b + 108ab2
= (4a)3 – (3b)3 – 3(4a)(3b)(4a – 3b)
= (4a – 3b)3
Using a3 – b3 – 3 ab(a – b) = (a – b)3
= (4a – 3b)(4a – 3b)(4a – 3b)

(v) 27p3– (1/216)−(9/2) p2+(1/4)p

(3p)3–(1/6)3–3(3p)2(1/6)+3(3p)(1/6)2

Using a3 – b3 – 3 ab(a – b) = (a – b)3

= (3p–1/6)3

= (3p–1/6)(3p–1/6)(3p–1/6)

EX 2.5 QUESTION 9.

Verify
(i) x3 + y3 = (x + y)-(x2 – xy + y2)
(ii) x3 – y3 = (x – y) (x2 + xy + y2)
Solution:
(i) ∵ (x + y)3 = x3 + y3 + 3xy(x + y)
⇒ (x + y)3 – 3(x + y)(xy) = x3 + y3
⇒ (x + y)[(x + y)2-3xy] = x3 + y3
⇒ (x + y)(x2 + y2 – xy) = x3 + y3
LHS = RHS

(ii) ∵ (x – y)3 = x3 – y3 – 3xy(x – y)
⇒ (x – y)3 + 3xy(x – y) = x3 – y3
⇒ (x – y)[(x – y)2 + 3xy)] = x3 – y3
⇒ (x – y)(x2 + y2 + xy) = x3 – y3
LHS = RHS

EX 2.5 QUESTION 10.

Factorise each of the following
(i) 27y3 + 125z3
(ii) 64m3 – 343n3
Solution:
(i) 27y3 + 125z3

= (3y)3 + (5z)3
= (3y + 5z)[(3y)2 – (3y)(5z) + (5z)2]                [x3 + y3 = (x + y)(x2 – xy + y2)]
= (3y + 5z)(9y2 – 15yz + 25z2)

(ii) 64m3 – 343n3

= (4m)3 – (7n)3
= (4m – 7n)[(4m)2 + (4m)(7n) + (7n)2]            [x3 + y3 = (x + y)(x2 – xy + y2)]
= (4m – 7n)(16m2 + 28mn + 49n2)

EX 2.5 QUESTION 11.

Factorise 27x3 +y3 +z3 -9xyz.
Solution:
27x3 + y3 + z3 – 9xyz = (3x)3 + (y)3 + (z)3 – 3(3x)(y)(z)
Using ,x3 + y3 + z3 – 3xyz = (x + y + z)(x2 + y2 + z2 – xy – yz – zx)
(3x)3 + (y)3 + (z)3 – 3(3x)(y)(z)
= (3x + y + z)[(3x)3 + y3 + z3 – (3x × y) – (y × 2) – (z × 3x)]
= (3x + y + z)(9x2 + y2 + z2 – 3xy – yz – 3zx)

EX 2.5 QUESTION 12.

Verify that
x3 +y3 +z3 – 3xyz =  (x + y+z)[(x-y)2 + (y – z)2 +(z – x)2]
Solution:
R.H.S
(x + y + z)[(x – y)2+(y – z)2+(z – x)2]
(x + y + 2)[(x2 + y2 – 2xy) + (y2 + z2 – 2yz) + (z2 + x2 – 2zx)]
(x + y + 2)(x2 + y2 + y2 + z2 + z2 + x2 – 2xy – 2yz – 2zx)
(x + y + z)[2(x2 + y2 + z2 – xy – yz – zx)]
= 2 x  x (x + y + z)(x2 + y2 + z2 – xy – yz – zx)
= (x + y + z)(x2 + y2 + z2 – xy – yz – zx)
= x3 + y3 + z3 – 3xyz = L.H.S.
R.H.S = L.H.S.

EX 2.5 QUESTION 13.

If x + y + z = 0, show that x3 + y3 + z3 = 3 xyz.
Solution:
x + y + z = 0
⇒ x + y = -z (x + y)3 = (-z)3
⇒ x3 + y3 + 3xy(x + y) = -z3
⇒ x3 + y3 + 3xy(-z) = -z3 [∵ x + y = -z]
⇒ x3 + y3 – 3xyz = -z3
⇒ x3 + y3 + z3 = 3xyz
if x + y + z = 0, then
x3 + y3 + z3 = 3xyz

EX 2.5 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:
(i) (-12)3 + (7)3 + (5)3
Let a = -12, b = 7 and c = 5.
Then, a + b + c = -12 + 7 + 5 = 0
We know that if a + b + c = 0, then, a3 + b3 + c3 = 3xyz
∴ (-12)3 + (7)3 + (5)3 = 3[(-12)(7)(5)]
= 3[-420] = -1260

(ii) (28)3 + (-15)3 + (-13)3
Let a = 28, b = -15 and c = -13.
Then, a + b + c = 28 – 15 – 13 = 0
We know that if a + b + c= 0, then a3 + b3 + c3 = 3xyz
∴ (28)3 + (-15)3 + (-13)3 = 3(28)(-15)(-13)
= 3(5460) = 16380

EX 2.5 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:
Area of a rectangle = (Length) x (Breadth)
(i) 25a2 – 35a + 12

=25a2 – 20a – 15a + 12

= 5a(5a – 4) – 3(5a – 4)

= (5a – 4)(5a – 3)
Thus, the possible length and breadth are (5a – 3) and (5a – 4).

(ii) 35y2+ 13y -12

= 35y2 + 28y – 15y -12

= 7y(5y + 4) – 3(5y + 4)

= (5 y + 4)(7y – 3)
Thus, the possible length and breadth are (5y + 4) and (7y – 3).

EX 2.5 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:
Volume of a cuboid = (Length) x (Breadth) x (Height)
(i) 3x2 – 12x

= 3(x2 – 4x)

= 3 x × x (x – 4)
∴ The possible dimensions of the cuboid are 3, x and (x – 4).

(ii) 12ky2 + 8ky – 20k
= 4[3ky2 + 2ky – 5k] = 4[k(3y2 + 2y – 5)]
= 4 x k x (3y2 + 2y – 5)
= 4k[3y2 – 3y + 5y – 5]
= 4k[3y(y – 1) + 5(y – 1)]
= 4k[(3y + 5) x (y – 1)]
= 4k x (3y + 5) x (y – 1)
Thus, the possible dimensions of the cuboid are 4k, (3y + 5) and (y -1).

UKPSC Forest Guard Exam 09 April 2023 – Answer Key

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