Polynomials – Exercise 2.5 – (MATHEMATICS) – 9th Class

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Exercise 2.5

1. Use suitable identities to find the fallowing products:
(i) (x + 4)(x + 10)

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Ans. –  (x + 4 ) (x + 10 ) = x² + (4 + 10) x + (4 × 10 )
[Using identity: (x + a ) (x + b ) = x² + (a +b ) x + ab ]
= x² + 14x + 40.

(ii) (x + 8)(x – 10)

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Ans. –  (x + 8)(x- 10) = x² + (8 – 10)x + 8 (-10 )
[using identity: ( x +a )(x -b)= x² + (a + b)x + ab]
= x²   – 2x – 80 .

(iii)(3x + 4)(3x – 5)

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Ans. – (3x + 4 ) (3x – 5) = (3x)² + (4 – 5 ) (3x) + 4 × (-5)
Using identity : (x + a) (x + b) = x² + (a + b)x + ab ]
= 9x² – 3x – 20.

(iv) \left( {{y^2} + {3 \over 2}} \right)\left( {{y^2} - {3 \over 2}} \right)

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Ans. – \left( {{y^2} + {3 \over 2}} \right)\left( {{y^2} - {3 \over 2}} \right) = {({y^2})^2} - {\left( {{3 \over 2}} \right)^2}
[ Using identity: (x +a) (x – a) = x² – a²]
 = {y^4} - {9 \over 4}

(v) (3 – 2x) (3 + 2x)

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Ans. – (3 - 2x)(3 + 2x) = {(3)^2} - {(2x)^2}
[ Using identity : ( x + a) (x-a)= x² – a²]
= 9 – 4x²

2. Evaluate the following products without multiplying directly:
(i) 103 × 107

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Ans. – 103 × 107 = (100 + 3)(100 + 7)
.                             = (100)2 + (3 + 7) × 100 + 3 × 7
[Using identity: (x + a)(x + b) = x2 + (a + b)x + ab]
= 10000 + 1000 + 21 =  11021.

(ii) 95 × 96

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Ans. –  95 × 96 = (90 + 5)(90 + 6)
.                          = (90)2 + (5 + 6) × 90 + 5 × 6
[Using identity: (x + a)(x + b) = x2 + (a + b)× ab]
= 8100 + 990 + 30 = 9120.

(iii) 104 × 96.

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Ans. – 104 × 96 = (100 + 4)(100 – 4) = (100)2 (4)2
[Using identity: (x + y)(x – y) = x2 – y2]
= 10000 – 16 = 9984.

3. Factorise the following using appropriate identities:

(i) 9x2  +  6xy + y2

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Ans. – 9x2 + 6xy  + y2 = (3x)2 + 2 × 3x  ×   y + (y)2   =  (3x + y)2
[Using identity: x2 + 2xy  + y2  = (x +y)2 ]
= (3x + y) (3x + y)

(ii) 4y2 – 4y + 1

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Ans. –    4y2 – 4y + 1 = (2y)2 – 2 × 2y × 1 + (1)2 = (2y – 1)2
[Using identity: x2 – 2xy + y2 =  (x – y)2]
=  (2y – 1)( 2y – 1).

(iii) x2– {{{y^2}} \over {100}}

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Ans. –   {x^2} - {{{y^2}} \over {100}} = {(x)^2} - {\left( {{y \over 2}} \right)^2} = \left( {x - {y \over {10}}} \right)\left( {x + {y \over {10}}} \right)

4. Expand each of the following, using suitable identities :
(i)   (x + 2y + 4z)2

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Ans.  –   (x + 2y +  4z)2 =  (x)2  + (2y)2 + (4z)2 + 2(x)(2y) + 2(2y)(4z) + 2(x)(4z)
[Using identity: (x + y + z)2 = x2 + y2 + z2 + 2xy + 2yz + 2zx]
= x2 + 4y2 + 16z2 + 4xy + 16yz + 8xz.

(ii) (2x – y + z)2

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Ans.  – (2x – y + x)² = (2x)² + (-y)² + (z)² + 2(2x) (-y) + 2(-y) (z) + 2(2x) (z)
[Using identity : (x + y + z )² =x² + y² + z² + 2xy + 2yz + 2zx ]
= 4x² + y² + z² – 4xy – 2yz + 4xz.

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

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Ans.  –
(- 2x + 3y + 2z)2 = (- 2x)2 + (3y)2 + (2z)+ 2 (- 2x)(3y) + 2(3y)(2z) + 2 (- 2x) (2z)
[Using identity: (x + y + z)2 = x2 + y2 + z2 + 2xy + 2yz + 2zx]
=4x² + 9y² +4z² – 12xy + 12yz – 8xz.

(iv) (3a – 7b — c)2

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Ans.  –
(3a – 7b – c)2 = (3a)2 + (- 7b)2 + (- c)2 + 2(3a)(- 7b) + 2 (- 7b)(-c) + 2(3a)(- c)
[Using identity: (x + y + z)2 = x2 + y2 + z2 + 2xy + 2yz + 2zx]
= 9a2 + 49b2 + c2 – 42ab + 14bc – 6ac.

(v) (- 2x + 5y — 3z)2

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Ans.  –
(- 2x + 5y – 3z)2 = (- 2x)2 (5y)2 + (- 3z)2 +  2(- 2x) ( 5y)+ 2(5y)(- 3z) + 2(- 2x)(- 3z)
[Using identity: (x + y + z)2 = x2 + y2 + z2 + 2xy + 2yz + 2zx]
= 4x2 + 25y2 + 9z2   20xy – 30yz + 12xz.

(vi) {\left[ {{1 \over 4}a - {1 \over 2}b + 1} \right]^2}

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Ans.  –  {\left[ {{1 \over 4}a - {1 \over 2}b + 1} \right]^2} = {\left( {{1 \over 2}a} \right)^2} + {\left( { - {1 \over 2}b} \right)^2} + {(1)^2} + 2\left( {{1 \over 4}a} \right)\left( { - {1 \over 2}b} \right) + 2\left( { - {1 \over 2}b} \right)(1) + 2\left( {{1 \over 4}a} \right)(1)
[ Using identity : (x + y + z)² = x² +  y² + z² + 2xy  + 2yz + 2zx ]
 = {1 \over {16}}{a^2} + {1 \over 4}{b^2} + 1 - {1 \over 4}ab - b + {1 \over 2}a

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

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Ans. – (i) 4x2 + 9y2 + 16z2 + 12xy – 24yz – 16xz = (2x)2
+ (3y)2 + (- 4z)+ 2(2x)(3y) + 2(3y)( -4z) + 2(2x)(- 4z)
[Using identity: x2 + y2 + z2 + 2xy + 2yz + 2zx = (x + y + z)2]
= (2x + 3y – 4x)2.
= (2x + 3y – 4z)(2x + 3y – 4z).

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

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Ans. –
2x2 + y2 + 8z2 – 2 √2 xy + 4√2 yz – 8xz
= (- √2 x)2 + ( y)2 + (2 √2 z)2 + 2(- √2x)(y) + 2( y) (2 √2  z) + 2 (- √2x)(2√2 z)
[Using identity: x2 + y2 + z2 + 2xy + 2yz + 2zx = (x + y + z)2]
 = {( - \sqrt 2 x + y + 2\sqrt 2 z)^2}
 = ( - \sqrt 2 x + y + 2\sqrt 2 z)( - \sqrt 2 x + y + 2\sqrt 2 z)

6. Write the following cubes in expanded form:
(i) (2x + 1)3

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Ans. –  (2x + 1)3 = (2x)3 + (1)3 + 3(2x)(1)(2x + 1)
[Using identity: (x + y)3 = x3 + y3 + 3xy (x + y)]
= 8x3 + 1 + 12x2 + 6x = 8x3 + 12x2 + 6x + 1.

(ii) (2a — 3b)3

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Ans. –  (2a – 3b)3 = (2a)3 – (3b)3 – 3(2a)(3b)(2a – 3b)
[Using identity: (x – y)3 = x3 – y3 – 3xy(x – y)]
= 8a3 – 27b3 – 36a2b + 54ab2.

(iii) {\left[ {{3 \over 2}x + 1} \right]^3}

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Ans. –{\left( {{3 \over 2}x + 1} \right)^3} = {\left( {{3 \over 2}x} \right)^3} + {(1)^3} + 3 \times {3 \over 2}x \times 1\left( {{3 \over 2}x + 1} \right)
[Using identity :  ( x + y )³ = x³ + y³ + 3xy (x + y )]
 = {{27} \over 8}{x^3} + 1 + {{27} \over 4}x + {9 \over 2}x = {{27} \over 8}x + {{27} \over 4}{x^2} + {9 \over 2}x + 1

(iv) {\left[ {x - {3 \over 2}y} \right]^3}

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Ans. – {\left( {x - {3 \over 2}y} \right)^3} = {\left( x \right)^3} + {\left( {{2 \over 3}y} \right)^3} - 3(x)\left( {{2 \over 3}y} \right)\left( {x - {2 \over 3}y} \right)
[Using identity : (x – y )³ = x³ -y³ – 3xy (x-y) ]
 = {x^3} - {8 \over {27}}{y^3} - 2{x^2}y + {4 \over 3}x{y^2}

7. Evaluate the following using suitable identities:
(i) (99)³

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Ans. –  (99)3 = (100 – 1)3
= (100)3 – (1)3 –  3 × 100 ×    1 (100 – 1)
[Using identity: (x –  y)3 = x3 – y3 – 3xy (x – y)]
= 1000000 – 1 – 30000 + 300 = 970299.

(ii) (102)³

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Ans. – (102)3 = (100 + 2)3
= (100)3 + (2)3 + 3 × 100 × 2 (100 + 2)
[Using identity: (x + y)³ = x³ + y³ + 3xy(x + y)]
= 1000000 + 8 + 60000 + 1200 = 1061208.

(iii) (998)³

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Ans. – (998)3 = (1000 – 2)3
= (1000)3 – (2)3 – 3 × 1000 × 2(1000 – 2)
[Using identity: (x – y)3 = x3 – y3 – 3xy(x – y)]
= 1000000000 – 8 – 6000000 + 12000 = 994011992.

8. Factorise each of the following:
(i) 8a3 + b3 + 12a2b + 6ab2

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Ans. –
8a3 + b3 + 12a2b + 6ab2
= (2a)3 + (b)3+3(2a)(b)(2a + b)
= (2a b)3
[Using identity: x3 + y3 + 3xy(x + y) = (x + y)3]
= (2a + b)(2a + b)(2a + b).

(ii) 8a3 – b3 – 12a2b + 6ab2

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Ans. –  8a3 – b3 – 12a2b + 6ab2
= (2a)3 – (b)3 – 3(2a)2b + 3(2a)b2
= (2a – b)3
[Using identity: x3 – y3 3x2y + 3xy2 = (x –  y)3]
= (2a – b)(2a – b)(2a – b). 

(iii) 27 – 125a3 – 135a + 225a2

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Ans. – 27 – 125a3 – 135a + 225a2
= (3)3 – (5a)3 – 3 × (3)2  × (5a) + 3(3)(5a)2
= (3 – 5a)³
[Using identity: x³ – y³ – 3x²y + 3xy² = (x – y)³]
= (3 – 5a)(3 – 5a)(3 – 5a).

(iv) 64a3 – 27b3 – 144a2b + 108ab2

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Ans. – 64a³ – 27b³ 144a²b + 108ab²
= (4a)³ – (3b)³ – 3(4a)²(3b) + 3(4a)(3b)²
= (4a – 3b)³
[Using identity: x³ – y³ – 3x²y + 3xy² = (x – y)³]
= (4a – 3b)(4a – 3b)(4a – 3b).

(v) 27{p^3} - {1 \over {216}} - {9 \over 2}{p^2} + {1 \over 4}p

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Ans. – 27{p^3} - {1 \over {216}} - {9 \over 2}{p^2} + {1 \over 4}p = {(3p)^3} - {\left( {{1 \over 6}} \right)^3} - 3 \times {(3p)^2} \times {1 \over 6} + 3 \times (3p) \times {\left( {{1 \over 6}} \right)^2}
 = {\left( {3p - {1 \over 6}} \right)^3}
 = \left( {3p - {1 \over 6}} \right)\left( {3p - {1 \over 6}} \right)\left( {3p - {1 \over 6}} \right)

9. Verify:
(i) x3 + y3 = (x + y)(x2 – xy + y2)

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Ans. –  Consider the identity (x + y)3 = x3 + y3 + 3xy(x + y)
⇒  x3 + y= (x + y)3 – 3xy(x + y)
= (x + y)((x + y)2 – 3xy)
⇒ x³ + y³ = (x + y) (x² + y² + 2xy – 3xy)
= (x + y)(x² – xy + y²).

(ii) x3 – y3 = (x – y)(x² + xy + y²)

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Ans. –
Consider the identity (x – y)² = x³ – y³ – 3xy(x – y)
⇒ x³ – y³ = (x – y)³ + 3xy (x – y)
= (x – y) {(x – y)² + 3xy}
⇒ x³ – y³ = (x – y) {x² + y² – 2xy + 3xy)
= (x – y)(x² + xy + y²).

10. Factorise each of the following:
(i) 27y³ + 125z³

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Ans. – Consider 27y³ + 125z³ = (3y)³ + (5z)²
= (3y + 5z) {(3y)² – (3y)(5z) + (5z)²}
[Using identity: x³ + y³ = (x + y)(x² – xy + y²)]
= (3y + 5z)(9y² – 15yz + 25z²)

(ii) 64m³ – 343n³.

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Ans. – Consider 64m³ – 343n³ = (4m)³ – (7n)³
= (4m – 7n) {(4m)² + (4m)(7n) + (7n)²}
[Using identity: x³ – y³ = – y)(x² + xy + y²)]
= (4m – 7n)(16m² + 28mn + 49n²).

11. Factorise: 27x³ + y³ + z³ – 9xyz.

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Sol. Consider 27x³ + y³ + z³ – 9xyz
= (3x)³ + ( y)³ + (z)³ – 3(3x)(y)(z)
[Using identity: x³ + y³ + z³ – 3xyz
= (x + y + z)(x² + y² + z² – xy – yz – zx)]
= (3x + y + z) [(3x)² + y² + z² – (3x)y – yz – (3x)z)
= (3x + y + z) (9x² + y² + z2 3xy – yz – 3xz).

12. Verify that:

x3 + y3 + z3 – 3xyz
 = {1 \over 2}(x + y + z)[{(x - y)^2} + {(y - z)^2} + {(z - x)^2}]

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Sol.  Consider identity x³ + y³ + z³ – 3xyz
= (x + y + z)(x² + y² + z² -xy – yz – zx )
 = {1 \over 2}(x + y + z)(2{x^2} + 2{y^2} + 2{z^2} - 2xy - 2yz - 2zx)
 = {1 \over 2}(x + y + z)\{ ({x^2} + {y^2} - 2xy) + ({y^2} + 2yz - 2zx) + ({z^2} + {x^2} - 2xz)\}
 = {1 \over 2}(x + y + z)\{ {(x - y)^2} + {(y - z)^2} + {(z - x)^2}\}

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

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Sol. Consider the identity x³ + y³ + z³ – 3xyz
= (x + y + z) (x² + y² + z² – xy – yz – zx)                               …….(i)
If x + y + z = 0, then
(x + y + z)(x² + y² + z² – xy – yz – zx) =    0
∴  x³ + y³ + z³ – 3xyz = 0                                              [From (i)]
⇒       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

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Ans. –  (i) Consider (- 12)³ + (7)³ + (5)³
Let x = – 12, y = 7, z = 5
Now, x + y + z = – 12 + 7 + 5 = 0
We know, if x + y + z = 0, then x³ + y³ + z³ = 3xyz.
∴ (- 12)³ + (7)³ + (5)³ = 3(- 12)(7)(5)
= – 21 × 60 = – 1260.

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

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Ans. –   Consider (28)³ + (- 15)³ +(+ 13)³
Let x = 28, y = – 15, z = – 13
Now, x + y + z = 28 – 15 – 13 = 0
We know, if x + y + z = 0, then x³ + y³ + z³ = 3xyz.
∴ (28)³ + (- 15)³ + (- 13)³
= 3(28) (- 15) (- 13)  = 16380.

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

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Ans. – Are2 = 25a² – 35a + 12 = 25a² – 20a –  15a + 12
= 5a(5a – 4) – 3 (5a – 4) = (5a – 3)(5a – 4).

Possible expressions for length and breadth are
(5a – 3) and (5a – 4) respectively


(ii) Area: 35y² + 13y – 12

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Ans. – Area = 35y² + 13y – 12 = 35y² + 28y – 15y – 12
= 7y (5y + 4) –  3(5y + 4) = (7y – 3)(5y + 4),
Possible expressions for length and breadth are (7y – 3) and (5y + 4).

16. What are the possible expressions for the dimensions of the cuboids whose volumes are given below?
(i) Volume  :  3x² – 12x

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Ans.-  Volume = 3x² – 12x = 3x(x – 4)
Possible dimensions are 3, x and (x – 4).

(ii) Volume  : 12ℜy² + 8ℜy – 20ℜ

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Ans.-
Volume = 12ℜy² + 8ℜy – 20ℜ = 4ℜ(3y² + 2y – 5)
= 4ℜ(3y² + 5y – 3y – 5)
= 4ℜ { y(3y + 5) – 1(3y + 5)}
= 4ℜ (y – 1)(3y + 5).
Possible dimensions are 4ℜ, y – 1 and 3y + 5.

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