GSEB Solutions Class 9 Maths Chapter 2 Polynomials Ex 2.5

   

Gujarat Board GSEB Solutions Class 9 Maths Chapter 2 Polynomials Ex 2.5 Textbook Questions and Answers.

Gujarat Board Textbook Solutions Class 9 Maths Chapter 2 Polynomials 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 + \(\frac {3}{2}\) )(y2 – \(\frac {3}{2}\))
(v) (3 – 2x)(3 + 2x)
Solution:
(i) Using identity
(x + a)(x + b) = x2 + (a + b) x + ab, we get
= (x + 4)(x + 10) = x2 + (4 + 10) x + 4 x 10
= x2 + 14x + 40

(ii) Using identity
(x + a)(x + b) = x2 + (a + b) x + ab,we get
(x + 8)(x – 10)
= (x + 8) {x + (-10)}
= x2 + {8 + (-10)} x + 8 x(-10)
= x2 + {8 – 10} x – 80
= x2 – 2x – 80

GSEB Solutions Class 9 Maths Chapter 2 Polynomials Ex 2.5

(iii) Using identity
(x + a) (x + b) = x2 + (a + b)x + ab, we get
(3x + 4) (3x – 5)
= (3x)2+{4 + (-5)(3x) + 4x(-5)
= 9x2 + (-1)3x – 20
= 9x2 – 3x – 20

(iv) Using identity
(a + b)(a – b) = a2 – b2,we get
(y2 + \(\frac {3}{2}\)) (y2 – \(\frac {3}{2}\)) = (y2)2 – (\(\frac {3}{2}\))2 = y – \(\frac {9}{4}\)

(v) Using identity
(a – b)(a + b) = a2 – b2,weget
(3 – 2x)(3 + 2x) = 32 – (2x)2
= 9 – 4x2

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)
Using identity
(x + a)(x + b) = x2 + (a + b)x + ab, we get (100 + 3) (100 + 7)
= 1002 + (3 + 7) x 100 + 3 x 7
= 10000 + 10 x 100 + 21
= 10000 + 1000 + 21
= 11021

GSEB Solutions Class 9 Maths Chapter 2 Polynomials Ex 2.5

(ii) 95 x 96 = (100 – 5)(100 – 4)
Using identity
(x + a)(x + b) = x2 + (a + b)x + ab,we get {100 + (-5)} {100 + (-4)}
= 1002 + {5 + (-4)} x 100 + (-5) (-4)
= 10000 + (-9) x 100 + 20
= 10000 – 900 + 20 = 9120

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

Question 3.
Factorise the following using appropriate identities:
(i) 9x2 + 6xy + y2
(ii) 4y2 – 4y + 1
(iii) x2 – \(\frac{y^{2}}{100}\)
Solution:
(i) 9x2 + 6xy + y2 = (3x)2 + 2 x 3x x y + (y)2
= (3x + y)2
[ ∴ a2 + 2ab + b2 = (a + b)2]

(ii) 4y2 – 4y + 1 = (2y)2 – 2 x 2y x 1 + 12
=(2y – 1)2
[∴ a2 – 2ab + b2 = (a – b)2]

(iii) x2 – \(\frac{y^{2}}{100}\) = x2 – \(\frac{y^{2}}{100}\)2
= (x – \(\frac {y}{10}\)) (x + \(\frac {y}{10}\))
[∴ a2 – b2 = (a – b) (a + b)]

GSEB Solutions Class 9 Maths Chapter 2 Polynomials Ex 2.5

Question 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
(vi) [\(\frac {1}{4}\) a – \(\frac {1}{2}\) b + 1]2
Solution:
Using identity
(a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca
(i) (x + 2y + 4z)2 = (x)2 + (2y)2 + (4z)2 + 2 x x x 2y + 2 + 2y + 4z + 2 + 4z + x
= x2 + 4y2 + 16z2 + 4xy + 16yz + 8zx

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

(iii) (- 2x + 3y + 2z)2
= (-2x)2 + (3y)2 + (2z)2 + (-2x) x 3y+ 2 x (3y) x (2z) + 2 x 2z x (-2x)
= 4x2 + 9y2 + 4z2 – 12xy + 12yz – 8zx

GSEB Solutions Class 9 Maths Chapter 2 Polynomials Ex 2.5

(iv) (3a – 7b – c)2 = {3a + (-7b) + (-c)}2
= (3a)2 + (-7b)2 + (-c)2 + 2 x 3a x (-7b) + 2 x (-7b)(-c) + 2x(-c) x 3a
= 9a2 + 49b2 + c2 – 42ab + 14bc – 6ca

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

(vi) [\(\frac {1}{4}\) a – \(\frac {1}{2}\) b + 1]2 =[\(\frac {1}{4}\) a + ( – \(\frac {1}{2}\)b) + 1]2
= (\(\frac {1}{4}\) a)2 + (-\(\frac {1}{4}\) b)2 + 12 + 2 + \(\frac {1}{4}\) a x (- \(\frac {1}{2}\)b) + 2 (- \(\frac {1}{2}\)b) + 1 + 2 x 1 x \(\frac {1}{4}\) a
= \(\frac{a^{2}}{16}\) + \(\frac{b^{2}}{4}\) + 1 – \(\frac {1}{4}\)ab – b + \(\frac {b}{2}\)

Question 5.
Factorise:
(i) 4x2 + 9y2 + 16z2 + l2xy – 24yz – 16xz
(ii) 2x2 + y2 + 8z2 – 2\(\sqrt{2}\)xy + 4 \(\sqrt{2}\)yz – 8zx
Solution:
(i) 4x2 + 9y2 + 16z2 + 12xy – 24yz – 16xz
= (2x)2 + (3y)2 + (-4z)2 + 2 x 2x x 3y + 2 x 3y x (-4z) + 2(- 4z) x (2x)
[∴ a2 + b2 + c2 + 2ab + 2bc + 2ca = (a + b + c)2]
= [2x + 3y + (-4z)]2 = (2x + 3y – 4z)2
= (2x + 3y – 4z) (2x + 3y – 4z)

GSEB Solutions Class 9 Maths Chapter 2 Polynomials Ex 2.5

(ii) 2x2 + y2 + 8z2 – 2\(\sqrt{2}\)xy + 4\(\sqrt{2}\)yz – 8zx
= (- \(\sqrt{2}\)x)2 + y2 + (2\(\sqrt{2}\)x)2 + 2 x (-\(\sqrt{2}\)x)y + 2y x 2 \(\sqrt{2}\) z + 2 x 2\(\sqrt{2}\)z x (-\(\sqrt{2}\)x)
= (- \(\sqrt{2}\) x + y + 2\(\sqrt{2}\)z)2
= (-2\(\sqrt{2}\) x + y + 2\(\sqrt{2}\)z) (-2\(\sqrt{2}\)x + y + 2\(\sqrt{2}\)z)

Question 6.
Write the following cubes in expanded form:
(i) (2x + 1)3
(ii) (2a – 3b)3
(iii) (\(\frac{3}{2}\) x + 1)3
(iv) (x – \(\frac{2}{3}\) y)3
Solution:
(i) Using identity
(a + b)3 = a3 + b3 + 3ab (a + b), we get
(2x + 1)3 = (2x)3 + 13 + 3 x 2x x 1(2x + 1)
= 8x3 + 1 + 6x(2x + 1)
= 8x3 + 1 + 12x3 + 6x
= 8x3 + 12x3 + 6x + 1

(ii) Using identity
(a – b)3 = a3 – b3 – 3ab (a – b), we get
(2a – 3b)3 = (2a)3 – (3b)3 – 3 x 2a x 3b(2a – 3b)
= 8a3 – 27b3 – l8ab(2a – 3b)
= 8a3 – 27b3 – 36a2b + 54ab2

GSEB Solutions Class 9 Maths Chapter 2 Polynomials Ex 2.5

(iii) Using identity
(a + b)3 = a3 + b3 + 3ab(a + b),weget
(\(\frac {3}{2}\) x + 1)3 = (\(\frac {3}{2}\)x)3 + 13 + 3 x \(\frac {3}{2}\) x x 1 (\(\frac {3}{2}\)x + 1)
= \(\frac {27}{8}\) x3 + 1 + \(\frac {9}{2}\)x (\(\frac {3}{2}\)x + 1)
= \(\frac {27}{8}\) x3 + 1 + \(\frac {9}{2}\)x(\(\frac {3}{2}\)x + 1)
= \(\frac {27}{8}\) x3 + 1 + \(\frac {27}{4}\) x2 + \(\frac {9}{2}\)x
= \(\frac {27}{8}\) x3 + 1 + \(\frac {27}{4}\) x2 + \(\frac {9}{2}\)x + 1

(iv) Using identity
(a – b)3 = a3 – b3 – 3ab(a – b), we get
(x – \(\frac {2}{3}\) y)3 = x3 – (\(\frac {2}{3}\) y)3 – 3x x \(\frac {2}{3}\) y (x – \(\frac {2}{3}\)y)
= x3 – \(\frac {8}{27}\)y3 – 2xy(x – \(\frac {2}{3}\) y)
= x3 – \(\frac {8}{27}\)y3 – 2x2y + \(\frac{4 x y^{2}}{3}\)
= x3 – 2x2y + \(\frac{4 x y^{2}}{3}\) – \(\frac {8}{27}\) y3

Question 7.
Evaluate the following using suitable identities:
(i) (99)3
(ii) (102)3
(iii) (998)3
Solution:
(i) (99)3 = (100 – 1)3
Using identity
(a – b)3 = a3 – b3 – 3ab(a – b), we get
(100 – 1) = 100 – 1 – 3 x 100 x 1(100 – 1)
= 1000000 – 1 – 300(100 – 1)
= 1000000 – 1 – 30000 + 300
= 970299

(ii) (102)3 = (100 + 2)
Using identity
(a + b)3 = a3 + b3 + 3ab(a + b),we get
(100+2)3 = 1003 + 23 + 3 x 100 x 2(100+2)
= 1000000 + 8 + 600 (100 + 2)
= 1000000 + 8 + 60000 + 1200
= 1061208

GSEB Solutions Class 9 Maths Chapter 2 Polynomials Ex 2.5

(iii) (998)3 = (1000 – 2)3
Using identity
(a – b)3 = a3 – b3 – 3ab(a – b), we get (1000 – 2)3
= 1oo03 – 23 – 3 x 1000 x 2(1000 – 2)
1000000000 – 8 – 6000(1000 – 2)
= 1000000000 – 8 – 6000000 + 12000
= 994011992

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 – \(\frac {1}{216}\) – \(\frac {9}{2}\) p2 + \(\frac {1}{4}\)p
Solution:
(i) 8a3 + b3 + 12a2b + 6ab2
= (2a)3 + b3 + 3 x 2a x b(2a + b)
= (2a + b)3
[∴ a3 + b3 + 3ab(a + b) = (a + b)3]
= (2a + b)(2a + b)(2a + b)

(ii) 8a3 – b3 – 12a2b +6ab2
= (2a)3 – b3 – 3 x 2a x b (2a – b)
= (2a – b)3
[∴ a3 – b3 – 3ab(a – b) = (a – b)3]
= (2a – b)(2a – b)(2a – b)

GSEB Solutions Class 9 Maths Chapter 2 Polynomials Ex 2.5

(iii) 27 – 125a3 – 135a + 225a2
= (3)3 – (5a)3 – 3 x 3 x 5a (3 – 5a)
= (3 – 5a)3
[∴ a3 – b3 – 3ab(a – b) = (a – b)3]
= (3 – 5a)(3 – 5a)(3 – 5a)

(iv) 64a3 – 27b3 – 14a2b + 108ab2
= (4a)3 – (3b)3 – 3 x 4a x 3b(4a – 3b)
=(4a – 3b)3
[∴ a3 – b3 – 3ab(a – b) = (a – b)3]
= (4a – 3b)(4a – 3b)(4a – 3b)

(v) 27p3 – \(\frac {1}{216}\) – \(\frac {9}{2}\) p2 + \(\frac {1}{4}\)P
= (3p)3 – (\(\frac {1}{6}\))3 – 3 x 3P x \(\frac {1}{6}\) (3P – \(\frac {1}{6}\).)
= (3p – \(\frac {1}{6}\))3
[∴ a3 – b3 – 3ab(a – b) = (a – b)3]
= (3p – \(\frac {1}{6}\))3
= [∴ a3 – b3 – 3ab(a – b) = (a – b)3]
= (3p – \(\frac {1}{6}\)) (3p – \(\frac {1}{6}\)) (3p – \(\frac {1}{6}\))

GSEB Solutions Class 9 Maths Chapter 2 Polynomials Ex 2.5

Question 9.
Verify:
(i) x3 + y3 = (x + y)(x2 – xy + y2)
(ii) x3 – y3 = (x – y)(x2 + xy + y2)
Solution:
(i) We know that
(x + y)3 = x3 + y3 + 3xy (x + y)
x3 + y3 (x + y)3 + 3xy(x+y)
= x3 + y3 = (x + y){(x + y)2 – 3xy)
= x3 + y3 = (x + y){x2 + y2 + 2xy – 3xy)
x3 + y3 = (x + y){x2 + y2 — xy)
=> x3 + y3 = (x + y)(x2 – xy + y2)

(ii) We know that
(x – y)3 =x3 – y3 – 3xy(x – y)
x3 – y3 = (x – y)3 + 3xy(x – y)
x3 – y = (x – y){(x – y)2 + 3xy}
⇒ x3 – y3 = (x – y)(x2 + y2 – 2xy + 3xy)
x3 – y3 = (x – y)(x2 + xy + y2)

GSEB Solutions Class 9 Maths Chapter 2 Polynomials Ex 2.5

Question 10.
Factorise each of the following:
(i) 27y3 + 125z3
(ii) 64m3 – 343n3
Solution:
(i) 27 y3 + 125 z3 = (3y)3 + (5z)3
Using identity
a3 + b3 (a + b) (a2 – ab + b2), we get (3y)3 + (5z)3
= (3y + 5z) {(3y)2 – 3y x 5z + (5z)2)
= 27 y3 + 125 z3
= (3y + 5z)(9y2 – 15yz + 25z2)

(ii) 64m3 – 343n3 = (4m)3 – (7n)3
Using identity
a3 – b3 = (a – b) (a2 + ab + b2), we get
(4m)3 – (7n)3
= (4m – 7n){(4m)2 + 4m x 7n + (7n)2}
64m3 – 343n3
= (4m – 7n) (16m2 + 28mn + 49n2)

GSEB Solutions Class 9 Maths Chapter 2 Polynomials Ex 2.5

Question 11.
Factorise: 27x3 + y3 + z3 – 9xyz
Solution:
27 x3 + y3 + z 3 – 9xyz
= (3x)3 + y3 + z3 – 3 x 3x x y x z
Using identity
a3 + b3 + c3 – 3abc
= (a + b + c)(a2 + b2 + c2 – ab – bc – ca)
27 x3 + y3 + z3 – 9xyz
= (3x +y – z){(3x)2 + y2 + z2 – 3x x y – yz – z x 3x)
= (3x + y + z)(9x2 + y2 + z2 – 3xy – yz – 3zx)

Question 12.
Verify that x3 + y3 + z3 – 3xyz = (x + y + z)((x – y)2 + (y – z)2+(z – x)3]
Solution:
Using identity
a3 + b3 + c3 – 3abc
= (a + b + c) (a2 + b2 + c2 -ab – bc – cu),we get x3 + y3 + z3 – 3xyz
= (x + y + z)[x2 + y2 + z2 – xy – yz – zx]
= \(\frac {1}{2}\) (x + y + z)[2x2 + 2y2 + 2z2 – 2xy – 2yz – 2zx]
= \(\frac {1}{2}\) (x + y + z)[x2 + y2 – 2xy + y2 + z2 – 2yz + z2 + x2 – 2zx]
\(\frac {1}{2}\) (x + y + z) [(x – y)2 + (y – z)2 + z – x)2]

GSEB Solutions Class 9 Maths Chapter 2 Polynomials Ex 2.5

Question 13.
If x + y + z = 0, show that x2 + y2 + z2 = 3xyz.
Solution:
We know that
x2 + y2 + z2 – 3xyz
= (x + y + z)(x2 + y2 + z2 – xy – yz – zx)
But we have x + y + z = 0
∴ x2 + y2 + z2 – 3xyz
= 0 x (x2 + y2 + z2 – xy – yz – zx)
= x3 + y3 + z3 – 3xyz = 0
∴ x3 + y3 + z3 = 3xyz

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) Let a = -12, b = 7 and c = 5
a + b + c = -12 + 7 + 5 = – 12 + 12
We know that ifa + b + c = 0, then
a3 + b3 + c3 = 3abc
=(-12) + (7)3 + (5)3 = 3 x(-12) x 7 x 5
= -1260

GSEB Solutions Class 9 Maths Chapter 2 Polynomials Ex 2.5

(ii) Let a = 28, b = -15 and c = -13
a + b + c = 28 + (-15) + (-13)
= 28 – 28 = 0
As a + b + c = 0
:.a3 + b3 + c3 = 3abc
= (28)3 + (-15)3 + (-13)3
= 3 x 28 x (-15) x (-13)
= 16380

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:
(i) 25a2 – 35a + 12 = 25a2 – 20a – 15a + 12
= 5a(5a – 4) – 3(5a – 4)
= (5a – 4)(5a – 3)
∴ The possible expressions for the length and breadth of the rectangle are 5a – 3 and 5a – 4.

(ii) 35y2 + 13y – 12
= 35y2 + 28y – 15y – l2
= 7y (5y + 4) – 3(5y + 4)
= (5y + 4)(7y – 3)
∴ The possible expressions for the length and breadth of the rectangle are 7y – 3 and 5y + 4.

GSEB Solutions Class 9 Maths Chapter 2 Polynomials 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:
(i) 3x2 – 12x = 3x (x – 4)
∴ The possible expressions for the dimensions of the cuboid are 3, x and x – 4.

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

GSEB Solutions Class 9 Maths Chapter 2 Polynomials Ex 2.5

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