{"id":24801,"date":"2023-04-11T11:38:04","date_gmt":"2023-04-11T06:08:04","guid":{"rendered":"https:\/\/ncertsolutions.guru\/?p=24801"},"modified":"2023-04-11T17:15:45","modified_gmt":"2023-04-11T11:45:45","slug":"ncert-solutions-for-class-9-maths-chapter-2-ex-2-5","status":"publish","type":"post","link":"https:\/\/ncertsolutions.guru\/ncert-solutions-for-class-9-maths-chapter-2-ex-2-5\/","title":{"rendered":"NCERT Solutions for Class 9 Maths Chapter 2 Polynomials Ex 2.5"},"content":{"rendered":"

These NCERT Solutions for Class 9 Maths<\/a> Chapter 2 Polynomials Ex 2.5 Questions and Answers are prepared by our highly skilled subject experts.<\/p>\n

NCERT Solutions for Class 9 Maths Chapter 2 Polynomials Exercise 2.5<\/h2>\n

Question 1.
\nUse suitable identities to find the following products:
\n(i) (x + 4) (x + 10)
\n(ii) (x + 8) (x – 10)
\n(iii) (3x + 4) (3x – 5)
\n(iv) (y2<\/sup> + \\(\\frac{3}{2}\\)) (y2<\/sup> – \\(\\frac{3}{2}\\))
\n(v) (3 – 2x) (3 + 2x)
\nSolution:
\n(i) We have given, (x + 4)(x + 10)
\nWe know that
\n(x + a) (x + b) = x2<\/sup> + (a + b)x + ab
\nTherefore,
\n(x + 4) (x + 10)
\n= x2<\/sup> + (4 + 10)x + 4 \u00d7 10
\n= x2<\/sup> + 14x + 40<\/p>\n

(ii) We have given (x + 8) (x – 10)
\nor, (x + 8) (x + (-10))
\nWe know that,
\n(x + a) (x + b) = x2<\/sup> + (a + b)x + ab
\n(x + 8) (x + (-10))
\n= x2<\/sup> + (8 + (-10))x + 8 \u00d7 (-10)
\n= x2<\/sup> – 2x – 80<\/p>\n

\"NCERT<\/p>\n

(iii) We have given, (3x + 4)(3x – 5)
\nor, (3x + 4) (3x + (- 5))
\nWe know that
\n(x + a) (x + b) = x2<\/sup> + (a + b)x + ab
\n(3x + 4)(3x – 5)
\n= (3x)2<\/sup> + (4 + (-5))3x + 4 \u00d7 (-5)
\n= 9x2<\/sup> – 3x – 20<\/p>\n

(iv) We have given (y2<\/sup> + \\(\\frac{3}{2}\\)) (y2<\/sup> – \\(\\frac{3}{2}\\))
\nWe know that (x + y)(x – y) = x2<\/sup> – y2<\/sup>
\n\\(\\left(y^{2}+\\frac{3}{2}\\right)\\left(y^{2}-\\frac{3}{2}\\right)=\\left(y^{2}\\right)^{2}-\\left(\\frac{3}{2}\\right)^{2}\\)
\n= \\(y^{4}-\\frac{9}{4}\\)<\/p>\n

(v) We have given (3 – 2x)(3 + 2x)
\nWe know that (x – y)(x + y) = x2<\/sup> – y2<\/sup>
\n(3 – 2x)(3 + 2x)
\n= (3)2<\/sup> – (2x)2<\/sup>
\n= 9 – 4x2<\/sup><\/p>\n

\"NCERT<\/p>\n

Question 2.
\nEvaluate the following products without multiplying directly.
\n(i) 103 \u00d7 107
\n(ii) 95 \u00d7 96
\n(iii) 104 \u00d7 96
\nSolution:
\n(i) We have given, 103 \u00d7 107
\nWe can also write
\n(100 + 3) \u00d7 (100 + 7)
\nWe know that
\n(x + a) (x + b) = x2<\/sup> + (a + b)x + ab
\n(100 + 3) (100 + 7)
\n= (100)2<\/sup> + (3 + 7) \u00d7 100 + 3 \u00d7 7
\n= 10000 + 1000 + 21
\n= 11021<\/p>\n

(ii) We have given, 95 \u00d7 96
\nWe can also write (90 + 5) \u00d7 (90 + 6)
\nWe know that
\n(x + a) (x + b) = x2<\/sup> + (a + b)x + ab
\n(90 + 5) \u00d7 (90 + 6)
\n= (90)2<\/sup> + (5 + 6) \u00d7 90 + 5 \u00d7 6
\n= 8100 + 990 + 30
\n= 9120<\/p>\n

\"NCERT<\/p>\n

(iii) We have given, 104 \u00d7 96
\nWe can also write (100 + 4) \u00d7 (100 – 4)
\nWe know that (x + y)(x – y) = x2<\/sup> – y2<\/sup>
\n(100 + 4)(100 – 4)
\n= (100)2<\/sup> – (4)2<\/sup>
\n= 10000 – 16
\n= 9984<\/p>\n

Question 3.
\nFactorise the following using appropriate identities:
\n(i) 9x2<\/sup> + 6xy + y2<\/sup>
\n(ii) 4y2<\/sup> – 4y + 1
\n(iii) \\(x^{2}-\\frac{y^{2}}{100}\\)
\nSolution:
\n(i) We have given, 9x2<\/sup> + 6xy + y2<\/sup>
\nWe can also write (3x)2<\/sup> + 2 . 3x . y + (y)2<\/sup>
\nWe know that
\na2<\/sup> + 2ab + b2<\/sup> = (a + b)2<\/sup>
\n(3x)2<\/sup> + 2(3x)(y) + (y)2<\/sup>
\n= (3x + y)2<\/sup>
\n= (3x + y)(3x + y)<\/p>\n

(ii) We have, 4y2<\/sup> – 4y + 1
\nWe can also write (2y)2<\/sup> – 2 . 2y . 1 + (1)2<\/sup>
\nWe know that
\nx2<\/sup> – 2xy + y2<\/sup> = (x – y)2<\/sup>
\n(2y)2<\/sup> – 2 . 2y . 1 + (1)2<\/sup>
\n= (2y – 1)2<\/sup>
\n= (2y – 1)(2y – 1)<\/p>\n

(iii) We have given, \\(x^{2}-\\frac{y^{2}}{100}\\)
\nWe can also write, \\((x)^{2}-\\left(\\frac{y}{100}\\right)^{2}\\)
\nWe know that
\na2<\/sup> – b2<\/sup> = (a + b)(a – b)
\n\\((x)^{2}-\\left(\\frac{y}{100}\\right)^{2}=\\left(x+\\frac{y}{10}\\right)\\left(x-\\frac{y}{10}\\right)\\)<\/p>\n

\"NCERT<\/p>\n

Question 4.
\nExpand each of the following using suitable identities:
\n(i) (x + 2y + 4z)2<\/sup>
\n(ii) (2x – y + z)2<\/sup>
\n(iii) (-2x + 3y + 2z)2<\/sup>
\n(iv) (3a – 7b – c)2<\/sup>
\n(iv) (-2x + 5y – 3z)2<\/sup>
\n(v) \\(\\left[\\frac{1}{4} a-\\frac{1}{2} b+1\\right]^{2}\\)
\nSolution:
\n(i) We have given that, (x + 2y + 4z)2<\/sup>
\nWe know that
\n(a + b + c)2<\/sup> = a2<\/sup> + b2<\/sup> + c2<\/sup> + 2ab<\/span>\u00a0+ 2bc + 2ca
\nTherefore,
\n(x + 2y + 4z)2<\/sup>
\n= (x)2<\/sup> + (2y)2<\/sup> + (4z)2<\/sup> + 2 . x . 2y + 2 . 2y . 4z + 2 . 4z . x
\n= x2<\/sup> + 4y2<\/sup> + 16z2<\/sup> + 4xy + 16yz + 8zx<\/p>\n

(ii) We have given that (2x – y + z)2<\/sup>
\nWe can also write, (2x + (-y) + z)2<\/sup>
\nNow, we know that
\n(a + b + c)2<\/sup> = a2<\/sup> + b2<\/sup> + c2<\/sup> + 2ab + 2bc + 2ca
\nTherefore,
\n(2x + (-y) + z)2<\/sup>
\n= (2x)2<\/sup> + (-y)2<\/sup> + (z)2<\/sup> + 2 . 2x . (-y) + 2 . (-y) . z + 2 . z . 2x
\n= 4x2<\/sup> + y2<\/sup> + z2<\/sup> – 4xy – 2yz + 4zx<\/p>\n

\"NCERT<\/p>\n

(iii) We have given that, (-2x + 3y + 2z)2<\/sup>
\nWe can also write, [(-2x) + 3y + 2z]2<\/sup>
\nNow, we know that
\n(a + b + c)2<\/sup> = a2<\/sup> + b2<\/sup> + c2<\/sup> + 2ab + 2bc + 2ca
\nTherefore,
\n((-2x) + 3y + 2z)2<\/sup> = (-2x)2<\/sup> + (3y)2<\/sup> + (2z)2<\/sup> + 2 . (-2x) . 3y + 2 . 3y . 2z + 2 . 2z . (-2x)
\n= 4x2<\/sup> + 9y2<\/sup> + 4z2<\/sup> – 12xy + 12yz – 8zx<\/p>\n

(iv) We have given that, (3a – 7b – c)2<\/sup>
\nWe can also write (3a + (-7b) + (-c)2<\/sup>
\nNow, we know that
\n(a + b + c)2<\/sup> = a2<\/sup> + b2<\/sup> + c2<\/sup> + 2ab + 2bc + 2ca
\nTherefore,
\n(3a + (-7b) + (-c))2<\/sup>
\n= (3a)2<\/sup> + (-7b)2<\/sup> + (-c)2<\/sup> + 2 . 3a . (-7b) + 2 . (-7b) . (-c) + 2 . (-c) . 3a
\n= 9a2<\/sup> + 49b2<\/sup> + c2<\/sup> – 42ab + 14bc – 6ca<\/p>\n

(v) We have given that, (-2x + 5y – 3z)2<\/sup>
\nWe can also write [(-2x) + 5y + (-3z)]2<\/sup>
\nNow, we know that
\n(a + b + c)2<\/sup> = a2<\/sup> + b2<\/sup> + c2<\/sup> + 2ab + 2bc + 2ca
\nTherefore,
\n[(-2x) + 5y + (-3z)]2<\/sup>
\n= (-2x)2<\/sup> + (5y)2<\/sup> + (-3z)2<\/sup> + 2 . (-2x). 5y + 2 . 5y . (-3z) + 2 . (-3z) . (-2x)
\n= 4x2<\/sup> + 25y2<\/sup> + 9z2<\/sup> – 20xy – 30yz + 12zx<\/p>\n

\"NCERT<\/p>\n

(vi) We have given that, \\(\\left[\\frac{1}{4} a-\\frac{1}{2} b+1\\right]^{2}\\)
\nWe can also write \\(\\left[\\frac{1}{4} a\\left(-\\frac{1}{2} b\\right)+1\\right]^{2}\\)
\nNow, we know that
\n(a + b + c)2<\/sup> = a2<\/sup> + b2<\/sup> + c2<\/sup> + 2ab + 2bc + 2ca
\nTherefore,
\n\"NCERT<\/p>\n

Question 5.
\nFactorise:
\n(i) 4x2<\/sup> + 9y2<\/sup> + 16z2<\/sup> + 12xy – 24yz – 16xz
\n(ii) 2x2<\/sup> + y2<\/sup> + 8z2<\/sup> – 2\u221a2xy + 4\u221a2yz – 8xz
\nSolution:
\n(i) We have given
\n4x2<\/sup> + 9y2<\/sup> + 16z2<\/sup> + 12xy – 24yz – 16xz
\nWe can also write
\n(2x)2<\/sup> + (3y)2<\/sup> + (-4z)2<\/sup> + 2 . 2x . 3y + 2 . 3y . (-4z) + 2 . (-4z) . 2x
\nNow we know that
\na2<\/sup> + b2<\/sup> + c2<\/sup> + 2ab + 2bc + 2ca = (a + b + c)2<\/sup>
\nTherefore,
\n(2x)2<\/sup> + (3y)2<\/sup> + (-4z)2<\/sup> + 2 . 2x . 3y + 2 . 3y (-4z) + 2 . (-4z). 2x
\n= (2x + 3y + (-4z))2<\/sup>
\n= (2x + 3y – 4z)2<\/sup>
\n= (2x + 3y – 4z) (2x + 3y – 4z)<\/p>\n

(ii) We have given that
\n2x2<\/sup> + y2<\/sup> + 8z2<\/sup> – 2\u221a2xy + 4\u221a2yz – 8xz
\nWe can also write
\n(-\u221a2x)2<\/sup> + (y)2<\/sup> + (2\u221a2z)2<\/sup> + 2 . (-\u221a2x) . y + 2 . y . (2\u221a2z) + 2 . (2\u221a2 z) . (-\u221a2x)
\nNow, we know that
\na2<\/sup> + b2<\/sup> + c2<\/sup> + 2ab + 2bc + 2ca = (a + b + c)2<\/sup>
\nTherefore,
\n(-\u221a2x)2<\/sup> + (y)2<\/sup> + (2\u221a2z)2<\/sup> + 2 . (-\u221a2x) . y + 2 . y . (2\u221a2z) + 2 . (2\u221a2z) . (-\u221a2x)
\n= ((-\u221a2x) + y + 2\u221a2z)2<\/sup>
\n= (-\u221a2x + y + 2\u221a2z) (-\u221a2x + y + 2\u221a2z)<\/p>\n

\"NCERT<\/p>\n

Question 6.
\nWrite the following cubes in expanded front:
\n(i) (2x + 1)3<\/sup>
\n(ii) (2a – 3b)3<\/sup>
\n(iii) \\(\\left[\\frac{3}{2} x+1\\right]^{3}\\)
\n(iv) \\(\\left[x-\\frac{2}{3} y\\right]\\)
\nSolution:
\n(i) We have given that, (2x + 1)3<\/sup>
\nWe know that
\n(a + b)3<\/sup> = a3<\/sup> + b3<\/sup> + 3ab(a + b)
\nTherefore,
\n(2x + 1)3<\/sup> = (2x)3<\/sup> + (1)3<\/sup> + 3 . (2x)(1)(2x + 1)
\n= 8x3<\/sup> + 1 + 6x(2x + 1)
\n= 8x3<\/sup> + 1 + 12x3<\/sup> + 6x
\n= 8x3<\/sup> + 12x3<\/sup> + 6x + 1<\/p>\n

(ii) We have given, (2a – 3b)3<\/sup>
\nWe know that (a – b)3<\/sup> = a3<\/sup> – b3<\/sup> – 3ab(a – b)
\n(2a – 3b)3<\/sup>
\n= 8a3<\/sup> – 27b3<\/sup> – 18ab(2a – 3b)
\n= 8a3<\/sup> – 27b3<\/sup> – 36a2<\/sup>b + 54ab2<\/sup><\/p>\n

(iii) We have given that, \\(\\left[\\frac{3}{2} x+1\\right]^{3}\\)
\nWe know that
\n(a + b)3<\/sup> = a3<\/sup> + b3<\/sup> + 3ab(a + b)
\nTherefore,
\n\"NCERT<\/p>\n

(iv) We have given that, \\(\\left[x-\\frac{2}{3} y\\right]\\)
\nWe know that
\n(a – b)3<\/sup> = a3<\/sup> – b3<\/sup> – 3ab(a – b)
\nTherefore,
\n\"NCERT<\/p>\n

Question 7.
\nEvaluate the following using suitable identities:
\n(i) (99)3<\/sup>
\n(ii) (102)3<\/sup>
\n(iii) (998)3<\/sup>
\nSolution:
\n(i) We have given, (99)3<\/sup>
\nWe can also write, (100 – 1)3<\/sup>
\nWe know that (a – b)3<\/sup> = a3<\/sup> – b3<\/sup> – 3ab(a – b)
\n(100 – 1)3<\/sup> = (100)3<\/sup> – (1)3<\/sup> – 3 . 100 . 1 (100 – 1)
\n= 10,00,000 – 1 – 30000 + 300
\n= 9,70,299<\/p>\n

(ii) We have given that, (102)3<\/sup>
\nWe can also write (100 + 2)3<\/sup>
\nNow, we know that
\n(a + b)3<\/sup> = a3<\/sup> + b3<\/sup> + 3ab(a + b)
\n(100 + 2)3<\/sup> = (100)3 <\/sup>+ (2)3<\/sup>+ 3 . 100 . 2 (100 + 2)
\n= 10,00,000 + 8 + 600(100 + 2)
\n= 10,00,000 + 8 + 60000 + 1200
\n= 10,61,208<\/p>\n

(iii) We have given that (998)3<\/sup>
\nWe can also write (1000 – 2)3<\/sup>
\nNow, we know that (a – b)3<\/sup> = a3<\/sup> – b3<\/sup> – 3ab(a – b)
\n(1000 – 2)3<\/sup> = (1000)3<\/sup> – (2)3<\/sup> – 3 . 1000 . 2(1000 – 2)
\n= 1,00,00,00,000 – 8 – 6000 (100 – 2)
\n= 99,40,11,992<\/p>\n

\"NCERT<\/p>\n

Question 8.
\nFactarise each of the following:
\n(i) 8a3<\/sup> + b3<\/sup> + 12a2<\/sup>b + 6ab2<\/sup>
\n(ii) 8a3<\/sup> – b3<\/sup> – 12a2<\/sup>b + 6ab2<\/sup>
\n(iii) 27 – 125a3<\/sup> – 135a + 225a3<\/sup>
\n(iv) 64a3<\/sup> – 27b3<\/sup> -144a2<\/sup>b + 108ab2<\/sup>
\n(v) \\(27 \\mathrm{P}^{3}-\\frac{1}{216}-\\frac{9}{2} \\mathrm{P}^{2}+\\frac{1}{4} \\mathrm{P}\\)
\nSolution:
\n(i) We have given that
\n8a3<\/sup> + b3<\/sup> + 12a2<\/sup>b + 6ab2<\/sup>
\nWe can also write
\n(2a)3<\/sup> + (b)3<\/sup> + 6ab(2a + b)
\nor, (2a)3<\/sup> + b3<\/sup> + 3 . 2a . b(2a + b)
\nWe know that
\na3<\/sup> + b3<\/sup> + 3ab(a + b) = (a + b)3<\/sup>
\n(2a)3<\/sup> + (b)3<\/sup> + 3 . 2a . b (2a + b) = (2a + b)3<\/sup><\/p>\n

(ii) We have given that
\n8a3<\/sup> – b3<\/sup>– 12a2<\/sup>b + 6ab2<\/sup>
\nWe can also write
\n(2a)3<\/sup> – (b)3<\/sup> – 6ab(2a – b)
\nor, (2a)3<\/sup> – (b)3<\/sup> – 3 . 2a . b(2a – b)
\nWe know that
\na3<\/sup> – b3<\/sup> – 3ab(a – b) = (a – b)3<\/sup>
\n(2a)3<\/sup> – (b)3<\/sup> – 3 . 2a . b(2a – b) = (2a – b)3<\/sup><\/p>\n

\"NCERT<\/p>\n

(iii) We have given that
\n27 – 125a3<\/sup> – 135a + 225a2<\/sup>
\nWe can also write (3)3<\/sup> – (5a)3<\/sup> – 45a(3 – 5a)
\nor, (3)3<\/sup> – (5a)3<\/sup> – 3 . 3 . 5a(3 – 5a)
\nNow, we know that,
\na3<\/sup> – b3<\/sup> – 3ab(a – b) = (a – b)3<\/sup>
\n(3)3<\/sup> – (5a)3<\/sup> – 3 . 3 . 5a(3 – 5a)
\n= (3 – 5a)3<\/sup>
\n= (3 – 5a) (3 – 5a) (3 – 5a)<\/p>\n

(iv) We have given that
\n64a3<\/sup> – 27b3<\/sup> – 144a2<\/sup>b + 108ab2<\/sup>
\nWe can also write
\n(4a)3<\/sup> – (3b)3<\/sup> – 36ab(4a – 3b)
\nor, (4a)3<\/sup> – (3b)3<\/sup> – 3 . 4a . 3b(4a – 3b)
\nNow, we know that
\na3<\/sup> – b3<\/sup> – 3ab(a – b) = (a – b)3<\/sup>
\n(4a)3<\/sup> – (3b)3<\/sup> – 3 . 4a . 3b(4a – 3b)
\n= (4a – 3b)3<\/sup><\/p>\n

(v) We have given that
\n\\(27 \\mathrm{P}^{3}-\\frac{1}{216}-\\frac{9}{2} \\mathrm{P}^{2}+\\frac{1}{4} \\mathbf{P}\\)
\nWe can also write,
\n\\((3 P)^{3}-\\left(\\frac{1}{6}\\right)^{3}-3.3 P \\cdot \\frac{1}{6}\\left(3 P-\\frac{1}{6}\\right)\\)
\nWe know that
\na3<\/sup> – b3<\/sup> – 3ab(a – b) = (a – b)3<\/sup>
\n\\((3 P)^{3}-\\left(\\frac{1}{6}\\right)^{3}-3.3 P \\cdot \\frac{1}{6}\\left(3 P-\\frac{1}{6}\\right)=\\left(3 P-\\frac{1}{6}\\right)^{3}\\)<\/p>\n

\"NCERT<\/p>\n

Question 9.
\nVerify
\n(i) x3<\/sup> + y3<\/sup> = (x + y) (x2<\/sup> – xy + y2<\/sup>)
\n(ii) x3<\/sup> – y3<\/sup> = (x – y) (x2<\/sup> + xy + y2<\/sup>)
\nSolution:
\n(i) R.H.S.
\n(x + y) (x2<\/sup> – xy + y2<\/sup>)
\nBy actual multiplication we have
\nx3<\/sup> – x2<\/sup>y + xy2<\/sup> + x2<\/sup>y – xy2<\/sup> + y3<\/sup>
\n= x3<\/sup> + y3<\/sup>
\n= L.H.S<\/p>\n

(ii) R.H.S.
\n(x – y)(x2<\/sup> + xy + y2<\/sup>)
\nBy actual multiplication we have
\nx3<\/sup> + x2<\/sup>y + xy2<\/sup> – x2<\/sup>y – xy2<\/sup> – y3<\/sup>
\n= x3<\/sup> – y3<\/sup>
\n= L.H.S.<\/p>\n

Question 10.
\nFactorise each of the following:
\n(i) 27y3<\/sup> + 125z3<\/sup>
\n(ii) 64m3<\/sup> – 343n3<\/sup>
\nSolution:
\n(i) We have given, 27y3<\/sup> + 125z3<\/sup>
\nWe can also write, (3y)3<\/sup> + (5z)3<\/sup>
\nWe know that
\na3<\/sup> + b3<\/sup> = (a + b)(a2<\/sup> – ab + b2<\/sup>)
\nTherefore,
\n(3y)3<\/sup> + (5z)3<\/sup> = (3y + 5z) ((3y)2<\/sup> – (3y).(5z) + (5z)2<\/sup>)
\n= (3y + 5z) (9y2<\/sup> – 15yz + 25z2<\/sup>)<\/p>\n

(ii) We have given, 64m3<\/sup> – 343n3<\/sup>
\nWe can also write (4m)3<\/sup> – (7n)3<\/sup>
\nWe know that,
\na3<\/sup> – b3<\/sup> = (a – b)(a2<\/sup> + ab + b2<\/sup>)
\nTherefore,
\n(4m)3<\/sup> – (7n)3<\/sup> = (4m – 7n)((4m)2<\/sup> + (4m)(7n) + (7n)2<\/sup>)
\n= (4m – 7n)(16m2<\/sup> + 28mn + 49n2<\/sup>)<\/p>\n

\"NCERT<\/p>\n

Question 11.
\nFactorise: 27x3<\/sup> + y3<\/sup> + z3<\/sup> – 9xyz
\nSolution:
\nWe have given,
\n27x3<\/sup> + y3<\/sup> + z3<\/sup> – 9xyz
\nWe can also write,
\n(3x)3<\/sup> + (y)3<\/sup> + (z)3<\/sup> – 3 . (3x) . (y) . (z)
\nWe know that
\na3<\/sup> + b3<\/sup> + c3<\/sup> – 3abc = (a + b + c) (a2<\/sup> + b2<\/sup> + c2<\/sup> – ab – bc – ca)
\nTherefore,
\n(3x)3<\/sup> + (y)3<\/sup> + (z)3<\/sup> – 3 . (3x) (y) (z)
\n= (3x + y + z) [(3x)2<\/sup> + (y)2<\/sup> + (z)2<\/sup> – (3x)(y) -(y)(z) – (z)(3x)]
\n= (3x + y + z) (9x2<\/sup> + y2<\/sup> + z2<\/sup> – 3xy – yz – 3zx)<\/p>\n

Question 12.
\nVerify that x3<\/sup> + y3<\/sup> + z3<\/sup> – 3xyz = 1\/2 (x + y + z) [(x – y)2<\/sup> + (y – z)2<\/sup> + (z – x)2<\/sup>]
\nSolution:
\nWe have given L.H.S. is x3<\/sup> + y3<\/sup> + z3<\/sup> – 3xyz
\nWe know that
\nx3<\/sup> + y3<\/sup> + z3<\/sup> – 3xyz = (x + y + z)(x2<\/sup> + y2<\/sup> + z2<\/sup> – xy – yz – zx)
\nMultiply and divide 2 we get
\n= (x + y + z)2<\/sup> \u00d7 \\(\\frac{1}{2}\\) (x2<\/sup> + y2<\/sup> + z2<\/sup> – xy – yz – zx)
\n= \\(\\frac{1}{2}\\) (x + y + z)(2x2<\/sup> + 2y2<\/sup> + 2z2<\/sup> – 2xy – 2yz – 2zx)
\n= \\(\\frac{1}{2}\\) (x + y + z) (x2<\/sup> + y2<\/sup> – 2xy + y2<\/sup> + z2<\/sup> – 2yz + z2<\/sup> + x2<\/sup> – 2zx)
\n= \\(\\frac{1}{2}\\) (x + y + z)((x – y)2<\/sup> + (y – z)2<\/sup> + (z – x)2<\/sup>)
\nHence verified.<\/p>\n

\"NCERT<\/p>\n

Question 13.
\nIf x + y + z = 0, show that x3<\/sup> + y3<\/sup> + z3<\/sup> = 3xyz.
\nSolution:
\nWe know that,
\nx3<\/sup> + y3<\/sup> + z3<\/sup> – 3xyz = (x + y + z) (x2<\/sup> + y2<\/sup> + z2<\/sup> – xy – yz – zx)
\nHere, we have given x + y + z = 0
\nx3<\/sup> + y3<\/sup> + z3<\/sup> – 3xyz = 0 \u00d7 (x2<\/sup> + y2<\/sup> + z2<\/sup> – xy – yz – zx)
\nor, x3<\/sup> + y3<\/sup> + z3<\/sup> – 3xyz = 0
\nx3<\/sup> + y3<\/sup> + z3<\/sup> = 3xyz<\/p>\n

Question 14.
\nWithout actually calculating the cubes, find the value of each of the following:
\n(i) (-12)3<\/sup> + (7)3<\/sup> + (5)3<\/sup>
\n(ii) (28)3<\/sup> + (-15)3<\/sup> + (-13)3<\/sup>
\nSolution:
\n(i) We have given that (-12)3<\/sup> + (7)3<\/sup> + (5)3<\/sup>
\nFirst we check the value of x + y + z
\nWe have -12 + 7 + 5 = 0
\nWe know that if x + y + z = 0 then x3<\/sup> + y3<\/sup> + z3<\/sup> = 3xyz
\n(-12)3<\/sup> + (7)3<\/sup> + (5)3<\/sup> = 3 \u00d7 (-12) \u00d7 7 \u00d7 5 = -1260<\/p>\n

(ii) We have given that
\n(28)3<\/sup> + (-15)3<\/sup> + (-13)3<\/sup>
\nFirst we check the value of x + y + z
\nWe have 28 + (-15) + (-13) = 0
\nWe know that if x + y + z = 0 then,
\nx3<\/sup> + y3<\/sup> + z3<\/sup> = 3xyz
\n(28)3<\/sup> + (-15)3<\/sup> + (-13)3<\/sup>
\n= 3 \u00d7 28 \u00d7 (-15) \u00d7 (-13)
\n= 16,380<\/p>\n

\"NCERT<\/p>\n

Question 15.
\nGive possible expressions of the length and breadth of each of the following rectangles in which their areas are given.
\n(i) Area: 25a2<\/sup> – 35a + 12
\n(ii) Area: 35y2<\/sup> + 13y – 12
\nSolution:
\n(i) We know that
\nArea of rectangle = length \u00d7 breadth …. (i)
\nBut, we have given that
\nArea of rectangle = 25a2<\/sup> – 35a + 12 …. (ii)
\nfrom equation (i) and (ii)
\nl \u00d7 b = 25a2<\/sup> – 35a + 12
\n= 25a2<\/sup> – 20a – 15a + 12
\n= 5a(5a – 4) – 3(5a – 4)
\nl \u00d7 b = (5a – 4)(5a – 3)
\nlength (l) = (5a – 4) and breadth (b) = (5a – 3)<\/p>\n

(ii) We know that
\nArea of rectangle = length \u00d7 breadth ……. (i)
\nBut we have given that
\nArea of rectangle = 35y2<\/sup> + 13y – 12 ……. (ii)
\nFrom equation (i) and (ii)
\nlength \u00d7 breadth = 35y2<\/sup> + 13y – 12
\n= 35y2<\/sup> + 28y – 15y – 12
\n= 7y(5y + 4) – 3(5y + 4)
\nor, length \u00d7 breadth = (5y + 4) (7y – 3)
\nTherefore,
\nlength = (5y + 4)
\nbreadth = (7y – 3)<\/p>\n

\"NCERT<\/p>\n

Question 16.
\nWhat are the possible expression for the dimensions of the cuboids whose volumes are given below:
\n(i) volume = 3x2<\/sup> – 12x
\n(ii) volume = 12ky2<\/sup> + 8ky – 20k
\nSolution:
\n(i) We know that
\nVolume of cuboid = length \u00d7 breadth \u00d7 height ……. (i)
\nBut we have given that
\nVolume of cuboid = 3x2<\/sup> – 12x …… (ii)
\nFrom equation (i) and (ii)
\nlength \u00d7 breadth \u00d7 height = 3x2<\/sup> – 12x = 3x(x – 4)
\nor, length \u00d7 breadth \u00d7 height = 3 \u00d7 x \u00d7 (x – 4)
\nlength of cuboid = 3
\nbreadth of cuboid = x
\nheight of cuboid = (x – 4)<\/p>\n

(ii) We know that,
\nVolume of cuboid = length \u00d7 breadth \u00d7 height …….. (i)
\nBut, we have given that Volume of cuboid = 12ky2<\/sup> + 8ky – 20k ……. (ii)
\nFrom equation (i) and (ii)
\nlength \u00d7 breadth \u00d7 height
\n= 12ky2<\/sup> + 8ky – 20k
\n= 4k (3y2<\/sup> + 2y – 5)
\n= 4k (3y2<\/sup> + 5y – 3y – 5)
\n= 4k[y(3y + 5) – 1(3y + 5)]
\nor, length \u00d7 breadth \u00d7 height = 4k(3y + 5)(y – 1)
\nlength of cuboid = 4k
\nbreadth of cuboid = (3y + 5)
\nand height of cuboid = (y – 1)<\/p>\n","protected":false},"excerpt":{"rendered":"

These NCERT Solutions for Class 9 Maths Chapter 2 Polynomials Ex 2.5 Questions and Answers are prepared by our highly skilled subject experts. NCERT Solutions for Class 9 Maths Chapter 2 Polynomials Exercise 2.5 Question 1. Use suitable identities to find the following products: (i) (x + 4) (x + 10) (ii) (x + 8) …<\/p>\n

NCERT Solutions for Class 9 Maths Chapter 2 Polynomials Ex 2.5<\/span> Read More »<\/a><\/p>\n","protected":false},"author":9,"featured_media":0,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"spay_email":""},"categories":[6],"tags":[],"yoast_head":"\nNCERT Solutions for Class 9 Maths Chapter 2 Polynomials Ex 2.5 - NCERT Solutions<\/title>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/ncertsolutions.guru\/ncert-solutions-for-class-9-maths-chapter-2-ex-2-5\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"NCERT Solutions for Class 9 Maths Chapter 2 Polynomials Ex 2.5 - NCERT Solutions\" \/>\n<meta property=\"og:description\" content=\"These NCERT Solutions for Class 9 Maths Chapter 2 Polynomials Ex 2.5 Questions and Answers are prepared by our highly skilled subject experts. 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