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>\nNCERT 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<\/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<\/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<\/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>\nQuestion 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<\/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<\/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<\/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<\/p>\nQuestion 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<\/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<\/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<\/p>\nQuestion 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<\/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 + 6ab