{"id":24668,"date":"2023-03-29T15:38:27","date_gmt":"2023-03-29T10:08:27","guid":{"rendered":"https:\/\/ncertsolutions.guru\/?p=24668"},"modified":"2023-03-29T17:23:23","modified_gmt":"2023-03-29T11:53:23","slug":"ncert-solutions-for-class-9-maths-chapter-2-ex-2-4","status":"publish","type":"post","link":"https:\/\/ncertsolutions.guru\/ncert-solutions-for-class-9-maths-chapter-2-ex-2-4\/","title":{"rendered":"NCERT Solutions for Class 9 Maths Chapter 2 Polynomials Ex 2.4"},"content":{"rendered":"

These NCERT Solutions for Class 9 Maths<\/a> Chapter 2 Polynomials Ex 2.4 Questions and Answers are prepared by our highly skilled subject experts.<\/p>\n

NCERT Solutions for Class 9 Maths Chapter 2 Polynomials Exercise 2.4<\/h2>\n

Question 1.
\nWhich of the following polynomials has (x + 1) a factor:
\n(i) x3<\/sup> + x2<\/sup> + x + 1
\n(ii) x4<\/sup> + x3<\/sup> + x2<\/sup> + x + 1
\n(iii) x4<\/sup> + 3x3<\/sup> + 3x2<\/sup> + x +1
\n(iv) x3<\/sup> – x2<\/sup> – (2 + \u221a2)x + \u221a2
\nSolution:
\n(i) Here, p(x) = x3<\/sup> + x2<\/sup> + x + 1
\nand the zero of x + 1 is -1.
\nSo, p(-1) = (-1)3<\/sup> + (-1)2<\/sup> + (-1) + 1
\n= -1 + 1 – 1 + 1
\n= 0
\nHere, remainder is zero, therefore x + 1 is the factor of x3<\/sup> + x2<\/sup> + x + 1.<\/p>\n

(ii) Here, p(x) = x4<\/sup> + x3<\/sup> + x2<\/sup> + x + 1
\nand the zero of x + 1 is -1.
\nSo p(-1) = (-1)4<\/sup> + (-1)3<\/sup> + (-1)2<\/sup> + (-1) + 1
\n= 1 – 1 + 1 – 1 + 1
\n= 1
\nHere, remainder is 1, therefore x + 1 is not the factor of x4<\/sup> + x3<\/sup> + x2<\/sup> + x + 1.<\/p>\n

\"NCERT<\/p>\n

(iii) Here, p(x) = x4<\/sup> + 3x3<\/sup> + 3x2<\/sup> + x + 1
\nand the zero of x + 1 is -1
\nSo, p(-1) = (-1)4<\/sup> + 3(-1)3<\/sup> + 3(-1)2<\/sup> + (-1) + 1
\n= 1 – 3 + 3 – 1 + 1
\n= 1
\nHere, remainder is 1, therefore x + 1 is not the factor of x4<\/sup> + 3x3<\/sup> + 3x2<\/sup> + x + 1.<\/p>\n

(iv) Here, p(x) = x3<\/sup> – x2<\/sup> – (2 + \u221a2)x + \u221a2 and the zero of x + 1 is -1
\nSo, p(-1) = (-1)3<\/sup> – (-1)2<\/sup> – (2 + \u221a2) (-1) + \u221a2
\n= -1 – 1 + 2 + \u221a2 + \u221a2
\n= 2\u221a2
\nHere, remainder is 2\u221a2 therefore x + 1 is not the factor of x3<\/sup> – x2<\/sup> – (2 + \u221a2 )x + \u221a2<\/p>\n

Question 2.
\nUse the factor theorem to determine whether g(x) is a factor of p(x) in each of the following cases:
\n(i) p(x) = 2x3<\/sup> + x2<\/sup> – 2x – 1, g(x) = x + 1
\n(ii) p(x) = x3<\/sup> + 3x2<\/sup> + 3x + 1, g(x) = x + 2
\n(iii) pix) = x3<\/sup> – 4x2<\/sup> + x + 6, g(x) = x – 3
\nSolution:
\n(i) Here, p(x) = 2x3<\/sup> + x2<\/sup> – 2x – 1 and the zero of g(x) = x + 1 is -1
\nSo p(-1) = 2(-1)3<\/sup> + (-1)2<\/sup> – 2(-1) – 1
\n= -2 + 1 + 2 – 1
\n= 0
\nHere, remainder is zero, therefore g(x) = x + 1 is the factor of p(x) = 2x3<\/sup> + x2<\/sup> – 2x – 1<\/p>\n

\"NCERT<\/p>\n

(ii) Here, p(x) = x3<\/sup> + 3x2<\/sup> + 3x + 1
\nand the zero of g(x) = x + 2 is -2.
\nSo, p(-2) = (-2)3<\/sup> + 3(-2)2<\/sup> + 3(-2) + 1
\n= -8 + 12 – 6 + 1
\n= -1
\nHere, remainder is -1, therefore g(x) = x + 2 is not the factor of p(x) = x3<\/sup> + 3x2<\/sup> + 3x + 1<\/p>\n

(iii) Here, p(x) = x3<\/sup> – 4x2<\/sup> + x + 6
\nand the zero of g(x) = x – 3 is 3.
\nSo, p(3) = (3)3<\/sup> – 4(3)2<\/sup> + 3 + 6
\n= 27 – 36 + 3 + 6
\n= 0
\nHere, remainder is 0, therefore g(x) = x – 3 is the factor of p(x) = x3<\/sup> – 4x2<\/sup> + x + 6.<\/p>\n

Question 3.
\nFind the value of k if x – 1 is a factor of p(x) in each of the following cases:
\n(i) p(x) = x2<\/sup> + x + k
\n(ii) p(x) = 2x2<\/sup> + kx + \u221a2
\n(iii) p(x) = kx2<\/sup> – \u221a2x + 1
\n(iv) p(x) = kx2<\/sup> – 3x + k
\nSolution:
\n(i) We have give that x – 1 is the factor of p(x) = x2<\/sup> + x + k
\nSo, (1)2<\/sup> + 1 + k = 0
\nor, 1 + 1 + k = 0
\n2 + k = 0
\nk = -2
\nTherefore, the value of k in p(x) = x2<\/sup> + x + k is -2.<\/p>\n

\"NCERT<\/p>\n

(ii) We have given that x – 1 is the factor of P(x) = 2x2<\/sup> + kx + \u221a2
\nP(1) = 0
\nNow, P(1) = 2(1)2<\/sup> + k(1) + \u221a2
\nSo, 0 = 2 + k + \u221a2
\nor, k = -2 – \u221a2
\nor, k = -(2 + \u221a2)
\nTherefore the value of k in p(x) = 2x2<\/sup> + kx + 2 is -(2 + \u221a2)<\/p>\n

(iii) We have given that x – 1 is a factor of
\np(x) = kx2<\/sup> – \u221a2x + 1
\np(1) = 0
\nNow, p(1) = k(1)2<\/sup> – \u221a2(1) + 1
\nSo, k – \u221a2 + 1 = 0
\nk = \u221a2 – 1
\nTherefore, the value of it in p(x) = kx2<\/sup> – \u221a2x + 1 is \u221a2 – 1.<\/p>\n

(iv) We have given that x – 1 is the factor of p(x) = kx2<\/sup> – 3x + k
\nP(1) = 0
\nNow, p(1) = k(1)2<\/sup> – 3(1) + k
\nSo, k – 3 + k = 0
\nor, 2k – 3 = 0
\nk = \\(\\frac{3}{2}\\)
\nTherefore the value of k in p(x) = kx2<\/sup> – 3x + k is \\(\\frac{3}{2}\\)<\/p>\n

\"NCERT<\/p>\n

Question 4.
\nFactorise
\n(i) 12x2<\/sup> – 7x + 1
\n(ii) 2x2<\/sup> + 7x + 3
\n(iii) 6x2<\/sup> + 5x – 6
\n(iv) 3x2<\/sup> – x – 4
\nSolution:
\n(i) we have given 12x2<\/sup> – 7x + 1
\nWe can also write 12x2<\/sup> – 4x – 3x – 1
\nor, 4x(3x – 1) – 1(3x – 1)
\nor, (3x – 1)(4x – 1)<\/p>\n

(ii) We have given 2x2<\/sup> + 7x + 3
\nWe can also write 2x2<\/sup> + 6x + x + 3
\nor, 2x(x + 3) + 1(x + 3)
\nor, (x + 3) (2x + 1)<\/p>\n

\"NCERT<\/p>\n

(iii) We have given 6x2<\/sup> + 5x – 6
\nWe can also write 6x2<\/sup> + 9x – 4x – 6
\nor, 3x(2x + 3) – 2(2x + 3)
\nor, (2x + 3) (3x – 2)<\/p>\n

(iv) We have given 3x2<\/sup> – x – 4
\nWe can also write 3x2<\/sup> – 4x + 3x – 4
\nor, x(3x – 4) + 1(3x – 4)
\nor, (3x – 4)(x + 1)<\/p>\n

Question 5.
\nFactorise:
\n(i) x3<\/sup> – 2x2<\/sup> – x + 2
\n(ii) x3<\/sup> – 3x2<\/sup> – 9x – 5
\n(iii) x3<\/sup> – 13x2<\/sup> + 20
\n(iv) 2y3<\/sup> + y2<\/sup> – 2y – 1
\nSolution:
\n(i) Let p(x) = x3<\/sup> – 2x2<\/sup> – x + 2
\nThe factors of 2 are \u00b11, \u00b12
\nBy trial, we find that p(2) = 0
\nSo, (x – 2) must be the factor of p(x)
\nTherefore, x3<\/sup> – 2x2<\/sup> – x + 2
\n= (x – 2)(x2<\/sup> – 1)
\n= (x – 2)(x + 1)(x – 1)
\n[\u2235 a2<\/sup> – b2<\/sup> = (a + b)(a – b)]
\n\u2234 x3<\/sup> – 2x2<\/sup> – x + 2 = (x – 2)(x – 1)(x + 1)<\/p>\n

\"NCERT<\/p>\n

(ii) Let P(x) = x3<\/sup> – 3x2<\/sup> – 9x – 5
\nThe factors of -5 are \u00b11, \u00b15
\nBy trial, we find that P(-1) = 0
\nSo, (x + 1) is the fact or of p(x)
\nTherefore,
\nx3<\/sup> – 3x2<\/sup> – 9x – 5
\n= (x + 1) (x2<\/sup> – 4x – 5)
\n= (x + 1) [x2<\/sup> – 5x + x – 5]
\n= (x + 1) [x(x – 5) + 1(x – 5)]
\n\u2234 x3<\/sup> – 3x2<\/sup> – 9x – 5 = (x + 1)(x + 1)(x – 5)<\/p>\n

(iii) Let P(x) = x3<\/sup> + 13x2<\/sup> + 32x + 20
\nThe factor of 20 are +1, \u00b12, \u00b14, \u00b15, \u00b110, \u00b120.
\nBy trial, we find that P(-1) = 0
\nSo, (x +1) must be the factor of P(x) Therefore,
\nx3<\/sup> + 12x2<\/sup> + 32x + 20
\n= (x + 1) (x2<\/sup> + 12x + 20)
\n= (x + 1) [x2<\/sup> + 10x + 2x + 20]
\n= (x + 1) [x(x +10) + 2(x + 10)]
\n\u2234 x3<\/sup> + 13x2<\/sup> + 32x + 20 = (x + 1)(x + 10) (x + 2)<\/p>\n

\"NCERT<\/p>\n

(iv) Let P(y) = 2y3<\/sup> + y2<\/sup> – 2y – 1
\nThe factor of -1 are \u00b11
\nBy trial, we find that P(1) = 0
\nSo, (y – 1) must be the factor of 2y3<\/sup> + y2<\/sup> – 2y – 1
\nTherefore,
\n2y3<\/sup> + y2<\/sup> – 2y – 1
\n= (y – 1) (2y2<\/sup> + 3y + 1)
\n= (y – 1) [2y2<\/sup> + 2y + y + 1]
\n= (y – 1) [2y(y + 1) + 1(y + 1)]
\n\u2234 2y3<\/sup> + y2<\/sup> – 2y – 1 = (y – 1)(y + 1)(2y + 1)<\/p>\n","protected":false},"excerpt":{"rendered":"

These NCERT Solutions for Class 9 Maths Chapter 2 Polynomials Ex 2.4 Questions and Answers are prepared by our highly skilled subject experts. NCERT Solutions for Class 9 Maths Chapter 2 Polynomials Exercise 2.4 Question 1. Which of the following polynomials has (x + 1) a factor: (i) x3 + x2 + x + 1 …<\/p>\n

NCERT Solutions for Class 9 Maths Chapter 2 Polynomials Ex 2.4<\/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 2.4 - 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-2-4\/\" \/>\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 2.4 - NCERT Solutions\" \/>\n<meta property=\"og:description\" content=\"These NCERT Solutions for Class 9 Maths Chapter 2 Polynomials Ex 2.4 Questions and Answers are prepared by our highly skilled subject experts. 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NCERT Solutions for Class 9 Maths Chapter 2 Polynomials Exercise 2.4 Question 1. 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