{"id":24648,"date":"2021-06-19T16:32:28","date_gmt":"2021-06-19T11:02:28","guid":{"rendered":"https:\/\/ncertsolutions.guru\/?p=24648"},"modified":"2021-06-19T16:37:36","modified_gmt":"2021-06-19T11:07:36","slug":"ncert-solutions-for-class-9-maths-chapter-2-ex-2-2","status":"publish","type":"post","link":"https:\/\/ncertsolutions.guru\/ncert-solutions-for-class-9-maths-chapter-2-ex-2-2\/","title":{"rendered":"NCERT Solutions for Class 9 Maths Chapter 2 Polynomials Ex 2.2"},"content":{"rendered":"

These NCERT Solutions for Class 9 Maths<\/a> Chapter 2 Polynomials Ex 2.2 Questions and Answers are prepared by our highly skilled subject experts.<\/p>\n

NCERT Solutions for Class 9 Maths Chapter 2 Polynomials Exercise 2.2<\/h2>\n

Question 1.
\nFind the value of the polynomial 5x – 4x2<\/sup> + 3 at
\n(i) x = 0
\n(ii) x = -1
\n(iii) x = 2
\nSolution:
\n(i) Let P(x) = 5x – 4x2<\/sup> + 3
\nTherefore, P(0) = 5(0) – 4(0)2<\/sup> + 3
\n= 0 – 0 + 3
\n= 3
\nSo, the value of P(x) at x = 0 is 3.<\/p>\n

(ii) Let P(x) = 5x – 4x2<\/sup> + 3
\nTherefore, P(-1) = 5(-1) – 4(-1)2<\/sup> + 3
\n= -5 – 4 + 3
\n= -6
\nSo, the value of P(x) at x = -1 is -6.<\/p>\n

(iii) Let P(x) = 5x – 4x2<\/sup> + 3
\nTherefore, P(2) = 5(2) – 4(2)2<\/sup> + 3
\n= 10 – 16 + 3
\n= -3
\nSo, the value of P(x) at x = 2 is -3.<\/p>\n

\"NCERT<\/p>\n

Question 2.
\nFind p(0), p(1) and p(2) for each of the following polynomials:
\n(i) p(y) = y2<\/sup> – y + 1
\n(ii) p(t) = 2 + t + 2t2<\/sup> – t3<\/sup>
\n(iii) p(x) = x3<\/sup>
\n(iv) p(x) = (x – 1) (x + 1)
\nSolution:
\n(i) We have given
\np(y) = y2<\/sup> – y + 1
\nTherefore, the value of polynomial p(y) at y = 0 is
\np(0) = 02<\/sup> – 0 + 1 = 1
\nAgain, the value of polynomial p(y) at y = 1 is
\np(1) = 12<\/sup> – 1 + 1 = 1
\nAgain, the value of polynomial p(y) at y = 2 is
\np(2) = 22<\/sup> – 2 + 1 = 3<\/p>\n

(ii) We have given,
\np(t) = 2 + t + 2t2<\/sup> – t3<\/sup>
\nTherefore, the value of polynomial p(t) at t = 0 is
\np(0) = 2 + 0 + 2(0)2<\/sup> – (0)3<\/sup> = 2
\nAgain, the value of polynomial p(t) at t = 1 is
\np(1) = 2 + 1 + 2(1)2<\/sup> – (1)3<\/sup> = 4
\nAgain, the value of polynomial p(t) at t = 2 is
\np(2) = 2 + 2 + 2(2)2<\/sup> – (2)3<\/sup> = 4<\/p>\n

(iii) We have given,
\nP(x) = x3<\/sup>,
\nTherefore, the value of polynomial p(x) at x = 0 is
\np(0) = (0)3<\/sup> = 0
\nAgain the value of polynomial p(x) at x = 1 is
\np(1) = (1)3<\/sup> = 1
\nAgain, the value of polynomial p(x) at x = 2 is
\nP(2) = (2)3<\/sup> = 8<\/p>\n

(iv) We have given
\np(x) = (x – 1)(x + 1)
\nTherefore, the value of polynomial p(x) at x = 0 is
\nP(0) = (0 – 1)(0 + 1)
\n= (-1) \u00d7 (+1)
\n= -1
\nAgain, the value of polynomial p(x) at x = 1 is
\np(1) = (1 – 1) \u00d7 (1 + 1)
\n= 0 \u00d7 2
\n= 0
\nAgain, the value of polynomial p(x), at x = 2 is
\np(2) = (2 – 1)(2 + 1)
\n= 1 \u00d7 3
\n= 3<\/p>\n

\"NCERT<\/p>\n

Question 3.
\nVerify whether the following are zeros of the polynomial indicated against them:
\n(i) p(x) = 3x + 1, x = \\(-\\frac{1}{3}\\)
\n(ii) p(x) = 5x – p, x = \\(\\frac{4}{5}\\)
\n(iii) p(x) = x2<\/sup> – 1, x = 1, -1
\n(iv) p(x) = (x + 1) (x – 2), x = -1, 2
\n(v) p(x) = x2<\/sup>, x = 0
\n(vi) p(x) = 1x + m, x = \\(-\\frac{m}{1}\\)
\n(vii) p(x) = 3x2<\/sup> – 1, x = \\(-\\frac{1}{\\sqrt{3}}, \\frac{2}{\\sqrt{3}}\\)
\n(viii) p(x) = 2x + 1, x = \\(\\frac{1}{2}\\)
\nSolution:
\n(i) We have given that, p(x) = 3x + 1.
\nTherefore, the value of polynomial p(x) at x = \\(-\\frac{1}{3}\\) is
\n\\(P\\left(-\\frac{1}{3}\\right)=3 \\times \\frac{(-1)}{3}+1\\) = 0
\nyes x = \\(-\\frac{1}{3}\\) is the zero of polynomial p(x).<\/p>\n

(ii) We have given that p(x) = 5x – p
\nTherefore, the value of polynomial p(x) at x = \\(\\frac{4}{5}\\) is
\n\\(P\\left(\\frac{4}{5}\\right)=5 \\times \\frac{4}{5}-\\frac{22}{7}\\) (\u2235 \u03c0 = \\(\\frac {22}{7}\\))
\n= 4 – \\(\\frac{22}{7}\\)
\n\\(P\\left(\\frac{4}{5}\\right)=\\frac{6}{7}\\)
\nNo, x = \\(\\frac{4}{5}\\) is not the zero of p(x) = 5x – \u03c0.<\/p>\n

(iii) We have given that p(x) = x2<\/sup> – 1
\nTherefore, the value of polynomial p(x) at x = 1 is
\np(1) = 12<\/sup> – 1 = 0
\nAgain, the value of polynomial p(x) at x = -1 is
\np(-1) = (-1)2<\/sup> – 1 = 0
\nyes x = 1, -1 are the zero of polynomial p(x) = x2<\/sup> – 1<\/p>\n

(iv) We have given that
\np(x) = (x + 1)(x – 2)
\nTherefore, the value of polynomial p(x) at x = -1 is
\np(-1) = (-1 + 1)(-1 – 2)
\n= 0 \u00d7 (-3)
\n= 0
\nAgain, the value of polynomial p(x) at x = 2 is
\nP(2) = (2 + 1)(2 – 2)
\n= 3 \u00d7 0
\n= 0
\nyes x = -1, 2 are the zero of polynomial p(x) = (x + 1)(x – 2)<\/p>\n

\"NCERT<\/p>\n

(v) We have given that p(x) = x2<\/sup>
\nTherefore, the value of polynomial p(x) at x = 0 is
\np(0) = 02<\/sup> = 0
\nyes x = 0 is the zero of polynomial p(x) = x2<\/sup>.<\/p>\n

(vi) We have given that, p(x) = lx + m
\nTherefore, the value of polynomial p(x) at x = \\(-\\frac{m}{\\ell}\\) is
\n\\(P\\left(-\\frac{m}{\\ell}\\right)=\\ell\\left(-\\frac{m}{\\ell}\\right)+m\\) = 0
\nyes x = \\(-\\frac{m}{\\ell}\\) is the zero of polynomial p(x) = lx + m.<\/p>\n

(vii) We have given that, p(x) = 3x2<\/sup> – 1
\nTherefore, the value of polynomial p(x) at x = \\(-\\frac{1}{\\sqrt{3}}\\) is
\n\\(P\\left(-\\frac{1}{\\sqrt{3}}\\right)=3 \\times\\left(-\\frac{1}{\\sqrt{3}}\\right)^{2}-1\\)
\n= \\(3 \\times \\frac{1}{3}-1\\)
\n= 0
\nAgain, the value of polynomial p(x) at x = \\(\\frac{2}{\\sqrt{3}}\\) is
\n\\(P\\left(\\frac{2}{\\sqrt{3}}\\right)=3 \\times\\left(\\frac{2}{\\sqrt{3}}\\right)^{2}-1\\)
\n= \\(3 \\times \\frac{4}{3}-1\\)
\n= 4 – 1
\n= 3
\nTherefore, x = \\(-\\frac{1}{\\sqrt{3}}\\) is the zero of polynomial p(x) and x = \\(\\frac{2}{\\sqrt{3}}\\) is not the zero of polynomial p(x).<\/p>\n

(viii) We have given that p(x) = 2x + 1
\nTherefore, the value of polynomial p(x) at x = \\(\\frac{1}{2}\\) is
\n\\(\\mathrm{P}\\left(\\frac{1}{2}\\right)=2 \\times \\frac{1}{2}+1\\) = 2
\nTherefore, x = \\(\\frac{1}{2}\\) is not the zero of polynomial p(x).<\/p>\n

\"NCERT<\/p>\n

Question 4.
\nFind the zero of the polynomial in each of the following cases:
\n(i) p(x) = x + 5
\n(ii) p(x) = x – 5
\n(iii) p(x) = 2x + 5
\n(iv) p(x) = 3x – 2
\n(v) p(x) = 3x
\n(vi) p(x) = ax, a \u2260 0
\n(vii) p(x) = cx + d, c \u2260 0; c, d are real numbers.
\nSolution:
\n(i) We have given that
\np(x) = x + 5 ……..(i)
\nTo find the zero of polynomial p(x), we can take
\nP(x) = 0 ………(ii)
\nfrom equation (i) and (ii)
\nx + 5 = 0
\n\u2234 x = -5
\nTherefore, x = -5 is the zero of polynomial p(x) = x + 5.<\/p>\n

(ii) We have given that,
\np(x) = x – 5 ……..(i)
\nTo find the zero of polynomial p(x) we can take
\np(x) = 0 ……..(ii)
\nfrom equation (i) and (ii)
\nx – 5 = 0
\n\u2234 x = 5
\nTherefore x = 5 is the zero of polynomial p(x) = x – 5.<\/p>\n

(iii) We have given that
\np(x) = 2x + 5 ………(i)
\nTo find the zero of polynomial p(x), we can take
\np(x) = 0 …….(ii)
\nfrom equation (i) and (ii)
\n2x + 5 = 0
\nor x = \\(-\\frac{5}{2}\\)
\nTherefore x = \\(-\\frac{5}{2}\\) is the zero of polynomial p(x) = 2x + 5.<\/p>\n

\"NCERT<\/p>\n

(iv) We have given that
\np(x) = 3x – 2 …….(i)
\nTo find the zero of polynomial p(x) we can take
\np(x) = 0 ……..(ii)
\n3x – 2 = 0
\nx = \\(\\frac{2}{3}\\)
\nTherefore, x = \\(\\frac{2}{3}\\) is the zero of polynomial p(x) = 3x – 2.<\/p>\n

(v) We have given that
\np(x) = 3x ……..(i)
\nTo find the zero of polynomial p(x) we can take
\nP(x) = 0 ……..(ii)
\nfrom equation (i) and (ii)
\n3x = 0
\nx = 0
\nTherefore, x = 0 is the zero of polynomial p(x) = 3x.<\/p>\n

(vi) We have given that,
\np(x) = ax, a \u2260 0 ……(i)
\nTo find the zero of polynomial p(x) we can take
\np(x) = 0 ………(ii)
\nfrom equation (i) and (ii)
\nax = 0
\nx = \\(\\frac{0}{a}\\) = 0
\nTherefore, x = 0 is the zero of polynomial p(x) = ax, a \u2260 0.<\/p>\n

\"NCERT<\/p>\n

(vii) We have given
\np(x) = cx + d …….(i)
\nc \u2260 0, c, d are real numbers.
\nTo find the zero of polynomial p(x) we can take
\np(x) = 0 ………(ii)
\nfrom equation (i) and (ii)
\ncx + d = 0
\nx = \\(-\\frac{d}{c}\\)
\nwhere c \u2260 0 and c, d are real numbers.
\nTherefore, x = \\(\\frac{d}{c}\\), where c \u2260 0 and c, d are real numbers is the solution of polynomial p(x) = cx + d.<\/p>\n","protected":false},"excerpt":{"rendered":"

These NCERT Solutions for Class 9 Maths Chapter 2 Polynomials Ex 2.2 Questions and Answers are prepared by our highly skilled subject experts. NCERT Solutions for Class 9 Maths Chapter 2 Polynomials Exercise 2.2 Question 1. Find the value of the polynomial 5x – 4×2 + 3 at (i) x = 0 (ii) x = …<\/p>\n

NCERT Solutions for Class 9 Maths Chapter 2 Polynomials Ex 2.2<\/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 2.2 - 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-2-2\/\" \/>\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 2.2 - NCERT Solutions\" \/>\n<meta property=\"og:description\" content=\"These NCERT Solutions for Class 9 Maths Chapter 2 Polynomials Ex 2.2 Questions and Answers are prepared by our highly skilled subject experts. NCERT Solutions for Class 9 Maths Chapter 2 Polynomials Exercise 2.2 Question 1. Find the value of the polynomial 5x – 4x2 + 3 at (i) x = 0 (ii) x = … NCERT Solutions for Class 9 Maths Chapter 2 Polynomials Ex 2.2 Read More »\" \/>\n<meta property=\"og:url\" content=\"https:\/\/ncertsolutions.guru\/ncert-solutions-for-class-9-maths-chapter-2-ex-2-2\/\" \/>\n<meta property=\"og:site_name\" content=\"NCERT Solutions\" \/>\n<meta property=\"article:publisher\" content=\"https:\/\/www.facebook.com\/NCERTSolutionsGuru\/\" \/>\n<meta property=\"article:published_time\" content=\"2021-06-19T11:02:28+00:00\" \/>\n<meta property=\"article:modified_time\" content=\"2021-06-19T11:07:36+00:00\" \/>\n<meta property=\"og:image\" content=\"https:\/\/ncertsolutions.guru\/wp-content\/uploads\/2021\/01\/NCERT-Solutions-Guru.png\" \/>\n<meta name=\"twitter:card\" content=\"summary_large_image\" \/>\n<meta name=\"twitter:creator\" content=\"@ncertsolguru\" \/>\n<meta name=\"twitter:site\" content=\"@ncertsolguru\" \/>\n<meta name=\"twitter:label1\" content=\"Written by\" \/>\n\t<meta name=\"twitter:data1\" content=\"Prasanna\" \/>\n\t<meta name=\"twitter:label2\" content=\"Est. reading time\" \/>\n\t<meta name=\"twitter:data2\" content=\"8 minutes\" \/>\n<script type=\"application\/ld+json\" class=\"yoast-schema-graph\">{\"@context\":\"https:\/\/schema.org\",\"@graph\":[{\"@type\":\"Organization\",\"@id\":\"https:\/\/ncertsolutions.guru\/#organization\",\"name\":\"NCERT Solutions\",\"url\":\"https:\/\/ncertsolutions.guru\/\",\"sameAs\":[\"https:\/\/www.facebook.com\/NCERTSolutionsGuru\/\",\"https:\/\/www.instagram.com\/ncertsolutionsguru\/\",\"https:\/\/www.linkedin.com\/in\/ncertsolutionsguru\/\",\"https:\/\/in.pinterest.com\/ncertsolutionsguru\/\",\"https:\/\/en.wikipedia.org\/wiki\/National_Council_of_Educational_Research_and_Training\",\"https:\/\/twitter.com\/ncertsolguru\"],\"logo\":{\"@type\":\"ImageObject\",\"@id\":\"https:\/\/ncertsolutions.guru\/#logo\",\"inLanguage\":\"en-US\",\"url\":\"https:\/\/i1.wp.com\/ncertsolutions.guru\/wp-content\/uploads\/2020\/05\/NCERT-Solutions.png?fit=460%2C45&ssl=1\",\"contentUrl\":\"https:\/\/i1.wp.com\/ncertsolutions.guru\/wp-content\/uploads\/2020\/05\/NCERT-Solutions.png?fit=460%2C45&ssl=1\",\"width\":460,\"height\":45,\"caption\":\"NCERT Solutions\"},\"image\":{\"@id\":\"https:\/\/ncertsolutions.guru\/#logo\"}},{\"@type\":\"WebSite\",\"@id\":\"https:\/\/ncertsolutions.guru\/#website\",\"url\":\"https:\/\/ncertsolutions.guru\/\",\"name\":\"NCERT Solutions\",\"description\":\"NCERT Solutions for Class 1 to 12\",\"publisher\":{\"@id\":\"https:\/\/ncertsolutions.guru\/#organization\"},\"potentialAction\":[{\"@type\":\"SearchAction\",\"target\":{\"@type\":\"EntryPoint\",\"urlTemplate\":\"https:\/\/ncertsolutions.guru\/?s={search_term_string}\"},\"query-input\":\"required name=search_term_string\"}],\"inLanguage\":\"en-US\"},{\"@type\":\"ImageObject\",\"@id\":\"https:\/\/ncertsolutions.guru\/ncert-solutions-for-class-9-maths-chapter-2-ex-2-2\/#primaryimage\",\"inLanguage\":\"en-US\",\"url\":\"https:\/\/i1.wp.com\/ncertsolutions.guru\/wp-content\/uploads\/2021\/01\/NCERT-Solutions-Guru.png?fit=170%2C17&ssl=1\",\"contentUrl\":\"https:\/\/i1.wp.com\/ncertsolutions.guru\/wp-content\/uploads\/2021\/01\/NCERT-Solutions-Guru.png?fit=170%2C17&ssl=1\",\"width\":170,\"height\":17,\"caption\":\"NCERT Solutions\"},{\"@type\":\"WebPage\",\"@id\":\"https:\/\/ncertsolutions.guru\/ncert-solutions-for-class-9-maths-chapter-2-ex-2-2\/#webpage\",\"url\":\"https:\/\/ncertsolutions.guru\/ncert-solutions-for-class-9-maths-chapter-2-ex-2-2\/\",\"name\":\"NCERT Solutions for Class 9 Maths Chapter 2 Polynomials Ex 2.2 - NCERT Solutions\",\"isPartOf\":{\"@id\":\"https:\/\/ncertsolutions.guru\/#website\"},\"primaryImageOfPage\":{\"@id\":\"https:\/\/ncertsolutions.guru\/ncert-solutions-for-class-9-maths-chapter-2-ex-2-2\/#primaryimage\"},\"datePublished\":\"2021-06-19T11:02:28+00:00\",\"dateModified\":\"2021-06-19T11:07:36+00:00\",\"breadcrumb\":{\"@id\":\"https:\/\/ncertsolutions.guru\/ncert-solutions-for-class-9-maths-chapter-2-ex-2-2\/#breadcrumb\"},\"inLanguage\":\"en-US\",\"potentialAction\":[{\"@type\":\"ReadAction\",\"target\":[\"https:\/\/ncertsolutions.guru\/ncert-solutions-for-class-9-maths-chapter-2-ex-2-2\/\"]}]},{\"@type\":\"BreadcrumbList\",\"@id\":\"https:\/\/ncertsolutions.guru\/ncert-solutions-for-class-9-maths-chapter-2-ex-2-2\/#breadcrumb\",\"itemListElement\":[{\"@type\":\"ListItem\",\"position\":1,\"name\":\"Home\",\"item\":\"https:\/\/ncertsolutions.guru\/\"},{\"@type\":\"ListItem\",\"position\":2,\"name\":\"NCERT Solutions for Class 9 Maths Chapter 2 Polynomials Ex 2.2\"}]},{\"@type\":\"Article\",\"@id\":\"https:\/\/ncertsolutions.guru\/ncert-solutions-for-class-9-maths-chapter-2-ex-2-2\/#article\",\"isPartOf\":{\"@id\":\"https:\/\/ncertsolutions.guru\/ncert-solutions-for-class-9-maths-chapter-2-ex-2-2\/#webpage\"},\"author\":{\"@id\":\"https:\/\/ncertsolutions.guru\/#\/schema\/person\/4ba9570f32f2057e70e670c7885e47f3\"},\"headline\":\"NCERT Solutions for Class 9 Maths Chapter 2 Polynomials Ex 2.2\",\"datePublished\":\"2021-06-19T11:02:28+00:00\",\"dateModified\":\"2021-06-19T11:07:36+00:00\",\"mainEntityOfPage\":{\"@id\":\"https:\/\/ncertsolutions.guru\/ncert-solutions-for-class-9-maths-chapter-2-ex-2-2\/#webpage\"},\"wordCount\":1256,\"publisher\":{\"@id\":\"https:\/\/ncertsolutions.guru\/#organization\"},\"image\":{\"@id\":\"https:\/\/ncertsolutions.guru\/ncert-solutions-for-class-9-maths-chapter-2-ex-2-2\/#primaryimage\"},\"thumbnailUrl\":\"https:\/\/ncertsolutions.guru\/wp-content\/uploads\/2021\/01\/NCERT-Solutions-Guru.png\",\"articleSection\":[\"CBSE Class 9\"],\"inLanguage\":\"en-US\"},{\"@type\":\"Person\",\"@id\":\"https:\/\/ncertsolutions.guru\/#\/schema\/person\/4ba9570f32f2057e70e670c7885e47f3\",\"name\":\"Prasanna\",\"image\":{\"@type\":\"ImageObject\",\"@id\":\"https:\/\/ncertsolutions.guru\/#personlogo\",\"inLanguage\":\"en-US\",\"url\":\"https:\/\/secure.gravatar.com\/avatar\/174540ad43736c7d1a4c4f83c775e74d?s=96&d=mm&r=g\",\"contentUrl\":\"https:\/\/secure.gravatar.com\/avatar\/174540ad43736c7d1a4c4f83c775e74d?s=96&d=mm&r=g\",\"caption\":\"Prasanna\"},\"url\":\"https:\/\/ncertsolutions.guru\/author\/prasanna\/\"}]}<\/script>\n<!-- \/ Yoast SEO plugin. -->","yoast_head_json":{"title":"NCERT Solutions for Class 9 Maths Chapter 2 Polynomials Ex 2.2 - NCERT Solutions","robots":{"index":"index","follow":"follow","max-snippet":"max-snippet:-1","max-image-preview":"max-image-preview:large","max-video-preview":"max-video-preview:-1"},"canonical":"https:\/\/ncertsolutions.guru\/ncert-solutions-for-class-9-maths-chapter-2-ex-2-2\/","og_locale":"en_US","og_type":"article","og_title":"NCERT Solutions for Class 9 Maths Chapter 2 Polynomials Ex 2.2 - NCERT Solutions","og_description":"These NCERT Solutions for Class 9 Maths Chapter 2 Polynomials Ex 2.2 Questions and Answers are prepared by our highly skilled subject experts. NCERT Solutions for Class 9 Maths Chapter 2 Polynomials Exercise 2.2 Question 1. Find the value of the polynomial 5x – 4x2 + 3 at (i) x = 0 (ii) x = … NCERT Solutions for Class 9 Maths Chapter 2 Polynomials Ex 2.2 Read More »","og_url":"https:\/\/ncertsolutions.guru\/ncert-solutions-for-class-9-maths-chapter-2-ex-2-2\/","og_site_name":"NCERT Solutions","article_publisher":"https:\/\/www.facebook.com\/NCERTSolutionsGuru\/","article_published_time":"2021-06-19T11:02:28+00:00","article_modified_time":"2021-06-19T11:07:36+00:00","og_image":[{"url":"https:\/\/ncertsolutions.guru\/wp-content\/uploads\/2021\/01\/NCERT-Solutions-Guru.png"}],"twitter_card":"summary_large_image","twitter_creator":"@ncertsolguru","twitter_site":"@ncertsolguru","twitter_misc":{"Written by":"Prasanna","Est. reading time":"8 minutes"},"schema":{"@context":"https:\/\/schema.org","@graph":[{"@type":"Organization","@id":"https:\/\/ncertsolutions.guru\/#organization","name":"NCERT Solutions","url":"https:\/\/ncertsolutions.guru\/","sameAs":["https:\/\/www.facebook.com\/NCERTSolutionsGuru\/","https:\/\/www.instagram.com\/ncertsolutionsguru\/","https:\/\/www.linkedin.com\/in\/ncertsolutionsguru\/","https:\/\/in.pinterest.com\/ncertsolutionsguru\/","https:\/\/en.wikipedia.org\/wiki\/National_Council_of_Educational_Research_and_Training","https:\/\/twitter.com\/ncertsolguru"],"logo":{"@type":"ImageObject","@id":"https:\/\/ncertsolutions.guru\/#logo","inLanguage":"en-US","url":"https:\/\/i1.wp.com\/ncertsolutions.guru\/wp-content\/uploads\/2020\/05\/NCERT-Solutions.png?fit=460%2C45&ssl=1","contentUrl":"https:\/\/i1.wp.com\/ncertsolutions.guru\/wp-content\/uploads\/2020\/05\/NCERT-Solutions.png?fit=460%2C45&ssl=1","width":460,"height":45,"caption":"NCERT Solutions"},"image":{"@id":"https:\/\/ncertsolutions.guru\/#logo"}},{"@type":"WebSite","@id":"https:\/\/ncertsolutions.guru\/#website","url":"https:\/\/ncertsolutions.guru\/","name":"NCERT Solutions","description":"NCERT Solutions for Class 1 to 12","publisher":{"@id":"https:\/\/ncertsolutions.guru\/#organization"},"potentialAction":[{"@type":"SearchAction","target":{"@type":"EntryPoint","urlTemplate":"https:\/\/ncertsolutions.guru\/?s={search_term_string}"},"query-input":"required name=search_term_string"}],"inLanguage":"en-US"},{"@type":"ImageObject","@id":"https:\/\/ncertsolutions.guru\/ncert-solutions-for-class-9-maths-chapter-2-ex-2-2\/#primaryimage","inLanguage":"en-US","url":"https:\/\/i1.wp.com\/ncertsolutions.guru\/wp-content\/uploads\/2021\/01\/NCERT-Solutions-Guru.png?fit=170%2C17&ssl=1","contentUrl":"https:\/\/i1.wp.com\/ncertsolutions.guru\/wp-content\/uploads\/2021\/01\/NCERT-Solutions-Guru.png?fit=170%2C17&ssl=1","width":170,"height":17,"caption":"NCERT Solutions"},{"@type":"WebPage","@id":"https:\/\/ncertsolutions.guru\/ncert-solutions-for-class-9-maths-chapter-2-ex-2-2\/#webpage","url":"https:\/\/ncertsolutions.guru\/ncert-solutions-for-class-9-maths-chapter-2-ex-2-2\/","name":"NCERT Solutions for Class 9 Maths Chapter 2 Polynomials Ex 2.2 - NCERT Solutions","isPartOf":{"@id":"https:\/\/ncertsolutions.guru\/#website"},"primaryImageOfPage":{"@id":"https:\/\/ncertsolutions.guru\/ncert-solutions-for-class-9-maths-chapter-2-ex-2-2\/#primaryimage"},"datePublished":"2021-06-19T11:02:28+00:00","dateModified":"2021-06-19T11:07:36+00:00","breadcrumb":{"@id":"https:\/\/ncertsolutions.guru\/ncert-solutions-for-class-9-maths-chapter-2-ex-2-2\/#breadcrumb"},"inLanguage":"en-US","potentialAction":[{"@type":"ReadAction","target":["https:\/\/ncertsolutions.guru\/ncert-solutions-for-class-9-maths-chapter-2-ex-2-2\/"]}]},{"@type":"BreadcrumbList","@id":"https:\/\/ncertsolutions.guru\/ncert-solutions-for-class-9-maths-chapter-2-ex-2-2\/#breadcrumb","itemListElement":[{"@type":"ListItem","position":1,"name":"Home","item":"https:\/\/ncertsolutions.guru\/"},{"@type":"ListItem","position":2,"name":"NCERT Solutions for Class 9 Maths Chapter 2 Polynomials Ex 2.2"}]},{"@type":"Article","@id":"https:\/\/ncertsolutions.guru\/ncert-solutions-for-class-9-maths-chapter-2-ex-2-2\/#article","isPartOf":{"@id":"https:\/\/ncertsolutions.guru\/ncert-solutions-for-class-9-maths-chapter-2-ex-2-2\/#webpage"},"author":{"@id":"https:\/\/ncertsolutions.guru\/#\/schema\/person\/4ba9570f32f2057e70e670c7885e47f3"},"headline":"NCERT Solutions for Class 9 Maths Chapter 2 Polynomials Ex 2.2","datePublished":"2021-06-19T11:02:28+00:00","dateModified":"2021-06-19T11:07:36+00:00","mainEntityOfPage":{"@id":"https:\/\/ncertsolutions.guru\/ncert-solutions-for-class-9-maths-chapter-2-ex-2-2\/#webpage"},"wordCount":1256,"publisher":{"@id":"https:\/\/ncertsolutions.guru\/#organization"},"image":{"@id":"https:\/\/ncertsolutions.guru\/ncert-solutions-for-class-9-maths-chapter-2-ex-2-2\/#primaryimage"},"thumbnailUrl":"https:\/\/ncertsolutions.guru\/wp-content\/uploads\/2021\/01\/NCERT-Solutions-Guru.png","articleSection":["CBSE Class 9"],"inLanguage":"en-US"},{"@type":"Person","@id":"https:\/\/ncertsolutions.guru\/#\/schema\/person\/4ba9570f32f2057e70e670c7885e47f3","name":"Prasanna","image":{"@type":"ImageObject","@id":"https:\/\/ncertsolutions.guru\/#personlogo","inLanguage":"en-US","url":"https:\/\/secure.gravatar.com\/avatar\/174540ad43736c7d1a4c4f83c775e74d?s=96&d=mm&r=g","contentUrl":"https:\/\/secure.gravatar.com\/avatar\/174540ad43736c7d1a4c4f83c775e74d?s=96&d=mm&r=g","caption":"Prasanna"},"url":"https:\/\/ncertsolutions.guru\/author\/prasanna\/"}]}},"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/ncertsolutions.guru\/wp-json\/wp\/v2\/posts\/24648"}],"collection":[{"href":"https:\/\/ncertsolutions.guru\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/ncertsolutions.guru\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/ncertsolutions.guru\/wp-json\/wp\/v2\/users\/9"}],"replies":[{"embeddable":true,"href":"https:\/\/ncertsolutions.guru\/wp-json\/wp\/v2\/comments?post=24648"}],"version-history":[{"count":0,"href":"https:\/\/ncertsolutions.guru\/wp-json\/wp\/v2\/posts\/24648\/revisions"}],"wp:attachment":[{"href":"https:\/\/ncertsolutions.guru\/wp-json\/wp\/v2\/media?parent=24648"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/ncertsolutions.guru\/wp-json\/wp\/v2\/categories?post=24648"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/ncertsolutions.guru\/wp-json\/wp\/v2\/tags?post=24648"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}