{"id":4221,"date":"2023-02-02T06:00:48","date_gmt":"2023-02-02T06:00:48","guid":{"rendered":"https:\/\/theexampillar.com\/ncert\/?p=4221"},"modified":"2023-02-02T06:03:55","modified_gmt":"2023-02-02T06:03:55","slug":"ncert-solutions-class-9-maths-chapter-2-polynomials-ex-2-3","status":"publish","type":"post","link":"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-9-maths-chapter-2-polynomials-ex-2-3\/","title":{"rendered":"NCERT Solutions Class 9 Maths Chapter 2 Polynomials Ex 2.3"},"content":{"rendered":"<h2 style=\"text-align: center;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\">NCERT Solutions Class 9 Maths\u00a0<\/span><br \/>\n<span style=\"color: #000000; font-family: georgia, palatino, serif;\">Chapter &#8211; 2 (Polynomials)\u00a0<\/span><\/h2>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\">The NCERT Solutions in English Language for Class 9 Mathematics\u00a0<strong>Chapter &#8211; 2 Polynomials <\/strong>Exercise 2.3 has been provided here to help the students in solving the questions from this exercise.\u00a0<\/span><\/p>\n<p><span style=\"color: #000000;\"><strong>Chapter 2: Polynomials\u00a0<\/strong><\/span><\/p>\n<ul>\n<li><a href=\"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-9-maths-chapter-2-polynomials-ex-2-1\/\" target=\"_blank\" rel=\"noopener\">NCERT Solution Class 9 Maths Ex &#8211; 2.1<\/a><\/li>\n<li><a href=\"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-9-maths-chapter-2-polynomials-ex-2-2\/\" target=\"_blank\" rel=\"noopener\">NCERT Solution Class 9 Maths Ex &#8211; 2.2<\/a><\/li>\n<li><a href=\"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-9-maths-chapter-2-polynomials-ex-2-4\/\" target=\"_blank\" rel=\"noopener\">NCERT Solution Class 9 Maths Ex &#8211; 2.4<\/a><\/li>\n<li><a href=\"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-9-maths-chapter-2-polynomials-ex-2-5\/\" target=\"_blank\" rel=\"noopener\">NCERT Solution Class 9 Maths Ex &#8211; 2.5<\/a><\/li>\n<\/ul>\n<h2 style=\"text-align: center;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\">Exercise &#8211; 2.3\u00a0<\/span><\/h2>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>1. Find the remainder when\u00a0x<sup>3 <\/sup>+ 3x<sup>2 <\/sup>+ 3x + 1 is divided by<br \/>\n<\/strong><strong>(i) x + 1<br \/>\n<\/strong><strong>(ii) x \u2212 1\/2<br \/>\n<\/strong><strong>(iii) x<br \/>\n<\/strong><strong>(iv) x + \u03c0<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>Answer &#8211;<br \/>\n<\/strong><strong>(i) x + 1<br \/>\n<\/strong>The zero of x + 1 = 0<br \/>\n\u21d2 x = \u22121<br \/>\nLet p(x) = x<sup>3<\/sup>\u00a0+ 3x<sup>2<\/sup> + 3x +1<br \/>\n\u2234 p(-1) = (-1)3 + 3(-1)2 + 3(-1) +1<br \/>\n\u21d2 -1 + 3- 3 + 1<br \/>\n\u21d2 0<br \/>\nThus, the required remainder = 0<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(ii) x\u22121\/2<br \/>\n<\/strong>The zero of x \u2212 1\/2 = 0<br \/>\n\u21d2 x = 1\/2<br \/>\nLet p(x) = x<sup>3<\/sup>\u00a0+ 3x<sup>2<\/sup> + 3x +1<br \/>\np(1\/2) = (1\/2)<sup>3 <\/sup>+ 3(1\/2)<sup>2 <\/sup>+ 3(1\/2) + 1<br \/>\n\u21d2 (1\/8) + (3\/4) + (3\/2) + 1<br \/>\n\u21d2 <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{1&amp;space;+&amp;space;3\\times&amp;space;2&amp;space;+&amp;space;3\\times&amp;space;4&amp;space;+&amp;space;1\\times8}{8}\" alt=\"\\frac{1 + 3\\times 2 + 3\\times 4 + 1\\times8}{8}\" align=\"absmiddle\" \/>\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 LCM of 8, 4 and 2 = 8 <\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\">\u21d2 <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{1+6+12+8}{8}\" alt=\"\\frac{1+6+12+8}{8}\" align=\"absmiddle\" \/> <\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\">\u21d2 <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{27}{8}\" alt=\"\\frac{27}{8}\" align=\"absmiddle\" \/><br \/>\nThus, the required remainder = <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{27}{8}\" alt=\"\\frac{27}{8}\" align=\"absmiddle\" \/><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(iii) x<br \/>\n<\/strong>The zero of x = 0<br \/>\n\u21d2 x = 0<br \/>\nLet p(x) = x<sup>3<\/sup>\u00a0+ 3x<sup>2<\/sup> + 3x +1<br \/>\np(0) = (0)<sup>3<\/sup>+3(0)<sup>2<\/sup>+3(0)+1<br \/>\n\u21d2 1<br \/>\nThus, the required remainder = 1.<br \/>\n<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(iv) x + \u03c0<br \/>\n<\/strong>The zero of x + \u03c0 = 0<br \/>\n\u21d2 x = &#8211; \u03c0<br \/>\nLet p(x) = x<sup>3<\/sup>\u00a0+ 3x<sup>2<\/sup> + 3x +1<br \/>\np(\u03c0) = (\u2212\u03c0)<sup>3\u00a0<\/sup>+3(\u2212\u03c0)<sup>2 <\/sup>+ 3(\u2212\u03c0) + 1<br \/>\n\u21d2 \u2212\u03c0<sup>3 <\/sup>+ 3\u03c0<sup>2 <\/sup>\u2212 3\u03c0 + 1<br \/>\nThus, the required remainder is -\u03c0<sup>3<\/sup>\u00a0+ 3\u03c0<sup>2<\/sup> \u2013 3\u03c0 + 1.<br \/>\n<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(v) 5 + 2x<br \/>\n<\/strong>The zero of 5 + 2x = 0<br \/>\n\u21d2 x = &#8211; 5\/2<br \/>\nLet p(x) = x<sup>3<\/sup>\u00a0+ 3x<sup>2<\/sup> + 3x +1<br \/>\np(-5\/2) = (-5\/2)<sup>3 <\/sup>+ 3(-5\/2)<sup>2 <\/sup>+ 3(-5\/2) + 1<br \/>\n\u21d2 (-125\/8) + (75\/4) &#8211; (15\/2) + 1<br \/>\n\u21d2 <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{-125&amp;space;+75\\times&amp;space;2&amp;space;-&amp;space;15\\times&amp;space;4&amp;space;+1\\times&amp;space;8}{8}\" alt=\"\\frac{-125 +75\\times 2 - 15\\times 4 +1\\times 8}{8}\" align=\"absmiddle\" \/> \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0LCM of 8, 4 and 2 = 8<\/span><\/p>\n<p><span style=\"color: #000000;\">\u21d2 <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{-125&amp;space;+&amp;space;150&amp;space;-60&amp;space;+8}{8}\" alt=\"\\frac{-125 + 150 -60 +8}{8}\" align=\"absmiddle\" \/> \u00a0<\/span><\/p>\n<p><span style=\"color: #000000;\">\u21d2 <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{-27}{8}\" alt=\"\\frac{-27}{8}\" align=\"absmiddle\" \/><br \/>\nThus, the required remainder is <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{-27}{8}\" alt=\"\\frac{-27}{8}\" align=\"absmiddle\" \/>.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>2. Find the remainder when\u00a0x<sup>3 <\/sup>\u2212 ax<sup>2 <\/sup>+ 6x \u2212 a is divided by x &#8211; a.<br \/>\n<\/strong><strong>Answer &#8211;<\/strong><br \/>\nLet\u00a0p(x) = x<sup>3 <\/sup>\u2212 ax<sup>2 <\/sup>+ 6x \u2212 a<br \/>\nx \u2212 a = 0<br \/>\n\u2234 x = a<br \/>\nThe remainder:<br \/>\np(a) = (a)<sup>3 <\/sup>\u2212 a(a<sup>2<\/sup>) + 6(a) \u2212 a<br \/>\n= a<sup>3 <\/sup>\u2212 a<sup>3 <\/sup>+ 6a \u2212 a<br \/>\n= 5a<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>3. Check whether 7 + 3x is a factor of 3x<sup>3 <\/sup>+ 7x.<br \/>\n<\/strong><strong>Answer &#8211;<\/strong><br \/>\nLet p(x) =\u00a03x<sup>3 <\/sup>+ 7x<br \/>\n7 + 3x = 0<br \/>\n\u21d2 3x = \u22127<br \/>\n\u21d2 x = -7\/3<br \/>\n\u2234 The remainder:<br \/>\np(-7\/3) = 3(-7\/3)<sup>3 <\/sup>+ 7(-7\/3)<br \/>\n\u21d2 -(343\/9) + (-49\/3) <\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\">\u21d2 <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{-343&amp;space;-&amp;space;49\\times&amp;space;3}{9}\" alt=\"\\frac{-343 - 49\\times 3}{9}\" align=\"absmiddle\" \/>\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 LCM of 9 and 3 = 9<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\">\u21d2 <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{-343&amp;space;-&amp;space;147}{9}\" alt=\"\\frac{-343 - 147}{9}\" align=\"absmiddle\" \/> <\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\">\u21d2 <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{-490}{9}\" alt=\"\\frac{-490}{9}\" align=\"absmiddle\" \/> \u2260 0<br \/>\n\u2234 7 + 3x is not a factor of 3x<sup>3 <\/sup>+ 7x.<\/span><\/p>\n<p>&nbsp;<\/p>\n<p>&nbsp;<\/p>\n","protected":false},"excerpt":{"rendered":"<p>NCERT Solutions Class 9 Maths\u00a0 Chapter &#8211; 2 (Polynomials)\u00a0 The NCERT Solutions in English Language for Class 9 Mathematics\u00a0Chapter &#8211; 2 Polynomials Exercise 2.3 has been provided here to help the students in solving the questions from this exercise.\u00a0 Chapter 2: Polynomials\u00a0 NCERT Solution Class 9 Maths Ex &#8211; 2.1 NCERT Solution Class 9 Maths [&hellip;]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[698],"tags":[703,702,186,704,701,5],"class_list":["post-4221","post","type-post","status-publish","format-standard","hentry","category-class-9-maths","tag-class-9th-chapter-2-polynomials-in-english","tag-ncert-solution-class-9","tag-ncert-solutions","tag-ncert-solutions-class-9-maths-chapter-2-in-english","tag-ncert-solutions-class-9-maths-in-english","tag-ncert-solutions-in-english"],"yoast_head":"<!-- This site is optimized with the Yoast SEO Premium plugin v22.9 (Yoast SEO v27.4) - 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