{"id":4220,"date":"2023-02-02T06:00:43","date_gmt":"2023-02-02T06:00:43","guid":{"rendered":"https:\/\/theexampillar.com\/ncert\/?p=4220"},"modified":"2023-02-02T06:04:02","modified_gmt":"2023-02-02T06:04:02","slug":"ncert-solutions-class-9-maths-chapter-2-polynomials-ex-2-2","status":"publish","type":"post","link":"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-9-maths-chapter-2-polynomials-ex-2-2\/","title":{"rendered":"NCERT Solutions Class 9 Maths Chapter 2 Polynomials Ex 2.2"},"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 2.2 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-2-3\/\" target=\"_blank\" rel=\"noopener\">NCERT Solution Class 9 Maths Ex &#8211; 2.3<\/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; 2.2\u00a0<\/span><\/h2>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>1. Find the value of the polynomial (x) = 5x \u2212 4x<sup>2 <\/sup>+ 3<br \/>\n<\/strong><\/span><span style=\"color: #000000;\"><strong>(i) x = 0<br \/>\n<\/strong><\/span><span style=\"color: #000000;\"><strong>(ii) x\u00a0= \u2013 1<br \/>\n<\/strong><\/span><span style=\"color: #000000;\"><strong>(iii) x\u00a0= 2<br \/>\n<\/strong><\/span><strong><span style=\"color: #000000;\">Answer &#8211;<br \/>\n<\/span><\/strong><span style=\"color: #000000;\">Let f(x) = 5x \u2212 4x<sup>2 <\/sup>+ 3<br \/>\n<\/span><strong><span style=\"color: #000000;\">(i) When x = 0<br \/>\n<\/span><\/strong><span style=\"color: #000000;\">f(0) = 5(0) &#8211; 4(0)<sup>2 <\/sup>+ 3<br \/>\n<\/span><span style=\"color: #000000;\">= 3<\/span><\/p>\n<p style=\"text-align: justify;\"><strong><span style=\"color: #000000;\">(ii) When x = -1<br \/>\n<\/span><\/strong><span style=\"color: #000000;\">f(x) = 5x \u2212 4x<sup>2 <\/sup>+ 3<br \/>\n<\/span><span style=\"color: #000000;\">f(\u22121) = 5(\u22121) \u2212 4(\u22121)<sup>2 <\/sup>+ 3<br \/>\n<\/span><span style=\"color: #000000;\">= \u22125 \u2013 4 + 3<br \/>\n<\/span><span style=\"color: #000000;\">= \u22126<\/span><\/p>\n<p style=\"text-align: justify;\"><strong><span style=\"color: #000000;\">(iii) When x = 2<br \/>\n<\/span><\/strong><span style=\"color: #000000;\">f(x) = 5x \u2212 4x<sup>2 <\/sup>+ 3<br \/>\n<\/span><span style=\"color: #000000;\">f(2) = 5(2) \u2212 4(2)<sup>2 <\/sup>+ 3<br \/>\n<\/span><span style=\"color: #000000;\">= 10 \u2013 16 + 3<br \/>\n<\/span><span style=\"color: #000000;\">= \u22123<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>2. Find p(0), p(1) and p(2) for each of the following polynomials.<br \/>\n<\/strong><\/span><span style=\"color: #000000;\"><strong>(i) p(y) = y<sup>2 <\/sup>\u2212 y + 1<br \/>\n(ii) p(t) = 2 + t + 2t<sup>2 <\/sup>\u2212 t<sup>3<br \/>\n<\/sup><\/strong><\/span><span style=\"color: #000000;\"><strong>(iii) p(x) = x<sup>3 <\/sup><br \/>\n<\/strong><\/span><span style=\"color: #000000;\"><strong>(iv) P(x) = (x \u2212 1)(x + 1)<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>Answer &#8211;<\/strong><br \/>\n<\/span><span style=\"color: #000000;\"><strong>(i) p(y) = y<sup>2 <\/sup>\u2212 y + 1<br \/>\n<\/strong><\/span><span style=\"color: #000000;\"><strong>p(0)<\/strong> = (0)<sup>2 <\/sup>\u2212 (0) + 1<br \/>\n= 1 <\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>p(1)<\/strong><br \/>\n= (1)<sup>2 <\/sup>\u2013 (1) + 1<br \/>\n=1<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>p(2)<\/strong><br \/>\n= (2)<sup>2 <\/sup>\u2013 (2) + 1<br \/>\n= 4 &#8211; 2 +1<br \/>\n= 3<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(ii) p(t) = 2 + t + 2t<sup>2 <\/sup>\u2212 t<sup>3<br \/>\n<\/sup><\/strong><\/span><span style=\"color: #000000;\"><strong>p(0)<\/strong> = 2 + 0 + 2(0)<sup>2 <\/sup>\u2013 (0)<sup>3<br \/>\n<\/sup>= 2<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>p(1)<\/strong><br \/>\n= 2 + 1 + 2(1)<sup>2 <\/sup>\u2013 (1)<sup>3<br \/>\n<\/sup>= 2 + 1 + 2 \u2013 1<br \/>\n= 4<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>p(2)<\/strong><br \/>\n= 2 + 2 + 2(2)<sup>2 <\/sup>\u2013 (2)<sup>3<br \/>\n<\/sup>= 2 + 2 + 8 \u2013 8<br \/>\n= 4<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(iii) p(x) = x<sup>3 <\/sup><\/strong><sup><br \/>\n<\/sup><\/span><span style=\"color: #000000;\"><strong>p(0)<\/strong> = (0)<sup>3<br \/>\n<\/sup>= 0<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>p(1) <\/strong>= (1)<sup>3<br \/>\n<\/sup>= 1<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>p(2)<\/strong> = (2)<sup>3<br \/>\n<\/sup>= 8<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(iv) P(x) = (x \u2212 1)(x + 1)<br \/>\n<\/strong><\/span><span style=\"color: #000000;\"><strong>p(0)<\/strong> = (0 \u2013 1)(0 + 1)<br \/>\n= (\u22121)(1)<br \/>\n= \u20131<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>p(1)<\/strong> = (1 \u2013 1)(1 + 1)<br \/>\n= 0(2)<br \/>\n= 0<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>p(2)<\/strong><br \/>\n= (2 \u2013 1)(2 + 1)<br \/>\n= 1(3)<br \/>\n= 3<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>3. Verify whether the following are zeroes of the polynomial indicated against them.<br \/>\n<\/strong><\/span><span style=\"color: #000000;\"><strong>(i) p(x) = 3x + 1,\u00a0 x =\u22121\/3<br \/>\n<\/strong><\/span><span style=\"color: #000000;\"><strong>(ii) p(x) = 5x \u2013 \u03c0,\u00a0 x = 4\/5<br \/>\n<\/strong><\/span><span style=\"color: #000000;\"><strong>(iii) p(x) = x<sup>2 <\/sup>\u2212 1,\u00a0 x=1, \u22121<br \/>\n<\/strong><\/span><span style=\"color: #000000;\"><strong>(iv) p(x) = (x + 1)(x \u2013 2),\u00a0 x =\u22121, 2<br \/>\n<\/strong><\/span><span style=\"color: #000000;\"><strong>(v) p(x) = x<sup>2<\/sup>, x = 0<br \/>\n<\/strong><\/span><span style=\"color: #000000;\"><strong>(vi) p(x) =<em> lx\u00a0<\/em>+ m,\u00a0 \u00a0x = \u2212m\/<em>l<br \/>\n<\/em><\/strong><\/span><span style=\"color: #000000;\"><strong>(vii) p(x) = 3x<sup>2 <\/sup>\u2212 1,\u00a0 \u00a0x = -1\/\u221a3 , 2\/\u221a3<br \/>\n<\/strong><\/span><span style=\"color: #000000;\"><strong>(viii) p(x) = 2x + 1, x = 1\/2 <\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>Answer &#8211;<\/strong><br \/>\n<\/span><span style=\"color: #000000;\"><strong>(i) p(x) = 3x + 1,\u00a0 x =\u22121\/3<br \/>\n<\/strong><\/span><span style=\"color: #000000;\"><strong>p(\u22121\/3)<\/strong> = 3(-1\/3) + 1<br \/>\n= \u22121 + 1<br \/>\n= 0<br \/>\n<\/span><span style=\"color: #000000;\">\u2234 &#8211; <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{1}{3}\" alt=\"\\frac{1}{3}\" align=\"absmiddle\" \/> is a zero of p(x).<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(ii) p(x) = 5x \u2013 \u03c0,\u00a0 \u00a0 \u00a0 \u00a0 x = 4\/5<br \/>\n<\/strong><\/span><span style=\"color: #000000;\"><strong>p(4\/5)<\/strong> = 5 (4\/5) &#8211; \u03c0<br \/>\n= 4 &#8211; \u03c0<br \/>\n<\/span><span style=\"color: #000000;\">\u2234 <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{4}{5}\" alt=\"\\frac{4}{5}\" align=\"absmiddle\" \/> is not a zero of p(x).<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(iii) p(x) = x<sup>2 <\/sup>\u2212 1,\u00a0 \u00a0 x = 1, \u22121<\/strong><br \/>\n<\/span><span style=\"color: #000000;\"><strong>p(1) <\/strong>= 1<sup>2 <\/sup>\u2212 1<br \/>\n= 1 \u2212 1<br \/>\n= 0<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>p(\u22121) <\/strong>= (-1)<sup>2 <\/sup>\u2212 1<br \/>\n= 1 \u2212 1<br \/>\n= 0<br \/>\n<\/span><span style=\"color: #000000;\">\u22341, \u22121\u00a0are zeros of p(x).<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(iv) p(x) = (x + 1)(x \u2013 2), x =\u22121, 2<\/strong><br \/>\n<\/span><span style=\"color: #000000;\"><strong>p(\u22121)<\/strong> = (\u22121 + 1)(\u22121 \u2013 2)<br \/>\n<\/span><span style=\"color: #000000;\">= (0)(\u22123)<br \/>\n= 0<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>p(2)<\/strong> = (2+1)(2\u20132)<br \/>\n= (3)(0)<br \/>\n= 0<br \/>\n<\/span><span style=\"color: #000000;\">\u2234 \u22121, 2 are zeros of p(x).<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(v) p(x) = x<sup>2<\/sup>, x = 0<br \/>\n<\/strong><\/span><span style=\"color: #000000;\"><strong>p(0)<\/strong> = 0<sup>2<br \/>\n<\/sup>= 0<br \/>\n<\/span><span style=\"color: #000000;\">\u2234 0\u00a0is a\u00a0zero of p(x).<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(vi) p(x) =<em> lx\u00a0<\/em>+ m,\u00a0 \u00a0 \u00a0 \u00a0x = \u2212m\/<em>l<br \/>\n<\/em><\/strong><\/span><span style=\"color: #000000;\"><strong>p(-m\/<em>l) <\/em><\/strong>=\u00a0<em>l<\/em>(-m\/<em>l<\/em>) + m<br \/>\n= \u2212m + m<br \/>\n= 0<br \/>\n<\/span><span style=\"color: #000000;\">\u2234 -m\/<em>l <\/em>\u00a0is a zero of p(x).<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(vii) p(x) = 3x<sup>2<\/sup>\u22121, x = -1\/\u221a3 , 2\/\u221a3<br \/>\n<\/strong><\/span><span style=\"color: #000000;\"><strong>p(-1\/\u221a3)<\/strong> = 3(-1\/\u221a3)<sup>2 <\/sup>&#8211; 1<br \/>\n= 3(1\/3) &#8211; 1<br \/>\n= 1 &#8211; 1<br \/>\n= 0<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>p(2\/\u221a3)<\/strong> = 3(2\/\u221a3)<sup>2 <\/sup>&#8211; 1<br \/>\n= 3(4\/3) &#8211; 1<br \/>\n= 4 \u2212 1<br \/>\n= 3 \u2260 0 <\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\">\u2234 -1\/\u221a3 is a zero of p(x) but 2\/\u221a3\u00a0\u00a0is not a zero of p(x).<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(viii) p(x) =2x + 1, x = 1\/2<br \/>\n<\/strong><\/span><span style=\"color: #000000;\"><strong>p(1\/2) <\/strong>= 2(1\/2) + 1<br \/>\n= 1 + 1<br \/>\n= 2 \u22600<br \/>\n<\/span><span style=\"color: #000000;\">\u2234 1\/2 is not a zero of p(x).<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>4. Find the zero of the polynomials in each of the following cases.<br \/>\n<\/strong><\/span><span style=\"color: #000000;\"><strong>(i) p(x) = x + 5<br \/>\n(ii) p (x) = x \u2013 5<br \/>\n(iii) p (x) = 2x + 5<br \/>\n(iv) p (x) = 3x \u2013 2<br \/>\n(v) p (x) = 3x<br \/>\n(vi) p (x)= ax, a \u2260 0<br \/>\n(vii) p (x) = cx + d, c \u2260 0 where c and d are real numbers.<br \/>\n<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>Answer &#8211;<\/strong><br \/>\n<\/span><span style=\"color: #000000;\"><strong>(i) p(x) = x+5<br \/>\n<\/strong><\/span><span style=\"color: #000000;\">\u21d2 x + 5 = 0<br \/>\n<\/span><span style=\"color: #000000;\">\u21d2 x = \u22125<br \/>\n<\/span><span style=\"color: #000000;\">\u2234 -5 is a zero polynomial of the polynomial p(x).<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(ii) p(x) =\u00a0x\u20135<br \/>\n<\/strong><\/span><span style=\"color: #000000;\">\u21d2 x\u22125 = 0<br \/>\n<\/span><span style=\"color: #000000;\">\u21d2 x = 5<br \/>\n<\/span><span style=\"color: #000000;\">\u2234 5 is a zero polynomial of the polynomial p(x).<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(iii) p(x) = 2x+5<br \/>\n<\/strong><\/span><span style=\"color: #000000;\">\u21d2 2x + 5 = 0<br \/>\n<\/span><span style=\"color: #000000;\">\u21d2 2x = \u22125<br \/>\n<\/span><span style=\"color: #000000;\">\u21d2 x = -5\/2<br \/>\n<\/span><span style=\"color: #000000;\">\u2234 x = -5\/2 is a zero polynomial of the polynomial p(x).<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(iv) p(x) = 3x\u20132<br \/>\n<\/strong><\/span><span style=\"color: #000000;\">\u21d2 3x \u2212 2 = 0<br \/>\n<\/span><span style=\"color: #000000;\">\u21d2 3x = 2<br \/>\n<\/span><span style=\"color: #000000;\">\u21d2 x = 2\/3<br \/>\n<\/span><span style=\"color: #000000;\">\u2234 x = 2\/3 is a zero polynomial of the polynomial p(x).<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(v) p(x) = 3x<br \/>\n<\/strong><\/span><span style=\"color: #000000;\">\u21d2 3x = 0<br \/>\n<\/span><span style=\"color: #000000;\">\u21d2 x = 0<br \/>\n<\/span><span style=\"color: #000000;\">\u2234 0 is a zero polynomial of the polynomial p(x).<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(vi) p(x) = ax, a<b>\u2260<\/b>0<br \/>\n<\/strong><\/span><span style=\"color: #000000;\">\u21d2 ax = 0<br \/>\n<\/span><span style=\"color: #000000;\">\u21d2 x = 0<br \/>\n<\/span><span style=\"color: #000000;\">\u2234 x = 0 is a zero polynomial of the polynomial p(x).<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(vii) p(x) = cx + d, c \u2260 0, c, d are real numbers.<br \/>\n<\/strong><\/span><span style=\"color: #000000;\">\u21d2 cx + d = 0<br \/>\n<\/span><span style=\"color: #000000;\">\u21d2 x = -d\/c<br \/>\n<\/span><span style=\"color: #000000;\">\u2234 x = -d\/c is a zero polynomial of the polynomial p(x).<\/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 2.2 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-4220","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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