{"id":6013,"date":"2023-07-19T06:52:24","date_gmt":"2023-07-19T06:52:24","guid":{"rendered":"https:\/\/theexampillar.com\/ncert\/?p=6013"},"modified":"2023-07-23T05:43:29","modified_gmt":"2023-07-23T05:43:29","slug":"ncert-solutions-class-10-maths-chapter-2-polynomials-ex-2-4","status":"publish","type":"post","link":"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-10-maths-chapter-2-polynomials-ex-2-4\/","title":{"rendered":"NCERT Solutions Class 10 Maths Chapter 2 Polynomials Ex 2.4"},"content":{"rendered":"<h2 style=\"text-align: center;\"><strong><span style=\"color: #000000; font-family: georgia, palatino, serif;\">NCERT Solutions Class 10 Maths\u00a0<\/span><\/strong><br \/>\n<strong><span style=\"color: #000000; font-family: georgia, palatino, serif;\">Chapter &#8211; 2 (Polynomials)\u00a0<\/span><\/strong><\/h2>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\">The NCERT Solutions in English Language for Class 10 Mathematics <strong>Chapter &#8211; 2 Polynomials <\/strong>Exercise 2.4 has been provided here to help the students in solving the questions from this exercise.\u00a0<\/span><\/p>\n<p><span style=\"color: #ff0000;\"><strong>Chapter : 2 Polynomials<\/strong> <\/span><\/p>\n<ul>\n<li><span style=\"color: #0000ff;\"><a style=\"color: #0000ff;\" href=\"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-10-maths-chapter-2-polynomials-ex-2-1\/\" target=\"_blank\" rel=\"noopener\">NCERT Class 10 Maths Solution Ex &#8211; 2.1<\/a><\/span><\/li>\n<li><span style=\"color: #0000ff;\"><a style=\"color: #0000ff;\" href=\"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-10-maths-chapter-2-polynomials-ex-2-2\/\" target=\"_blank\" rel=\"noopener\">NCERT Class 10 Maths Solution Ex &#8211; 2.2<\/a><\/span><\/li>\n<li><span style=\"color: #0000ff;\"><a style=\"color: #0000ff;\" href=\"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-10-maths-chapter-2-polynomials-ex-2-3\/\" target=\"_blank\" rel=\"noopener\">NCERT Class 10 Maths Solution Ex &#8211; 2.3<\/a><\/span><\/li>\n<\/ul>\n<h2 style=\"text-align: center;\"><strong><span style=\"color: #000000; font-family: georgia, palatino, serif;\">Exercise &#8211; 2.4\u00a0<\/span><\/strong><\/h2>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>1. Verify that the numbers given alongside of the cubic polynomials below are their zeroes. Also verify the relationship between the zeroes and the coefficients in each case:<br \/>\n<\/strong><strong>(i) 2x<sup>3 <\/sup>+ x<sup>2 <\/sup>&#8211; 5x + 2;\u00a0 \u00a0 <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\boldsymbol{-\\frac{1}{2}}\" alt=\"\\boldsymbol{-\\frac{1}{2}}\" align=\"absmiddle\" \/>, 1, -2<br \/>\n<\/strong><strong>(ii) x<sup>3 <\/sup>&#8211; 4x<sup>2 <\/sup>+ 5x &#8211; 2<\/strong>;\u00a0 \u00a0<strong>2,\u00a0 1,\u00a0 1 <\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>Solution &#8211;<br \/>\n<\/strong><strong>(i) 2x<sup>3 <\/sup>+ x<sup>2 <\/sup>&#8211; 5x + 2;\u00a0 \u00a0<img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\boldsymbol{\\frac{1}{2}}\" alt=\"\\boldsymbol{\\frac{1}{2}}\" align=\"absmiddle\" \/>, 1, -2<\/strong><br \/>\np(x) <strong>= <\/strong>2x<sup>3 <\/sup>+ x<sup>2 <\/sup>&#8211; 5x + 2<br \/>\nGiven zeroes are <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{1}{2}\" alt=\"\\frac{1}{2}\" align=\"absmiddle\" \/>, 1, &#8211; 2<\/span><br \/>\n<span style=\"color: #000000;\">Substitute x = <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{1}{2}\" alt=\"\\frac{1}{2}\" align=\"absmiddle\" \/> in p (x) = 2x<sup>3<\/sup>\u00a0+ x<sup>2<\/sup>\u00a0&#8211; 5x + 2<\/span><br \/>\n<span style=\"color: #000000;\">\u2234 p(1\/2) = 2(1\/2)<sup>3 <\/sup>+ (1\/2)<sup>2 <\/sup>&#8211; 5(1\/2) + 2 <\/span><br \/>\n<span style=\"color: #000000;\">= (1\/4) + (1\/4) &#8211; (5\/2) + 2 <\/span><br \/>\n<span style=\"color: #000000;\">= 0<\/span><br \/>\n<span style=\"color: #000000;\">Substitute x = 1\u00a0in p (x) = 2x<sup>3<\/sup>\u00a0+ x<sup>2<\/sup>\u00a0&#8211; 5x + 2<br \/>\np(1) = 2(1)<sup>3 <\/sup>+ (1)<sup>2 <\/sup>&#8211; 5(1) + 2<br \/>\n= 2 + 1 &#8211; 5 + 2<br \/>\n= 0<br \/>\nSubstitute x = -2 in p (x) = 2x<sup>3<\/sup>\u00a0+ x<sup>2<\/sup>\u00a0&#8211; 5x + 2<br \/>\np(-2) = 2(-2)<sup>3 <\/sup>+ (-2)<sup>2 <\/sup>&#8211; 5(-2) + 2<br \/>\n= 2 \u00d7 (-8) + 4 + 10 +2<br \/>\n= -16 + 16<br \/>\n= 0<br \/>\nHence, proved 1\/2, 1, -2 are the zeroes of 2x<sup>3<\/sup>+x<sup>2<\/sup>-5x+2.<br \/>\nNow, comparing the given polynomial with general expression, we get;<br \/>\n\u2234 ax<sup>3 <\/sup>+ bx<sup>2 <\/sup>+ cx + d = 2x<sup>3 <\/sup>+ x<sup>2 <\/sup>&#8211; 5x + 2<br \/>\na = 2, b = 1, c = -5 and d = 2<br \/>\nAs we know, if \u03b1, \u03b2, \u03b3 are the zeroes of the cubic polynomial ax<sup>3 <\/sup>+ bx<sup>2 <\/sup>+ cx + d , then;<br \/>\n\u03b1 + \u03b2 + \u03b3 = \u2013b\/a<br \/>\n\u03b1\u03b2 + \u03b2\u03b3 + \u03b3\u03b1 = c\/a<br \/>\n\u03b1 \u03b2\u03b3 = \u2013 d\/a.<br \/>\nTherefore, putting the values of zeroes of the polynomial,<br \/>\n\u03b1 + \u03b2 + \u03b3<br \/>\n= \u00bd + 1 + (-2)<br \/>\n= -\u00bd = \u2013b\/a<br \/>\n\u03b1\u03b2 + \u03b2\u03b3 + \u03b3\u03b1<br \/>\n= (\u00bd \u00d7 1) + (1 \u00d7 -2) + (-2 \u00d7 \u00bd)<br \/>\n= -5\/2<br \/>\n= c\/a<br \/>\n\u03b1 \u03b2 \u03b3<br \/>\n= \u00bd \u00d7 1 \u00d7 (-2)<br \/>\n= -2\/2<br \/>\n= -d\/a<br \/>\nHence, the relationship between the zeroes and the coefficients are satisfied.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(ii) x<sup>3 <\/sup>&#8211; 4x<sup>2 <\/sup>+ 5x &#8211; 2<\/strong>;\u00a0 \u00a0<strong>2,\u00a0 1,\u00a0 1<br \/>\n<\/strong>Given, p(x) = x<sup>3 <\/sup>&#8211; 4x<sup>2 <\/sup>+ 5x &#8211; 2<br \/>\nGiven zeroes are 2, 1, 1<\/span><br \/>\n<span style=\"color: #000000;\">Substitute x = 2 in p (x) = x<sup>3<\/sup>\u00a0&#8211; 4x<sup>2<\/sup>\u00a0+ 5x &#8211; 2<\/span><br \/>\n<span style=\"color: #000000;\">p (2) = (2)<sup>3<\/sup>\u00a0&#8211; 4(2)<sup>2<\/sup>\u00a0+ 5(2) &#8211; 2<\/span><br \/>\n<span style=\"color: #000000;\">p (2) = 8 &#8211; 16 + 10 &#8211; 2<\/span><br \/>\n<span style=\"color: #000000;\">p (2) = 18\u00a0&#8211; 18<\/span><br \/>\n<span style=\"color: #000000;\">p (2) = 0<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\">Substitute x = 1 in x<sup>3<\/sup>\u00a0&#8211; 4x<sup>2<\/sup>\u00a0+ 5x &#8211; 2<\/span><br \/>\n<span style=\"color: #000000;\">p (1) = (1)<sup>3<\/sup>\u00a0&#8211; 4(1)<sup>2<\/sup>\u00a0+ 5(1) &#8211; 2<\/span><br \/>\n<span style=\"color: #000000;\">p (1) = 1 &#8211; 4 + 5 &#8211; 2<\/span><br \/>\n<span style=\"color: #000000;\">p (1) = &#8211; 3 + 3<\/span><br \/>\n<span style=\"color: #000000;\">p (1) = 0<\/span><br \/>\n<span style=\"color: #000000;\">Therefore, 2,1 and 1 are the zeroes of the polynomial.<\/span><br \/>\n<span style=\"color: #000000;\">\u2234 ax<sup>3 <\/sup>+ bx<sup>2 <\/sup>+ cx + d = x<sup>3 <\/sup>&#8211; 4x<sup>2 <\/sup>+5x &#8211; 2<\/span><br \/>\n<span style=\"color: #000000;\">a = 1, b = -4, c = 5 and d = -2<\/span><br \/>\n<span style=\"color: #000000;\">Now let \u03b1 = 2, \u03b2 = 1 and \u03b3 = 1<\/span><br \/>\n<span style=\"color: #000000;\">As we know, if \u03b1, \u03b2, \u03b3 are the zeroes of the cubic polynomial ax<sup>3 <\/sup>+ bx<sup>2 <\/sup>+ cx + d , then;<br \/>\n\u03b1 + \u03b2 + \u03b3 = \u2013b\/a<br \/>\n\u03b1\u03b2 + \u03b2\u03b3 + \u03b3\u03b1 = c\/a<br \/>\n\u03b1 \u03b2 \u03b3 = \u2013 d\/a.<br \/>\nTherefore, putting the values of zeroes of the polynomial,<br \/>\n\u03b1 + \u03b2 + \u03b3<br \/>\n= 2 + 1 + 1<br \/>\n= 4<br \/>\n= -(-4)\/1 = \u2013b\/a<br \/>\n\u03b1\u03b2 + \u03b2\u03b3 + \u03b3\u03b1<br \/>\n= 2 \u00d7 1 + 1 \u00d7 1 + 1 \u00d7 2<br \/>\n= 5<br \/>\n= 5\/1 = c\/a<br \/>\n\u03b1\u03b2\u03b3<br \/>\n= 2 \u00d7 1 \u00d7 1<br \/>\n= 2<br \/>\n= -(-2)\/1 = -d\/a<br \/>\nHence, the relationship between the zeroes and the coefficients are satisfied.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>2. Find a cubic polynomial with the sum, sum of the product of its zeroes taken two at a time, and the product of its zeroes as 2, \u20137, \u201314 respectively.<br \/>\n<\/strong><strong>Solution &#8211;<br \/>\n<\/strong>Let the polynomial be ax<sup>3 <\/sup>+ bx<sup>2 <\/sup>+ cx + d and the zeroes be \u03b1, \u03b2 and \u03b3.<br \/>\nAs per the given question,<br \/>\n\u03b1 + \u03b2 + \u03b3<br \/>\n= -b\/a<br \/>\n= 2\/1<br \/>\n\u03b1\u03b2 + \u03b2\u03b3 + \u03b3\u03b1<br \/>\n= c\/a<br \/>\n= -7\/1<br \/>\n\u03b1\u03b2\u03b3<br \/>\n= -d\/a<br \/>\n= -14\/1<br \/>\nThus, from above three expressions we get the values of coefficient of polynomial.<br \/>\na = 1, b = -2, c = -7, d = 14<br \/>\nNow, substitute the values of a, b, c, and d in the cubic polynomial\u00a0ax<sup>3<\/sup>\u00a0+ bx<sup>2<\/sup>\u00a0+ cx + d.<br \/>\nHence, the cubic polynomial is x<sup>3 <\/sup>&#8211; 2x<sup>2 <\/sup>&#8211; 7x + 14<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>3. If the zeroes of the polynomial x<sup>3<\/sup>-3x<sup>2<\/sup>+x+1<\/strong> <strong>are a \u2013 b, a, a + b, find a and b.<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>Solution &#8211;<br \/>\n<\/strong>We are given with the polynomial here,<br \/>\np(x) = x<sup>3 <\/sup>&#8211; 3x<sup>2 <\/sup>+ x + 1<br \/>\nzeroes are given as a \u2013 b, a, a + b<br \/>\nNow, comparing the given polynomial with general expression, we get;<br \/>\n\u2234 px<sup>3 <\/sup>+ qx<sup>2 <\/sup>+ rx + s<br \/>\n= x<sup>3 <\/sup>&#8211; 3x<sup>2 <\/sup>+ x + 1<br \/>\np = 1, q = -3, r = 1 and s = 1<br \/>\nSum of zeroes = -q\/p<br \/>\n= a \u2013 b + a + a + b<br \/>\n= 3a<br \/>\nPutting the values q and p.<br \/>\n-q\/p = -(-3)\/1<br \/>\n3a = 3<br \/>\na = 1<br \/>\nThus, the zeroes are 1 &#8211; b, 1, 1 + b.<br \/>\nNow, product of zeroes = 1(1 &#8211; b)(1 + b)<br \/>\n-s\/p = 1 &#8211; b<sup>2<\/sup><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\">-1\/1 = 1 &#8211; b<sup>2<\/sup><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\">-1 = 1 &#8211; b<sup>2<\/sup><br \/>\nb<sup>2 <\/sup>= 1 + 1 = 2<br \/>\nb = \u00b1\u221a2<br \/>\nHence,1 &#8211; \u221a2, 1 ,1 + \u221a2 are the zeroes of x<sup>3 <\/sup>&#8211; 3x<sup>2 <\/sup>+ x + 1.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>4. If two zeroes of the polynomial x<sup>4 <\/sup>&#8211; 6x<sup>3 <\/sup>&#8211; 26x<sup>2 <\/sup>+ 138x &#8211; 35<\/strong> <strong>are 2 \u00b1 <\/strong>\u221a<strong>3,<\/strong> <strong>find other zeroes.<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>Solution &#8211;<br \/>\n<\/strong>Given polynomial is x<sup>4<\/sup>\u00a0&#8211; 6x<sup>3<\/sup>\u00a0&#8211; 26x<sup>2<\/sup>\u00a0+ 138x &#8211; 35 and the zeroes of the polynomial are 2 \u00b1 \u221a3<br \/>\nSince this is a polynomial equation of degree 4, hence there will be total 4 roots.<br \/>\nLet f(x) = x<sup>4 <\/sup>&#8211; 6x<sup>3 <\/sup>&#8211; 26x<sup>2 <\/sup>+ 138x &#8211; 35<br \/>\nSince 2 + \u221a3 and 2 &#8211; \u221a3 are zeroes of given polynomial f(x).<br \/>\n\u2234 [x \u2212 (2 + \u221a3)] [x \u2212 (2 &#8211; \u221a3)] = 0<br \/>\n(x \u2212 2 \u2212 \u221a3)(x \u2212 2 + \u221a3) = 0<br \/>\nx<sup>2 <\/sup>&#8211; 4x + 1, this is a factor of a given polynomial f(x).<br \/>\nNow, if we will divide f(x) by g(x), the quotient will also be a factor of f(x) and the remainder will be 0.<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-6159\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/06\/NCERT-Class-10-Maths-Ex2.4-Que4.png\" alt=\"NCERT Class 10 Maths Solution\" width=\"277\" height=\"279\" srcset=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/06\/NCERT-Class-10-Maths-Ex2.4-Que4.png 277w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/06\/NCERT-Class-10-Maths-Ex2.4-Que4-150x150.png 150w\" sizes=\"auto, (max-width: 277px) 100vw, 277px\" \/><br \/>\nSo, x<sup>4 <\/sup>&#8211; 6x<sup>3 <\/sup>&#8211; 26x<sup>2 <\/sup>+ 138x &#8211; 35 = (x<sup>2 <\/sup>&#8211; 4x + 1)(x<sup>2<\/sup> \u2013 2x \u2212 35)<br \/>\nNow, on further factorizing (x<sup>2 <\/sup>\u2013 2x \u2212 35) we get,<br \/>\nx<sup>2 <\/sup>\u2013 (7 \u2212 5)x \u2212 35<br \/>\n\u21d2\u00a0x<sup>2 <\/sup>\u2013 7x + 5x + 35 = 0<br \/>\n\u21d2\u00a0x(x \u22127)+5(x\u22127) = 0<br \/>\n\u21d2\u00a0(x+5)(x\u22127) = 0<br \/>\nSo, its zeroes are given by:<br \/>\nx= \u22125 and x = 7.<br \/>\nTherefore, all four zeroes of given polynomial equation are: 2 + \u221a3 , 2 &#8211; \u221a3, \u22125 and 7<strong>.<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><b>5. If the polynomial x<\/b><sup><b>4<\/b><\/sup><b> \u2013 6x<\/b><sup><b>3<\/b><\/sup><b> + 16x<\/b><sup><b>2<\/b><\/sup><b> \u2013 25x + 10 is divided by another polynomial x<\/b><sup><b>2<\/b><\/sup><b> \u2013 2x + k, the remainder comes out to be x + a, find k and a.<\/b><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><b>Solution &#8211;<br \/>\n<\/b>By division algorithm,<\/span><br \/>\n<span style=\"color: #000000;\">Dividend = Divisor \u00d7 Quotient + Remainder<\/span><br \/>\n<span style=\"color: #000000;\">Dividend \u2212 Remainder = Divisor \u00d7 Quotient<\/span><br \/>\n<span style=\"color: #000000;\">Let\u2019s divide x<sup>4<\/sup> \u2013 6x<sup>3<\/sup> + 16x<sup>2<\/sup> \u2013 25x + 10 by x<sup>2<\/sup> \u2013 2x + k.<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-6160\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/06\/NCERT-Class-10-Maths-Ex2.4-Que5.png\" alt=\"NCERT Class 10 Maths Solution\" width=\"431\" height=\"301\" srcset=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/06\/NCERT-Class-10-Maths-Ex2.4-Que5.png 431w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/06\/NCERT-Class-10-Maths-Ex2.4-Que5-300x210.png 300w\" sizes=\"auto, (max-width: 431px) 100vw, 431px\" \/><br \/>\nIt can be observed that (-10 + 2k)x + (10 &#8211; a &#8211; 8k + k<sup>2<\/sup>) will be 0.<br \/>\nTherefore, (-10 + 2k) = 0 and (10 &#8211; a &#8211; 8k + k<sup>2<\/sup>) = 0<br \/>\nFor (-10 + 2k) = 0,<\/span><br \/>\n<span style=\"color: #000000;\">2<i>k<\/i>\u00a0 = 10<\/span><br \/>\n<span style=\"color: #000000;\"><i>k<\/i> = 5<\/span><\/p>\n<p><span style=\"color: #000000;\">For (10 &#8211; a &#8211; 8k + k<sup>2<\/sup>) = 0\u00a0<\/span><br \/>\n<span style=\"color: #000000;\">10 \u2212 <i>a<\/i> \u2212 8 \u00d7 5 + 25 = 0<\/span><br \/>\n<span style=\"color: #000000;\">10 \u2212 <i>a<\/i> \u2212 40 + 25 = 0<\/span><br \/>\n<span style=\"color: #000000;\">\u2212 5 \u2212 <i>a<\/i> = 0<\/span><br \/>\n<span style=\"color: #000000;\"><i>a<\/i> = \u22125<\/span><br \/>\n<span style=\"color: #000000;\"><i><\/i>Therefore, k = 5 and a = -5.<\/span><\/p>\n<p><a href=\"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-10-maths\/\"><span style=\"color: #0000ff;\"><b><i>Go Back To Chapters<\/i><\/b><\/span><\/a><\/p>\n","protected":false},"excerpt":{"rendered":"<p>NCERT Solutions Class 10 Maths\u00a0 Chapter &#8211; 2 (Polynomials)\u00a0 The NCERT Solutions in English Language for Class 10 Mathematics Chapter &#8211; 2 Polynomials Exercise 2.4 has been provided here to help the students in solving the questions from this exercise.\u00a0 Chapter : 2 Polynomials NCERT Class 10 Maths Solution Ex &#8211; 2.1 NCERT Class 10 [&hellip;]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1043],"tags":[1045,1053,1057,1044,1049,1048],"class_list":["post-6013","post","type-post","status-publish","format-standard","hentry","category-class-10-maths","tag-class-10-ncert-mathematics-solutions","tag-ncert-class-10-mathematics-chapter-2-polynomials-solutions","tag-ncert-class-10-mathematics-exercise-2-4-solutions","tag-ncert-class-10-mathematics-solutions-class-10-ncert-solutions","tag-ncert-solutions-class-10-mathematics","tag-ncert-solutions-class-10-maths"],"yoast_head":"<!-- This site is optimized with the Yoast SEO Premium plugin v22.9 (Yoast SEO v27.3) - 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