{"id":6012,"date":"2023-07-19T06:52:11","date_gmt":"2023-07-19T06:52:11","guid":{"rendered":"https:\/\/theexampillar.com\/ncert\/?p=6012"},"modified":"2023-07-23T05:43:20","modified_gmt":"2023-07-23T05:43:20","slug":"ncert-solutions-class-10-maths-chapter-2-polynomials-ex-2-3","status":"publish","type":"post","link":"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-10-maths-chapter-2-polynomials-ex-2-3\/","title":{"rendered":"NCERT Solutions Class 10 Maths Chapter 2 Polynomials Ex 2.3"},"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 2.3 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-2-4\/\" target=\"_blank\" rel=\"noopener\">NCERT Class 10 Maths Solution Ex &#8211; 2.4<\/a><\/span><\/li>\n<\/ul>\n<h2 style=\"text-align: center;\"><strong><span style=\"color: #000000; font-family: georgia, palatino, serif;\">Exercise &#8211; 2.3\u00a0<\/span><\/strong><\/h2>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>1. Divide the polynomial p(x) by the polynomial g(x) and find the quotient and remainder in each of the following:<br \/>\n<\/strong><strong>(i)<\/strong> <strong>p(x) = x<sup>3 <\/sup>\u2013\u00a03x<sup>2 <\/sup>+ 5x \u2013 3,\u00a0 \u00a0 \u00a0 \u00a0g(x) = x<sup>2 <\/sup>\u2013 2<br \/>\n(ii) p(x) = x<sup>4 <\/sup>\u2013 3x<sup>2 <\/sup>+ 4x + 5,\u00a0 \u00a0 \u00a0 g(x) = x<sup>2 <\/sup>+ 1 \u2013 x<br \/>\n(iii) p(x) =x<sup>4 <\/sup>\u2013 5x + 6,\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 g(x) = 2 \u2013 x<sup>2<\/sup><\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>Solution &#8211;<br \/>\n<\/strong><strong>(i)<\/strong> <strong>p(x) = x<sup>3 <\/sup>\u2013\u00a03x<sup>2 <\/sup>+ 5x \u2013 3,\u00a0 \u00a0 \u00a0 \u00a0g(x) = x<sup>2 <\/sup>\u2013 2<br \/>\n<\/strong>Given,<br \/>\nDividend = p(x) = x<sup>3 <\/sup>\u2013\u00a03x<sup>2 <\/sup>+ 5x \u2013 3<br \/>\nDivisor = g(x) = x<sup>2 <\/sup>\u2013 2<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-6146\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/06\/NCERT-Class-10-Maths-Ex2.3-Q1-i.png\" alt=\"NCERT Class 10 Maths Solution \" width=\"163\" height=\"203\" \/><br \/>\nQuotient = x \u2013 3<br \/>\nRemainder = 7x \u2013 9<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(ii) p(x) = x<sup>4 <\/sup>\u2013 3x<sup>2 <\/sup>+ 4x + 5,\u00a0 \u00a0 \u00a0 g(x) = x<sup>2 <\/sup>+ 1 \u2013 x<br \/>\n<\/strong>Dividend = p(x) = x<sup>4 <\/sup>\u2013 3x<sup>2 <\/sup>+ 4x +5<br \/>\nDivisor = g(x) = x<sup>2<\/sup> + 1 \u2013 x<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-6147\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/06\/NCERT-Class-10-Maths-Ex2.3-Q1-ii.png\" alt=\"NCERT Class 10 Maths Solution \" width=\"224\" height=\"277\" \/><br \/>\nQuotient = x<sup>2 <\/sup>+ x\u20133<br \/>\nRemainder = 8<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(iii) p(x) =x<sup>4 <\/sup>\u2013 5x + 6,\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 g(x) = 2 \u2013 x<sup>2<br \/>\n<\/sup><\/strong>Dividend = p(x) =x<sup>4<\/sup> \u2013 5x + 6 = x<sup>4 <\/sup>+ 0x<sup>2 <\/sup>\u2013 5x + 6<br \/>\nDivisor = g(x) = 2 \u2013 x<sup>2<\/sup> = \u2013 x<sup>2 <\/sup>+ 2<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-6148\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/06\/NCERT-Class-10-Maths-Ex2.3-Q1-iii.png\" alt=\"NCERT Class 10 Maths Solution \" width=\"241\" height=\"201\" \/><br \/>\nQuotient = -x<sup>2 <\/sup>&#8211; 2<br \/>\nRemainder = &#8211; 5x + 10<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>2. Check whether the first polynomial is a factor of the second polynomial by dividing the second polynomial by the first polynomial:<br \/>\n<\/strong><strong>(i) t<sup>2 <\/sup>&#8211; 3, 2t<sup>4 <\/sup>+ 3t<sup>3 <\/sup>&#8211; 2t<sup>2 <\/sup>&#8211; 9t &#8211; 12<br \/>\n(ii) x<sup>2 <\/sup>+ 3x + 1 , 3x<sup>4 <\/sup>+ 5x<sup>3 <\/sup>&#8211; 7x<sup>2 <\/sup>+ 2x + 2<br \/>\n(iii) x<sup>3 <\/sup>&#8211; 3x + 1, x<sup>5 <\/sup>&#8211; 4x<sup>3 <\/sup>+ x<sup>2 <\/sup>+ 3x + 1 <\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>Solutions &#8211;<br \/>\n<\/strong><strong>(i) t<sup>2 <\/sup>&#8211; 3, 2t<sup>4 <\/sup>+ 3t<sup>3 <\/sup>&#8211; 2t<sup>2 <\/sup>&#8211; 9t &#8211; 12<br \/>\n<\/strong>First polynomial = t<sup>2 <\/sup>&#8211; 3<br \/>\nSecond polynomial = 2t<sup>4 <\/sup>+ 3t<sup>3 <\/sup>&#8211; 2t<sup>2 <\/sup>&#8211; 9t &#8211; 12<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-6150\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/06\/NCERT-Class-10-Maths-Ex2.3-Q2-i.png\" alt=\"NCERT Class 10 Maths Solution\" width=\"311\" height=\"286\" srcset=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/06\/NCERT-Class-10-Maths-Ex2.3-Q2-i.png 488w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/06\/NCERT-Class-10-Maths-Ex2.3-Q2-i-300x275.png 300w\" sizes=\"auto, (max-width: 311px) 100vw, 311px\" \/><br \/>\nAs we can see, the remainder is left as 0. Therefore, we say that, t<sup>2 <\/sup>&#8211; 3 is a factor of 2t<sup>4\u00a0 <\/sup>+ 3t<sup>3 <\/sup>&#8211; 2t<sup>2 <\/sup>&#8211; 9t &#8211; 12.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(ii) x<sup>2 <\/sup>+ 3x + 1 , 3x<sup>4 <\/sup>+ 5x<sup>3 <\/sup>&#8211; 7x<sup>2 <\/sup>+ 2x + 2<br \/>\n<\/strong>First polynomial = x<sup>2 <\/sup>+ 3x + 1<br \/>\nSecond polynomial = 3x<sup>4 <\/sup>+ 5x<sup>3 <\/sup>&#8211; 7x<sup>2 <\/sup>+ 2x + 2<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-6151\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/06\/NCERT-Class-10-Maths-Ex2.3-Q2-ii.png\" alt=\"NCERT Class 10 Maths Solution\" width=\"317\" height=\"282\" srcset=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/06\/NCERT-Class-10-Maths-Ex2.3-Q2-ii.png 404w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/06\/NCERT-Class-10-Maths-Ex2.3-Q2-ii-300x267.png 300w\" sizes=\"auto, (max-width: 317px) 100vw, 317px\" \/><br \/>\nAs we can see, the remainder is left as 0. Therefore, we say that, x<sup>2<\/sup> + 3x + 1 is a factor of 3x<sup>4 <\/sup>+ 5x<sup>3 <\/sup>&#8211; 7x<sup>2 <\/sup>+ 2x + 2.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(iii) x<sup>3 <\/sup>&#8211; 3x + 1, x<sup>5 <\/sup>&#8211; 4x<sup>3 <\/sup>+ x<sup>2 <\/sup>+ 3x + 1<br \/>\n<\/strong>First polynomial = x<sup>3 <\/sup>&#8211; 3x + 1<br \/>\nSecond polynomial = x<sup>5 <\/sup>&#8211; 4x<sup>3 <\/sup>+ x<sup>2 <\/sup>+ 3x + 1<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-6152\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/06\/NCERT-Class-10-Maths-Ex2.3-Q2-iii.png\" alt=\"NCERT Class 10 Maths Solution\" width=\"314\" height=\"183\" srcset=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/06\/NCERT-Class-10-Maths-Ex2.3-Q2-iii.png 500w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/06\/NCERT-Class-10-Maths-Ex2.3-Q2-iii-300x175.png 300w\" sizes=\"auto, (max-width: 314px) 100vw, 314px\" \/><br \/>\nAs we can see, the remainder is not equal to 0. Therefore, we say that, x<sup>3 <\/sup>&#8211; 3x + 1 is not a factor of x<sup>5 <\/sup>&#8211; 4x<sup>3 <\/sup>+ x<sup>2 <\/sup>+ 3x + 1 .<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>3. Obtain all other zeroes of 3x<sup>4 <\/sup>+ 6x<sup>3 <\/sup>&#8211; 2x<sup>2 <\/sup>&#8211; 10x &#8211; 5, if two of its zeroes are <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\mathbf{\\sqrt{\\frac{5}{3}}}\" alt=\"\\mathbf{\\sqrt{\\frac{5}{3}}}\" align=\"absmiddle\" \/> and \u2013 <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\mathbf{\\sqrt{\\frac{5}{3}}}\" alt=\"\\mathbf{\\sqrt{\\frac{5}{3}}}\" align=\"absmiddle\" \/>.<br \/>\n<\/strong><strong>Solutions &#8211;<br \/>\n<\/strong>p (x) = 3x<sup>4 <\/sup>+ 6x<sup>3 <\/sup>&#8211; 2x<sup>2 <\/sup>&#8211; 10x &#8211; 5<br \/>\nSince the two zeroes are <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\mathbf{\\sqrt{\\frac{5}{3}}}\" alt=\"\\mathbf{\\sqrt{\\frac{5}{3}}}\" align=\"absmiddle\" \/> and \u2013 <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\mathbf{\\sqrt{\\frac{5}{3}}}\" alt=\"\\mathbf{\\sqrt{\\frac{5}{3}}}\" align=\"absmiddle\" \/>.<br \/>\n\u2234 <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\left&amp;space;(&amp;space;x-\\sqrt{\\frac{5}{3}}&amp;space;\\right&amp;space;)\\left&amp;space;(&amp;space;x+\\sqrt{\\frac{5}{3}}&amp;space;\\right&amp;space;)\" alt=\"\\left ( x-\\sqrt{\\frac{5}{3}} \\right )\\left ( x+\\sqrt{\\frac{5}{3}} \\right )\" align=\"absmiddle\" \/> = <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\left&amp;space;(&amp;space;x^2&amp;space;-&amp;space;\\frac{5}{3}\\right&amp;space;)\" alt=\"\\left ( x^2 - \\frac{5}{3}\\right )\" align=\"absmiddle\" \/> is a factor of 3x<sup>4 <\/sup>+ 6x<sup>3 <\/sup>&#8211; 2x<sup>2 <\/sup>&#8211; 10x &#8211; 5.<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-6153\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/06\/NCERT-Class-10-Maths-Ex2.3-Que3.png\" alt=\"NCERT Class 10 Maths Solution\" width=\"310\" height=\"311\" srcset=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/06\/NCERT-Class-10-Maths-Ex2.3-Que3.png 401w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/06\/NCERT-Class-10-Maths-Ex2.3-Que3-300x301.png 300w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/06\/NCERT-Class-10-Maths-Ex2.3-Que3-150x150.png 150w\" sizes=\"auto, (max-width: 310px) 100vw, 310px\" \/><\/span><br \/>\n<span style=\"color: #000000;\">Therefore, 3x<sup>4<\/sup>\u00a0+ 6x<sup>3<\/sup>\u00a0&#8211; 2x<sup>2<\/sup> &#8211; 10x &#8211; 5 = <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\left&amp;space;(&amp;space;x^2&amp;space;-&amp;space;\\frac{5}{3}\\right&amp;space;)\" alt=\"\\left ( x^2 - \\frac{5}{3}\\right )\" align=\"absmiddle\" \/> (3x<sup>2<\/sup>\u00a0+ 6x + 3) + 0<\/span><br \/>\n<span style=\"color: #000000;\">= 3\u00a0(x<sup>2<\/sup> &#8211; <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{5}{3}\" alt=\"\\frac{5}{3}\" align=\"absmiddle\" \/>) (x<sup>2<\/sup>\u00a0+ 2x + 1)<\/span><br \/>\n<span style=\"color: #000000;\">On factorizing\u00a0x<sup>2<\/sup>\u00a0+ 2x + 1, we get (x + 1)<sup>2<br \/>\n<\/sup>Therefore, its zero is given by<\/span><br \/>\n<span style=\"color: #000000;\">x + 1 = 0<\/span><br \/>\n<span style=\"color: #000000;\">x = &#8211; 1<\/span><br \/>\n<span style=\"color: #000000;\">As it has the term (x + 1)<sup>2<\/sup>,<\/span><br \/>\n<span style=\"color: #000000;\">Therefore, there will be two identical zeroes at x = &#8211; 1<\/span><br \/>\n<span style=\"color: #000000;\">Hence the zeroes of the given polynomial are <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\mathbf{\\sqrt{\\frac{5}{3}}}\" alt=\"\\mathbf{\\sqrt{\\frac{5}{3}}}\" align=\"absmiddle\" \/> and \u2013 <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\mathbf{\\sqrt{\\frac{5}{3}}}\" alt=\"\\mathbf{\\sqrt{\\frac{5}{3}}}\" align=\"absmiddle\" \/>, &#8211; 1 and &#8211; 1.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>4. On dividing x<sup>3<\/sup>-3x<sup>2<\/sup>+x+2<\/strong> <strong>by a polynomial g(x), the quotient and remainder were x\u20132 and \u20132x+4, respectively. Find g(x).<br \/>\n<\/strong><strong>Solution &#8211;<br \/>\n<\/strong>Dividend, p(x) = x<sup>3 <\/sup>&#8211; 3x<sup>2 <\/sup>+ x + 2<br \/>\nQuotient = x &#8211; 2<br \/>\nRemainder = \u20132x + 4<br \/>\nWe have to find the value of Divisor, g(x) =?<br \/>\nDividend = Divisor \u00d7 Quotient + Remainder<br \/>\n\u2234 x<sup>3 <\/sup>&#8211; 3x<sup>2 <\/sup>+ x + 2 = g(x) \u00d7 (x &#8211; 2) + (-2x + 4)<br \/>\nx<sup>3 <\/sup>&#8211; 3x<sup>2 <\/sup>+ x + 2 &#8211; (-2x + 4) = g(x) \u00d7 (x &#8211; 2)<br \/>\nTherefore, g(x) \u00d7 (x &#8211; 2) = x<sup>3 <\/sup>&#8211; 3x<sup>2 <\/sup>+ 3x &#8211; 2<br \/>\nNow, for finding g(x) we will divide x<sup>3 <\/sup>&#8211; 3x<sup>2 <\/sup>+ 3x &#8211; 2 with (x &#8211; 2)<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-6154\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/06\/NCERT-Class-10-Maths-Ex2.3-Que4.png\" alt=\"NCERT Class 10 Maths Solution\" width=\"230\" height=\"303\" srcset=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/06\/NCERT-Class-10-Maths-Ex2.3-Que4.png 303w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/06\/NCERT-Class-10-Maths-Ex2.3-Que4-228x300.png 228w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/06\/NCERT-Class-10-Maths-Ex2.3-Que4-300x395.png 300w\" sizes=\"auto, (max-width: 230px) 100vw, 230px\" \/><br \/>\nTherefore, g(x) = (x<sup>2 <\/sup>\u2013 x + 1)<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>5. Give examples of polynomials p(x), g(x), q(x) and r(x), which satisfy the division algorithm and<br \/>\n<\/strong><strong>(i) deg p(x) = deg q(x)<br \/>\n<\/strong><strong>(ii) deg q(x) = deg r(x)<br \/>\n<\/strong><strong>(iii) deg r(x) = 0<br \/>\n<\/strong><strong>Solutions &#8211;\u00a0 <\/strong>According to the division algorithm, dividend p(x) and divisor g(x) are two polynomials, where g(x)\u22600. Then we can find the value of quotient q(x) and remainder r(x), with the help of below given formula;<br \/>\nDividend = Divisor \u00d7 Quotient + Remainder<br \/>\n\u2234 p(x) = g(x)\u00d7q(x)+r(x)<br \/>\nWhere r(x) = 0 or degree of r(x)&lt; degree of g(x).<br \/>\nNow let us proof the three given cases as per division algorithm by taking examples for each.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(i) deg p(x) = deg q(x)<br \/>\n<\/strong>Degree of dividend is equal to degree of quotient, only when the divisor is a constant term.<br \/>\nLet us take an example, p(x) = 3x<sup>2<\/sup>+3x+3 is a polynomial to be divided by g(x) = 3.<br \/>\nSo, (3x<sup>2<\/sup>+3x+3)\/3 = x<sup>2<\/sup>+x+1 = q(x)<br \/>\nThus, you can see, the degree of quotient q(x) = 2, which also equal to the degree of dividend p(x).<br \/>\nHence, division algorithm is satisfied here.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(ii) deg q(x) = deg r(x)<br \/>\n<\/strong>Let us take an example, p(x) = x<sup>2\u00a0<\/sup>+ 3 is a polynomial to be divided by g(x) = x \u2013 1.<br \/>\nSo,\u00a0x<sup>2\u00a0<\/sup>+ 3 = (x \u2013 1)\u00d7(x) + (x + 3)<br \/>\nHence, quotient q(x) = x<br \/>\nAlso, remainder r(x) = x + 3<br \/>\nThus, you can see, the degree of quotient q(x) = 1, which is also equal to the degree of remainder r(x).<br \/>\nHence, division algorithm is satisfied here.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(iii) deg r(x) = 0<br \/>\n<\/strong>The degree of remainder is 0 only when the remainder left after division algorithm is constant.<br \/>\nLet us take an example, p(x) = x<sup>2\u00a0<\/sup>+ 1 is a polynomial to be divided by g(x) = x.<br \/>\nSo, x<sup>2\u00a0<\/sup>+ 1 = (x)\u00d7(x) + 1<br \/>\nHence, quotient q(x) = x<br \/>\nAnd, remainder r(x) = 1<br \/>\nClearly, the degree of remainder here is 0.<br \/>\nHence, division algorithm is satisfied here.<\/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 2.3 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":[1045,1053,1056,1044,1049,1048],"class_list":["post-6012","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-2-3-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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