{"id":3495,"date":"2021-07-08T11:10:35","date_gmt":"2021-07-08T11:10:35","guid":{"rendered":"https:\/\/themindpalace.in\/?p=3495"},"modified":"2021-08-26T06:56:56","modified_gmt":"2021-08-26T06:56:56","slug":"polynomials","status":"publish","type":"post","link":"https:\/\/themindpalace.in\/index.php\/2021\/07\/08\/polynomials\/","title":{"rendered":"Polynomials"},"content":{"rendered":"\n<ul class=\"wp-block-list\"><li>Polynomials<\/li><li>Introduction<\/li><li>Polynomials In One Variable<\/li><li>Zeroes Of A Polynomial<\/li><li>Remainder Theorem<\/li><li>Factorisation Of Polynomials<\/li><li>Algebraic Identities<\/li><li>Summary<\/li><\/ul>\n\n\n\n<p class=\"wp-block-paragraph\">Polynomial is an algebraic expression that includes constants, variables, and exponents. It is the expression in which the variables have only positive integral.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Example<\/strong><\/h3>\n\n\n\n<ol class=\"wp-block-list\"><li> 4x<sup>3<\/sup>&nbsp;+ 3x<sup>2<\/sup>&nbsp;+ x +3 is a polynomial in variable x.<\/li><li> 4x<sup>2<\/sup>&nbsp;+ 3x<sup>-1<\/sup>&nbsp;&#8211; 4 is not a polynomial as it has negative power.<\/li><li> 3x<sup>3\/2&nbsp;<\/sup>+ 2x \u2013 3&nbsp;is not a polynomial.<\/li><\/ol>\n\n\n\n<div class=\"wp-block-image\"><figure class=\"aligncenter size-large is-resized\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/themindpalace.in\/wp-content\/uploads\/2021\/06\/polynomials-9-pic1.png\" alt=\"Polynomial\" class=\"wp-image-3496\" width=\"671\" height=\"395\" srcset=\"https:\/\/themindpalace.in\/wp-content\/uploads\/2021\/06\/polynomials-9-pic1.png 895w, https:\/\/themindpalace.in\/wp-content\/uploads\/2021\/06\/polynomials-9-pic1-300x176.png 300w, https:\/\/themindpalace.in\/wp-content\/uploads\/2021\/06\/polynomials-9-pic1-768x451.png 768w\" sizes=\"auto, (max-width: 671px) 100vw, 671px\" \/><figcaption>Polynomial <\/figcaption><\/figure><\/div>\n\n\n\n<ul class=\"wp-block-list\"><li>Polynomials are denoted by p(x), q(x) etc.<\/li><li>In the above polynomial 2x<sup>2<\/sup>, 3y and 2 are the terms of the polynomial.<\/li><li>2 and 3 are the coefficient of the x<sup>2<\/sup>&nbsp;and y respectively.<\/li><li>x and y are the variables.<\/li><li>2 is the constant term which has no variable.<\/li><\/ul>\n\n\n\n<div class=\"wp-block-kadence-spacer aligncenter kt-block-spacer-_fb706e-17\"><div class=\"kt-block-spacer kt-block-spacer-halign-center\" style=\"height:60px\"><hr class=\"kt-divider\" style=\"border-top-color:rgba(238, 238, 238, 1);border-top-width:1px;width:80%;border-top-style:solid\"\/><\/div><\/div>\n\n\n\n<h3 class=\"wp-block-heading\">Polynomials in One Variable<\/h3>\n\n\n\n<p class=\"wp-block-paragraph\">If there is only one variable in the expression, then this is called the polynomial in one variable.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>Example<\/strong><br><\/p>\n\n\n\n<ul class=\"wp-block-list\"><li>x<sup>3<\/sup>&nbsp;+ x \u2013 4 is polynomial in variable x and is denoted by p(x).<\/li><li>r<sup>2<\/sup>&nbsp;+ 2 is polynomial in variable r and is denoted by p(r).<\/li><\/ul>\n\n\n\n<h3 class=\"wp-block-heading\">Types of polynomials on the basis of the number of terms<\/h3>\n\n\n\n<figure class=\"wp-block-table aligncenter is-style-stripes\"><table><thead><tr><th>Number of non zero terms<\/th><th>Name<\/th><th>Example<\/th><\/tr><\/thead><tbody><tr><td>0<\/td><td>Zero Polinomial<\/td><td>0<\/td><\/tr><tr><td>1<\/td><td>Monomial<\/td><td>X<sup>2<\/sup><\/td><\/tr><tr><td>2<\/td><td>Binomial<\/td><td>X<sup>2<\/sup>+1<\/td><\/tr><tr><td>3<\/td><td>Trinomial<\/td><td>X<sup>3<\/sup>+1<\/td><\/tr><\/tbody><\/table><figcaption>Types of polynomials on the basis of the number of terms<\/figcaption><\/figure>\n\n\n\n<div class=\"wp-block-kadence-spacer aligncenter kt-block-spacer-_760a58-e6\"><div class=\"kt-block-spacer kt-block-spacer-halign-center\" style=\"height:60px\"><hr class=\"kt-divider\" style=\"border-top-color:rgba(238, 238, 238, 1);border-top-width:1px;width:80%;border-top-style:solid\"\/><\/div><\/div>\n\n\n\n<h3 class=\"wp-block-heading\">Types of polynomials on the basis of the number of degrees<\/h3>\n\n\n\n<p class=\"wp-block-paragraph\"><\/p>\n\n\n\n<p class=\"wp-block-paragraph\">The highest value of the power of the variable in the polynomial is the degree of the polynomial.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\"><\/p>\n\n\n\n<figure class=\"wp-block-image size-large\"><img loading=\"lazy\" decoding=\"async\" width=\"1024\" height=\"521\" src=\"https:\/\/themindpalace.in\/wp-content\/uploads\/2021\/07\/Types-of-Polynomials-1024x521.png\" alt=\"Types of polynomials \" class=\"wp-image-3588\" srcset=\"https:\/\/themindpalace.in\/wp-content\/uploads\/2021\/07\/Types-of-Polynomials-1024x521.png 1024w, https:\/\/themindpalace.in\/wp-content\/uploads\/2021\/07\/Types-of-Polynomials-300x153.png 300w, https:\/\/themindpalace.in\/wp-content\/uploads\/2021\/07\/Types-of-Polynomials-768x391.png 768w, https:\/\/themindpalace.in\/wp-content\/uploads\/2021\/07\/Types-of-Polynomials-1536x782.png 1536w, https:\/\/themindpalace.in\/wp-content\/uploads\/2021\/07\/Types-of-Polynomials.png 1880w\" sizes=\"auto, (max-width: 1024px) 100vw, 1024px\" \/><figcaption>Types of polynomials<\/figcaption><\/figure>\n\n\n\n<p class=\"wp-block-paragraph\"><\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Zeroes of a Polynomial<\/h3>\n\n\n\n<p class=\"wp-block-paragraph\">If p(x) is a polynomial, then the number \u2018a\u2019 will be the zero of the polynomial with p(a) = 0. We can find the zero of the polynomial by&nbsp;<strong>equating it to zero<\/strong>.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>Example: 1<\/strong><\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Given polynomial is p(x) = x &#8211; 4<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">To find the zero of the polynomial we will equate it to zero.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">x &#8211; 4 = 0<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">x = 4<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">p(4) = x \u2013 4 = 4 \u2013 4 = 0<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">This shows that if we place 4 in place of x, we got the value of the polynomial as zero. So 4 is the zero of this polynomial. And also we are getting the value 4 by equating the polynomial by 0.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">So 4 is the zero of the polynomial or root of the polynomial.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">The\u00a0<strong>root of the polynomial<\/strong>\u00a0is basically the\u00a0<strong>x-intercept<\/strong>\u00a0of the polynomial.<\/p>\n\n\n\n<div class=\"wp-block-image\"><figure class=\"aligncenter size-large\"><img loading=\"lazy\" decoding=\"async\" width=\"282\" height=\"127\" src=\"https:\/\/themindpalace.in\/wp-content\/uploads\/2021\/07\/image.png\" alt=\"\" class=\"wp-image-3592\"\/><figcaption>the root of the polynomial<\/figcaption><\/figure><\/div>\n\n\n\n<p class=\"wp-block-paragraph\">If the polynomial has one root, it will intersect the x-axis at one point only and, if it has two roots then it will intersect at two points and so on.<\/p>\n\n\n\n<div class=\"wp-block-kadence-spacer aligncenter kt-block-spacer-_dc7e34-6e\"><div class=\"kt-block-spacer kt-block-spacer-halign-center\" style=\"height:60px\"><hr class=\"kt-divider\" style=\"border-top-color:rgba(238, 238, 238, 1);border-top-width:1px;width:80%;border-top-style:solid\"\/><\/div><\/div>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>Example: 2<\/strong><\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Find p (1) for the polynomial p (t) = t<sup>2<\/sup>&nbsp;\u2013 t + 1<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">p (1) = (1)<sup>2<\/sup>&nbsp;\u2013 1 + 1<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">= 1 \u2013 1 + 1<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">= 1<\/p>\n\n\n\n<div class=\"wp-block-kadence-spacer aligncenter kt-block-spacer-_9a0df5-cb\"><div class=\"kt-block-spacer kt-block-spacer-halign-center\" style=\"height:60px\"><hr class=\"kt-divider\" style=\"border-top-color:rgba(238, 238, 238, 1);border-top-width:1px;width:80%;border-top-style:solid\"\/><\/div><\/div>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>Example 2<\/strong>.<strong>1 <\/strong><\/p>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>Which of the following expressions are polynomials in one variable and which are not? State reasons for your answer.<\/strong><br><strong><br><\/strong>(i) 4x<sup>2<\/sup>\u00a0\u2013 3x + 7<br>(ii) y<sup>2<\/sup>\u00a0+ \u221a2<br>(iii) 3 \u221at + t\u221a2<br>(iv) y+\u00a02\/y<br>(v) x<sup>10<\/sup>+ y<sup>3<\/sup>+t<sup>50<\/sup><\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Solution:<br>(i) We have 4x<sup>2<\/sup>&nbsp;\u2013 3x + 7 = 4x<sup>2<\/sup>&nbsp;\u2013 3x + 7x<sup>0<\/sup><br>It is a polynomial in one variable i.e., x<br>because each exponent of x is a whole number.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">(ii) We have y<sup>2<\/sup>&nbsp;+ \u221a2 = y<sup>2<\/sup>&nbsp;+ \u221a2y<sup>0<\/sup><br>It is a polynomial in one variable i.e., y<br>because each exponent of y is a whole number.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">(iii) We have 3 \u221at + t\u221a2 = 3 \u221at<sup>1\/2<\/sup>&nbsp;+ \u221a2.t<br>It is not a polynomial, because one of the exponents of t is&nbsp;1\/2,<br>which is not a whole number.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">(iv) We have y +&nbsp;y+2\/y&nbsp;= y + 2.y<sup>-1<\/sup><br>It is not a polynomial, because one of the exponents of y is -1,<br>which is not a whole number.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">(v) We have x<sup>10<\/sup>+\u00a0 y<sup>3\u00a0<\/sup>+ t<sup>50<\/sup><br>Here, the exponent of every variable is a whole number, but x<sup>10<\/sup>\u00a0+ y<sup>3<\/sup>\u00a0+ t<sup>50<\/sup>\u00a0is a polynomial in x, y, and t, i.e., in three variables.<br>So, it is not a polynomial in one variable.<\/p>\n\n\n\n<div class=\"wp-block-kadence-spacer aligncenter kt-block-spacer-_5f934c-55\"><div class=\"kt-block-spacer kt-block-spacer-halign-center\" style=\"height:60px\"><hr class=\"kt-divider\" style=\"border-top-color:rgba(238, 238, 238, 1);border-top-width:1px;width:80%;border-top-style:solid\"\/><\/div><\/div>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>2. Write the coefficients of x<sup>2<\/sup>\u00a0in each of the following<br><\/strong><br>(i) 2 + x<sup>2<\/sup>\u00a0+ x<br>(ii) 2 \u2013 x<sup>2<\/sup>\u00a0+ x<sup>3<\/sup><br>(iii)\u00a0\u03c0\/2\u00a0x<sup>2<\/sup>\u00a0+ x<br>(iv) \u221a2 x \u2013 1<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Solution:<br>(i) The given polynomial is 2 + x<sup>2<\/sup>&nbsp;+ x.<br>The coefficient of x<sup>2<\/sup>&nbsp;is 1.<br>(ii) The given polynomial is 2 \u2013 x<sup>2<\/sup>&nbsp;+ x<sup>3<\/sup>.<br>The coefficient of x<sup>2<\/sup>&nbsp;is -1.<br>(iii) The given polynomial is&nbsp;\u03c0\/2 x 2&nbsp;+ x.<br>The coefficient of x<sup>2<\/sup>&nbsp;is&nbsp;\u03c0\/2.<br>(iv) The given polynomial is \u221a2 x \u2013 1.<br>The coefficient of x<sup>2<\/sup>&nbsp;is 0.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>3. Give one example each of a binomial of degree 35, and of a monomial of degree 100.<br><\/strong><br>Solution:<br>(i) A binomial of degree 35 can be 3x<sup>35<\/sup>\u00a0-4.<br>(ii) A monomial of degree 100 can be \u221a2y<sup>100<\/sup>.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>4. Write the degree of each of the following polynomials.<\/strong><br><strong><br><\/strong>(i) 5x<sup>3<\/sup>+4x<sup>2<\/sup>\u00a0+ 7x<br>(ii) 4 \u2013 y<sup>2<\/sup><br>(iii) 5t \u2013 \u221a7<br>(iv) 3<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Solution:<br>(i) The given polynomial is 5x<sup>3<\/sup>&nbsp;+ 4x<sup>2<\/sup>&nbsp;+ 7x.<br>The highest power of the variable x is 3.So, the degree of the polynomial is 3.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\"><br>(ii) The given polynomial is 4- y<sup>2<\/sup>. The highest power of the variable y is 2.<br>So, the degree of the polynomial is 2.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\"><br>(iii) The given polynomial is 5t \u2013 \u221a7 . The highest power of variable t is 1. So, the degree of the polynomial is 1.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\"><br>(iv) Since, 3 = 3x\u00b0 [\u2235 x\u00b0=1]<br>So, the degree of the polynomial is 0.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>5. Classify the following as linear, quadratic and cubic polynomials.<\/strong><br><strong><br><\/strong>(i) x<sup>2<\/sup>+ x<br>(ii) x \u2013 x<sup>3<\/sup><br>(iii) y + y<sup>2<\/sup>+4<br>(iv) 1 + x<br>(v) 3t<br>(vi) r<sup>2<\/sup><br>(vii) 7x<sup>3<\/sup><\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Solution:<br>(i) The degree of x<sup>2<\/sup>\u00a0+ x is 2. So, it is a quadratic polynomial.<br>(ii) The degree of x \u2013 x<sup>3<\/sup>\u00a0is 3. So, it is a cubic polynomial.<br>(iii) The degree of y + y<sup>2<\/sup>\u00a0+ 4 is 2. So, it is a quadratic polynomial.<br>(iv) The degree of 1 + x is 1. So, it is a linear polynomial.<br>(v) The degree of 3t is 1. So, it is a linear polynomial.<br>(vi) The degree of r<sup>2<\/sup>\u00a0is 2. So, it is a quadratic polynomial.<br>(vii) The degree of 7x<sup>3<\/sup>\u00a0is 3. So, it is a cubic polynomial.<\/p>\n\n\n\n<div class=\"wp-block-kadence-spacer aligncenter kt-block-spacer-_95e4eb-93\"><div class=\"kt-block-spacer kt-block-spacer-halign-center\" style=\"height:60px\"><hr class=\"kt-divider\" style=\"border-top-color:rgba(238, 238, 238, 1);border-top-width:1px;width:80%;border-top-style:solid\"\/><\/div><\/div>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>Example 2.2<\/strong><\/p>\n\n\n\n<ol class=\"wp-block-list\"><li><strong>Find the value of the polynomial 5x \u2013 4x<sup>2<\/sup>\u00a0+ 3 at<br><\/strong>(i) x = 0<br>(ii) x = \u2013 1<br>(iii) x = 2<\/li><\/ol>\n\n\n\n<p class=\"wp-block-paragraph\">Solution:<br>1et p(x) = 5x \u2013 4x<sup>2<\/sup>&nbsp;+ 3<br>(i) p(0) = 5(0) \u2013 4(0)<sup>2<\/sup>&nbsp;+ 3<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; = 0 \u2013 0 + 3<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; = 3<br>Thus, the value of 5x \u2013 4x<sup>2<\/sup>&nbsp;+ 3 at x = 0 is 3.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\"><br>(ii) p(-1) = 5(-1) \u2013 4(-1)<sup>2<\/sup>&nbsp;+ 3<br>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; = \u2013 5x \u2013 4x<sup>2<\/sup>&nbsp;+ 3<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; = -9 + 3<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; = -6<br>Thus, the value of 5x \u2013 4x<sup>2<\/sup>&nbsp;+ 3 at x = -1 is -6.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\"><br>(iii) p(2) = 5(2) \u2013 4(2)<sup>2<\/sup>&nbsp;+ 3<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; = 10 \u2013 4(4) + 3<br>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; = 10 \u2013 16 + 3<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; = -3<br>Thus, the value of 5x \u2013 4x<sup>2<\/sup>&nbsp;+ 3 at x = 2 is \u2013 3.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>2. Find p (0), p (1) and p (2) for each of the following polynomials.<br><\/strong>(i) p(y) = y<sup>2<\/sup>\u00a0\u2013 y +1<br>(ii) p (t) = 2 +1 + 2t<sup>2<\/sup>\u00a0-t<sup>3<\/sup><br>(iii) P (x) = x<sup>3<\/sup><br>(iv) p (x) = (x-1) (x+1)<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Solution:<br>(i) Given that p(y) = y<sup>2<\/sup>&nbsp;\u2013 y + 1.<br>\u2234 P(0) = (0)<sup>2<\/sup>&nbsp;\u2013 0 + 1 = 0 \u2013 0 + 1 = 1<br>p(1) = (1)<sup>2<\/sup>&nbsp;\u2013 1 + 1 = 1 \u2013 1 + 1 = 1<br>p(2) = (2)<sup>2<\/sup>&nbsp;\u2013 2 + 1 = 4 \u2013 2 + 1 = 3<br><br><\/p>\n\n\n\n<p class=\"wp-block-paragraph\">(ii) Given that p(t) = 2 + t + 2t<sup>2&nbsp;<\/sup>\u2013 t<sup>3<\/sup><br>\u2234p(0) = 2 + 0 + 2(0)<sup>2&nbsp;<\/sup>\u2013 (0)<sup>3<\/sup><br>= 2 + 0 + 0 \u2013 0=2<br>P(1) = 2 + 1 + 2(1)<sup>2&nbsp;<\/sup>\u2013 (1)<sup>3<\/sup><br>= 2 + 1 + 2 \u2013 1 = 4<br>p( 2) = 2 + 2 + 2(2)<sup>2&nbsp;<\/sup>\u2013 (2)<sup>3<\/sup><br>= 2 + 2 + 8 \u2013 8 = 4<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">(iii) Given that p(x) = x<sup>3<\/sup><br>\u2234 p(0) = (0)<sup>3<\/sup>&nbsp;= 0, p(1) = (1)<sup>3<\/sup>&nbsp;= 1<br>p(2) = (2)<sup>3<\/sup>&nbsp;= 8<br><br><\/p>\n\n\n\n<p class=\"wp-block-paragraph\">(iv) Given that p(x) = (x \u2013 1)(x + 1)<br>\u2234 p(0) = (0 \u2013 1)(0 + 1) = (-1)(1) = -1<br>p(1) = (1 \u2013 1)(1 +1) = (0)(2) = 0<br>P(2) = (2 \u2013 1)(2 + 1) = (1)(3) = 3<\/p>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>3. Verify whether the following are zeroes of the polynomial, indicated against them.<br><\/strong>(i) p(x) = 3x + 1,x = \u20131\/3<br>(ii) p (x) = 5x \u2013 \u03c0, x =\u00a04\/5<br>(iii) p (x) = x<sup>2<\/sup>\u00a0\u2013 1, x = x \u2013 1<br>(iv) p (x) = (x + 1) (x \u2013 2), x = \u2013 1,2<br>(v) p (x) = x<sup>2<\/sup>, x = 0<br>(vi) p (x) = 1x + m, x = \u2013\u00a0m\/1<br>(vii) P (x) = 3x<sup>2<\/sup>\u00a0\u2013 1, x = \u2013\u00a01\/\u221a3,2\/\u221a3<br>(viii) p (x) = 2x + 1, x =\u00a0\u00bd<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Solution:<br>(i) We have , p(x) = 3x + 1<\/p>\n\n\n\n<div class=\"wp-block-image\"><figure class=\"aligncenter size-large\"><img loading=\"lazy\" decoding=\"async\" width=\"386\" height=\"132\" src=\"https:\/\/themindpalace.in\/wp-content\/uploads\/2021\/07\/image-1.png\" alt=\"\" class=\"wp-image-3594\" srcset=\"https:\/\/themindpalace.in\/wp-content\/uploads\/2021\/07\/image-1.png 386w, https:\/\/themindpalace.in\/wp-content\/uploads\/2021\/07\/image-1-300x103.png 300w\" sizes=\"auto, (max-width: 386px) 100vw, 386px\" \/><\/figure><\/div>\n\n\n\n<p class=\"wp-block-paragraph\">(ii) We have, p(x) = 5x \u2013 \u03c0<br>\u2234\u00a0p(\u22121\/3)=3(\u22121\/3)+1=\u22121+1=0<\/p>\n\n\n\n<p class=\"wp-block-paragraph\"><br>(iii) We have, p(x) = x<sup>2<\/sup>\u00a0\u2013 1<br>\u2234 p(1) = (1)<sup>2<\/sup>\u00a0\u2013 1 = 1 \u2013 1=0<br>Since, p(1) = 0, so x = 1 is a zero of x<sup>2<\/sup>\u00a0-1.<br>Also, p(-1) = (-1)<sup>2<\/sup>\u00a0-1 = 1 \u2013 1 = 0<br>Since p(-1) = 0, so, x = -1, is also a zero of x<sup>2<\/sup>\u00a0\u2013 1.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">(iv) We have, p(x) = (x + 1)(x \u2013 2)<br>\u2234 p(-1) = (-1 +1) (-1 \u2013 2) = (0)(- 3) = 0<br>Since, p(-1) = 0, so, x = -1 is a zero of (x + 1)(x \u2013 2).<br>Also, p( 2) = (2 + 1)(2 \u2013 2) = (3)(0) = 0<br>Since, p(2) = 0, so, x = 2 is also a zero of (x + 1)(x \u2013 2).<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">(v) We have, p(x) = x<sup>2<\/sup><br>\u2234 p(o) = (0)<sup>2<\/sup>\u00a0= 0<br>Since, p(0) = 0, so, x = 0 is a zero of x<sup>2<\/sup>. (vi) We have, p(x) = lx + m<br><\/p>\n\n\n\n<div class=\"wp-block-image\"><figure class=\"aligncenter size-large\"><img loading=\"lazy\" decoding=\"async\" width=\"388\" height=\"122\" src=\"https:\/\/themindpalace.in\/wp-content\/uploads\/2021\/07\/image-2.png\" alt=\"\" class=\"wp-image-3595\" srcset=\"https:\/\/themindpalace.in\/wp-content\/uploads\/2021\/07\/image-2.png 388w, https:\/\/themindpalace.in\/wp-content\/uploads\/2021\/07\/image-2-300x94.png 300w\" sizes=\"auto, (max-width: 388px) 100vw, 388px\" \/><\/figure><\/div>\n\n\n\n<p class=\"wp-block-paragraph\">(vii) We have, p(x) = 3x<sup>2<\/sup>\u00a0\u2013 1<\/p>\n\n\n\n<div class=\"wp-block-image\"><figure class=\"aligncenter size-large\"><img loading=\"lazy\" decoding=\"async\" width=\"403\" height=\"304\" src=\"https:\/\/themindpalace.in\/wp-content\/uploads\/2021\/07\/image-3.png\" alt=\"\" class=\"wp-image-3596\" srcset=\"https:\/\/themindpalace.in\/wp-content\/uploads\/2021\/07\/image-3.png 403w, https:\/\/themindpalace.in\/wp-content\/uploads\/2021\/07\/image-3-300x226.png 300w\" sizes=\"auto, (max-width: 403px) 100vw, 403px\" \/><\/figure><\/div>\n\n\n\n<p class=\"wp-block-paragraph\"><br>(viii) We have, p(x) = 2x + 1<br>\u2234\u00a0p(1\/2)=2(1\/2)+1=1+1=2<br>Since,\u00a0p(1\/2)\u00a0\u2260 0, so, x =\u00a012\u00a0is not a zero of 2x + 1.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">4. <strong>Find the zero of the polynomial in each of the following cases<br><\/strong>(i) p(x)=x+5<br>(ii) p (x) = x \u2013 5<br>(iii) p (x) = 2x + 5<br>(iv) p (x) = 3x \u2013 2<br>(v) p (x) = 3x<br>(vi) p (x)= ax, a\u22600<br>(vii) p (x) = cx + d, c \u2260 0 where c and d are real numbers.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Solution:<br>(i) We have, p(x) = x + 5. Since, p(x) = 0<br>\u21d2 x + 5 = 0<br>\u21d2 x = -5.<br>Thus, zero of x + 5 is -5.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">(ii) We have, p(x) = x \u2013 5.<br>Since, p(x) = 0 \u21d2 x \u2013 5 = 0 \u21d2 x = -5<br>Thus, zero of x \u2013 5 is 5.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">(iii) We have, p(x) = 2x + 5. Since, p(x) = 0<br>\u21d2 2x + 5 =0<br>\u21d2 2x = -5<br>\u21d2 x =&nbsp;\u22125\/2<br>Thus, zero of 2x + 5 is&nbsp;\u22125\/2&nbsp;.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">(iv) We have, p(x) = 3x \u2013 2. Since, p(x) = 0<br>\u21d2 3x \u2013 2 = 0<br>\u21d2 3x = 2<br>\u21d2 x =&nbsp;2\/3<br>Thus, zero of 3x \u2013 2 is&nbsp;2\/3<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">(v) We have, p(x) = 3x. Since, p(x) = 0<br>\u21d2 3x = 0 \u21d2 x = 0<br>Thus, zero of 3x is 0.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">(vi) We have, p(x) = ax, a \u2260 0.<br>Since, p(x) = 0 =&gt; ax = 0 =&gt; x-0<br>Thus, zero of ax is 0.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">(vii) We have, p(x) = cx + d. Since, p(x) = 0<br>\u21d2 cx + d = 0 \u21d2 cx = -d \u21d2\u00a0x=\u2212d\/c<br>Thus, zero of cx + d is\u00a0\u2013d\/c<\/p>\n\n\n\n<div class=\"wp-block-kadence-spacer aligncenter kt-block-spacer-_139d33-30\"><div class=\"kt-block-spacer kt-block-spacer-halign-center\" style=\"height:60px\"><hr class=\"kt-divider\" style=\"border-top-color:rgba(238, 238, 238, 1);border-top-width:1px;width:80%;border-top-style:solid\"\/><\/div><\/div>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>Example 2.3<br><\/strong><br><\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Find the remainder when x<sup>3<\/sup>&nbsp;+ 3x<sup>2<\/sup>&nbsp;+ 3x + 1 is divided by<br>(i) x + 1<br>(ii) x \u2013&nbsp;1\/2<br>(iii) x<br>(iv) x + \u03c0<br>(v) 5 + 2x<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Solution:<br>Let p(x) = x<sup>3<\/sup>\u00a0+ 3x<sup>2<\/sup>\u00a0+ 3x +1<br>(i) The zero of x + 1 is -1.<br>\u2234 p(-1) = (-1)3 + 3(-1)2 + 3(-1) +1<br>= -1 + 3- 3 + 1 = 0<br>Thus, the required remainder = 0<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">(ii) The zero of\u00a0x\u221212\u00a0is\u00a012<br><\/p>\n\n\n\n<div class=\"wp-block-image\"><figure class=\"aligncenter size-large\"><img loading=\"lazy\" decoding=\"async\" width=\"359\" height=\"106\" src=\"https:\/\/themindpalace.in\/wp-content\/uploads\/2021\/07\/image-4.png\" alt=\"\" class=\"wp-image-3598\" srcset=\"https:\/\/themindpalace.in\/wp-content\/uploads\/2021\/07\/image-4.png 359w, https:\/\/themindpalace.in\/wp-content\/uploads\/2021\/07\/image-4-300x89.png 300w\" sizes=\"auto, (max-width: 359px) 100vw, 359px\" \/><\/figure><\/div>\n\n\n\n<p class=\"wp-block-paragraph\">Thus, the required remainder =\u00a027\/8<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">(iii) The zero of x is 0.<br>\u2234 p(0) = (0)<sup>3<\/sup>&nbsp;+ 3(0)<sup>2<\/sup>&nbsp;+ 3(0) + 1<br>= 0 + 0 + 0 + 1 = 1<br>Thus, the required remainder = 1.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">(iv) The zero of x + \u03c0 is -\u03c0.<br>p(-\u03c0) = (-\u03c0)<sup>3<\/sup>\u00a0+ 3(- \u03c0)<sup>2<\/sup>2 + 3(- \u03c0) +1<br>= -\u03c0<sup>3<\/sup>\u00a0+ 3\u03c0<sup>2<\/sup>\u00a0+ (-3\u03c0) + 1<br>= \u2013 \u03c0<sup>3<\/sup>\u00a0+ 3\u03c0<sup>2<\/sup>\u00a0\u2013 3\u03c0 +1<br>Thus, the required remainder is -\u03c0<sup>3<\/sup>\u00a0+ 3\u03c0<sup>2<\/sup>\u00a0\u2013 3\u03c0+1. <\/p>\n\n\n\n<p class=\"wp-block-paragraph\">(v) The zero of 5 + 2x is\u00a0\u22125\/2\u00a0.<br><\/p>\n\n\n\n<div class=\"wp-block-image\"><figure class=\"aligncenter size-large\"><img loading=\"lazy\" decoding=\"async\" width=\"342\" height=\"118\" src=\"https:\/\/themindpalace.in\/wp-content\/uploads\/2021\/07\/image-5.png\" alt=\"\" class=\"wp-image-3599\" srcset=\"https:\/\/themindpalace.in\/wp-content\/uploads\/2021\/07\/image-5.png 342w, https:\/\/themindpalace.in\/wp-content\/uploads\/2021\/07\/image-5-300x104.png 300w\" sizes=\"auto, (max-width: 342px) 100vw, 342px\" \/><\/figure><\/div>\n\n\n\n<div class=\"wp-block-image\"><figure class=\"aligncenter size-large\"><img loading=\"lazy\" decoding=\"async\" width=\"236\" height=\"51\" src=\"https:\/\/themindpalace.in\/wp-content\/uploads\/2021\/07\/image-6.png\" alt=\"\" class=\"wp-image-3600\"\/><\/figure><\/div>\n\n\n\n<p class=\"wp-block-paragraph\">Thus, the required remainder is&nbsp;\u221227\/8&nbsp;.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>2. Find the remainder when x<sup>3<\/sup>\u00a0\u2013 ax<sup>2<\/sup>\u00a0+ 6x \u2013 a is divided by x \u2013 a.<\/strong><\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Solution:<br>We have, p(x) = x<sup>3<\/sup>\u00a0\u2013 ax<sup>2<\/sup>\u00a0+ 6x \u2013 a and zero of x \u2013 a is a.<br>\u2234 p(a) = (a)<sup>3<\/sup>\u00a0\u2013 a(a)<sup>2<\/sup>\u00a0+ 6(a) \u2013 a<br>= a<sup>3<\/sup>\u00a0\u2013 a<sup>3<\/sup>\u00a0+ 6a \u2013 a = 5a<br>Thus, the required remainder is 5a. 3. Check whether 7 + 3x is a factor of 3x<sup>3<\/sup>+7x.<br>Solution:<br>We have, p(x) = 3x<sup>3<\/sup>+7x. and zero of 7 + 3x is\u00a0\u22127\/3.<\/p>\n\n\n\n<div class=\"wp-block-image\"><figure class=\"aligncenter size-large\"><img loading=\"lazy\" decoding=\"async\" width=\"375\" height=\"104\" src=\"https:\/\/themindpalace.in\/wp-content\/uploads\/2021\/07\/image-7.png\" alt=\"\" class=\"wp-image-3601\" srcset=\"https:\/\/themindpalace.in\/wp-content\/uploads\/2021\/07\/image-7.png 375w, https:\/\/themindpalace.in\/wp-content\/uploads\/2021\/07\/image-7-300x83.png 300w\" sizes=\"auto, (max-width: 375px) 100vw, 375px\" \/><\/figure><\/div>\n\n\n\n<p class=\"wp-block-paragraph\">Since,(\u00a0\u2212490\/9) \u2260 0<br>i.e. the remainder is not 0.<br>\u2234 3x<sup>3<\/sup>\u00a0+ 7x is not divisib1e by 7 + 3x.<br>Thus, 7 + 3x is not a factor of 3x<sup>3<\/sup>\u00a0+ 7x.<\/p>\n\n\n\n<div class=\"wp-block-kadence-spacer aligncenter kt-block-spacer-_a85374-59\"><div class=\"kt-block-spacer kt-block-spacer-halign-center\" style=\"height:60px\"><hr class=\"kt-divider\" style=\"border-top-color:rgba(238, 238, 238, 1);border-top-width:1px;width:80%;border-top-style:solid\"\/><\/div><\/div>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>Example 2.4<\/strong><\/p>\n\n\n\n<ol class=\"wp-block-list\" type=\"1\"><li><strong>Determine which of the following polynomials has (x +1) a factor.<br><\/strong>(i) x<sup>3<\/sup>+x<sup>2<\/sup>+x +1<br>(ii) x<sup>4<\/sup>\u00a0+ x<sup>3<\/sup>\u00a0+ x<sup>2<\/sup>\u00a0+ x + 1<br>(iii) x<sup>4<\/sup>\u00a0+ 3x<sup>3<\/sup>\u00a0+ 3x<sup>2<\/sup>\u00a0+ x + 1<br>(iv) x<sup>3<\/sup>\u00a0\u2013 x<sup>2<\/sup>\u00a0\u2013 (2 +\u221a2 )x + \u221a2<\/li><\/ol>\n\n\n\n<p class=\"wp-block-paragraph\">Solution:<br>The zero of x + 1 is -1.<br>(i) Let p (x) = x<sup>3<\/sup>&nbsp;+ x<sup>2<\/sup>&nbsp;+ x + 1<br>\u2234 p (-1) = (-1)<sup>3<\/sup>&nbsp;+ (-1)<sup>2<\/sup>&nbsp;+ (-1) + 1 .<br>= -1 + 1 \u2013 1 + 1<br>\u21d2 p (- 1) = 0<br>So, (x+ 1) is a factor of x<sup>3<\/sup>&nbsp;+ x<sup>2<\/sup>&nbsp;+ x + 1.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">(ii) Let p (x) = x<sup>4<\/sup>&nbsp;+ x<sup>3<\/sup>&nbsp;+ x<sup>2<\/sup>&nbsp;+ x + 1<br>\u2234 P(-1) = (-1)<sup>4<\/sup>&nbsp;+ (-1)<sup>3<\/sup>&nbsp;+ (-1)<sup>2<\/sup>&nbsp;+ (-1)+1<br>= 1 \u2013 1 + 1 \u2013 1 + 1<br>\u21d2 P (-1) \u2260 1<br>So, (x + 1) is not a factor of x<sup>4<\/sup>&nbsp;+ x<sup>3<\/sup>&nbsp;+ x<sup>2<\/sup>&nbsp;+ x+ 1<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">(iii) Let p (x) = x<sup>4<\/sup>&nbsp;+ 3x<sup>3<\/sup>&nbsp;+ 3x<sup>2<\/sup>&nbsp;+ x + 1 .<br>\u2234 p (-1)= (-1)<sup>4<\/sup>&nbsp;+ 3 (-1)<sup>3<\/sup>&nbsp;+ 3 (-1)<sup>2<\/sup>&nbsp;+ (- 1) + 1<br>= 1 \u2013 3 + 3 \u2013 1 + 1 = 1<br>\u21d2 p (-1) \u2260 0<br>So, (x + 1) is not a factor of x<sup>4<\/sup>&nbsp;+ 3x<sup>3<\/sup>&nbsp;+ 3x<sup>2<\/sup>&nbsp;+ x+ 1.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">(iv) Let p (x) = x<sup>3<\/sup>&nbsp;\u2013 x<sup>2<\/sup>&nbsp;\u2013 (2 + \u221a2) x + \u221a2<br>\u2234 p (- 1) =(- 1)3- (-1)2 \u2013 (2 + \u221a2)(-1) + \u221a2<br>= -1 \u2013 1 + 2 + \u221a2 + \u221a2<br>= 2\u221a2<br>\u21d2 p (-1) \u2260 0<br>So, (x + 1) is not a factor of x<sup>3<\/sup>&nbsp;\u2013 x<sup>2<\/sup>&nbsp;\u2013 (2 + \u221a2) x + \u221a2.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>2. Use the Factor Theorem to determine whether g (x) is a factor of p (x) in each of the following cases<\/strong><br><strong><br><\/strong>(i) p (x)= 2x<sup>3<\/sup>\u00a0+ x<sup>2<\/sup>\u00a0\u2013 2x \u2013 1, g (x) = x + 1<br>(ii) p(x)= x<sup>3<\/sup>\u00a0+ 3x<sup>2<\/sup>\u00a0+ 3x + 1, g (x) = x + 2<br>(iii) p (x) = x<sup>3<\/sup>\u00a0\u2013 4x<sup>2<\/sup>\u00a0+ x + 6, g (x) = x \u2013 3<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Solution:<br>(i) We have, p (x)= 2x<sup>3<\/sup>&nbsp;+ x<sup>2<\/sup>&nbsp;\u2013 2x \u2013 1 and g (x) = x + 1<br>\u2234 p(-1) = 2(-1)<sup>3<\/sup>&nbsp;+ (-1)<sup>2<\/sup>&nbsp;\u2013 2(-1) \u2013 1<br>= 2(-1) + 1 + 2 \u2013 1<br>= -2 + 1 + 2 -1 = 0<br>\u21d2 p(-1) = 0, so g(x) is a factor of p(x).<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">(ii) We have, p(x) x<sup>3<\/sup>&nbsp;+ 3x<sup>2<\/sup>&nbsp;+ 3x + 1 and g(x) = x + 2<br>\u2234 p(-2) = (-2)<sup>3<\/sup>&nbsp;+ 3(-2)<sup>2<\/sup>+ 3(-2) + 1<br>= -8 + 12 \u2013 6 + 1<br>= -14 + 13<br>= -1<br>\u21d2 p(-2) \u2260 0, so g(x) is not a factor of p(x).<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">(iii) We have, = x<sup>3<\/sup>&nbsp;\u2013 4x<sup>2<\/sup>&nbsp;+ x + 6 and g (x) = x \u2013 3<br>\u2234 p(3) = (3)<sup>3<\/sup>&nbsp;\u2013 4(3)<sup>2<\/sup>&nbsp;+ 3 + 6<br>= 27 \u2013 4(9) + 3 + 6<br>= 27 \u2013 36 + 3 + 6 = 0<br>\u21d2 p(3) = 0, so g(x) is a factor of p(x).<\/p>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>3. Find the value of k, if x \u2013 1 is a factor of p (x) in each of the following cases<\/strong><br><strong><br><\/strong>(i) p (x) = x<sup>2<\/sup>\u00a0+ x + k<br>(ii) p (x) = 2x<sup>2<\/sup>\u00a0+ kx + \u221a2<br>(iii) p (x) = kx<sup>2<\/sup>\u00a0\u2013 \u221a2 x + 1<br>(iv) p (x) = kx<sup>2<\/sup>\u00a0\u2013 3x + k<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Solution:<br>For (x \u2013 1) to be a factor of p(x), p(1) should be equal to 0.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">(i) Here, p(x) = x<sup>2<\/sup>&nbsp;+ x + k<br>Since, p(1) = (1)<sup>2<\/sup>&nbsp;+1 + k<br>\u21d2 p(1) = k + 2 = 0<br>\u21d2 k = -2.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">(ii) Here, p (x) = 2x<sup>2<\/sup>&nbsp;+ kx + \u221a2<br>Since, p(1) = 2(1)<sup>2<\/sup>&nbsp;+ k(1) + \u221a2<br>= 2 + k + \u221a2 =0<br>k = -2 \u2013 \u221a2 = -(2 + \u221a2)<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">(iii) Here, p (x) = kx<sup>2<\/sup>&nbsp;\u2013 \u221a2 x + 1<br>Since, p(1) = k(1)<sup>2<\/sup>&nbsp;\u2013 (1) + 1<br>= k \u2013 \u221a2 + 1 = 0<br>\u21d2 k = \u221a2 -1<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">(iv) Here, p(x) = kx<sup>2<\/sup>&nbsp;\u2013 3x + k<br>p(1) = k(1)<sup>2<\/sup>&nbsp;\u2013 3(1) + k<br>= k \u2013 3 + k<br>= 2k \u2013 3 = 0<br>\u21d2 k =&nbsp;\u00be<\/p>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>4. Factorise<br><\/strong><br>(i) 12x<sup>2<\/sup>\u00a0\u2013 7x +1<br>(ii) 2x<sup>2<\/sup>\u00a0+ 7x + 3<br>(iii) 6x<sup>2<\/sup>\u00a0+ 5x \u2013 6<br>(iv) 3x<sup>2<\/sup>\u00a0\u2013 x \u2013 4<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Solution:<br>(i) We have,<br>12x<sup>2<\/sup>&nbsp;\u2013 7x + 1 = 12x<sup>2<\/sup>&nbsp;\u2013 4x- 3x + 1<br>= 4x (3x \u2013 1 ) -1 (3x \u2013 1)<br>= (3x -1) (4x -1)<br>Thus, 12x<sup>2<\/sup>&nbsp;-7x + 3 = (2x \u2013 1) (x + 3)<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">(ii) We have, 2x<sup>2<\/sup>&nbsp;+ 7x + 3 = 2x<sup>2<\/sup>&nbsp;+ x + 6x + 3<br>= x(2x + 1) + 3(2x + 1)<br>= (2x + 1)(x + 3)<br>Thus, 2\u00d72 + 7x + 3 = (2x + 1)(x + 3)<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">(iii) We have, 6x<sup>2<\/sup>&nbsp;+ 5x \u2013 6 = 6x<sup>2<\/sup>&nbsp;+ 9x \u2013 4x \u2013 6<br>= 3x(2x + 3) \u2013 2(2x + 3)<br>= (2x + 3)(3x \u2013 2)<br>Thus, 6x<sup>2<\/sup>&nbsp;+ 5x \u2013 6 = (2x + 3)(3x \u2013 2)<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">(iv) We have, 3x<sup>2<\/sup>&nbsp;\u2013 x \u2013 4 = 3x<sup>2<\/sup>&nbsp;\u2013 4x + 3x \u2013 4<br>= x(3x \u2013 4) + 1(3x \u2013 4) = (3x \u2013 4)(x + 1)<br>Thus, 3x<sup>2<\/sup>&nbsp;\u2013 x \u2013 4 = (3x \u2013 4)(x + 1)<\/p>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>5. Factorise<\/strong><br><strong><br><\/strong>(i) x<sup>3<\/sup>\u00a0\u2013 2x<sup>2<\/sup>\u00a0\u2013 x + 2<br>(ii) x<sup>3<\/sup>\u00a0\u2013 3x<sup>2<\/sup>\u00a0\u2013 9x \u2013 5<br>(iii) x<sup>3<\/sup>\u00a0+ 13x<sup>2<\/sup>\u00a0+ 32x + 20<br>(iv) 2y<sup>3<\/sup>\u00a0+ y<sup>2<\/sup>\u00a0\u2013 2y \u2013 1<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Solution:<br>(i) We have, x<sup>3<\/sup>&nbsp;\u2013 2x<sup>2<\/sup>&nbsp;\u2013 x + 2<br>Rearranging the terms, we have x<sup>3<\/sup>&nbsp;\u2013 x \u2013 2x<sup>2<\/sup>&nbsp;+ 2<br>= x(x<sup>2<\/sup>&nbsp;\u2013 1) \u2013 2(x<sup>2<\/sup>&nbsp;-1) = (x<sup>2<\/sup>&nbsp;\u2013 1)(x \u2013 2)<br>= [(x)<sup>2<\/sup>&nbsp;\u2013 (1)<sup>2<\/sup>](x \u2013 2)<br>= (x \u2013 1)(x + 1)(x \u2013 2)<br>[\u2235 (a<sup>2<\/sup>&nbsp;\u2013 b<sup>2<\/sup>) = (a + b)(a-b)]<br>Thus, x<sup>3<\/sup>&nbsp;\u2013 2x<sup>2<\/sup>&nbsp;\u2013 x + 2 = (x \u2013 1)(x + 1)(x \u2013 2)<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">(ii) We have, x<sup>3<\/sup>&nbsp;\u2013 3x<sup>2<\/sup>&nbsp;\u2013 9x \u2013 5<br>= x<sup>3<\/sup>&nbsp;+ x<sup>2<\/sup>&nbsp;\u2013 4x<sup>2<\/sup>&nbsp;\u2013 4x \u2013 5x \u2013 5 ,<br>= x<sup>2<\/sup>&nbsp;(x + 1) \u2013 4x(x + 1) \u2013 5(x + 1)<br>= (x + 1)(x<sup>2<\/sup>&nbsp;\u2013 4x \u2013 5)<br>= (x + 1)(x<sup>2<\/sup>&nbsp;\u2013 5x + x \u2013 5)<br>= (x + 1)[x(x \u2013 5) + 1(x \u2013 5)]<br>= (x + 1)(x \u2013 5)(x + 1)<br>Thus, x<sup>3<\/sup>&nbsp;\u2013 3x<sup>2<\/sup>&nbsp;\u2013 9x \u2013 5 = (x + 1)(x \u2013 5)(x +1)<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">(iii) We have, x<sup>3<\/sup>&nbsp;+ 13x<sup>2<\/sup>&nbsp;+ 32x + 20<br>= x<sup>3<\/sup>&nbsp;+ x<sup>2<\/sup>&nbsp;+ 12x<sup>2<\/sup>&nbsp;+ 12x + 20x + 20<br>= x<sup>2<\/sup>(x + 1) + 12x(x +1) + 20(x + 1)<br>= (x + 1)(x<sup>2<\/sup>&nbsp;+ 12x + 20)<br>= (x + 1)(x<sup>2<\/sup>&nbsp;+ 2x + 10x + 20)<br>= (x + 1)[x(x + 2) + 10(x + 2)]<br>= (x + 1)(x + 2)(x + 10)<br>Thus, x<sup>3<\/sup>&nbsp;+ 13x<sup>2<\/sup>&nbsp;+ 32x + 20<br>= (x + 1)(x + 2)(x + 10)<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">(iv) We have, 2y<sup>3<\/sup>\u00a0+ y<sup>2<\/sup>\u00a0\u2013 2y \u2013 1<br>= 2y<sup>3<\/sup>\u00a0\u2013 2y<sup>2<\/sup>\u00a0+ 3y<sup>2<\/sup>\u00a0\u2013 3y + y \u2013 1<br>= 2y<sup>2<\/sup>(y \u2013 1) + 3y(y \u2013 1) + 1(y \u2013 1)<br>= (y \u2013 1)(2y<sup>2<\/sup>\u00a0+ 3y + 1)<br>= (y \u2013 1)(2y<sup>2<\/sup>\u00a0+ 2y + y + 1)<br>= (y \u2013 1)[2y(y + 1) + 1(y + 1)]<br>= (y \u2013 1)(y + 1)(2y + 1)<br>Thus, 2y<sup>3<\/sup>\u00a0+ y<sup>2<\/sup>\u00a0\u2013 2y \u2013 1<br>= (y \u2013 1)(y + 1)(2y +1)<\/p>\n\n\n\n<div class=\"wp-block-kadence-spacer aligncenter kt-block-spacer-_97d6e7-18\"><div class=\"kt-block-spacer kt-block-spacer-halign-center\" style=\"height:60px\"><hr class=\"kt-divider\" style=\"border-top-color:rgba(238, 238, 238, 1);border-top-width:1px;width:80%;border-top-style:solid\"\/><\/div><\/div>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>Example 2.5<\/strong><\/p>\n\n\n\n<ol class=\"wp-block-list\" type=\"1\"><li><strong>Use suitable identities to find the following products<br><\/strong>(i) (x + 4)(x + 10)<br>(ii) (x+8) (x -10)<br>(iii) (3x + 4) (3x \u2013 5)<br>(iv) (y<sup>2<\/sup>+\u00a03\/2) (y<sup>2<\/sup>\u2013\u00a03\/2)<br>(v) (3 \u2013 2x) (3 + 2x)<\/li><\/ol>\n\n\n\n<p class=\"wp-block-paragraph\">Solution:<br>(i) We have, (x+ 4) (x + 10)<br>Using identity,<br>(x+ a) (x+ b) = x<sup>2<\/sup>&nbsp;+ (a + b) x+ ab.<br>We have, (x + 4) (x + 10) = x<sup>2<\/sup>+(4 + 10) x + (4 x 10)<br>= x<sup>2<\/sup>&nbsp;+ 14x+40<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">(ii) We have, (x+ 8) (x -10)<br>Using identity,<br>(x + a) (x + b) = x<sup>2<\/sup>\u00a0+ (a + b) x + ab<br>We have, (x + 8) (x \u2013 10) = x<sup>2<\/sup>\u00a0+ [8 + (-10)] x + (8) (- 10)<br>= x<sup>2<\/sup>\u00a0\u2013 2x \u2013 80<\/p>\n\n\n\n<p class=\"wp-block-paragraph\"> (iii) We have, (3x + 4) (3x \u2013 5)<br>Using identity,<br>(x + a) (x + b) = x<sup>2<\/sup>\u00a0+ (a + b) x + ab<br>We have, (3x + 4) (3x \u2013 5) = (3x)<sup>2<\/sup>\u00a0+ (4 \u2013 5) x + (4) (- 5)<br>= 9x<sup>2<\/sup>\u00a0\u2013 x \u2013 20<br><\/p>\n\n\n\n<div class=\"wp-block-image\"><figure class=\"aligncenter size-large\"><img loading=\"lazy\" decoding=\"async\" width=\"392\" height=\"269\" src=\"https:\/\/themindpalace.in\/wp-content\/uploads\/2021\/07\/image-9.png\" alt=\"\" class=\"wp-image-3606\" srcset=\"https:\/\/themindpalace.in\/wp-content\/uploads\/2021\/07\/image-9.png 392w, https:\/\/themindpalace.in\/wp-content\/uploads\/2021\/07\/image-9-300x206.png 300w\" sizes=\"auto, (max-width: 392px) 100vw, 392px\" \/><\/figure><\/div>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>2. Evaluate the following products without multiplying directly<br><\/strong>(i) 103 x 107<br>(ii) 95 x 96<br>(iii) 104 x 96<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Solution:<br>(i)We have, 103 x 107 = (100 + 3) (100 + 7)<br>= ( 100)<sup>2<\/sup>&nbsp;+ (3 + 7) (100)+ (3 x 7)<br>[Using (x + a)(x + b) = x<sup>2<\/sup>&nbsp;+ (a + b)x + ab]<br>= 10000 + (10) x 100 + 21<br>= 10000 + 1000 + 21=11021<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">(ii) We have, 95 x 96 = (100 \u2013 5) (100 \u2013 4)<br>= ( 100)<sup>2<\/sup>&nbsp;+ [(- 5) + (- 4)] 100 + (- 5 x \u2013 4)<br>[Using (x + a)(x + b) = x<sup>2<\/sup>&nbsp;+ (a + b)x + ab]<br>= 10000 + (-9) + 20 = 9120<br>= 10000 + (-900) + 20 = 9120<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">(iii) We have 104 x 96 = (100 + 4) (100 \u2013 4)<br>= (100)<sup>2<\/sup>-4<sup>2<\/sup><br>[Using (a + b)(a -b) = a<sup>2<\/sup>\u2013 b<sup>2<\/sup>]<br>= 10000 \u2013 16 = 9984<\/p>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>3. Factorise the following using appropriate identities<br><\/strong>(i) 9x<sup>2\u00a0<\/sup>+ 6xy + y<sup>2<\/sup><br>(ii) 4y<sup>2<\/sup>-4y + 1<br>(iii) x<sup>2<\/sup>\u00a0\u2013\u00a0y2\/100<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Solution:<br>(i) We have, 9x<sup>2<\/sup>\u00a0+ 6xy + y<sup>2<\/sup><br>= (3x)<sup>2<\/sup>\u00a0+ 2(3x)(y) + (y)<sup>2<\/sup><br>= (3x + y)<sup>2<\/sup><br>[Using a<sup>2<\/sup>\u00a0+ 2ab + b<sup>2<\/sup>\u00a0= (a + b)<sup>2<\/sup>]<br>= (3x + y)(3x + y)<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">(ii) We have, 4y<sup>2<\/sup>\u00a0\u2013 4y + 1<sup>2<\/sup><br>= (2y)<sup>2<\/sup>\u00a0+ 2(2y)(1) + (1)<sup>2<\/sup><br>= (2y -1)<sup>2<\/sup><br>[Using a<sup>2<\/sup>\u00a0\u2013 2ab + b<sup>2<\/sup>\u00a0= (a- b)<sup>2<\/sup>]<br>= (2y \u2013 1)(2y \u2013 1 )<br><\/p>\n\n\n\n<div class=\"wp-block-image\"><figure class=\"aligncenter size-large\"><img loading=\"lazy\" decoding=\"async\" width=\"384\" height=\"131\" src=\"https:\/\/themindpalace.in\/wp-content\/uploads\/2021\/07\/image-11.png\" alt=\"\" class=\"wp-image-3608\" srcset=\"https:\/\/themindpalace.in\/wp-content\/uploads\/2021\/07\/image-11.png 384w, https:\/\/themindpalace.in\/wp-content\/uploads\/2021\/07\/image-11-300x102.png 300w\" sizes=\"auto, (max-width: 384px) 100vw, 384px\" \/><\/figure><\/div>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>4. Expand each of the following, using suitable identity<br><\/strong>(i) (x+2y+ 4z)<sup>2<\/sup><br>(ii) (2x \u2013 y + z)<sup>2<\/sup><br>(iii) (- 2x + 3y + 2z)<sup>2<\/sup><br>(iv) (3a -7b \u2013 c)<sup>z<\/sup><br>(v) (- 2x + 5y \u2013 3z)<sup>2<\/sup><br>(vi) [\u00a01\/4a \u20131\/4b + 1]\u00a0<sup>2<\/sup><\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Solution:<br>We know that<br>(x + y + z)<sup>2<\/sup>&nbsp;= x<sup>2<\/sup>&nbsp;+ y<sup>2<\/sup>&nbsp;+ z<sup>2<\/sup>&nbsp;+ 2xy + 2yz + 2zx<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">(i) (x + 2y + 4z)<sup>2<\/sup><br>= x<sup>2<\/sup>&nbsp;+ (2y)<sup>2<\/sup>&nbsp;+ (4z)<sup>2<\/sup>&nbsp;+ 2 (x) (2y) + 2 (2y) (4z) + 2(4z) (x)<br>= x<sup>2<\/sup>&nbsp;+ 4y<sup>2<\/sup>&nbsp;+ 16z<sup>2<\/sup>&nbsp;+ 4xy + 16yz + 8 zx<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">(ii) (2x \u2013 y + z)<sup>2<\/sup>&nbsp;= (2x)<sup>2<\/sup>&nbsp;+ (- y)<sup>2<\/sup>&nbsp;+ z<sup>2<\/sup>&nbsp;+ 2 (2x) (- y)+ 2 (- y) (z) + 2 (z) (2x)<br>= 4x<sup>2<\/sup>&nbsp;+ y<sup>2<\/sup>&nbsp;+ z<sup>2<\/sup>&nbsp;\u2013 4xy \u2013 2yz + 4zx<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">(iii) (- 2x + 3y + 2z)<sup>2<\/sup>&nbsp;= (- 2x)<sup>2<\/sup>&nbsp;+ (3y)<sup>2<\/sup>&nbsp;+ (2z)<sup>2<\/sup>&nbsp;+ 2 (- 2x) (3y)+ 2 (3y) (2z) + 2 (2z) (- 2x)<br>= 4x<sup>2<\/sup>&nbsp;+ 9y<sup>2<\/sup>&nbsp;+ 4z<sup>2<\/sup>&nbsp;\u2013 12xy + 12yz \u2013 8zx<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">(iv) (3a -7b- c)<sup>2&nbsp;<\/sup>= (3a)<sup>2<\/sup>&nbsp;+ (- 7b)<sup>2<\/sup>&nbsp;+ (- c)<sup>2<\/sup>&nbsp;+ 2 (3a) (- 7b) + 2 (- 7b) (- c) + 2 (- c) (3a)<br>= 9a<sup>2<\/sup>&nbsp;+ 49b<sup>2<\/sup>&nbsp;+ c<sup>2<\/sup>&nbsp;\u2013 42ab + 14bc \u2013 6ac<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">(v)(- 2x + 5y- 3z)<sup>2<\/sup>\u00a0= (- 2x)<sup>2<\/sup>\u00a0+ (5y)<sup>2<\/sup>\u00a0+ (- 3z)<sup>2<\/sup>\u00a0+ 2 (- 2x) (5y) + 2 (5y) (- 3z) + 2 (- 3z) (- 2x)<br>= 4x<sup>2<\/sup>\u00a0+ 25y<sup>2<\/sup>\u00a0+ 9z<sup>2<\/sup>\u00a0\u2013 20xy \u2013 30yz + 12zx<\/p>\n\n\n\n<div class=\"wp-block-image\"><figure class=\"aligncenter size-large\"><img loading=\"lazy\" decoding=\"async\" width=\"360\" height=\"148\" src=\"https:\/\/themindpalace.in\/wp-content\/uploads\/2021\/07\/image-13.png\" alt=\"\" class=\"wp-image-3610\" srcset=\"https:\/\/themindpalace.in\/wp-content\/uploads\/2021\/07\/image-13.png 360w, https:\/\/themindpalace.in\/wp-content\/uploads\/2021\/07\/image-13-300x123.png 300w\" sizes=\"auto, (max-width: 360px) 100vw, 360px\" \/><\/figure><\/div>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>5. Factorise<br><\/strong>(i) 4 x<sup>2<\/sup>\u00a0+ 9y<sup>2<\/sup>\u00a0+ 16z<sup>2<\/sup>\u00a0+ 12xy \u2013 24yz \u2013 16xz<br>(ii) 2x<sup>2<\/sup>\u00a0+ y<sup>2<\/sup>\u00a0+ 8z<sup>2<\/sup>\u00a0\u2013 2\u221a2xy + 4\u221a2yz \u2013 8xz<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Solution:<br>(i) 4x<sup>2<\/sup>&nbsp;+ 9y<sup>2<\/sup>&nbsp;+ 16z<sup>2<\/sup>&nbsp;+ 12xy \u2013 24yz \u2013 16xz<br>= (2x)<sup>2<\/sup>&nbsp;+ (3y)<sup>2<\/sup>&nbsp;+ (- 4z)<sup>2<\/sup>&nbsp;+ 2 (2x) (3y) + 2 (3y) (- 4z) + 2 (- 4z) (2x)<br>= (2x + 3y \u2013 4z)<sup>2<\/sup>&nbsp;= (2x + 3y + 4z) (2x + 3y \u2013 4z)<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">(ii) 2x<sup>2<\/sup>\u00a0+ y<sup>2<\/sup>\u00a0+ 8z<sup>2<\/sup>\u00a0\u2013 2\u221a2xy + 4\u221a2yz \u2013 8xz<br>= (- \u221a2x)<sup>2<\/sup>\u00a0+ (y)<sup>2<\/sup>\u00a0+ (2 \u221a2z)<sup>2<\/sup>y + 2(- \u221a2x) (y)+ 2 (y) (2\u221a2z) + 2 (2\u221a2z) (- \u221a2x)<br>= (- \u221a2x + y + 2 \u221a2z)<sup>2<\/sup><br>= (- \u221a2x + y + 2 \u221a2z) (- \u221a2x + y + 2 \u221a2z) <\/p>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>6. Write the following cubes in expanded form<br><\/strong><\/p>\n\n\n\n<div class=\"wp-block-image\"><figure class=\"aligncenter size-large\"><img loading=\"lazy\" decoding=\"async\" width=\"355\" height=\"77\" src=\"https:\/\/themindpalace.in\/wp-content\/uploads\/2021\/07\/image-15.png\" alt=\"\" class=\"wp-image-3612\" srcset=\"https:\/\/themindpalace.in\/wp-content\/uploads\/2021\/07\/image-15.png 355w, https:\/\/themindpalace.in\/wp-content\/uploads\/2021\/07\/image-15-300x65.png 300w\" sizes=\"auto, (max-width: 355px) 100vw, 355px\" \/><\/figure><\/div>\n\n\n\n<p class=\"wp-block-paragraph\">Solution:<br>We have, (x + y)<sup>3<\/sup>&nbsp;= x<sup>3<\/sup>&nbsp;+ y<sup>3<\/sup>&nbsp;+ 3xy(x + y) \u2026(1)<br>and (x \u2013 y)<sup>3<\/sup>&nbsp;= x<sup>3<\/sup>&nbsp;\u2013 y<sup>3<\/sup>&nbsp;\u2013 3xy(x \u2013 y) \u2026(2)<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">(i) (2x + 1)<sup>3<\/sup>&nbsp;= (2x)<sup>3<\/sup>&nbsp;+ (1)<sup>3<\/sup>&nbsp;+ 3(2x)(1)(2x + 1) [By (1)]<br>= 8x<sup>3<\/sup>&nbsp;+ 1 + 6x(2x + 1)<br>= 8x<sup>3<\/sup>&nbsp;+ 12x<sup>2<\/sup>&nbsp;+ 6x + 1<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">(ii) (2a \u2013 3b)<sup>3<\/sup>\u00a0= (2a)<sup>3<\/sup>\u00a0\u2013 (3b)<sup>3<\/sup>\u00a0\u2013 3(2a)(3b)(2a \u2013 3b) [By (2)]<br>= 8a<sup>3<\/sup>\u00a0\u2013 27b<sup>3<\/sup>\u00a0\u2013 18ab(2a \u2013 3b)<br>= 8a<sup>3<\/sup>\u00a0\u2013 27b<sup>3<\/sup>\u00a0\u2013 36a<sup>2<\/sup>b + 54ab<sup>2<\/sup><\/p>\n\n\n\n<div class=\"wp-block-image\"><figure class=\"aligncenter size-large\"><img loading=\"lazy\" decoding=\"async\" width=\"381\" height=\"84\" src=\"https:\/\/themindpalace.in\/wp-content\/uploads\/2021\/07\/image-16.png\" alt=\"\" class=\"wp-image-3613\" srcset=\"https:\/\/themindpalace.in\/wp-content\/uploads\/2021\/07\/image-16.png 381w, https:\/\/themindpalace.in\/wp-content\/uploads\/2021\/07\/image-16-300x66.png 300w\" sizes=\"auto, (max-width: 381px) 100vw, 381px\" \/><\/figure><\/div>\n\n\n\n<div class=\"wp-block-image\"><figure class=\"aligncenter size-large\"><img loading=\"lazy\" decoding=\"async\" width=\"382\" height=\"311\" src=\"https:\/\/themindpalace.in\/wp-content\/uploads\/2021\/07\/image-18.png\" alt=\"\" class=\"wp-image-3615\" srcset=\"https:\/\/themindpalace.in\/wp-content\/uploads\/2021\/07\/image-18.png 382w, https:\/\/themindpalace.in\/wp-content\/uploads\/2021\/07\/image-18-300x244.png 300w\" sizes=\"auto, (max-width: 382px) 100vw, 382px\" \/><\/figure><\/div>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>7. Evaluate the following using suitable identities<\/strong><br><strong><br><\/strong>(i) (99)<sup>3<\/sup><br>(ii) (102)<sup>3<\/sup><br>(iii) (998)<sup>3<\/sup><\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Solution:<br>(i) We have, 99 = (100 -1)<br>\u2234 99<sup>3<\/sup>&nbsp;= (100 \u2013 1)<sup>3<\/sup><br>= (100)<sup>3<\/sup>&nbsp;\u2013 1<sup>3<\/sup>&nbsp;\u2013 3(100)(1)(100 -1)<br>[Using (a \u2013 b)<sup>3<\/sup>&nbsp;= a<sup>3<\/sup>&nbsp;\u2013 b<sup>3<\/sup>&nbsp;\u2013 3ab (a \u2013 b)]<br>= 1000000 \u2013 1 \u2013 300(100 \u2013 1)<br>= 1000000 -1 \u2013 30000 + 300<br>= 1000300 \u2013 30001 = 970299<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">(ii) We have, 102 =100 + 2<br>\u2234 102<sup>3<\/sup>&nbsp;= (100 + 2)<sup>3<\/sup><br>= (100)<sup>3<\/sup>&nbsp;+ (2)<sup>3<\/sup>&nbsp;+ 3(100)(2)(100 + 2)<br>[Using (a + b)<sup>3<\/sup>&nbsp;= a<sup>3<\/sup>&nbsp;+ b<sup>3<\/sup>&nbsp;+ 3ab (a + b)]<br>= 1000000 + 8 + 600(100 + 2)<br>= 1000000 + 8 + 60000 + 1200 = 1061208<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">(iii) We have, 998 = 1000 \u2013 2<br>\u2234 (998)<sup>3<\/sup>\u00a0= (1000-2)<sup>3<\/sup><br>= (1000)<sup>3<\/sup>\u2013 (2)<sup>3<\/sup>\u00a0\u2013 3(1000)(2)(1000 \u2013 2)<br>[Using (a \u2013 b)<sup>3<\/sup>\u00a0= a<sup>3<\/sup>\u00a0\u2013 b<sup>3<\/sup>\u00a0\u2013 3ab (a \u2013 b)]<br>= 1000000000 \u2013 8 \u2013 6000(1000 \u2013 2)<br>= 1000000000 \u2013 8 \u2013 6000000 +12000<br>= 994011992 <\/p>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>8.Factorise each of the following<br><\/strong>(i) 8a<sup>3<\/sup>\u00a0+b<sup>3<\/sup>\u00a0+ 12a<sup>2<\/sup>b+6ab<sup>2<\/sup><br>(ii) 8a<sup>3<\/sup>\u00a0-b<sup>3<\/sup>-12a<sup>2<\/sup>b+6ab<sup>2<br><\/sup>(iii) 27-125a<sup>3<\/sup>\u00a0-135a+225a<sup>2<br><\/sup>(iv) 64a<sup>3<\/sup>\u00a0-27b<sup>3<\/sup>\u00a0-144a<sup>2<\/sup>b + 108ab<sup>2<\/sup><\/p>\n\n\n\n<div class=\"wp-block-image\"><figure class=\"aligncenter size-large\"><img loading=\"lazy\" decoding=\"async\" width=\"208\" height=\"43\" src=\"https:\/\/themindpalace.in\/wp-content\/uploads\/2021\/07\/image-19.png\" alt=\"\" class=\"wp-image-3616\"\/><\/figure><\/div>\n\n\n\n<p class=\"wp-block-paragraph\"><br><\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Solution:<br>(i) 8a<sup>3<\/sup>&nbsp;+b<sup>3<\/sup>&nbsp;+12a<sup>2<\/sup>b+6ab<sup>2<\/sup><br>= (2a)<sup>3<\/sup>&nbsp;+ (b)<sup>3<\/sup>&nbsp;+ 6ab(2a + b)<br>= (2a)<sup>3<\/sup>&nbsp;+ (b)<sup>3<\/sup>&nbsp;+ 3(2a)(b)(2a + b)<br>= (2 a + b)<sup>3<\/sup><br>[Using a<sup>3<\/sup>&nbsp;+ b<sup>3<\/sup>&nbsp;+ 3 ab(a + b) = (a + b)<sup>3<\/sup>]<br>= (2a + b)(2a + b)(2a + b)<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">(ii) 8a<sup>3<\/sup>&nbsp;\u2013 b<sup>3<\/sup>&nbsp;\u2013 12o<sup>2<\/sup>b + 6ab<sup>2<\/sup><br>= (2a)<sup>3<\/sup>&nbsp;\u2013 (b)<sup>3<\/sup>&nbsp;\u2013 3(2a)(b)(2a \u2013 b)<br>= (2a \u2013 b)<sup>3<\/sup><br>[Using a<sup>3<\/sup>&nbsp;+ b<sup>3<\/sup>&nbsp;+ 3 ab(a + b) = (a + b)<sup>3<\/sup>]<br>= (2a \u2013 b) (2a \u2013 b) (2a \u2013 b)<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">(iii) 27 \u2013 125a<sup>3<\/sup>&nbsp;\u2013 135a + 225a<sup>2<\/sup><br>= (3)<sup>3<\/sup>&nbsp;\u2013 (5a)<sup>3<\/sup>&nbsp;\u2013 3(3)(5a)(3 \u2013 5a)<br>= (3 \u2013 5a)<sup>3<\/sup><br>[Using a<sup>3<\/sup>&nbsp;+ b<sup>3<\/sup>&nbsp;+ 3 ab(a + b) = (a + b)<sup>3<\/sup>]<br>= (3 \u2013 5a) (3 \u2013 5a) (3 \u2013 5a)<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">(iv) 64a<sup>3<\/sup>\u00a0-27b<sup>3<\/sup>\u00a0-144a<sup>2<\/sup>b + 108ab<sup>2<\/sup><br>= (4a)<sup>3<\/sup>\u00a0\u2013 (3b)<sup>3<\/sup>\u00a0\u2013 3(4a)(3b)(4a \u2013 3b)<br>= (4a \u2013 3b)<sup>3<\/sup><br>[Using a<sup>3<\/sup>\u00a0\u2013 b<sup>3<\/sup>\u00a0\u2013 3 ab(a \u2013 b) = (a \u2013 b)<sup>3<\/sup>]<br>= (4a \u2013 3b)(4a \u2013 3b)(4a \u2013 3b)<\/p>\n\n\n\n<div class=\"wp-block-image\"><figure class=\"aligncenter size-large\"><img loading=\"lazy\" decoding=\"async\" width=\"356\" height=\"219\" src=\"https:\/\/themindpalace.in\/wp-content\/uploads\/2021\/07\/image-21.png\" alt=\"\" class=\"wp-image-3618\" srcset=\"https:\/\/themindpalace.in\/wp-content\/uploads\/2021\/07\/image-21.png 356w, https:\/\/themindpalace.in\/wp-content\/uploads\/2021\/07\/image-21-300x185.png 300w\" sizes=\"auto, (max-width: 356px) 100vw, 356px\" \/><\/figure><\/div>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>9. Verify<br><\/strong>(i) x<sup>3<\/sup>\u00a0+ y<sup>3<\/sup>\u00a0= (x + y)-(x<sup>2<\/sup>\u00a0\u2013 xy + y<sup>2<\/sup>)<br>(ii) x<sup>3<\/sup>\u00a0\u2013 y<sup>3<\/sup>\u00a0= (x \u2013 y) (x<sup>2<\/sup>\u00a0+ xy + y<sup>2<\/sup>)<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Solution:<br>(i) \u2235 (x + y)<sup>3<\/sup>&nbsp;= x<sup>3<\/sup>&nbsp;+ y<sup>3<\/sup>&nbsp;+ 3xy(x + y)<br>\u21d2 (x + y)<sup>3<\/sup>&nbsp;\u2013 3(x + y)(xy) = x<sup>3<\/sup>&nbsp;+ y<sup>3<\/sup><br>\u21d2 (x + y)[(x + y)2-3xy] = x<sup>3<\/sup>&nbsp;+ y<sup>3<\/sup><br>\u21d2 (x + y)(x<sup>2<\/sup>&nbsp;+ y<sup>2<\/sup>&nbsp;\u2013 xy) = x<sup>3<\/sup>&nbsp;+ y<sup>3<\/sup><br>Hence, verified.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">(ii) \u2235 (x \u2013 y)<sup>3<\/sup>&nbsp;= x<sup>3<\/sup>&nbsp;\u2013 y<sup>3<\/sup>&nbsp;\u2013 3xy(x \u2013 y)<br>\u21d2 (x \u2013 y)<sup>3<\/sup>&nbsp;+ 3xy(x \u2013 y) = x<sup>3<\/sup>&nbsp;\u2013 y<sup>3<\/sup><br>\u21d2 (x \u2013 y)[(x \u2013 y)<sup>2<\/sup>&nbsp;+ 3xy)] = x<sup>3<\/sup>&nbsp;\u2013 y<sup>3<\/sup><br>\u21d2 (x \u2013 y)(x<sup>2<\/sup>&nbsp;+ y<sup>2<\/sup>&nbsp;+ xy) = x<sup>3<\/sup>&nbsp;\u2013 y<sup>3<\/sup><br>Hence, verified.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>10.Factorise each of the following<br><\/strong>(i) 27y<sup>3<\/sup>\u00a0+ 125z<sup>3<\/sup><br>(ii) 64m<sup>3<\/sup>\u00a0\u2013 343n<sup>3<\/sup><br>[Hint See question 9]<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Solution:<br>(i) We know that<br>x<sup>3<\/sup>&nbsp;+ y<sup>3<\/sup>&nbsp;= (x + y)(x<sup>2<\/sup>&nbsp;\u2013 xy + y<sup>2<\/sup>)<br>We have, 27y<sup>3<\/sup>&nbsp;+ 125z<sup>3<\/sup>&nbsp;= (3y)<sup>3<\/sup>&nbsp;+ (5z)<sup>3<\/sup><br>= (3y + 5z)[(3y)<sup>2<\/sup>&nbsp;\u2013 (3y)(5z) + (5z)<sup>2<\/sup>]<br>= (3y + 5z)(9y<sup>2<\/sup>&nbsp;\u2013 15yz + 25z<sup>2<\/sup>)<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">(ii) We know that<br>x<sup>3<\/sup>&nbsp;\u2013 y<sup>3<\/sup>&nbsp;= (x \u2013 y)(x<sup>2<\/sup>&nbsp;+ xy + y<sup>2<\/sup>)<br>We have, 64m<sup>3<\/sup>&nbsp;\u2013 343n<sup>3<\/sup>&nbsp;= (4m)<sup>3<\/sup>&nbsp;\u2013 (7n)<sup>3<\/sup><br>= (4m \u2013 7n)[(4m)<sup>2<\/sup>&nbsp;+ (4m)(7n) + (7n)<sup>2<\/sup>]<br>= (4m \u2013 7n)(16m<sup>2<\/sup>&nbsp;+ 28mn + 49n<sup>2<\/sup>)<\/p>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>10.Factorise each of the following<br><\/strong>(i) 27y<sup>3<\/sup>\u00a0+ 125z<sup>3<\/sup><br>(ii) 64m<sup>3<\/sup>\u00a0\u2013 343n<sup>3<\/sup><br>[Hint See question 9]<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Solution:<br>(i) We know that<br>x<sup>3<\/sup>&nbsp;+ y<sup>3<\/sup>&nbsp;= (x + y)(x<sup>2<\/sup>&nbsp;\u2013 xy + y<sup>2<\/sup>)<br>We have, 27y<sup>3<\/sup>&nbsp;+ 125z<sup>3<\/sup>&nbsp;= (3y)<sup>3<\/sup>&nbsp;+ (5z)<sup>3<\/sup><br>= (3y + 5z)[(3y)<sup>2<\/sup>&nbsp;\u2013 (3y)(5z) + (5z)<sup>2<\/sup>]<br>= (3y + 5z)(9y<sup>2<\/sup>&nbsp;\u2013 15yz + 25z<sup>2<\/sup>)<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">(ii) We know that<br>x<sup>3<\/sup>&nbsp;\u2013 y<sup>3<\/sup>&nbsp;= (x \u2013 y)(x<sup>2<\/sup>&nbsp;+ xy + y<sup>2<\/sup>)<br>We have, 64m<sup>3<\/sup>&nbsp;\u2013 343n<sup>3<\/sup>&nbsp;= (4m)<sup>3<\/sup>&nbsp;\u2013 (7n)<sup>3<\/sup><br>= (4m \u2013 7n)[(4m)<sup>2<\/sup>&nbsp;+ (4m)(7n) + (7n)<sup>2<\/sup>]<br>= (4m \u2013 7n)(16m<sup>2<\/sup>&nbsp;+ 28mn + 49n<sup>2<\/sup>)<\/p>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>11.Factorise 27x<sup>3<\/sup>\u00a0+y<sup>3<\/sup>\u00a0+z<sup>3<\/sup>\u00a0-9xyz<\/strong><\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Solution:<br>We have,<br>27x<sup>3<\/sup>&nbsp;+ y<sup>3<\/sup>&nbsp;+ z<sup>3<\/sup>&nbsp;\u2013 9xyz = (3x)<sup>3<\/sup>&nbsp;+ (y)<sup>3<\/sup>&nbsp;+ (z)<sup>3<\/sup>&nbsp;\u2013 3(3x)(y)(z)<br>Using the identity,<br>x<sup>3<\/sup>&nbsp;+ y<sup>3<\/sup>&nbsp;+ z<sup>3<\/sup>&nbsp;\u2013 3xyz = (x + y + z)(x<sup>2<\/sup>&nbsp;+ y<sup>2<\/sup>&nbsp;+ z<sup>2<\/sup>&nbsp;\u2013 xy \u2013 yz \u2013 zx)<br>We have, (3x)<sup>3<\/sup>&nbsp;+ (y)<sup>3<\/sup>&nbsp;+ (z)<sup>3<\/sup>&nbsp;\u2013 3(3x)(y)(z)<br>= (3x + y + z)[(3x)<sup>3<\/sup>&nbsp;+ y<sup>3<\/sup>&nbsp;+ z<sup>3<\/sup>&nbsp;\u2013 (3x \u00d7 y) \u2013 (y \u00d7 2) \u2013 (z \u00d7 3x)]<br>= (3x + y + z)(9x<sup>2<\/sup>&nbsp;+ y<sup>2<\/sup>&nbsp;+ z<sup>2<\/sup>&nbsp;\u2013 3xy \u2013 yz \u2013 3zx)<\/p>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>12.Verify that<br><\/strong>x<sup>3<\/sup>\u00a0+y<sup>3<\/sup>\u00a0+z<sup>3<\/sup>\u00a0\u2013 3xyz =\u00a01\/2\u00a0(x + y+z)[(x-y)<sup>2<\/sup>\u00a0+ (y \u2013 z)<sup>2<\/sup>\u00a0+(z \u2013 x)<sup>2<\/sup>]<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Solution:<br>R.H.S<br>=&nbsp;1\/2(x + y + z)[(x \u2013 y)<sup>2<\/sup>+(y \u2013 z)<sup>2<\/sup>+(z \u2013 x)<sup>2<\/sup>]<br>=&nbsp;1\/2&nbsp;(x + y + 2)[(x<sup>2<\/sup>&nbsp;+ y<sup>2<\/sup>&nbsp;\u2013 2xy) + (y<sup>2<\/sup>&nbsp;+ z<sup>2<\/sup>&nbsp;\u2013 2yz) + (z<sup>2<\/sup>&nbsp;+ x<sup>2<\/sup>&nbsp;\u2013 2zx)]<br>=&nbsp;1\/2&nbsp;(x + y + 2)(x<sup>2<\/sup>&nbsp;+ y<sup>2<\/sup>&nbsp;+ y<sup>2<\/sup>&nbsp;+ z<sup>2<\/sup>&nbsp;+ z<sup>2<\/sup>&nbsp;+ x<sup>2<\/sup>&nbsp;\u2013 2xy \u2013 2yz \u2013 2zx)<br>=&nbsp;1\/2&nbsp;(x + y + z)[2(x<sup>2<\/sup>&nbsp;+ y<sup>2<\/sup>&nbsp;+ z<sup>2<\/sup>&nbsp;\u2013 xy \u2013 yz \u2013 zx)]<br>= 2 x&nbsp;1\/2&nbsp;x (x + y + z)(x<sup>2<\/sup>&nbsp;+ y<sup>2<\/sup>&nbsp;+ z<sup>2<\/sup>&nbsp;\u2013 xy \u2013 yz \u2013 zx)<br>= (x + y + z)(x<sup>2<\/sup>&nbsp;+ y<sup>2<\/sup>&nbsp;+ z<sup>2<\/sup>&nbsp;\u2013 xy \u2013 yz \u2013 zx)<br>= x<sup>3<\/sup>&nbsp;+ y<sup>3<\/sup>&nbsp;+ z<sup>3<\/sup>&nbsp;\u2013 3xyz = L.H.S.<br>Hence, verified.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>13. If x + y + z = 0, show that x<sup>3<\/sup>\u00a0+ y<sup>3<\/sup>\u00a0+ z<sup>3<\/sup>\u00a0= 3 xyz.<\/strong><\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Solution:<br>Since, x + y + z = 0<br>\u21d2 x + y = -z (x + y)<sup>3<\/sup>&nbsp;= (-z)<sup>3<\/sup><br>\u21d2 x<sup>3<\/sup>&nbsp;+ y<sup>3<\/sup>&nbsp;+ 3xy(x + y) = -z<sup>3<\/sup><br>\u21d2 x<sup>3<\/sup>&nbsp;+ y<sup>3<\/sup>&nbsp;+ 3xy(-z) = -z<sup>3<\/sup>&nbsp;[\u2235 x + y = -z]<br>\u21d2 x<sup>3<\/sup>&nbsp;+ y<sup>3<\/sup>&nbsp;\u2013 3xyz = -z<sup>3<\/sup><br>\u21d2 x<sup>3<\/sup>&nbsp;+ y<sup>3<\/sup>&nbsp;+ z<sup>3<\/sup>&nbsp;= 3xyz<br>Hence, if x + y + z = 0, then<br>x<sup>3<\/sup>&nbsp;+ y<sup>3<\/sup>&nbsp;+ z<sup>3<\/sup>&nbsp;= 3xyz<\/p>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>14. Without actually calculating the cubes, find the value of each of the following<br><\/strong>(i) (- 12)<sup>3<\/sup>\u00a0+ (7)<sup>3<\/sup>\u00a0+ (5)<sup>3<br><\/sup>(ii) (28)<sup>3<\/sup>\u00a0+ (- 15)<sup>3<\/sup>\u00a0+ (- 13)<sup>3<\/sup><br>Solution:<br>(i) We have, (-12)<sup>3<\/sup>\u00a0+ (7)<sup>3<\/sup>\u00a0+ (5)<sup>3<\/sup><br>Let x = -12, y = 7 and z = 5.<br>Then, x + y + z = -12 + 7 + 5 = 0<br>We know that if x + y + z = 0, then, x<sup>3<\/sup>\u00a0+ y<sup>3<\/sup>\u00a0+ z<sup>3<\/sup>\u00a0= 3xyz<br>\u2234 (-12)<sup>3<\/sup>\u00a0+ (7)<sup>3<\/sup>\u00a0+ (5)<sup>3<\/sup>\u00a0= 3[(-12)(7)(5)]<br>= 3[-420] = -1260<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">(ii) We have, (28)<sup>3<\/sup>&nbsp;+ (-15)<sup>3<\/sup>&nbsp;+ (-13)<sup>3<\/sup><br>Let x = 28, y = -15 and z = -13.<br>Then, x + y + z = 28 \u2013 15 \u2013 13 = 0<br>We know that if x + y + z = 0, then x<sup>3<\/sup>&nbsp;+ y<sup>3<\/sup>&nbsp;+ z<sup>3<\/sup>&nbsp;= 3xyz<br>\u2234 (28)<sup>3<\/sup>&nbsp;+ (-15)<sup>3<\/sup>&nbsp;+ (-13)<sup>3<\/sup>&nbsp;= 3(28)(-15)(-13)<br>= 3(5460) = 16380<\/p>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>15. Give possible expressions for the length and breadth of each of the following rectangles, in which their areas are given<br><\/strong>(i) Area 25a<sup>2<\/sup>\u00a0\u2013 35a + 12<br>(ii) Area 35y<sup>2<\/sup>\u00a0+ 13y \u2013 12<br>Solution:<br>Area of a rectangle = (Length) x (Breadth)<br>(i) 25a<sup>2<\/sup>\u00a0\u2013 35a + 12 = 25a<sup>2<\/sup>\u00a0\u2013 20a \u2013 15a + 12 = 5a(5a \u2013 4) \u2013 3(5a \u2013 4) = (5a \u2013 4)(5a \u2013 3)<br>Thus, the possible length and breadth are (5a \u2013 3) and (5a \u2013 4).<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">(ii) 35y<sup>2<\/sup>+ 13y -12 = 35y<sup>2<\/sup>&nbsp;+ 28y \u2013 15y -12<br>= 7y(5y + 4) \u2013 3(5y + 4) = (5 y + 4)(7y \u2013 3)<br>Thus, the possible length and breadth are (7y \u2013 3) and (5y + 4).<\/p>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>16. What are the possible expressions for the dimensions of the cuboids whose volumes are given below?<br><\/strong>(i) Volume 3x<sup>2<\/sup>\u00a0\u2013 12x<br>(ii) Volume 12ky<sup>2<\/sup>\u00a0+ 8ky \u2013 20k<\/p>\n\n\n\n<p class=\"wp-block-paragraph\"><br>Solution:<br>Volume of a cuboid = (Length) x (Breadth) x (Height)<br>(i) We have, 3x<sup>2<\/sup>\u00a0\u2013 12x = 3(x<sup>2<\/sup>\u00a0\u2013 4x)<br>= 3 x x x (x \u2013 4)<br>\u2234 The possible dimensions of the cuboid are 3, x and (x \u2013 4).<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">(ii) We have, 12ky<sup>2<\/sup>&nbsp;+ 8ky \u2013 20k<br>= 4[3ky<sup>2<\/sup>&nbsp;+ 2ky \u2013 5k] = 4[k(3y<sup>2<\/sup>&nbsp;+ 2y \u2013 5)]<br>= 4 x k x (3y<sup>2<\/sup>&nbsp;+ 2y \u2013 5)<br>= 4k[3y<sup>2<\/sup>&nbsp;\u2013 3y + 5y \u2013 5]<br>= 4k[3y(y \u2013 1) + 5(y \u2013 1)]<br>= 4k[(3y + 5) x (y \u2013 1)]<br>= 4k x (3y + 5) x (y \u2013 1)<br>Thus, the possible dimensions of the cuboid are 4k, (3y + 5) and (y -1).<\/p>\n","protected":false},"excerpt":{"rendered":"<p>Polynomials Introduction Polynomials In One Variable Zeroes Of A Polynomial Remainder Theorem Factorisation Of Polynomials Algebraic Identities Summary Polynomial is an algebraic expression that includes constants, variables, and exponents. It is the expression in&#46;&#46;&#46;<\/p>\n","protected":false},"author":2,"featured_media":3496,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_jetpack_newsletter_access":"","_jetpack_dont_email_post_to_subs":false,"_jetpack_newsletter_tier_id":0,"_jetpack_memberships_contains_paywalled_content":false,"_jetpack_memberships_contains_paid_content":false,"footnotes":""},"categories":[111,475,14],"tags":[409,410],"class_list":["post-3495","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-9th-class-ncert-syllabus","category-posts-in-english","category-mathematics","tag-9th-class-maths","tag-polynomials"],"cp_meta_data":{"_jetpack_related_posts_cache":["a:1:{s:32:\"8f6677c9d6b0f903e98ad32ec61f8deb\";a:2:{s:7:\"expires\";i:1783399964;s:7:\"payload\";a:3:{i:0;a:1:{s:2:\"id\";i:2659;}i:1;a:1:{s:2:\"id\";i:3625;}i:2;a:1:{s:2:\"id\";i:5978;}}}}"],"_edit_lock":["1629961016:2"],"_last_editor_used_jetpack":["block-editor"],"_edit_last":["2"],"_layout":["inherit"],"_heateor_sss_meta":["a:2:{s:7:\"sharing\";i:0;s:16:\"vertical_sharing\";i:0;}"],"_thumbnail_id":["3496"]},"jetpack_featured_media_url":"https:\/\/themindpalace.in\/wp-content\/uploads\/2021\/06\/polynomials-9-pic1.png","jetpack-related-posts":[{"id":2659,"url":"https:\/\/themindpalace.in\/index.php\/2021\/01\/08\/algebraic-expression-and-identities\/","url_meta":{"origin":3495,"position":0},"title":"Algebraic Expression and Identities","author":"Nancy Diana","date":"January 8, 2021","format":false,"excerpt":"Any mathematical expression which consists of numbers, variables and operations are called Algebraic Expression.","rel":"","context":"In &quot;8th Class&quot;","block_context":{"text":"8th Class","link":"https:\/\/themindpalace.in\/index.php\/category\/ncert-school-syllabus\/8th-class-ncert-syllabus\/"},"img":{"alt_text":"","src":"https:\/\/i0.wp.com\/themindpalace.in\/wp-content\/uploads\/2021\/01\/table.jpg?resize=350%2C200&ssl=1","width":350,"height":200,"srcset":"https:\/\/i0.wp.com\/themindpalace.in\/wp-content\/uploads\/2021\/01\/table.jpg?resize=350%2C200&ssl=1 1x, https:\/\/i0.wp.com\/themindpalace.in\/wp-content\/uploads\/2021\/01\/table.jpg?resize=525%2C300&ssl=1 1.5x"},"classes":[]},{"id":3625,"url":"https:\/\/themindpalace.in\/index.php\/2021\/07\/09\/linear-equations-in-two-variables\/","url_meta":{"origin":3495,"position":1},"title":"Linear Equations in Two Variables","author":"The Mind","date":"July 9, 2021","format":false,"excerpt":"Linear Equations The equation of a straight line is the linear equation. It could be in one variable or two variables. Linear Equation in One Variable The equation with one variable in it is known as a\u00a0Linear Equation in One Variable. The general form is px + q = s,\u2026","rel":"","context":"In &quot;9th Class&quot;","block_context":{"text":"9th Class","link":"https:\/\/themindpalace.in\/index.php\/category\/ncert-school-syllabus\/9th-class-ncert-syllabus\/"},"img":{"alt_text":"","src":"https:\/\/i0.wp.com\/themindpalace.in\/wp-content\/uploads\/2021\/07\/Slope-Intercept-form.png?resize=350%2C200&ssl=1","width":350,"height":200,"srcset":"https:\/\/i0.wp.com\/themindpalace.in\/wp-content\/uploads\/2021\/07\/Slope-Intercept-form.png?resize=350%2C200&ssl=1 1x, https:\/\/i0.wp.com\/themindpalace.in\/wp-content\/uploads\/2021\/07\/Slope-Intercept-form.png?resize=525%2C300&ssl=1 1.5x, https:\/\/i0.wp.com\/themindpalace.in\/wp-content\/uploads\/2021\/07\/Slope-Intercept-form.png?resize=700%2C400&ssl=1 2x"},"classes":[]},{"id":5978,"url":"https:\/\/themindpalace.in\/index.php\/2022\/09\/15\/how-to-read-greek-alphabets\/","url_meta":{"origin":3495,"position":2},"title":"How to read Greek Alphabets","author":"The Mind","date":"September 15, 2022","format":false,"excerpt":"The Greek alphabet has been used to write the Greek language since the late 9th or early 8th century BCE.The modern Greek alphabet has 24 letters. It is used to write the Greek language. The Greek alphabets are frequently used in science and mathematics to represent various values or variables.\u2026","rel":"","context":"In &quot;English Medium&quot;","block_context":{"text":"English Medium","link":"https:\/\/themindpalace.in\/index.php\/category\/posts-in-english\/"},"img":{"alt_text":"Greek Alphabets","src":"https:\/\/i0.wp.com\/themindpalace.in\/wp-content\/uploads\/2022\/09\/Greek-Alphabets.jpg?resize=350%2C200&ssl=1","width":350,"height":200},"classes":[]},{"id":490,"url":"https:\/\/themindpalace.in\/index.php\/2020\/08\/16\/matter-in-our-surroundings\/","url_meta":{"origin":3495,"position":3},"title":"Matter in Our Surroundings","author":"Nancy Diana","date":"August 16, 2020","format":false,"excerpt":"Summary of matter in our surrounding matter which occupies space,table , chair. Summary Matter Matter is anything which occupies space and has mass. Everything in this universe is made of materials which scientist has named \u2018matter\u2019. Example: Chair, bed, river, tree, building, etc. Classification of Matter fire earth sky sky\u2026","rel":"","context":"In &quot;9th Class&quot;","block_context":{"text":"9th Class","link":"https:\/\/themindpalace.in\/index.php\/category\/ncert-school-syllabus\/9th-class-ncert-syllabus\/"},"img":{"alt_text":"","src":"https:\/\/i0.wp.com\/themindpalace.in\/wp-content\/uploads\/2020\/08\/matter.jpg?resize=350%2C200&ssl=1","width":350,"height":200,"srcset":"https:\/\/i0.wp.com\/themindpalace.in\/wp-content\/uploads\/2020\/08\/matter.jpg?resize=350%2C200&ssl=1 1x, https:\/\/i0.wp.com\/themindpalace.in\/wp-content\/uploads\/2020\/08\/matter.jpg?resize=525%2C300&ssl=1 1.5x, https:\/\/i0.wp.com\/themindpalace.in\/wp-content\/uploads\/2020\/08\/matter.jpg?resize=700%2C400&ssl=1 2x, https:\/\/i0.wp.com\/themindpalace.in\/wp-content\/uploads\/2020\/08\/matter.jpg?resize=1050%2C600&ssl=1 3x, https:\/\/i0.wp.com\/themindpalace.in\/wp-content\/uploads\/2020\/08\/matter.jpg?resize=1400%2C800&ssl=1 4x"},"classes":[]},{"id":3453,"url":"https:\/\/themindpalace.in\/index.php\/2021\/06\/07\/coordinate-geometry\/","url_meta":{"origin":3495,"position":4},"title":"Coordinate Geometry","author":"The Mind","date":"June 7, 2021","format":false,"excerpt":"Cartesian System If we take two number lines, one horizontal and one vertical, and then combine them in such a way that they intersect each other at their zeroes, and then they form a\u00a0Cartesian Plane. 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Energy is measured by the work that the body can do. 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