{"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><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>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><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=\"(max-width: 671px) 100vw, 671px\" \/><figcaption>Polynomial <\/figcaption><\/figure><\/div>\n\n\n\n<ul><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>If there is only one variable in the expression, then this is called the polynomial in one variable.<\/p>\n\n\n\n<p><strong>Example<\/strong><br><\/p>\n\n\n\n<ul><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><\/p>\n\n\n\n<p>The highest value of the power of the variable in the polynomial is the degree of the polynomial.<\/p>\n\n\n\n<p><\/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=\"(max-width: 1024px) 100vw, 1024px\" \/><figcaption>Types of polynomials<\/figcaption><\/figure>\n\n\n\n<p><\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Zeroes of a Polynomial<\/h3>\n\n\n\n<p>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><strong>Example: 1<\/strong><\/p>\n\n\n\n<p>Given polynomial is p(x) = x &#8211; 4<\/p>\n\n\n\n<p>To find the zero of the polynomial we will equate it to zero.<\/p>\n\n\n\n<p>x &#8211; 4 = 0<\/p>\n\n\n\n<p>x = 4<\/p>\n\n\n\n<p>p(4) = x \u2013 4 = 4 \u2013 4 = 0<\/p>\n\n\n\n<p>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>So 4 is the zero of the polynomial or root of the polynomial.<\/p>\n\n\n\n<p>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>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><strong>Example: 2<\/strong><\/p>\n\n\n\n<p>Find p (1) for the polynomial p (t) = t<sup>2<\/sup>&nbsp;\u2013 t + 1<\/p>\n\n\n\n<p>p (1) = (1)<sup>2<\/sup>&nbsp;\u2013 1 + 1<\/p>\n\n\n\n<p>= 1 \u2013 1 + 1<\/p>\n\n\n\n<p>= 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><strong>Example 2<\/strong>.<strong>1 <\/strong><\/p>\n\n\n\n<p><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>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>(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>(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>(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>(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><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>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><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><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>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><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><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><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><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>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><strong>Example 2.2<\/strong><\/p>\n\n\n\n<ol><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>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>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; = 0 \u2013 0 + 3<\/p>\n\n\n\n<p>&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><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>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; = -9 + 3<\/p>\n\n\n\n<p>&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><br>(iii) p(2) = 5(2) \u2013 4(2)<sup>2<\/sup>&nbsp;+ 3<\/p>\n\n\n\n<p>&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>&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><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>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>(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>(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>(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><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>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=\"(max-width: 386px) 100vw, 386px\" \/><\/figure><\/div>\n\n\n\n<p>(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><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>(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>(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=\"(max-width: 388px) 100vw, 388px\" \/><\/figure><\/div>\n\n\n\n<p>(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=\"(max-width: 403px) 100vw, 403px\" \/><\/figure><\/div>\n\n\n\n<p><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>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>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>(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>(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>(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>(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>(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>(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><strong>Example 2.3<br><\/strong><br><\/p>\n\n\n\n<p>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>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>(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=\"(max-width: 359px) 100vw, 359px\" \/><\/figure><\/div>\n\n\n\n<p>Thus, the required remainder =\u00a027\/8<\/p>\n\n\n\n<p>(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>(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>(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=\"(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>Thus, the required remainder is&nbsp;\u221227\/8&nbsp;.<\/p>\n\n\n\n<p><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>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=\"(max-width: 375px) 100vw, 375px\" \/><\/figure><\/div>\n\n\n\n<p>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><strong>Example 2.4<\/strong><\/p>\n\n\n\n<ol 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>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>(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>(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>(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><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>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>(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>(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><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>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>(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>(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>(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>(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><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>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>(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>(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>(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><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>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>(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>(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>(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><strong>Example 2.5<\/strong><\/p>\n\n\n\n<ol 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>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>(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> (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=\"(max-width: 392px) 100vw, 392px\" \/><\/figure><\/div>\n\n\n\n<p><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>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>(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>(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><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>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>(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=\"(max-width: 384px) 100vw, 384px\" \/><\/figure><\/div>\n\n\n\n<p><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>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>(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>(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>(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>(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>(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=\"(max-width: 360px) 100vw, 360px\" \/><\/figure><\/div>\n\n\n\n<p><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>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>(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><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=\"(max-width: 355px) 100vw, 355px\" \/><\/figure><\/div>\n\n\n\n<p>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>(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>(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=\"(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=\"(max-width: 382px) 100vw, 382px\" \/><\/figure><\/div>\n\n\n\n<p><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>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>(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>(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><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><br><\/p>\n\n\n\n<p>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>(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>(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>(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=\"(max-width: 356px) 100vw, 356px\" \/><\/figure><\/div>\n\n\n\n<p><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>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>(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><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>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>(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><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>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>(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><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>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><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>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><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>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><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>(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><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>(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><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><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>(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_memberships_contains_paid_content":false,"footnotes":""},"categories":[111,475,14],"tags":[409,410],"cp_meta_data":{"_jetpack_related_posts_cache":["a:1:{s:32:\"8f6677c9d6b0f903e98ad32ec61f8deb\";a:2:{s:7:\"expires\";i:1776245664;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_sharing_enabled":true,"jetpack_featured_media_url":"https:\/\/themindpalace.in\/wp-content\/uploads\/2021\/06\/polynomials-9-pic1.png","jetpack-related-posts":[],"_links":{"self":[{"href":"https:\/\/themindpalace.in\/index.php\/wp-json\/wp\/v2\/posts\/3495"}],"collection":[{"href":"https:\/\/themindpalace.in\/index.php\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/themindpalace.in\/index.php\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/themindpalace.in\/index.php\/wp-json\/wp\/v2\/users\/2"}],"replies":[{"embeddable":true,"href":"https:\/\/themindpalace.in\/index.php\/wp-json\/wp\/v2\/comments?post=3495"}],"version-history":[{"count":13,"href":"https:\/\/themindpalace.in\/index.php\/wp-json\/wp\/v2\/posts\/3495\/revisions"}],"predecessor-version":[{"id":3620,"href":"https:\/\/themindpalace.in\/index.php\/wp-json\/wp\/v2\/posts\/3495\/revisions\/3620"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/themindpalace.in\/index.php\/wp-json\/wp\/v2\/media\/3496"}],"wp:attachment":[{"href":"https:\/\/themindpalace.in\/index.php\/wp-json\/wp\/v2\/media?parent=3495"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/themindpalace.in\/index.php\/wp-json\/wp\/v2\/categories?post=3495"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/themindpalace.in\/index.php\/wp-json\/wp\/v2\/tags?post=3495"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}