{"id":34998,"date":"2023-10-23T13:47:32","date_gmt":"2023-10-23T08:17:32","guid":{"rendered":"https:\/\/ccssanswers.com\/?p=34998"},"modified":"2023-11-10T10:47:58","modified_gmt":"2023-11-10T05:17:58","slug":"factorization-by-using-identities","status":"publish","type":"post","link":"https:\/\/ccssanswers.com\/factorization-by-using-identities\/","title":{"rendered":"Factorization by Using Identities | How to Factorise Polynomials using Algebraic Identities?"},"content":{"rendered":"<p>Know the process to find <a href=\"https:\/\/ccssanswers.com\/factorization\/\">Factorization<\/a> by using identities. Based on the identities, we can simply factorize an algebraic equation. That means, depending on the identities or identity values, we can easily reduce the number of expressions into n number of terms. Generally, for the simplest factorization process, we have to follow some basic expressions. They are<\/p>\n<p>(i) (x + y)^2 = (x)^2 + 2xy + (y)^2.<br \/>\n(ii) (x \u2013 y)^2 = (x)^2 \u2013 2xy + (y)^2.<br \/>\n(iii) x^2 \u2013 y^2 = (x + y) (x \u2013 y).<\/p>\n<h2>Factoring Polynomials Identities Steps<\/h2>\n<p>Go through the below-listed guidelines on how to factor polynomials using algebraic identities. They are in the following fashion<\/p>\n<ul>\n<li>Note down the given expression and compare the expression with the basic expressions.<\/li>\n<li>If there are three terms and all are identified with the positive sign, then that is related to (x + y)^2.<\/li>\n<li>If the first and last terms are with the positive sign and the middle term is identified with the negative sign, then that is related to (x \u2013 y)^2.<\/li>\n<li>If there are only two terms with one positive sign and one negative sign, then that is related to x^2 \u2013 y^2.<\/li>\n<li>Now, compare the co-efficient values with the basic expressions.<\/li>\n<li>Find the values of coefficients.<\/li>\n<li>Based on the values, reduce the given expression into simple terms.<\/li>\n<\/ul>\n<h3>Factorization Using Identities Examples with Answers<\/h3>\n<p>1. Factorize using the formula of a square of the sum of two terms<\/p>\n<p>(i) a<sup>2<\/sup> + 6a + 9<\/p>\n<p>Solution: Given expression is a<sup>2<\/sup> + 6a + 9<br \/>\nThere are three terms are identified with the positive sign. Then it is related to the expression (x)^2 + 2xy + (y)^2 = (x + y)^2.<br \/>\nNow, compare the co-efficient values of a<sup>2<\/sup> + 6a + 9 with the expression (x)^2 + 2xy + (y)^2.<br \/>\nHere, x = a, 2y = 6 then y = 3.<br \/>\nSo, (x + y)^2 = (a + 3)^2.<\/p>\n<p>Finally, our expression a^2 + 6a + 9 is reduced to (a + 3)^2.<\/p>\n<p>(ii) a^2 + 20a + 100<\/p>\n<p>Solution:<br \/>\nGiven expression is a^2 + 20a + 100.<br \/>\nThere are three terms that are identified with a positive sign. Then it is related to the expression (x)^2 + 2xy + (y)^2 = (x + y)^2.<br \/>\nNow, compare the co-efficient values of a^2 + 20a + 100 with the expression(x)^2 + 2xy + (y)^2.<br \/>\nHere, x = a, 2y = 20 then y = 10.<br \/>\nSo, (x + y)^2 = (a + 10)^2.<\/p>\n<p>Finally, our expression a^2 + 20a + 100 is reduced to (a + 10)^2.<\/p>\n<p>2. Factorize using the formula of the square of the difference of two terms<\/p>\n<p>(i) 4p^2 \u2013 12pq + 9q^2<\/p>\n<p>Solution:<br \/>\nGiven expression is 4p^2 \u2013 12pq + 9q^2.<br \/>\nThere are three terms that are identified with the positive sign of the first and last terms and the negative sign of the middle term.<br \/>\nThen it is related to the expression (x)^2 &#8211; 2xy + (y)^2 = (x \u2013 y)^2.<br \/>\nNow, compare the co-efficient values of 4p^2 \u2013 12pq + 9q^2with the expression (x)^2 &#8211; 2xy + (y)^2.<br \/>\nHere, x^2 = 4p^2 then x =2p, y^2 = 9q^2 then y = 3q.<br \/>\nSo, (x &#8211; y)^2 = (2p \u2013 3q)^2.<\/p>\n<p>Finally, our expression 4p^2 \u2013 12pq + 9q^2is reduced to (2p \u2013 3q)^2.<\/p>\n<p>(ii) a^2 \u2013 10a + 25<\/p>\n<p>Solution:<br \/>\nGiven expression is a^2 \u2013 10a + 25.<br \/>\nThere are three terms that are identified with the positive sign of the first and last terms and the negative sign of the middle term.<br \/>\nThen it is related to the expression (x)^2 &#8211; 2xy + (y)^2 = (x \u2013 y)^2.<br \/>\nNow, compare the co-efficient values of a^2 \u2013 10a + 25 with the expression (x)^2 &#8211; 2xy + (y)^2.<br \/>\nHere, x^2 = a^2 then x =a, y^2 = 25 then y = 5.<br \/>\nSo, (x &#8211; y)^2 = (a \u2013 5)^2.<\/p>\n<p>Finally, our expression a^2 \u2013 10a + 25 is reduced to (a &#8211; 5)^2.<\/p>\n<p>3. Factorize using the formula of a difference of two squares:<\/p>\n<p>(i) 49a^2 \u2013 64<\/p>\n<p>Solution:<br \/>\nGiven expression is 49a^2 \u2013 64<br \/>\nThere are two terms identified with the positive sign of the first term and the negative sign of the last term.<br \/>\nNow, compare the co-efficient values of 49a^2 &#8211; 64 with the expression (x)^2 &#8211; (y)^2.<br \/>\nHere, x^2 = 49a^2 = (7a)^2 then x =7a, y^2 = 64 = (8)^2 then y = 8.<br \/>\nSo, (x)^2 \u2013 (y)^2 = (7a)^2 \u2013 (8)^2.<br \/>\n(x + y) (x \u2013 y) = (7a + 8) (7a \u2013 8)<\/p>\n<p>Finally, our expression 49a^2 &#8211; 64is reduced to (7a + 8) (7a \u2013 8).<\/p>\n<p>(ii) 16a^2 \u2013 36b^2<\/p>\n<p>Solution:<br \/>\nGiven expression is 16a^2 \u2013 36b^2<br \/>\nThere are two terms identified with the positive sign of the first term and the negative sign of the last term.<br \/>\nNow, compare the co-efficient values of 16a^2 \u2013 36b^2 with the (x)^2 &#8211; (y)^2.<br \/>\nHere, x^2 = 16a^2 = (4a)^2 then x =4a, y^2 = 36b^2 = (6b)^2 then y = 6b.<br \/>\nSo, (x)^2 \u2013 (y)^2 = (4a)^2 \u2013 (6b)^2.<br \/>\n(x + y) (x \u2013 y) = (4a + 6b) (4a \u2013 6b)<\/p>\n<p>Finally, our expression 16a^2 \u2013 36b^2 is reduced to (4a + 6b) (4a \u2013 6b).<\/p>\n<p>(iii) 1 \u2013 [5(2p \u2013 5q)]^2<\/p>\n<p>Solution:<br \/>\nGiven expression is 1 \u2013 [5(2p \u2013 5q)]^2<br \/>\nWe can write it as (1)^2 \u2013 [5(2p \u2013 5q)]^2<br \/>\nThere are two terms identified with the positive sign of the first term and the negative sign of the last term.<br \/>\nNow, compare the co-efficient values of (1)^2 \u2013 [5(2p \u2013 5q)]^2 with the (x)^2 &#8211; (y)^2.<br \/>\nHere, x^2 = (1)^2 then x =1,<br \/>\ny^2 = [5(2p \u2013 5q)]^2then y = 5(2p \u2013 5q).<br \/>\nSo, (x)^2 \u2013 (y)^2 = (1)^2 \u2013 [5(2p \u2013 5q)]^2.<br \/>\n(x + y) (x \u2013 y) = [1 + 5(2p \u2013 5q)] [1 \u2013 5(2p \u2013 5q)].<\/p>\n<p>Finally, our expression 1 \u2013 25(2p \u2013 5q)^2 is reduced to [1 + 5(2p \u2013 5q)] [1 \u2013 5(2p \u2013 5q)].<\/p>\n<p>4. Factor completely using the formula of a difference of two squares<\/p>\n<p>(i) p^4 \u2013q^4<\/p>\n<p>Solution:<br \/>\nGiven expression is p^4 \u2013q^4<br \/>\nWe can write it as (p^2)^2 \u2013 (q^2)^2<br \/>\nThere are two terms identified with the positive sign of the first term and the negative sign of the last term.<br \/>\nNow, compare the co-efficient values of (p^2)^2 \u2013 (q^2)^2with the (x)^2 &#8211; (y)^2.<br \/>\nHere, x^2 = (p^2)^2 then x =p^2,<br \/>\ny^2 = (q^2)^2 then y = q^2.<br \/>\nSo, (x)^2 \u2013 (y)^2 = (p^2)^2 \u2013 (q^2)^2.<br \/>\n(x + y) (x \u2013 y) = (p^2 + q^2) (p^2 \u2013q^2).<\/p>\n<p>Finally, our expression p^4 \u2013 q^4 is reduced to (p^2 + q^2) (p^2 \u2013 q^2).<\/p>\n","protected":false},"excerpt":{"rendered":"<p>Know the process to find Factorization by using identities. Based on the identities, we can simply factorize an algebraic equation. That means, depending on the identities or identity values, we can easily reduce the number of expressions into n number of terms. Generally, for the simplest factorization process, we have to follow some basic expressions. &#8230; <a title=\"Factorization by Using Identities | How to Factorise Polynomials using Algebraic Identities?\" class=\"read-more\" href=\"https:\/\/ccssanswers.com\/factorization-by-using-identities\/\" aria-label=\"More on Factorization by Using Identities | How to Factorise Polynomials using Algebraic Identities?\">Read more<\/a><\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[5],"tags":[],"jetpack_sharing_enabled":true,"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/posts\/34998"}],"collection":[{"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/comments?post=34998"}],"version-history":[{"count":1,"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/posts\/34998\/revisions"}],"predecessor-version":[{"id":138488,"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/posts\/34998\/revisions\/138488"}],"wp:attachment":[{"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/media?parent=34998"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/categories?post=34998"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/tags?post=34998"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}