{"id":259,"date":"2023-01-30T18:21:08","date_gmt":"2023-01-30T18:21:08","guid":{"rendered":"https:\/\/cwassignments.com\/blog\/?p=259"},"modified":"2024-05-10T07:49:14","modified_gmt":"2024-05-10T07:49:14","slug":"factoring-a-quadratic-trinomial-by-grouping","status":"publish","type":"post","link":"https:\/\/cwassignments.com\/blog\/factoring-a-quadratic-trinomial-by-grouping\/","title":{"rendered":"Factoring a Quadratic Trinomial by Grouping"},"content":{"rendered":"<p style=\"font-weight: 400;\">Another method for factoring these kinds of quadratic trinomials is called factoring by grouping. Factoring by grouping can be a bit more tedious and is often not worth the trouble if you can find the correct factors by some quick trial and error. However, it works quite well when the factors are not immediately obvious, such as when you have a very large number of candidate factors. When this happens, the trial and error method becomes very tedious.<\/p>\n<p style=\"font-weight: 400;\">Factoring by grouping is best demonstrated with a few examples.<\/p>\n<p style=\"font-weight: 400;\"><b><strong>Example:<\/strong><\/b><\/p>\n<table style=\"font-weight: 400;\" width=\"638\">\n<tbody>\n<tr>\n<td width=\"50%\">Given:<\/td>\n<td width=\"50%\">5<em>x<\/em><sup>2<\/sup>\u00a0+\u00a011<em>x<\/em>\u00a0+\u00a02<\/td>\n<\/tr>\n<tr>\n<td width=\"50%\">Find the product\u00a0<em>ac<\/em>:<\/td>\n<td width=\"50%\">(5)(2)\u00a0=\u00a010<\/td>\n<\/tr>\n<tr>\n<td width=\"50%\">Think of two factors of 10 that add up to 11:<\/td>\n<td width=\"50%\">1 and 10<\/td>\n<\/tr>\n<tr>\n<td width=\"50%\">Write the 11<em>x<\/em>\u00a0as the sum of 1<em>x\u00a0<\/em>and 10<em>x<\/em>:<\/td>\n<td width=\"50%\">5<em>x<\/em><sup>2<\/sup>\u00a0+\u00a01<em>x<\/em>\u00a0+\u00a010<em>x<\/em>\u00a0+\u00a02<\/td>\n<\/tr>\n<tr>\n<td width=\"50%\">Group the two pairs of terms:<\/td>\n<td width=\"50%\">(5<em>x<\/em><sup>2<\/sup>\u00a0+\u00a01<em>x<\/em>)\u00a0+\u00a0(10<em>x<\/em>\u00a0+\u00a02)<\/td>\n<\/tr>\n<tr>\n<td width=\"50%\">Remove common factors from each group:<\/td>\n<td width=\"50%\"><em>x<\/em>(5<em>x<\/em>\u00a0+\u00a01)\u00a0+\u00a02(5<em>x<\/em>\u00a0+\u00a01)<\/td>\n<\/tr>\n<tr>\n<td width=\"50%\">Notice that the two quantities in parentheses are now identical. That means we can factor out a common factor of (5<em>x<\/em>\u00a0+\u00a01):<\/td>\n<td width=\"50%\">(5<em>x<\/em>\u00a0+\u00a01)(<em>x<\/em>\u00a0+\u00a02)<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p style=\"font-weight: 400;\"><b><strong>Example:<\/strong><\/b><\/p>\n<table style=\"font-weight: 400;\" width=\"638\">\n<tbody>\n<tr>\n<td width=\"50%\">Given:<\/td>\n<td width=\"50%\">4<em>x<\/em><sup>2<\/sup>\u00a0+\u00a07<em>x<\/em>\u00a0\u2013\u00a015<\/td>\n<\/tr>\n<tr>\n<td width=\"50%\">Find the product\u00a0<em>ac<\/em>:<\/td>\n<td width=\"50%\">(4)(\u201315)\u00a0=\u00a0\u201360<\/td>\n<\/tr>\n<tr>\n<td width=\"50%\">Think of two factors of\u00a0\u201360 that add up to 7:<\/td>\n<td width=\"50%\">\u20135 and 12<\/td>\n<\/tr>\n<tr>\n<td width=\"50%\">Write the 7<em>x<\/em>\u00a0as the sum of\u00a0\u20135<em>x<\/em>\u00a0and 12<em>x<\/em>:<\/td>\n<td width=\"50%\">4<em>x<\/em><sup>2<\/sup>\u00a0\u2013\u00a05<em>x<\/em>\u00a0+\u00a012<em>x<\/em>\u00a0\u2013\u00a015<\/td>\n<\/tr>\n<tr>\n<td width=\"50%\">Group the two pairs of terms:<\/td>\n<td width=\"50%\">(4<em>x<\/em><sup>2<\/sup>\u00a0\u2013\u00a05<em>x<\/em>)\u00a0+\u00a0(12<em>x<\/em>\u00a0\u2013\u00a015)<\/td>\n<\/tr>\n<tr>\n<td width=\"50%\">Remove common factors from each group:<\/td>\n<td width=\"50%\"><em>x<\/em>(4<em>x<\/em>\u00a0\u2013\u00a05)\u00a0+\u00a03(4<em>x<\/em>\u00a0\u2013\u00a05)<\/td>\n<\/tr>\n<tr>\n<td width=\"50%\">Notice that the two quantities in<br \/>\nparentheses are now identical. That means we can factor out a common factor<br \/>\nof (4<em>x<\/em>\u00a0\u2013\u00a05):<\/td>\n<td width=\"50%\">(4<em>x<\/em>\u00a0\u2013\u00a05)(<em>x<\/em>\u00a0+\u00a03)<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<h4><a name=\"_Toc470866381\"><\/a>The Procedure<\/h4>\n<p style=\"font-weight: 400;\">Given a general quadratic trinomial\u00a0<em>ax<\/em><sup>2<\/sup>\u00a0+\u00a0<em>bx<\/em>\u00a0+\u00a0<em>c<\/em><\/p>\n<ol>\n<li style=\"font-weight: 400;\">Find the product<em>ac<\/em>.<\/li>\n<li style=\"font-weight: 400;\">Find two numbers<em>h<\/em>and\u00a0<em>k<\/em>\u00a0such that<br \/>\n<em>hk<\/em>\u00a0=\u00a0<em>ac\u00a0<\/em>(<em>h<\/em>\u00a0and\u00a0<em>k<\/em>\u00a0are factors of the product of the coefficient of\u00a0<em>x<\/em><sup>2<\/sup>\u00a0and the constant term)<br \/>\nAND<br \/>\n<em>h<\/em>\u00a0+\u00a0<em>k<\/em>\u00a0=\u00a0<em>b\u00a0<\/em>(<em>h<\/em>\u00a0and\u00a0<em>k<\/em>\u00a0add to give the coefficient of\u00a0<em>x<\/em>)<\/li>\n<li style=\"font-weight: 400;\">Rewrite the quadratic as<em>ax<\/em><sup>2<\/sup>+\u00a0<em>hx<\/em>\u00a0+\u00a0<em>kx<\/em>\u00a0+\u00a0<em>c<\/em><\/li>\n<li style=\"font-weight: 400;\">Group the two pairs of terms that have common factors and simplify.<br \/>\n(<em>ax<\/em><sup>2<\/sup>+\u00a0<em>hx<\/em>)\u00a0+\u00a0(<em>kx<\/em>+\u00a0<em>c<\/em>)<br \/>\n<em>x<\/em>(<em>ax<\/em>\u00a0+\u00a0<em>h<\/em>)\u00a0+\u00a0(<em>kx<\/em>\u00a0+\u00a0<em>c<\/em>)<br \/>\n(note: because of the way you chose\u00a0<em>h<\/em>\u00a0and\u00a0<em>k<\/em>, you will be able to factor a constant out of the second parentheses, leaving you with two identical expressions in parentheses as in the examples).<\/li>\n<\/ol>\n<ul>\n<li style=\"font-weight: 400;\">Remember that this won\u2019t work for all quadratic<br \/>\ntrinomials, because not all quadratic trinomials can be factored into products<br \/>\nof binomials with integer coefficients. If you have a non-factorable trinomial,<br \/>\nyou will not be able to do step 2 above.<\/li>\n<\/ul>\n<table style=\"font-weight: 400;\">\n<tbody>\n<tr>\n<td width=\"590\">\n<h4>Why this works<\/h4>\n<p>Suppose the quadratic trinomial in question came from multiplying two arbitrary binomials:<br \/>\n(<em>px<\/em>\u00a0+\u00a0<em>n<\/em>)(<em>qx<\/em>\u00a0+\u00a0<em>m<\/em>)<br \/>\nIf we multiply this out we will get<br \/>\n<em>\u00a0 \u00a0 \u00a0 pqx<sup>2<\/sup>\u00a0+\u00a0pmx\u00a0+\u00a0qnx\u00a0+\u00a0nm<br \/>\n<\/em>\u00a0or<br \/>\n<em>\u00a0 \u00a0 \u00a0 pqx<\/em><sup>2<\/sup>\u00a0+\u00a0(<em>pm<\/em>\u00a0+\u00a0<em>qn<\/em>)<em>x<\/em>\u00a0+\u00a0<em>nm.<br \/>\n<\/em>Notice that the coefficient of\u00a0<em>x<\/em>\u00a0consists of a sum of two terms,\u00a0<em>pm\u00a0<\/em>and\u00a0<em>qn<\/em>. These are the two numbers we called\u00a0<em>h<\/em>\u00a0and\u00a0<em>k\u00a0<\/em>above.<br \/>\n<em>\u00a0 \u00a0 \u00a0pm<\/em>\u00a0=\u00a0<em>h<br \/>\n<\/em><em>\u00a0 \u00a0 \u00a0qn<\/em>\u00a0=\u00a0<em>k<br \/>\n<\/em>Now we see that the two numbers\u00a0<em>h<\/em>\u00a0and\u00a0<em>k<\/em>\u00a0add up to the<br \/>\ncoefficient of\u00a0<em>x<\/em>, which we called\u00a0<em>b<\/em>:<\/p>\n<p><em>h<\/em>\u00a0+\u00a0<em>k<\/em>\u00a0=\u00a0<em>b<\/em><\/p>\n<p>Obviously they are factors of their own product\u00a0<em>pmqn<\/em>, but we notice that\u00a0<em>pq<\/em>\u00a0=\u00a0<em>a<\/em>, and\u00a0<em>mn<\/em>\u00a0=\u00a0<em>c<\/em>, so<\/p>\n<p>(<em>pm<\/em>)(<em>qn<\/em>)\u00a0=\u00a0(<em>pq<\/em>)(<em>nm<\/em>)<\/p>\n<p>which is equivalent to<\/p>\n<p><em>hk<\/em>\u00a0=\u00a0<em>ac<\/em><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>Having trouble with factoring by grouping in a System of Equations? Get reliable <a href=\"https:\/\/cwassignments.com\/assignment-help.html\">assignment help online<\/a> to ace it with confidence!<\/p>\n","protected":false},"excerpt":{"rendered":"<p>Another method for factoring these kinds of quadratic trinomials is called factoring by grouping. Factoring by grouping can be a bit more tedious and is often not worth the trouble if you can find the correct factors by some quick trial and error. However, it works quite well when the factors are not immediately obvious, [&hellip;]<\/p>\n","protected":false},"author":1,"featured_media":692,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"inline_featured_image":false,"_swt_meta_header_display":false,"_swt_meta_footer_display":false,"_swt_meta_site_title_display":false,"_swt_meta_sticky_header":false,"_swt_meta_transparent_header":false,"footnotes":""},"categories":[7],"tags":[],"class_list":["post-259","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-algebra"],"_links":{"self":[{"href":"https:\/\/cwassignments.com\/blog\/wp-json\/wp\/v2\/posts\/259","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/cwassignments.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/cwassignments.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/cwassignments.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/cwassignments.com\/blog\/wp-json\/wp\/v2\/comments?post=259"}],"version-history":[{"count":3,"href":"https:\/\/cwassignments.com\/blog\/wp-json\/wp\/v2\/posts\/259\/revisions"}],"predecessor-version":[{"id":540,"href":"https:\/\/cwassignments.com\/blog\/wp-json\/wp\/v2\/posts\/259\/revisions\/540"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/cwassignments.com\/blog\/wp-json\/wp\/v2\/media\/692"}],"wp:attachment":[{"href":"https:\/\/cwassignments.com\/blog\/wp-json\/wp\/v2\/media?parent=259"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/cwassignments.com\/blog\/wp-json\/wp\/v2\/categories?post=259"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/cwassignments.com\/blog\/wp-json\/wp\/v2\/tags?post=259"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}