{"id":868,"date":"2024-04-30T00:50:26","date_gmt":"2024-04-30T00:50:26","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/?post_type=chapter&#038;p=868"},"modified":"2024-11-20T02:43:37","modified_gmt":"2024-11-20T02:43:37","slug":"factoring-polynomials-learn-it-3","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/chapter\/factoring-polynomials-learn-it-3\/","title":{"raw":"Factoring Polynomials: Learn It 3","rendered":"Factoring Polynomials: Learn It 3"},"content":{"raw":"<h2>Factoring a Perfect Square Trinomial<\/h2>\r\nA perfect square trinomial is a trinomial that can be written as the square of a binomial. Recall that when a binomial is squared, the result is the square of the first term added to twice the product of the two terms and the square of the last term.\r\n<div style=\"text-align: center;\">[latex]\\begin{array}{ccc}\\hfill {a}^{2}+2ab+{b}^{2}&amp; =&amp; {\\left(a+b\\right)}^{2}\\hfill \\\\ &amp; \\text{and}&amp; \\\\ \\hfill {a}^{2}-2ab+{b}^{2}&amp; =&amp; {\\left(a-b\\right)}^{2}\\hfill \\end{array}[\/latex]<\/div>\r\n<div style=\"text-align: left;\">We can use this equation to factor any perfect square trinomial.<\/div>\r\n<section class=\"textbox keyTakeaway\" aria-label=\"Key Takeaway\">\r\n<h3>perfect square trinomial<\/h3>\r\nA perfect square trinomial can be written as the square of a binomial:\r\n<div style=\"text-align: center;\">[latex]{a}^{2}+2ab+{b}^{2}={\\left(a+b\\right)}^{2}[\/latex]<\/div>\r\n<\/section><section class=\"textbox questionHelp\" aria-label=\"Question Help\"><strong>How To: Given a perfect square trinomial, factor it into the square of a binomial\r\n<\/strong>\r\n<ol>\r\n \t<li>Confirm that the first and last term are perfect squares.<\/li>\r\n \t<li>Confirm that the middle term is twice the product of [latex]ab[\/latex].<\/li>\r\n \t<li>Write the factored form as [latex]{\\left(a+b\\right)}^{2}[\/latex].<\/li>\r\n<\/ol>\r\n<\/section><section class=\"textbox example\">Factor [latex]25{x}^{2}+20x+4[\/latex].<strong>Solution<\/strong>\r\n\r\nTo factor the quadratic expression [latex]25{x}^{2}+20x+4[\/latex], recognizing it as a perfect square trinomial will streamline the process. This type of expression comes from squaring a binomial and has a special format, [latex]a^2 +2ab+b^2[\/latex], where it can be rewritten as [latex](a+b)^2[\/latex].[latex]\\begin{align*} \\text{Original expression:} &amp; \\quad 25x^2 + 20x + 4 \\\\ \\text{Identify square terms:} &amp; \\quad 25x^2 = (5x)^2 \\quad \\text{and} \\quad 4 = 2^2 \\\\ \\text{Check middle term:} &amp; \\quad 2 \\times 5x \\times 2 = 20x \\\\ \\text{Write as a square of a binomial:} &amp; \\quad (5x + 2)^2 \\end{align*}[\/latex]\r\n\r\n<\/section><section class=\"textbox example\" aria-label=\"Example\">Factor [latex]49{x}^{2}-14x+1[\/latex].[reveal-answer q=\"530221\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"530221\"][latex]{\\left(7x - 1\\right)}^{2}[\/latex][\/hidden-answer]<\/section><section class=\"textbox tryIt\">[ohm2_question hide_question_numbers=1]18884[\/ohm2_question]<\/section><section class=\"textbox tryIt\">[ohm2_question hide_question_numbers=1]18885[\/ohm2_question]<\/section>\r\n<h2>Factoring a Difference of Squares<\/h2>\r\nA difference of squares is a perfect square subtracted from a perfect square. Recall that a difference of squares can be rewritten as factors containing the same terms but opposite signs because the middle terms cancel each other out when the two factors are multiplied.\r\n<div style=\"text-align: center;\">[latex]{a}^{2}-{b}^{2}=\\left(a+b\\right)\\left(a-b\\right)[\/latex]<\/div>\r\nWe can use this equation to factor any differences of squares.\r\n\r\n<section class=\"textbox keyTakeaway\" aria-label=\"Key Takeaway\">\r\n<h3>difference of squares<\/h3>\r\nA difference of squares can be rewritten as two factors containing the same terms but opposite signs.\r\n<div style=\"text-align: center;\">[latex]{a}^{2}-{b}^{2}=\\left(a+b\\right)\\left(a-b\\right)[\/latex]<\/div>\r\n<\/section><section class=\"textbox questionHelp\" aria-label=\"Question Help\"><strong>How To: Given a difference of squares, factor it into binomials<\/strong>\r\n<ol>\r\n \t<li>Confirm that the first and last term are perfect squares.<\/li>\r\n \t<li>Write the factored form as [latex]\\left(a+b\\right)\\left(a-b\\right)[\/latex].<\/li>\r\n<\/ol>\r\n<\/section><section class=\"textbox example\">Factor [latex]9{x}^{2}-25[\/latex].<strong>Solution<\/strong>\r\n\r\nTo factor the quadratic expression [latex]9{x}^{2}-25[\/latex], we recognize that it is a difference of squares.[latex]\\begin{align*} \\text{Original expression:} &amp; \\quad 9x^2 - 25 \\\\ \\text{Identify square terms:} &amp; \\quad 9x^2 = (3x)^2 \\quad \\text{and} \\quad 25 = 5^2 \\\\ \\text{Apply difference of squares formula:} &amp; \\quad (3x + 5)(3x - 5) \\end{align*}[\/latex]\r\n\r\n<\/section><section class=\"textbox example\" aria-label=\"Example\">Factor [latex]81{y}^{2}-100[\/latex].[reveal-answer q=\"979538\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"979538\"][latex]\\left(9y+10\\right)\\left(9y - 10\\right)[\/latex][\/hidden-answer]<\/section><section class=\"textbox tryIt\">[ohm2_question hide_question_numbers=1]18886[\/ohm2_question]<\/section><section class=\"textbox tryIt\">[ohm2_question hide_question_numbers=1]18887[\/ohm2_question]<\/section>","rendered":"<h2>Factoring a Perfect Square Trinomial<\/h2>\n<p>A perfect square trinomial is a trinomial that can be written as the square of a binomial. Recall that when a binomial is squared, the result is the square of the first term added to twice the product of the two terms and the square of the last term.<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{array}{ccc}\\hfill {a}^{2}+2ab+{b}^{2}& =& {\\left(a+b\\right)}^{2}\\hfill \\\\ & \\text{and}& \\\\ \\hfill {a}^{2}-2ab+{b}^{2}& =& {\\left(a-b\\right)}^{2}\\hfill \\end{array}[\/latex]<\/div>\n<div style=\"text-align: left;\">We can use this equation to factor any perfect square trinomial.<\/div>\n<section class=\"textbox keyTakeaway\" aria-label=\"Key Takeaway\">\n<h3>perfect square trinomial<\/h3>\n<p>A perfect square trinomial can be written as the square of a binomial:<\/p>\n<div style=\"text-align: center;\">[latex]{a}^{2}+2ab+{b}^{2}={\\left(a+b\\right)}^{2}[\/latex]<\/div>\n<\/section>\n<section class=\"textbox questionHelp\" aria-label=\"Question Help\"><strong>How To: Given a perfect square trinomial, factor it into the square of a binomial<br \/>\n<\/strong><\/p>\n<ol>\n<li>Confirm that the first and last term are perfect squares.<\/li>\n<li>Confirm that the middle term is twice the product of [latex]ab[\/latex].<\/li>\n<li>Write the factored form as [latex]{\\left(a+b\\right)}^{2}[\/latex].<\/li>\n<\/ol>\n<\/section>\n<section class=\"textbox example\">Factor [latex]25{x}^{2}+20x+4[\/latex].<strong>Solution<\/strong><\/p>\n<p>To factor the quadratic expression [latex]25{x}^{2}+20x+4[\/latex], recognizing it as a perfect square trinomial will streamline the process. This type of expression comes from squaring a binomial and has a special format, [latex]a^2 +2ab+b^2[\/latex], where it can be rewritten as [latex](a+b)^2[\/latex].[latex]\\begin{align*} \\text{Original expression:} & \\quad 25x^2 + 20x + 4 \\\\ \\text{Identify square terms:} & \\quad 25x^2 = (5x)^2 \\quad \\text{and} \\quad 4 = 2^2 \\\\ \\text{Check middle term:} & \\quad 2 \\times 5x \\times 2 = 20x \\\\ \\text{Write as a square of a binomial:} & \\quad (5x + 2)^2 \\end{align*}[\/latex]<\/p>\n<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">Factor [latex]49{x}^{2}-14x+1[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q530221\">Show Solution<\/button><\/p>\n<div id=\"q530221\" class=\"hidden-answer\" style=\"display: none\">[latex]{\\left(7x - 1\\right)}^{2}[\/latex]<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox tryIt\"><iframe loading=\"lazy\" id=\"ohm18884\" class=\"resizable\" src=\"https:\/\/ohm.one.lumenlearning.com\/multiembedq.php?id=18884&theme=lumen&iframe_resize_id=ohm18884&source=tnh\" width=\"100%\" height=\"150\"><\/iframe><\/section>\n<section class=\"textbox tryIt\"><iframe loading=\"lazy\" id=\"ohm18885\" class=\"resizable\" src=\"https:\/\/ohm.one.lumenlearning.com\/multiembedq.php?id=18885&theme=lumen&iframe_resize_id=ohm18885&source=tnh\" width=\"100%\" height=\"150\"><\/iframe><\/section>\n<h2>Factoring a Difference of Squares<\/h2>\n<p>A difference of squares is a perfect square subtracted from a perfect square. Recall that a difference of squares can be rewritten as factors containing the same terms but opposite signs because the middle terms cancel each other out when the two factors are multiplied.<\/p>\n<div style=\"text-align: center;\">[latex]{a}^{2}-{b}^{2}=\\left(a+b\\right)\\left(a-b\\right)[\/latex]<\/div>\n<p>We can use this equation to factor any differences of squares.<\/p>\n<section class=\"textbox keyTakeaway\" aria-label=\"Key Takeaway\">\n<h3>difference of squares<\/h3>\n<p>A difference of squares can be rewritten as two factors containing the same terms but opposite signs.<\/p>\n<div style=\"text-align: center;\">[latex]{a}^{2}-{b}^{2}=\\left(a+b\\right)\\left(a-b\\right)[\/latex]<\/div>\n<\/section>\n<section class=\"textbox questionHelp\" aria-label=\"Question Help\"><strong>How To: Given a difference of squares, factor it into binomials<\/strong><\/p>\n<ol>\n<li>Confirm that the first and last term are perfect squares.<\/li>\n<li>Write the factored form as [latex]\\left(a+b\\right)\\left(a-b\\right)[\/latex].<\/li>\n<\/ol>\n<\/section>\n<section class=\"textbox example\">Factor [latex]9{x}^{2}-25[\/latex].<strong>Solution<\/strong><\/p>\n<p>To factor the quadratic expression [latex]9{x}^{2}-25[\/latex], we recognize that it is a difference of squares.[latex]\\begin{align*} \\text{Original expression:} & \\quad 9x^2 - 25 \\\\ \\text{Identify square terms:} & \\quad 9x^2 = (3x)^2 \\quad \\text{and} \\quad 25 = 5^2 \\\\ \\text{Apply difference of squares formula:} & \\quad (3x + 5)(3x - 5) \\end{align*}[\/latex]<\/p>\n<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">Factor [latex]81{y}^{2}-100[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q979538\">Show Solution<\/button><\/p>\n<div id=\"q979538\" class=\"hidden-answer\" style=\"display: none\">[latex]\\left(9y+10\\right)\\left(9y - 10\\right)[\/latex]<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox tryIt\"><iframe loading=\"lazy\" id=\"ohm18886\" class=\"resizable\" src=\"https:\/\/ohm.one.lumenlearning.com\/multiembedq.php?id=18886&theme=lumen&iframe_resize_id=ohm18886&source=tnh\" width=\"100%\" height=\"150\"><\/iframe><\/section>\n<section class=\"textbox tryIt\"><iframe loading=\"lazy\" id=\"ohm18887\" class=\"resizable\" src=\"https:\/\/ohm.one.lumenlearning.com\/multiembedq.php?id=18887&theme=lumen&iframe_resize_id=ohm18887&source=tnh\" width=\"100%\" height=\"150\"><\/iframe><\/section>\n","protected":false},"author":12,"menu_order":13,"template":"","meta":{"_candela_citation":"[]","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"part":55,"module-header":"learn_it","content_attributions":[],"internal_book_links":[],"video_content":null,"cc_video_embed_content":{"cc_scripts":"","media_targets":[]},"try_it_collection":null,"_links":{"self":[{"href":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/wp-json\/pressbooks\/v2\/chapters\/868"}],"collection":[{"href":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/wp-json\/wp\/v2\/users\/12"}],"version-history":[{"count":12,"href":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/wp-json\/pressbooks\/v2\/chapters\/868\/revisions"}],"predecessor-version":[{"id":6228,"href":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/wp-json\/pressbooks\/v2\/chapters\/868\/revisions\/6228"}],"part":[{"href":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/wp-json\/pressbooks\/v2\/parts\/55"}],"metadata":[{"href":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/wp-json\/pressbooks\/v2\/chapters\/868\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/wp-json\/wp\/v2\/media?parent=868"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/wp-json\/pressbooks\/v2\/chapter-type?post=868"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/wp-json\/wp\/v2\/contributor?post=868"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/wp-json\/wp\/v2\/license?post=868"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}