{"id":3016,"date":"2024-08-28T14:50:53","date_gmt":"2024-08-28T14:50:53","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/?post_type=chapter&#038;p=3016"},"modified":"2024-11-20T02:40:26","modified_gmt":"2024-11-20T02:40:26","slug":"polynomial-basics-learn-it-4","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/chapter\/polynomial-basics-learn-it-4\/","title":{"raw":"Polynomial Basics: Learn It 4","rendered":"Polynomial Basics: Learn It 4"},"content":{"raw":"<h2>Special Cases of Polynomials<\/h2>\r\nSpecial cases of polynomials are specific types of polynomial expressions that have distinct characteristics or follow certain patterns, making it easier to work with them, especially when you're multiplying or factoring. Knowing these special cases is really helpful because it can simplify and speed up the process of solving polynomial equations and carrying out algebraic operations.\r\n<h3>Perfect Square Trinomials<\/h3>\r\nCertain binomial products have special forms. When a binomial is squared, the result is called a <strong>perfect square trinomial<\/strong>. We can find the square by multiplying the binomial by itself. However, there is a special form that each of these perfect square trinomials takes, and memorizing the form makes squaring binomials much easier. Let\u2019s look at a few perfect square trinomials to familiarize ourselves with the form.\r\n<div style=\"text-align: center;\">[latex]\\begin{array}{ccc}\\hfill \\text{ }{\\left(x+5\\right)}^{2}&amp; =&amp; \\text{ }{x}^{2}+10x+25\\hfill \\\\ \\hfill {\\left(x - 3\\right)}^{2}&amp; =&amp; \\text{ }{x}^{2}-6x+9\\hfill \\\\ \\hfill {\\left(4x - 1\\right)}^{2}&amp; =&amp; 4{x}^{2}-8x+1\\hfill \\end{array}[\/latex]<\/div>\r\nNotice that the first term of each trinomial is the square of the first term of the binomial and, similarly, the last term of each trinomial is the square of the last term of the binomial. The middle term is double the product of the two terms. Lastly, we see that the first sign of the trinomial is the same as the sign of the binomial.\r\n\r\n<section class=\"textbox keyTakeaway\">\r\n<h3>Perfect Square Trinomials<\/h3>\r\nA <strong>perfect square trinomial<\/strong> is a type of polynomial that results from squaring a binomial. It is called a \"perfect square\" because it is the exact square of a binomial expression.\r\n\r\n&nbsp;\r\n\r\nThe general form of a perfect square trinomial is:\r\n<p style=\"text-align: center;\">[latex](a+b)^2 = a^2+2ab+b^2[\/latex]<\/p>\r\n\r\n<\/section><section class=\"textbox proTip\" aria-label=\"Pro Tip\">Characteristics of perfect square trinomials:\r\n<ul>\r\n \t<li>First and Last Terms: The first and last terms are perfect squares of the terms in the original binomial.<\/li>\r\n \t<li>Middle Term: The middle term is twice the product of the two terms in the binomial.<\/li>\r\n<\/ul>\r\n<\/section><section class=\"textbox example\">Expand [latex]{\\left(3x - 8\\right)}^{2}[\/latex].<strong>Solution<\/strong>\r\n\r\n[latex]\\begin{align*} \\text{Original expression:} &amp; \\quad (3x - 8)^2 \\\\ \\text{Apply the formula } (a+b)^2 = a^2 + 2ab + b^2, &amp; \\quad \\text{where } a = 3x \\text{ and } b = -8: \\\\ \\text{Square the first term (}a^2\\text{):} &amp; \\quad (3x)^2 = 9x^2 \\\\ \\text{Double the product of the two terms (}2ab\\text{):} &amp; \\quad 2 \\cdot (3x) \\cdot (-8) = -48x \\\\ \\text{Square the last term (}b^2\\text{):} &amp; \\quad (-8)^2 = 64 \\\\ \\text{Combine all terms:} &amp; \\quad 9x^2 - 48x + 64 \\\\ \\end{align*}[\/latex]\r\n\r\n[reveal-answer q=\"753990\"]FOIL method[\/reveal-answer]\r\n[hidden-answer a=\"753990\"][latex]\\begin{align*} \\text{Original expression:} &amp; \\quad (3x - 8)^2 &amp; \\text{Starting point} \\\\ &amp; = (3x - 8)(3x - 8) &amp; \\text{Write out the square as a product} \\\\ &amp; = 3x(3x) + 3x(-8) + (-8)(3x) + (-8)(-8) &amp; \\text{Apply the FOIL method: First, Outer, Inner, Last} \\\\ &amp; = 9x^2 - 24x - 24x + 64 &amp; \\text{Multiply each term} \\\\ &amp; = 9x^2 - 48x + 64 &amp; \\text{Combine like terms} \\\\ \\end{align*}[\/latex][\/hidden-answer]\r\n\r\n<\/section><section class=\"textbox example\" aria-label=\"Example\">Expand [latex]{\\left(4x - 1\\right)}^{2}[\/latex].[reveal-answer q=\"278544\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"278544\"][latex]16{x}^{2}-8x+1[\/latex][\/hidden-answer]<\/section><section class=\"textbox tryIt\">[ohm2_question hide_question_numbers=1]18874[\/ohm2_question]<\/section>\r\n<h3>Difference of Squares<\/h3>\r\nAnother special product is called the <strong>difference of squares\u00a0<\/strong>which occurs when we multiply a binomial by another binomial with the same terms but the opposite sign. Let\u2019s see what happens when we multiply [latex]\\left(x+1\\right)\\left(x - 1\\right)[\/latex] using the FOIL method.\r\n<div style=\"text-align: center;\">[latex]\\begin{array}{ccc}\\hfill \\left(x+1\\right)\\left(x - 1\\right)&amp; =&amp; {x}^{2}-x+x - 1\\hfill \\\\ &amp; =&amp; {x}^{2}-1\\hfill \\end{array}[\/latex]<\/div>\r\nThe middle term drops out resulting in a difference of squares. Just as we did with the perfect squares, let\u2019s look at a few examples.\r\n<div style=\"text-align: center;\">[latex]\\begin{array}{ccc}\\hfill \\left(x+5\\right)\\left(x - 5\\right)&amp; =&amp; {x}^{2}-25\\hfill \\\\ \\hfill \\left(x+11\\right)\\left(x - 11\\right)&amp; =&amp; {x}^{2}-121\\hfill \\\\ \\hfill \\left(2x+3\\right)\\left(2x - 3\\right)&amp; =&amp; 4{x}^{2}-9\\hfill \\end{array}[\/latex]<\/div>\r\nBecause the sign changes in the second binomial, the outer and inner terms cancel each other out, and we are left only with the square of the first term minus the square of the last term.\r\n\r\n<section class=\"textbox keyTakeaway\">\r\n<h3><strong>difference of squares<\/strong><\/h3>\r\nWhen you multiply a binomial by another binomial that contains the same terms but with opposite signs, the result is known as the <strong>difference of squares<\/strong>.\r\n<p style=\"text-align: center;\">[latex](a+b)(a-b) = a^2 - b^2[\/latex]<\/p>\r\n\r\n<\/section><section class=\"textbox questionHelp\" aria-label=\"Question Help\"><strong>Is there a special form for the sum of squares?<\/strong><em>No. The difference of squares occurs because the opposite signs of the binomials cause the middle terms to disappear. There are no two binomials that multiply to equal a sum of squares.<\/em><\/section><section class=\"textbox example\">Multiply [latex]\\left(9x+4\\right)\\left(9x - 4\\right)[\/latex].<strong>Solution<\/strong>\r\n\r\n[latex]\\begin{align*} \\text{Original expression:} &amp; \\quad (9x + 4)(9x - 4) \\\\ \\text{Apply the difference of squares formula, where } a = 9x, \\text{ and } b = 4: &amp; \\\\ \\text{Use the formula } (a+b)(a-b) = a^2 - b^2: &amp; \\quad a^2 - b^2 \\\\ \\text{Substitute } a = 9x \\text{ and } b = 4: &amp; \\quad (9x)^2 - (4)^2 \\\\ \\text{Simplify each term:} &amp; \\quad 81x^2 - 16 \\end{align*}[\/latex]\r\n\r\n[reveal-answer q=\"96962\"]FOIL Method[\/reveal-answer]\r\n[hidden-answer a=\"96962\"][latex]\\begin{align*} \\text{Original expression:} &amp; \\quad (9x + 4)(9x - 4) &amp; \\text{Identify as a difference of squares} \\\\ &amp; = 9x(9x) + 9x(-4) + 4(9x) + 4(-4) &amp; \\text{Apply the FOIL method} \\\\ &amp; = 81x^2 - 36x + 36x - 16 &amp; \\text{Multiply each term} \\\\ &amp; = 81x^2 - 16 &amp; \\text{Combine like terms (middle terms cancel out)} \\end{align*}[\/latex][\/hidden-answer]\r\n\r\n<\/section><section class=\"textbox tryIt\">[ohm2_question hide_question_numbers=1]18875[\/ohm2_question]<\/section><section class=\"textbox tryIt\">[ohm2_question hide_question_numbers=1]18876[\/ohm2_question]<\/section>","rendered":"<h2>Special Cases of Polynomials<\/h2>\n<p>Special cases of polynomials are specific types of polynomial expressions that have distinct characteristics or follow certain patterns, making it easier to work with them, especially when you&#8217;re multiplying or factoring. Knowing these special cases is really helpful because it can simplify and speed up the process of solving polynomial equations and carrying out algebraic operations.<\/p>\n<h3>Perfect Square Trinomials<\/h3>\n<p>Certain binomial products have special forms. When a binomial is squared, the result is called a <strong>perfect square trinomial<\/strong>. We can find the square by multiplying the binomial by itself. However, there is a special form that each of these perfect square trinomials takes, and memorizing the form makes squaring binomials much easier. Let\u2019s look at a few perfect square trinomials to familiarize ourselves with the form.<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{array}{ccc}\\hfill \\text{ }{\\left(x+5\\right)}^{2}& =& \\text{ }{x}^{2}+10x+25\\hfill \\\\ \\hfill {\\left(x - 3\\right)}^{2}& =& \\text{ }{x}^{2}-6x+9\\hfill \\\\ \\hfill {\\left(4x - 1\\right)}^{2}& =& 4{x}^{2}-8x+1\\hfill \\end{array}[\/latex]<\/div>\n<p>Notice that the first term of each trinomial is the square of the first term of the binomial and, similarly, the last term of each trinomial is the square of the last term of the binomial. The middle term is double the product of the two terms. Lastly, we see that the first sign of the trinomial is the same as the sign of the binomial.<\/p>\n<section class=\"textbox keyTakeaway\">\n<h3>Perfect Square Trinomials<\/h3>\n<p>A <strong>perfect square trinomial<\/strong> is a type of polynomial that results from squaring a binomial. It is called a &#8220;perfect square&#8221; because it is the exact square of a binomial expression.<\/p>\n<p>&nbsp;<\/p>\n<p>The general form of a perfect square trinomial is:<\/p>\n<p style=\"text-align: center;\">[latex](a+b)^2 = a^2+2ab+b^2[\/latex]<\/p>\n<\/section>\n<section class=\"textbox proTip\" aria-label=\"Pro Tip\">Characteristics of perfect square trinomials:<\/p>\n<ul>\n<li>First and Last Terms: The first and last terms are perfect squares of the terms in the original binomial.<\/li>\n<li>Middle Term: The middle term is twice the product of the two terms in the binomial.<\/li>\n<\/ul>\n<\/section>\n<section class=\"textbox example\">Expand [latex]{\\left(3x - 8\\right)}^{2}[\/latex].<strong>Solution<\/strong><\/p>\n<p>[latex]\\begin{align*} \\text{Original expression:} & \\quad (3x - 8)^2 \\\\ \\text{Apply the formula } (a+b)^2 = a^2 + 2ab + b^2, & \\quad \\text{where } a = 3x \\text{ and } b = -8: \\\\ \\text{Square the first term (}a^2\\text{):} & \\quad (3x)^2 = 9x^2 \\\\ \\text{Double the product of the two terms (}2ab\\text{):} & \\quad 2 \\cdot (3x) \\cdot (-8) = -48x \\\\ \\text{Square the last term (}b^2\\text{):} & \\quad (-8)^2 = 64 \\\\ \\text{Combine all terms:} & \\quad 9x^2 - 48x + 64 \\\\ \\end{align*}[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q753990\">FOIL method<\/button><\/p>\n<div id=\"q753990\" class=\"hidden-answer\" style=\"display: none\">[latex]\\begin{align*} \\text{Original expression:} & \\quad (3x - 8)^2 & \\text{Starting point} \\\\ & = (3x - 8)(3x - 8) & \\text{Write out the square as a product} \\\\ & = 3x(3x) + 3x(-8) + (-8)(3x) + (-8)(-8) & \\text{Apply the FOIL method: First, Outer, Inner, Last} \\\\ & = 9x^2 - 24x - 24x + 64 & \\text{Multiply each term} \\\\ & = 9x^2 - 48x + 64 & \\text{Combine like terms} \\\\ \\end{align*}[\/latex]<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">Expand [latex]{\\left(4x - 1\\right)}^{2}[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q278544\">Show Solution<\/button><\/p>\n<div id=\"q278544\" class=\"hidden-answer\" style=\"display: none\">[latex]16{x}^{2}-8x+1[\/latex]<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox tryIt\"><iframe loading=\"lazy\" id=\"ohm18874\" class=\"resizable\" src=\"https:\/\/ohm.one.lumenlearning.com\/multiembedq.php?id=18874&theme=lumen&iframe_resize_id=ohm18874&source=tnh\" width=\"100%\" height=\"150\"><\/iframe><\/section>\n<h3>Difference of Squares<\/h3>\n<p>Another special product is called the <strong>difference of squares\u00a0<\/strong>which occurs when we multiply a binomial by another binomial with the same terms but the opposite sign. Let\u2019s see what happens when we multiply [latex]\\left(x+1\\right)\\left(x - 1\\right)[\/latex] using the FOIL method.<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{array}{ccc}\\hfill \\left(x+1\\right)\\left(x - 1\\right)& =& {x}^{2}-x+x - 1\\hfill \\\\ & =& {x}^{2}-1\\hfill \\end{array}[\/latex]<\/div>\n<p>The middle term drops out resulting in a difference of squares. Just as we did with the perfect squares, let\u2019s look at a few examples.<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{array}{ccc}\\hfill \\left(x+5\\right)\\left(x - 5\\right)& =& {x}^{2}-25\\hfill \\\\ \\hfill \\left(x+11\\right)\\left(x - 11\\right)& =& {x}^{2}-121\\hfill \\\\ \\hfill \\left(2x+3\\right)\\left(2x - 3\\right)& =& 4{x}^{2}-9\\hfill \\end{array}[\/latex]<\/div>\n<p>Because the sign changes in the second binomial, the outer and inner terms cancel each other out, and we are left only with the square of the first term minus the square of the last term.<\/p>\n<section class=\"textbox keyTakeaway\">\n<h3><strong>difference of squares<\/strong><\/h3>\n<p>When you multiply a binomial by another binomial that contains the same terms but with opposite signs, the result is known as the <strong>difference of squares<\/strong>.<\/p>\n<p style=\"text-align: center;\">[latex](a+b)(a-b) = a^2 - b^2[\/latex]<\/p>\n<\/section>\n<section class=\"textbox questionHelp\" aria-label=\"Question Help\"><strong>Is there a special form for the sum of squares?<\/strong><em>No. The difference of squares occurs because the opposite signs of the binomials cause the middle terms to disappear. There are no two binomials that multiply to equal a sum of squares.<\/em><\/section>\n<section class=\"textbox example\">Multiply [latex]\\left(9x+4\\right)\\left(9x - 4\\right)[\/latex].<strong>Solution<\/strong><\/p>\n<p>[latex]\\begin{align*} \\text{Original expression:} & \\quad (9x + 4)(9x - 4) \\\\ \\text{Apply the difference of squares formula, where } a = 9x, \\text{ and } b = 4: & \\\\ \\text{Use the formula } (a+b)(a-b) = a^2 - b^2: & \\quad a^2 - b^2 \\\\ \\text{Substitute } a = 9x \\text{ and } b = 4: & \\quad (9x)^2 - (4)^2 \\\\ \\text{Simplify each term:} & \\quad 81x^2 - 16 \\end{align*}[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q96962\">FOIL Method<\/button><\/p>\n<div id=\"q96962\" class=\"hidden-answer\" style=\"display: none\">[latex]\\begin{align*} \\text{Original expression:} & \\quad (9x + 4)(9x - 4) & \\text{Identify as a difference of squares} \\\\ & = 9x(9x) + 9x(-4) + 4(9x) + 4(-4) & \\text{Apply the FOIL method} \\\\ & = 81x^2 - 36x + 36x - 16 & \\text{Multiply each term} \\\\ & = 81x^2 - 16 & \\text{Combine like terms (middle terms cancel out)} \\end{align*}[\/latex]<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox tryIt\"><iframe loading=\"lazy\" id=\"ohm18875\" class=\"resizable\" src=\"https:\/\/ohm.one.lumenlearning.com\/multiembedq.php?id=18875&theme=lumen&iframe_resize_id=ohm18875&source=tnh\" width=\"100%\" height=\"150\"><\/iframe><\/section>\n<section class=\"textbox tryIt\"><iframe loading=\"lazy\" id=\"ohm18876\" class=\"resizable\" src=\"https:\/\/ohm.one.lumenlearning.com\/multiembedq.php?id=18876&theme=lumen&iframe_resize_id=ohm18876&source=tnh\" width=\"100%\" height=\"150\"><\/iframe><\/section>\n","protected":false},"author":15,"menu_order":7,"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\/3016"}],"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\/15"}],"version-history":[{"count":13,"href":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/wp-json\/pressbooks\/v2\/chapters\/3016\/revisions"}],"predecessor-version":[{"id":6224,"href":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/wp-json\/pressbooks\/v2\/chapters\/3016\/revisions\/6224"}],"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\/3016\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/wp-json\/wp\/v2\/media?parent=3016"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/wp-json\/pressbooks\/v2\/chapter-type?post=3016"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/wp-json\/wp\/v2\/contributor?post=3016"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/wp-json\/wp\/v2\/license?post=3016"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}