{"id":2719,"date":"2016-07-20T16:26:54","date_gmt":"2016-07-20T16:26:54","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/intermediatealgebra\/?post_type=chapter&#038;p=2719"},"modified":"2019-08-06T19:02:51","modified_gmt":"2019-08-06T19:02:51","slug":"read-simple-polynomial-equations","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/intermediatealgebra\/chapter\/read-simple-polynomial-equations\/","title":{"raw":"Solve Polynomial Equations","rendered":"Solve Polynomial Equations"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Outcomes<\/h3>\r\n<ul>\r\n \t<li>Use factoring methods to factor polynomial equations<\/li>\r\n<\/ul>\r\n<\/div>\r\nNot all of the techniques we use for solving linear equations will apply to solving polynomial equations. In this section, we will introduce a method for solving polynomial equations that combines factoring and the zero product principle.\r\n<h2>The Principle of Zero Products<\/h2>\r\n<img class=\"wp-image-4778\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/121\/2016\/07\/20161944\/Screen-Shot-2016-06-10-at-10.04.31-AM-172x300.png\" alt=\"The number zero\" width=\"55\" height=\"94\" \/>\r\n\r\nWhat if we told you that we multiplied two numbers together and got an answer of zero? What could you say about the two numbers? Could they be\u00a0[latex]2[\/latex] and\u00a0[latex]5[\/latex]? Could they be\u00a0[latex]9[\/latex] and\u00a0[latex]1[\/latex]? No! When the result (answer) from multiplying two numbers is zero, that means that one of them\u00a0<em>had\u00a0<\/em>to be zero. This idea is called the zero product principle, and it is useful for solving polynomial\u00a0equations that can be factored.\r\n<div class=\"textbox shaded\">\r\n<h3>Principle of Zero Products<\/h3>\r\nThe Principle of Zero Products states that if the product of two numbers is\u00a0[latex]0[\/latex], then at least one of the factors is\u00a0[latex]0[\/latex].\u00a0If [latex]ab=0[\/latex], then either [latex]a=0[\/latex] or [latex]b=0[\/latex], or both <em>a<\/em> and <em>b<\/em> are\u00a0[latex]0[\/latex].\r\n\r\n<\/div>\r\nLet us start with a simple example. We will factor a GCF from a binomial and apply the principle of zero products to solve a polynomial equation.\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nSolve:\r\n\r\n[latex]-t^2+t=0[\/latex]\r\n[reveal-answer q=\"612316\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"612316\"]\r\n\r\nEach term has a common factor of\u00a0[latex]t[\/latex],\u00a0so we can factor and\u00a0use the zero product principle.\u00a0Rewrite each term as the product of the GCF and the remaining terms.\r\n\r\n[latex]\\begin{array}{c}-t^2=t\\left(-t\\right)\\\\t=t\\left(1\\right)\\end{array}[\/latex]\r\n\r\nRewrite the polynomial equation\u00a0using the factored terms in place of the original terms.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{c}-t^2+t=0\\\\t\\left(-t\\right)+t\\left(1\\right)\\\\t\\left(-t+1\\right)=0\\end{array}[\/latex]<\/p>\r\n<p style=\"text-align: left;\">Now we have a product on one side and zero on the other, so we can set each factor equal to zero using the zero product principle.<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{c}t=0\\,\\,\\,\\,\\,\\,\\,\\,\\text{ OR }\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,-t+1=0\\\\\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\underline{-1}\\,\\,\\,\\underline{-1}\\\\\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,-t=-1\\\\\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\frac{-t}{-1}=\\frac{-1}{-1}\\\\\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,t=1\\end{array}[\/latex]<\/p>\r\nTherefore, [latex]t=0\\text{ OR }t=1[\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nIn the following video, we show two more examples of using both factoring and the principle of zero products to solve a polynomial equation.\r\n\r\nhttps:\/\/youtu.be\/gIwMkTAclw8\r\n\r\nIn the next video, we show that you can use previously learned methods to factor a trinomial in order to solve a quadratic equation.\r\n\r\nhttps:\/\/youtu.be\/bi7i_RuIGl0\r\n\r\nWe all know that it is rare to be given an equation to solve that has zero on one side, so let us try another example.\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nSolve: [latex]s^2-4s=5[\/latex]\r\n[reveal-answer q=\"165196\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"165196\"]\r\n\r\nFirst, move all the terms to one side. The goal is to try and see if we can use the zero product principle since that is the only tool we know for solving polynomial equations.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{c}\\,\\,\\,\\,\\,\\,\\,s^2-4s=5\\\\\\,\\,\\,\\,\\,\\,\\,s^2-4s-5=0\\\\\\end{array}[\/latex]<\/p>\r\n<p style=\"text-align: left;\">We now have all the terms on the left side and zero on the other side. The polynomial [latex]s^2-4s-5[\/latex] factors nicely which makes this equation a good candidate for the zero product principle.<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{c}s^2-4s-5=0\\\\\\left(s+1\\right)\\left(s-5\\right)=0\\end{array}[\/latex]<\/p>\r\n<p style=\"text-align: left;\">We separate our factors into two linear equations using the principle of zero products.<\/p>\r\n[latex]\\begin{array}{c}\\left(s-5\\right)=0\\\\s-5=0\\\\\\,\\,\\,\\,\\,\\,\\,\\,\\,s=5\\end{array}[\/latex]\r\n\r\nOR\r\n\r\n[latex]\\begin{array}{c}\\left(s+1\\right)=0\\\\s+1=0\\\\s=-1\\end{array}[\/latex]\r\n\r\nTherefore, [latex]s=-1\\text{ OR }s=5[\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nWe will work through one more example that is similar to the one above, except this example has fractions, yay!\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nSolve [latex]y^2-5=-\\frac{7}{2}y+\\frac{5}{2}[\/latex]\r\n[reveal-answer q=\"164090\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"164090\"]\r\n\r\nWe can solve this in one of two ways. \u00a0One way is to eliminate the fractions like you may have done when\u00a0solving linear equations, and the second is to find a common denominator and factor fractions.\u00a0Eliminating fractions is easier, so we will show that way.\r\n\r\nStart by multiplying the whole equation by\u00a0[latex]2[\/latex] to eliminate fractions:\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{ccc}2\\left(y^2-5=-\\frac{7}{2}y+\\frac{5}{2}\\right)\\\\\\,\\,\\,\\,\\,\\,2(y^2)+2(-5)=2\\left(-\\frac{7}{2}y\\right)+2\\left(\\frac{5}{2}\\right)\\\\2y^2-10=-7y+5\\end{array}[\/latex]<\/p>\r\nNow we can move all the terms to one side and see if this will factor so we can use the principle of zero products.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{c}2y^2-10=-7y+5\\\\2y^2-10+7y-5=0\\\\2y^2-15+7y=0\\\\2y^2+7y-15=0\\end{array}[\/latex]<\/p>\r\nWe can now check whether this polynomial will factor. Using a table we can list factors until we find two numbers with a product of [latex]2\\cdot-15=-30[\/latex] and a sum of 7.\r\n<table style=\"width: 20%;\" summary=\"A table with 7 rows and 2 columns. The first column is labeled: Factors of -30 while the second is labeled: Sum of Factors. The entries in the first column are: 1, -30; -1, 30; 2, -15; -2, 15; 3, -10; and -3, 10. The entries in the second column are: -29, 29, -13, 13, -7, and 7.\">\r\n<thead>\r\n<tr>\r\n<th>Factors of [latex]2\\cdot-15=-30[\/latex]<\/th>\r\n<th>Sum of Factors<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr>\r\n<td>[latex]1,-30[\/latex]<\/td>\r\n<td>[latex]-29[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]-1,30[\/latex]<\/td>\r\n<td>[latex]29[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]2,-15[\/latex]<\/td>\r\n<td>[latex]-13[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]-2,15[\/latex]<\/td>\r\n<td>[latex]13[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]3,-10[\/latex]<\/td>\r\n<td>[latex]-7[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]-3,10[\/latex]<\/td>\r\n<td>[latex]7[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nWe have found the factors that will produce the middle term we want,[latex]-3,10[\/latex]. We need to place the factors in a way that will lead to a middle term of [latex]7y[\/latex]:\r\n\r\n[latex]\\left(2y-3\\right)\\left(y+5\\right)=0[\/latex]\r\n\r\nNow we can set each factor equal to zero and solve:\r\n\r\n[latex]\\begin{array}{ccc}\\left(2y-3\\right)=0\\text{ OR }\\left(y+5\\right)=0\\\\2y=3\\text{ OR }y=-5\\\\y=\\frac{3}{2}\\text{ OR }y=-5\\end{array}[\/latex]\r\n\r\nYou can always check your work to make sure your solutions are correct:\r\n\r\nCheck [latex]y=\\frac{3}{2}[\/latex]\r\n\r\n[latex]\\begin{array}{ccc}\\left(\\frac{3}{2}\\right)^2-5=-\\frac{7}{2}\\left(\\frac{3}{2}\\right)+\\frac{5}{2}\\\\\\frac{9}{4}-5=-\\frac{21}{4}+\\frac{5}{2}\\\\\\text{ common denominator = 4}\\\\\\frac{9}{4}-\\frac{20}{4}=-\\frac{21}{4}+\\frac{10}{4}\\\\-\\frac{11}{4}=-\\frac{11}{4}\\end{array}[\/latex]\r\n\r\n[latex]y=\\frac{3}{2}[\/latex] is indeed a solution, now check\u00a0[latex]y=-5[\/latex]\r\n\r\n[latex]\\begin{array}{ccc}\\left(-5\\right)^2-5=-\\frac{7}{2}\\left(-5\\right)+\\frac{5}{2}\\\\25-5=\\frac{35}{2}+\\frac{5}{2}\\\\20=\\frac{40}{2}\\\\20=20\\end{array}[\/latex]\r\n\r\n[latex]y=-5[\/latex] is also a solution, so we must have done something right!\r\n<p style=\"text-align: left;\">Therefore, [latex]y=\\frac{3}{2}\\text{ OR }y=-5[\/latex].<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nIn our last video, we show how to solve another quadratic equation that contains fractions.\r\n\r\nhttps:\/\/youtu.be\/kDj_qdKW-ls","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li>Use factoring methods to factor polynomial equations<\/li>\n<\/ul>\n<\/div>\n<p>Not all of the techniques we use for solving linear equations will apply to solving polynomial equations. In this section, we will introduce a method for solving polynomial equations that combines factoring and the zero product principle.<\/p>\n<h2>The Principle of Zero Products<\/h2>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"wp-image-4778\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/121\/2016\/07\/20161944\/Screen-Shot-2016-06-10-at-10.04.31-AM-172x300.png\" alt=\"The number zero\" width=\"55\" height=\"94\" \/><\/p>\n<p>What if we told you that we multiplied two numbers together and got an answer of zero? What could you say about the two numbers? Could they be\u00a0[latex]2[\/latex] and\u00a0[latex]5[\/latex]? Could they be\u00a0[latex]9[\/latex] and\u00a0[latex]1[\/latex]? No! When the result (answer) from multiplying two numbers is zero, that means that one of them\u00a0<em>had\u00a0<\/em>to be zero. This idea is called the zero product principle, and it is useful for solving polynomial\u00a0equations that can be factored.<\/p>\n<div class=\"textbox shaded\">\n<h3>Principle of Zero Products<\/h3>\n<p>The Principle of Zero Products states that if the product of two numbers is\u00a0[latex]0[\/latex], then at least one of the factors is\u00a0[latex]0[\/latex].\u00a0If [latex]ab=0[\/latex], then either [latex]a=0[\/latex] or [latex]b=0[\/latex], or both <em>a<\/em> and <em>b<\/em> are\u00a0[latex]0[\/latex].<\/p>\n<\/div>\n<p>Let us start with a simple example. We will factor a GCF from a binomial and apply the principle of zero products to solve a polynomial equation.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Solve:<\/p>\n<p>[latex]-t^2+t=0[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q612316\">Show Solution<\/span><\/p>\n<div id=\"q612316\" class=\"hidden-answer\" style=\"display: none\">\n<p>Each term has a common factor of\u00a0[latex]t[\/latex],\u00a0so we can factor and\u00a0use the zero product principle.\u00a0Rewrite each term as the product of the GCF and the remaining terms.<\/p>\n<p>[latex]\\begin{array}{c}-t^2=t\\left(-t\\right)\\\\t=t\\left(1\\right)\\end{array}[\/latex]<\/p>\n<p>Rewrite the polynomial equation\u00a0using the factored terms in place of the original terms.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{c}-t^2+t=0\\\\t\\left(-t\\right)+t\\left(1\\right)\\\\t\\left(-t+1\\right)=0\\end{array}[\/latex]<\/p>\n<p style=\"text-align: left;\">Now we have a product on one side and zero on the other, so we can set each factor equal to zero using the zero product principle.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{c}t=0\\,\\,\\,\\,\\,\\,\\,\\,\\text{ OR }\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,-t+1=0\\\\\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\underline{-1}\\,\\,\\,\\underline{-1}\\\\\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,-t=-1\\\\\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\frac{-t}{-1}=\\frac{-1}{-1}\\\\\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,t=1\\end{array}[\/latex]<\/p>\n<p>Therefore, [latex]t=0\\text{ OR }t=1[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>In the following video, we show two more examples of using both factoring and the principle of zero products to solve a polynomial equation.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-1\" title=\"Ex:  Factor and Solve Quadratic Equation - Greatest Common Factor Only\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/gIwMkTAclw8?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<p>In the next video, we show that you can use previously learned methods to factor a trinomial in order to solve a quadratic equation.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-2\" title=\"Ex: Factor and Solve Quadratic Equations When A equals 1\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/bi7i_RuIGl0?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<p>We all know that it is rare to be given an equation to solve that has zero on one side, so let us try another example.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Solve: [latex]s^2-4s=5[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q165196\">Show Solution<\/span><\/p>\n<div id=\"q165196\" class=\"hidden-answer\" style=\"display: none\">\n<p>First, move all the terms to one side. The goal is to try and see if we can use the zero product principle since that is the only tool we know for solving polynomial equations.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{c}\\,\\,\\,\\,\\,\\,\\,s^2-4s=5\\\\\\,\\,\\,\\,\\,\\,\\,s^2-4s-5=0\\\\\\end{array}[\/latex]<\/p>\n<p style=\"text-align: left;\">We now have all the terms on the left side and zero on the other side. The polynomial [latex]s^2-4s-5[\/latex] factors nicely which makes this equation a good candidate for the zero product principle.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{c}s^2-4s-5=0\\\\\\left(s+1\\right)\\left(s-5\\right)=0\\end{array}[\/latex]<\/p>\n<p style=\"text-align: left;\">We separate our factors into two linear equations using the principle of zero products.<\/p>\n<p>[latex]\\begin{array}{c}\\left(s-5\\right)=0\\\\s-5=0\\\\\\,\\,\\,\\,\\,\\,\\,\\,\\,s=5\\end{array}[\/latex]<\/p>\n<p>OR<\/p>\n<p>[latex]\\begin{array}{c}\\left(s+1\\right)=0\\\\s+1=0\\\\s=-1\\end{array}[\/latex]<\/p>\n<p>Therefore, [latex]s=-1\\text{ OR }s=5[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>We will work through one more example that is similar to the one above, except this example has fractions, yay!<\/p>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Solve [latex]y^2-5=-\\frac{7}{2}y+\\frac{5}{2}[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q164090\">Show Solution<\/span><\/p>\n<div id=\"q164090\" class=\"hidden-answer\" style=\"display: none\">\n<p>We can solve this in one of two ways. \u00a0One way is to eliminate the fractions like you may have done when\u00a0solving linear equations, and the second is to find a common denominator and factor fractions.\u00a0Eliminating fractions is easier, so we will show that way.<\/p>\n<p>Start by multiplying the whole equation by\u00a0[latex]2[\/latex] to eliminate fractions:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{ccc}2\\left(y^2-5=-\\frac{7}{2}y+\\frac{5}{2}\\right)\\\\\\,\\,\\,\\,\\,\\,2(y^2)+2(-5)=2\\left(-\\frac{7}{2}y\\right)+2\\left(\\frac{5}{2}\\right)\\\\2y^2-10=-7y+5\\end{array}[\/latex]<\/p>\n<p>Now we can move all the terms to one side and see if this will factor so we can use the principle of zero products.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{c}2y^2-10=-7y+5\\\\2y^2-10+7y-5=0\\\\2y^2-15+7y=0\\\\2y^2+7y-15=0\\end{array}[\/latex]<\/p>\n<p>We can now check whether this polynomial will factor. Using a table we can list factors until we find two numbers with a product of [latex]2\\cdot-15=-30[\/latex] and a sum of 7.<\/p>\n<table style=\"width: 20%;\" summary=\"A table with 7 rows and 2 columns. The first column is labeled: Factors of -30 while the second is labeled: Sum of Factors. The entries in the first column are: 1, -30; -1, 30; 2, -15; -2, 15; 3, -10; and -3, 10. The entries in the second column are: -29, 29, -13, 13, -7, and 7.\">\n<thead>\n<tr>\n<th>Factors of [latex]2\\cdot-15=-30[\/latex]<\/th>\n<th>Sum of Factors<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr>\n<td>[latex]1,-30[\/latex]<\/td>\n<td>[latex]-29[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]-1,30[\/latex]<\/td>\n<td>[latex]29[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]2,-15[\/latex]<\/td>\n<td>[latex]-13[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]-2,15[\/latex]<\/td>\n<td>[latex]13[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]3,-10[\/latex]<\/td>\n<td>[latex]-7[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]-3,10[\/latex]<\/td>\n<td>[latex]7[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>We have found the factors that will produce the middle term we want,[latex]-3,10[\/latex]. We need to place the factors in a way that will lead to a middle term of [latex]7y[\/latex]:<\/p>\n<p>[latex]\\left(2y-3\\right)\\left(y+5\\right)=0[\/latex]<\/p>\n<p>Now we can set each factor equal to zero and solve:<\/p>\n<p>[latex]\\begin{array}{ccc}\\left(2y-3\\right)=0\\text{ OR }\\left(y+5\\right)=0\\\\2y=3\\text{ OR }y=-5\\\\y=\\frac{3}{2}\\text{ OR }y=-5\\end{array}[\/latex]<\/p>\n<p>You can always check your work to make sure your solutions are correct:<\/p>\n<p>Check [latex]y=\\frac{3}{2}[\/latex]<\/p>\n<p>[latex]\\begin{array}{ccc}\\left(\\frac{3}{2}\\right)^2-5=-\\frac{7}{2}\\left(\\frac{3}{2}\\right)+\\frac{5}{2}\\\\\\frac{9}{4}-5=-\\frac{21}{4}+\\frac{5}{2}\\\\\\text{ common denominator = 4}\\\\\\frac{9}{4}-\\frac{20}{4}=-\\frac{21}{4}+\\frac{10}{4}\\\\-\\frac{11}{4}=-\\frac{11}{4}\\end{array}[\/latex]<\/p>\n<p>[latex]y=\\frac{3}{2}[\/latex] is indeed a solution, now check\u00a0[latex]y=-5[\/latex]<\/p>\n<p>[latex]\\begin{array}{ccc}\\left(-5\\right)^2-5=-\\frac{7}{2}\\left(-5\\right)+\\frac{5}{2}\\\\25-5=\\frac{35}{2}+\\frac{5}{2}\\\\20=\\frac{40}{2}\\\\20=20\\end{array}[\/latex]<\/p>\n<p>[latex]y=-5[\/latex] is also a solution, so we must have done something right!<\/p>\n<p style=\"text-align: left;\">Therefore, [latex]y=\\frac{3}{2}\\text{ OR }y=-5[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>In our last video, we show how to solve another quadratic equation that contains fractions.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-3\" title=\"Solve a Quadratic Equations with Fractions by Factoring (a not 1)\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/kDj_qdKW-ls?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n\n\t\t\t <section class=\"citations-section\" role=\"contentinfo\">\n\t\t\t <h3>Candela Citations<\/h3>\n\t\t\t\t\t <div>\n\t\t\t\t\t\t <div id=\"citation-list-2719\">\n\t\t\t\t\t\t\t <div class=\"licensing\"><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Original<\/div><ul class=\"citation-list\"><li>Solve a Quadratic Equations with Fractions by Factoring (a not 1). <strong>Authored by<\/strong>: James Sousa (Mathispower4u.com) for Lumen Learning. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/kDj_qdKW-ls\">https:\/\/youtu.be\/kDj_qdKW-ls<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>Revision and Adaptation. <strong>Provided by<\/strong>: Lumen Learning. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><\/ul><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Shared previously<\/div><ul class=\"citation-list\"><li>Ex: Factor and Solve Quadratic Equation - Greatest Common Factor Only. <strong>Authored by<\/strong>: James Sousa (Mathispower4u.com) . <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/gIwMkTAclw8\">https:\/\/youtu.be\/gIwMkTAclw8<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>Ex: Factor and Solve Quadratic Equations When A equals 1. <strong>Authored by<\/strong>: James Sousa (Mathispower4u.com) . <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/bi7i_RuIGl0\">https:\/\/youtu.be\/bi7i_RuIGl0<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>Unit 12: Factoring, from Developmental Math: An Open Program. <strong>Provided by<\/strong>: Monterey Institute of Technology. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"http:\/\/nrocnetwork.org\/dm-opentext\">http:\/\/nrocnetwork.org\/dm-opentext<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><\/ul><\/div>\n\t\t\t\t\t\t <\/div>\n\t\t\t\t\t <\/div>\n\t\t\t <\/section>","protected":false},"author":21,"menu_order":11,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc\",\"description\":\"Ex: Factor and Solve Quadratic Equation - 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