{"id":16302,"date":"2019-10-03T03:06:08","date_gmt":"2019-10-03T03:06:08","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/wm-developmentalemporium\/chapter\/read-simple-polynomial-equations\/"},"modified":"2020-10-22T09:31:38","modified_gmt":"2020-10-22T09:31:38","slug":"read-simple-polynomial-equations","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/suny-rockland-developmentalemporium\/chapter\/read-simple-polynomial-equations\/","title":{"raw":"13.1.d - Solving a Polynomial in Factored Form","rendered":"13.1.d &#8211; Solving a Polynomial in Factored Form"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Outcomes<\/h3>\r\n<ul>\r\n \t<li>Factor the greatest common monomial out of a polynomial<\/li>\r\n \t<li>Solve a polynomial in factored form by setting it equal to zero<\/li>\r\n<\/ul>\r\n<\/div>\r\nIn this section we will apply factoring a monomial from a polynomial to solving polynomial equations. Recall that not all of the techniques we use for solving linear equations will apply to solving polynomial equations, so we will be using the zero product principle to solve for a variable.\r\n\r\nWe will begin with an example where the polynomial is already equal to zero.\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 t,\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\n\r\n<h4 style=\"text-align: left\">Answer<\/h4>\r\n[latex]t=0\\text{ OR }t=1[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nNotice how we were careful with signs in the last example. \u00a0Even though one of the terms was\u00a0negative, we factored out the positive common term of t. \u00a0In the next example we will see what to do when the polynomial you are working with is not set equal to zero. In the following video, we present more examples of solving quadratic equations by factoring.\r\n\r\nhttps:\/\/youtu.be\/Hpb8DVYBDzA\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nSolve: [latex]6t=3t^2-12t[\/latex]\r\n[reveal-answer q=\"162196\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"162196\"]\r\n\r\nFirst, move all the terms to one side. \u00a0The 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}\\,\\,\\,\\,\\,\\,\\,6t=3t^2-12t\\\\\\underline{-6t}\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\underline{-6t}\\\\\\,\\,\\,\\,\\,\\,\\,0=3t^2-18t\\\\\\end{array}[\/latex]<\/p>\r\n<p style=\"text-align: left\">We now have all the terms on the right side, and zero on the other side. Each term has a common factor of [latex]3t[\/latex], so we can factor and use the zero product principle. If you are uncertain how we found the common factor [latex]3t[\/latex], review the section on greatest common factor.<\/p>\r\n<p style=\"text-align: left\">Rewrite each term as the product of the GCF and the remaining terms. Note how we leave the negative sign on the [latex]6[\/latex] when we factor [latex]3t[\/latex] out of [latex]-18t[\/latex].<\/p>\r\n<p style=\"text-align: left\">[latex]\\begin{array}{c}3t^2=3t\\left(t\\right)\\\\-18t=3t\\left(-6\\right)\\end{array}[\/latex]<\/p>\r\n<p style=\"text-align: left\">Rewrite the polynomial equation\u00a0using the factored terms in place of the original terms.<\/p>\r\n<p style=\"text-align: center\">[latex]\\begin{array}{c}0=3t^2-18t\\\\\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,0=3t\\left(t\\right)+3t\\left(-6\\right)\\\\0=3t\\left(t-6\\right)\\end{array}[\/latex]<\/p>\r\n<p style=\"text-align: left\">Solve the two equations.<\/p>\r\n<p style=\"text-align: left\">[latex]\\begin{array}{l}\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,0=t-6\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\text{ OR }\\,\\,\\,\\,\\,\\,\\,\\,\\,0=3t\\\\\\,\\,\\,\\,\\,\\,\\,\\,\\,\\underline{+6}\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\underline{+6}\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\frac{0}{3}=\\frac{3t}{3}\\\\\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,6=t\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\text{ OR }\\,\\,\\,\\,\\,\\,\\,\\,\\,0=t\\end{array}[\/latex]<\/p>\r\n\r\n<h4 style=\"text-align: left\">Answer<\/h4>\r\n[latex]t=6\\text{ OR }t=0[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nThe video that follows provides another example of solving a polynomial equation using the zero product principle and factoring.\r\n\r\nhttps:\/\/youtu.be\/oYytjgbd6Q0\r\n\r\nWe will work through one more example that is similar to the ones above, except this example has fractions and the greatest common monomial is negative.\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nSolve [latex]\\frac{1}{2}y=-4y-\\frac{1}{2}y^2[\/latex]\r\n[reveal-answer q=\"164090\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"164090\"]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. We will work through the second way.\r\n\r\nFirst, we will find a common denominator, then factor fractions.\r\n<p style=\"text-align: center\">[latex]\\begin{array}{l}\\frac{1}{2}y=-4y-\\frac{1}{2}y^2\\\\0=-\\frac{1}{2}y-4y-\\frac{1}{2}y^2\\end{array}[\/latex]<\/p>\r\n[latex]2[\/latex] is\u00a0a common denominator for \u00a0[latex]-\\frac{1}{2}y\\text{ and }-4y[\/latex]\r\n<p style=\"text-align: center\">[latex]\\frac{2}{2}\\cdot{-4y}=-\\frac{8y}{2}[\/latex]<\/p>\r\nRewrite the equation with the common denominator, then combine like terms.\r\n<p style=\"text-align: center\">[latex]\\begin{array}{l}0=-\\frac{1}{2}y-4y-\\frac{1}{2}y^2\\\\\\text{ }\\\\0=-\\frac{1}{2}y-\\frac{8y}{2}-\\frac{1}{2}y^2\\\\\\text{}\\\\0=-\\frac{9}{2}y-\\frac{1}{2}y^2\\end{array}[\/latex]<\/p>\r\n<p style=\"text-align: left\">Find the greatest common factor of the terms of the polynomial:<\/p>\r\n<p style=\"text-align: left\">Factors of [latex]-\\frac{9}{2}y[\/latex]<\/p>\r\n<p style=\"text-align: center\">[latex]-\\frac{1}{2}\\cdot{3}\\cdot{3}\\cdot{y}[\/latex]<\/p>\r\n<p style=\"text-align: left\">Factors of [latex]-\\frac{1}{2}y^2[\/latex]<\/p>\r\n<p style=\"text-align: center\">[latex]-\\frac{1}{2}\\cdot{y}\\cdot{y}[\/latex]<\/p>\r\n<p style=\"text-align: left\">Both terms have [latex]-\\frac{1}{2}\\text{ and }y[\/latex] in common.<\/p>\r\n<p style=\"text-align: left\">Rewrite each term as the product of the GCF and the remaining terms.<\/p>\r\n<p style=\"text-align: left\">[latex]-\\frac{9}{2}y=-\\frac{1}{2}y\\left(3\\cdot{3}\\right)=-\\frac{1}{2}y\\left(9\\right)[\/latex]<\/p>\r\n<p style=\"text-align: left\">[latex]-\\frac{1}{2}y^2=-\\frac{1}{2}y\\left(y\\right)[\/latex]<\/p>\r\n<p style=\"text-align: left\">\u00a0Rewrite the polynomial equation\u00a0using the factored terms in place of the original terms. Pay attention to signs when we factor. Notice that we end up with a sum as a factor because the common factor is a negative number. [latex]\\left(9+y\\right)[\/latex]<\/p>\r\n<p style=\"text-align: center\">[latex]\\begin{array}{l}0=-\\frac{9}{2}y-\\frac{1}{2}y^2\\\\\\text{}\\\\-\\frac{1}{2}y\\left(9\\right)-\\frac{1}{2}y\\left(y\\right)\\\\\\text{}\\\\-\\frac{1}{2}y\\left(9+y\\right)\\end{array}[\/latex]<\/p>\r\n<p style=\"text-align: left\">Solve the two equations.<\/p>\r\n<p style=\"text-align: center\">[latex]\\begin{array}{l}\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,0=9+y\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\text{ OR }\\,\\,\\,\\,\\,\\,\\,\\,\\,0=-\\frac{1}{2}y\\\\\\,\\,\\,\\,\\,\\,\\,\\,\\,\\underline{-9}\\,\\,\\,\\,\\,\\,\\underline{-9}\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\\\\\,\\,\\,\\,\\,\\,\\,\\,\\,-9=y\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\text{ OR }\\,\\,\\,\\,\\,\\,\\,\\,\\,0=y\\end{array}[\/latex]<\/p>\r\n\r\n<h4 style=\"text-align: left\">Answer<\/h4>\r\n<p style=\"text-align: left\">[latex]y=-9\\text{ OR }y=0[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nWow! In the last example, we used <em>many<\/em> skills to solve one equation. \u00a0Let's summarize them:\r\n<ul>\r\n \t<li>We needed a common denominator to combine the like terms [latex]-4y\\text{ and }-\\frac{1}{2}y[\/latex], after we moved all the terms to one side of the equation<\/li>\r\n \t<li>We found the GCF\u00a0of the terms\u00a0[latex]-\\frac{9}{2}y\\text{ and }-\\frac{1}{2}y^2[\/latex]<\/li>\r\n \t<li>We used the GCF to factor the polynomial [latex]-\\frac{9}{2}y-\\frac{1}{2}y^2[\/latex]<\/li>\r\n \t<li>We used the zero product principle to solve the polynomial equation [latex]0=-\\frac{1}{2}y\\left(9+y\\right)[\/latex]<\/li>\r\n<\/ul>\r\nSometimes solving an equation requires the combination of many algebraic principles and techniques. \u00a0The last facet of solving the polynomial equation [latex]\\frac{1}{2}y=-4y-\\frac{1}{2}y^2[\/latex] that we should talk about is negative signs.\r\n\r\nWe found that the GCF [latex]-\\frac{1}{2}y[\/latex] contained a negative coefficient. \u00a0This meant that when we factored it out of all the terms in the polynomial, we were left with two positive factors, 9 and y. \u00a0This explains why we were left with \u00a0[latex]\\left(9+y\\right)[\/latex] as one of the factors of our final product.\r\n\r\nIn the following video we present another example of solving a quadratic polynomial with fractional coefficients using factoring and the zero product principle.\r\n\r\nhttps:\/\/youtu.be\/wm6DJ1bnaJs\r\n<div class=\"textbox shaded\"><img class=\"wp-image-2132 alignleft\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1468\/2016\/03\/22011815\/traffic-sign-160659-300x265.png\" alt=\"Caution\" width=\"44\" height=\"39\" \/>If the GCF of a polynomial is\u00a0negative, pay attention to the signs that are left when you factor it from the terms of a polynomial.<\/div>\r\nIn the next unit, we will learn more factoring techniques that will allow you to be able to solve a wider variety of polynomial equations such as [latex]3x^2-x=2[\/latex].\r\n<h2>Summary<\/h2>\r\nIn this section we practiced using the zero product principle as a method for solving polynomial equations.\u00a0 We also found that a polynomial can be rewritten as a product by factoring out the greatest common factor. We used both factoring and the zero product principle to solve second degree polynomials.","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li>Factor the greatest common monomial out of a polynomial<\/li>\n<li>Solve a polynomial in factored form by setting it equal to zero<\/li>\n<\/ul>\n<\/div>\n<p>In this section we will apply factoring a monomial from a polynomial to solving polynomial equations. Recall that not all of the techniques we use for solving linear equations will apply to solving polynomial equations, so we will be using the zero product principle to solve for a variable.<\/p>\n<p>We will begin with an example where the polynomial is already equal to zero.<\/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 t,\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<h4 style=\"text-align: left\">Answer<\/h4>\n<p>[latex]t=0\\text{ OR }t=1[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>Notice how we were careful with signs in the last example. \u00a0Even though one of the terms was\u00a0negative, we factored out the positive common term of t. \u00a0In the next example we will see what to do when the polynomial you are working with is not set equal to zero. In the following video, we present more examples of solving quadratic equations by factoring.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-1\" title=\"Ex 1:  Factor and Solve a Quadratic Equation - GCF\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/Hpb8DVYBDzA?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Solve: [latex]6t=3t^2-12t[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q162196\">Show Solution<\/span><\/p>\n<div id=\"q162196\" class=\"hidden-answer\" style=\"display: none\">\n<p>First, move all the terms to one side. \u00a0The 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}\\,\\,\\,\\,\\,\\,\\,6t=3t^2-12t\\\\\\underline{-6t}\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\underline{-6t}\\\\\\,\\,\\,\\,\\,\\,\\,0=3t^2-18t\\\\\\end{array}[\/latex]<\/p>\n<p style=\"text-align: left\">We now have all the terms on the right side, and zero on the other side. Each term has a common factor of [latex]3t[\/latex], so we can factor and use the zero product principle. If you are uncertain how we found the common factor [latex]3t[\/latex], review the section on greatest common factor.<\/p>\n<p style=\"text-align: left\">Rewrite each term as the product of the GCF and the remaining terms. Note how we leave the negative sign on the [latex]6[\/latex] when we factor [latex]3t[\/latex] out of [latex]-18t[\/latex].<\/p>\n<p style=\"text-align: left\">[latex]\\begin{array}{c}3t^2=3t\\left(t\\right)\\\\-18t=3t\\left(-6\\right)\\end{array}[\/latex]<\/p>\n<p style=\"text-align: left\">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}0=3t^2-18t\\\\\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,0=3t\\left(t\\right)+3t\\left(-6\\right)\\\\0=3t\\left(t-6\\right)\\end{array}[\/latex]<\/p>\n<p style=\"text-align: left\">Solve the two equations.<\/p>\n<p style=\"text-align: left\">[latex]\\begin{array}{l}\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,0=t-6\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\text{ OR }\\,\\,\\,\\,\\,\\,\\,\\,\\,0=3t\\\\\\,\\,\\,\\,\\,\\,\\,\\,\\,\\underline{+6}\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\underline{+6}\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\frac{0}{3}=\\frac{3t}{3}\\\\\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,6=t\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\text{ OR }\\,\\,\\,\\,\\,\\,\\,\\,\\,0=t\\end{array}[\/latex]<\/p>\n<h4 style=\"text-align: left\">Answer<\/h4>\n<p>[latex]t=6\\text{ OR }t=0[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>The video that follows provides another example of solving a polynomial equation using the zero product principle and factoring.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-2\" title=\"Factor and Solve a Quadratic Equation - GCF Only\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/oYytjgbd6Q0?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<p>We will work through one more example that is similar to the ones above, except this example has fractions and the greatest common monomial is negative.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Solve [latex]\\frac{1}{2}y=-4y-\\frac{1}{2}y^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\">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. We will work through the second way.<\/p>\n<p>First, we will find a common denominator, then factor fractions.<\/p>\n<p style=\"text-align: center\">[latex]\\begin{array}{l}\\frac{1}{2}y=-4y-\\frac{1}{2}y^2\\\\0=-\\frac{1}{2}y-4y-\\frac{1}{2}y^2\\end{array}[\/latex]<\/p>\n<p>[latex]2[\/latex] is\u00a0a common denominator for \u00a0[latex]-\\frac{1}{2}y\\text{ and }-4y[\/latex]<\/p>\n<p style=\"text-align: center\">[latex]\\frac{2}{2}\\cdot{-4y}=-\\frac{8y}{2}[\/latex]<\/p>\n<p>Rewrite the equation with the common denominator, then combine like terms.<\/p>\n<p style=\"text-align: center\">[latex]\\begin{array}{l}0=-\\frac{1}{2}y-4y-\\frac{1}{2}y^2\\\\\\text{ }\\\\0=-\\frac{1}{2}y-\\frac{8y}{2}-\\frac{1}{2}y^2\\\\\\text{}\\\\0=-\\frac{9}{2}y-\\frac{1}{2}y^2\\end{array}[\/latex]<\/p>\n<p style=\"text-align: left\">Find the greatest common factor of the terms of the polynomial:<\/p>\n<p style=\"text-align: left\">Factors of [latex]-\\frac{9}{2}y[\/latex]<\/p>\n<p style=\"text-align: center\">[latex]-\\frac{1}{2}\\cdot{3}\\cdot{3}\\cdot{y}[\/latex]<\/p>\n<p style=\"text-align: left\">Factors of [latex]-\\frac{1}{2}y^2[\/latex]<\/p>\n<p style=\"text-align: center\">[latex]-\\frac{1}{2}\\cdot{y}\\cdot{y}[\/latex]<\/p>\n<p style=\"text-align: left\">Both terms have [latex]-\\frac{1}{2}\\text{ and }y[\/latex] in common.<\/p>\n<p style=\"text-align: left\">Rewrite each term as the product of the GCF and the remaining terms.<\/p>\n<p style=\"text-align: left\">[latex]-\\frac{9}{2}y=-\\frac{1}{2}y\\left(3\\cdot{3}\\right)=-\\frac{1}{2}y\\left(9\\right)[\/latex]<\/p>\n<p style=\"text-align: left\">[latex]-\\frac{1}{2}y^2=-\\frac{1}{2}y\\left(y\\right)[\/latex]<\/p>\n<p style=\"text-align: left\">\u00a0Rewrite the polynomial equation\u00a0using the factored terms in place of the original terms. Pay attention to signs when we factor. Notice that we end up with a sum as a factor because the common factor is a negative number. [latex]\\left(9+y\\right)[\/latex]<\/p>\n<p style=\"text-align: center\">[latex]\\begin{array}{l}0=-\\frac{9}{2}y-\\frac{1}{2}y^2\\\\\\text{}\\\\-\\frac{1}{2}y\\left(9\\right)-\\frac{1}{2}y\\left(y\\right)\\\\\\text{}\\\\-\\frac{1}{2}y\\left(9+y\\right)\\end{array}[\/latex]<\/p>\n<p style=\"text-align: left\">Solve the two equations.<\/p>\n<p style=\"text-align: center\">[latex]\\begin{array}{l}\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,0=9+y\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\text{ OR }\\,\\,\\,\\,\\,\\,\\,\\,\\,0=-\\frac{1}{2}y\\\\\\,\\,\\,\\,\\,\\,\\,\\,\\,\\underline{-9}\\,\\,\\,\\,\\,\\,\\underline{-9}\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\\\\\,\\,\\,\\,\\,\\,\\,\\,\\,-9=y\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\text{ OR }\\,\\,\\,\\,\\,\\,\\,\\,\\,0=y\\end{array}[\/latex]<\/p>\n<h4 style=\"text-align: left\">Answer<\/h4>\n<p style=\"text-align: left\">[latex]y=-9\\text{ OR }y=0[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>Wow! In the last example, we used <em>many<\/em> skills to solve one equation. \u00a0Let&#8217;s summarize them:<\/p>\n<ul>\n<li>We needed a common denominator to combine the like terms [latex]-4y\\text{ and }-\\frac{1}{2}y[\/latex], after we moved all the terms to one side of the equation<\/li>\n<li>We found the GCF\u00a0of the terms\u00a0[latex]-\\frac{9}{2}y\\text{ and }-\\frac{1}{2}y^2[\/latex]<\/li>\n<li>We used the GCF to factor the polynomial [latex]-\\frac{9}{2}y-\\frac{1}{2}y^2[\/latex]<\/li>\n<li>We used the zero product principle to solve the polynomial equation [latex]0=-\\frac{1}{2}y\\left(9+y\\right)[\/latex]<\/li>\n<\/ul>\n<p>Sometimes solving an equation requires the combination of many algebraic principles and techniques. \u00a0The last facet of solving the polynomial equation [latex]\\frac{1}{2}y=-4y-\\frac{1}{2}y^2[\/latex] that we should talk about is negative signs.<\/p>\n<p>We found that the GCF [latex]-\\frac{1}{2}y[\/latex] contained a negative coefficient. \u00a0This meant that when we factored it out of all the terms in the polynomial, we were left with two positive factors, 9 and y. \u00a0This explains why we were left with \u00a0[latex]\\left(9+y\\right)[\/latex] as one of the factors of our final product.<\/p>\n<p>In the following video we present another example of solving a quadratic polynomial with fractional coefficients using factoring and the zero product principle.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-3\" title=\"Factor and Solve a Quadratic Equation with Fractions - GCF Only\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/wm6DJ1bnaJs?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<div class=\"textbox shaded\"><img loading=\"lazy\" decoding=\"async\" class=\"wp-image-2132 alignleft\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1468\/2016\/03\/22011815\/traffic-sign-160659-300x265.png\" alt=\"Caution\" width=\"44\" height=\"39\" \/>If the GCF of a polynomial is\u00a0negative, pay attention to the signs that are left when you factor it from the terms of a polynomial.<\/div>\n<p>In the next unit, we will learn more factoring techniques that will allow you to be able to solve a wider variety of polynomial equations such as [latex]3x^2-x=2[\/latex].<\/p>\n<h2>Summary<\/h2>\n<p>In this section we practiced using the zero product principle as a method for solving polynomial equations.\u00a0 We also found that a polynomial can be rewritten as a product by factoring out the greatest common factor. We used both factoring and the zero product principle to solve second degree polynomials.<\/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-16302\">\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>Factor and Solve a Quadratic Equation - GCF Only. <strong>Authored by<\/strong>: James Sousa (Mathispower4u.com) for Lumen Learning. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/oYytjgbd6Q0\">https:\/\/youtu.be\/oYytjgbd6Q0<\/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>Factor and Solve a Quadratic Equation with Fractions - GCF Only Mathispower4u . <strong>Authored by<\/strong>: James Sousa (Mathispower4u.com) for Lumen Learning. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/wm6DJ1bnaJs\">https:\/\/youtu.be\/wm6DJ1bnaJs<\/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 1: Factor and Solve a Quadratic Equation - GCF Mathispower4u . <strong>Authored by<\/strong>: James Sousa (Mathispower4u.com) . <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/Hpb8DVYBDzA\">https:\/\/youtu.be\/Hpb8DVYBDzA<\/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\/resources\/downloads\/nroc-math-open-textbook-units-1-12-pdf-and-word-formats\/\">http:\/\/nrocnetwork.org\/resources\/downloads\/nroc-math-open-textbook-units-1-12-pdf-and-word-formats\/<\/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":169554,"menu_order":6,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc\",\"description\":\"Ex 1: Factor and Solve a Quadratic Equation - 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