{"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":"2018-05-17T00:57:46","modified_gmt":"2018-05-17T00:57:46","slug":"read-simple-polynomial-equations","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/odessa-coreq-collegealgebra\/chapter\/read-simple-polynomial-equations\/","title":{"raw":"Polynomial Equations","rendered":"Polynomial Equations"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Objectives<\/h3>\r\n<ul>\r\n \t<li>Polynomial Equations\r\n<ul>\r\n \t<li>Use factoring methods to factor polynomial equations<\/li>\r\n \t<li>Use the principle of zero products to solve polynomial equations<\/li>\r\n<\/ul>\r\n<\/li>\r\n \t<li>Applications of Polynomial Equations\r\n<ul>\r\n \t<li>Projectiles<\/li>\r\n \t<li>Use the Pythagorean theorem to find the lengths of a right triangle<\/li>\r\n \t<li>Geometric Applications<\/li>\r\n \t<li>Cost, Revenue, and Profit Polynomials<\/li>\r\n<\/ul>\r\n<\/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[caption id=\"attachment_4778\" align=\"alignleft\" width=\"55\"]<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\" \/> Zero[\/caption]\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 2 and 5? Could they be 9 and 1? 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\r\n&nbsp;\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 0, then at least one of the factors is 0.\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 0.\r\n\r\n<\/div>\r\nLet's start with a simple example. \u00a0We 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 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\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 factor a trinomial using methods previously learned to solve a quadratic equation.\r\n\r\nhttps:\/\/youtu.be\/bi7i_RuIGl0\r\n\r\nYou and I both know that it is rare to be given an equation to solve that has zero on one side, so let's try another one.\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. \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}\\,\\,\\,\\,\\,\\,\\,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. (imagine that)<\/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<h4 style=\"text-align: left\">Answer<\/h4>\r\n[latex]s=-1\\text{ OR }s=5[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n&nbsp;\r\n\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\"]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.\r\n\r\nStart by multiplying the whole equation by 2 to eliminate fractions:\r\n\r\n[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}\\right)+2\\left(-\\frac{5}{2}\\right)\\\\2y^2-10=-7y+5\\end{array}[\/latex]\r\n\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\r\n[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]\r\n\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>29<\/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>13<\/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>7<\/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 term of [latex]10y[\/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<h4>\u00a0Answer<\/h4>\r\n<p style=\"text-align: left\">[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\r\n<h2>\u00a0Projectile Motion<\/h2>\r\nProjectile motion happens when you throw a ball into the air and it comes back down because of gravity. \u00a0A projectile will follow a curved path that behaves in a predictable way. \u00a0This predictable motion has been studied for centuries, and in simple cases it's height from the ground\u00a0at a given time, t, can be modeled with a polynomial function of the form [latex]h(t)=at^2+bt+c[\/latex], where h(t) = height of an object at a given time, t. \u00a0Projectile motion is also called a parabolic trajectory because of the shape of the path of a projectile's motion, as in the image of water in the fountain below.\r\n\r\n[caption id=\"attachment_4890\" align=\"aligncenter\" width=\"436\"]<img class=\"wp-image-4890\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/121\/2016\/07\/20161949\/ParabolicWaterTrajectory-225x300.jpg\" alt=\"Water from a fountain shoing classic parabolic motion.\" width=\"436\" height=\"581\" \/> Parabolic WaterTrajectory[\/caption]\r\n\r\nParabolic motion and it's related functions\u00a0allow us to launch satellites for telecommunications, and rockets for space exploration. Recently, police departments have even begun using projectiles with GPS to track fleeing suspects in vehicles, rather than pursuing them by high-speed chase\u00a0[footnote]\"Cops' Latest Tool in High-speed Chases: GPS Projectiles.\" <i>CBSNews<\/i>. CBS Interactive, n.d. Web. 14 June 2016.[\/footnote].\r\n\r\nIn this section we will use\u00a0polynomial functions\u00a0to answer questions about the parabolic motion of a projectile. The real mathematical model for the path of a rocket or a police GPS projectile may have different coefficients or more variables, but the concept remains the same. We will also learn to interpret the meaning of the variables in a polynomial function that models projectile motion.\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Example<\/h3>\r\nA small toy rocket is launched from a 4-foot pedestal. The height (<i>h, <\/i>in feet) of the rocket <i>t<\/i> seconds after taking off is given by the function [latex]h(t)=\u22122t^{2}+7t+4[\/latex]. How long will it take the rocket to hit the ground?\r\n[reveal-answer q=\"679533\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"679533\"]\r\n\r\n<strong><strong>\u00a0<\/strong><\/strong>The rocket will be on the ground when the [latex]h(t)=0[\/latex]. We want to know how long, t, \u00a0the rocket is in the air.\r\n\r\n[latex]\\begin{array}{l}h(t)=\u22122t^{2}+7t+4=0\\\\0=\u22122t^{2}+7t+4\\end{array}[\/latex]\r\n\r\nWe can factor the polynomial [latex]\u22122t^{2}+7t+4[\/latex] more easily by first factoring out a [latex]-1[\/latex]\r\n\r\n[latex]\\begin{array}{c}0=-1(2t^{2}-7t-4)\\\\0=-1\\left(2t+1\\right)\\left(t-4\\right)\\end{array}[\/latex]\r\n\r\nUse the Zero Product Property. There is no need to set the constant factor [latex]-1[\/latex] to zero, because [latex]-1[\/latex] will never equal zero.\r\n<p style=\"text-align: center\">[latex]2t+1=0\\,\\,\\,\\,\\,\\,\\text{or}\\,\\,\\,\\,\\,\\,t-4=0[\/latex]<\/p>\r\nSolve each equation.\r\n<p style=\"text-align: center\">[latex]t=-\\frac{1}{2}\\,\\,\\,\\,\\,\\,\\text{or}\\,\\,\\,\\,\\,\\,t=4[\/latex]<\/p>\r\nInterpret the answer. Since <i>t<\/i> represents time, it cannot be a negative number; only [latex]t=4[\/latex]\u00a0makes sense in this context.\r\n\r\n[latex]t=4[\/latex]\r\n\r\nWe can check our answer: [latex]\\begin{array}{c}h(4)=\u22122(4)^{2}+7(4)+4=0\\\\h(4)=-2(16)+28+4=0\\\\h(4)-32+32=0\\\\h(4)=0\\end{array}[\/latex]\r\n<h4>Answer<\/h4>\r\nThe rocket will hit the ground 4 seconds after being launched.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nIn the next example we will solve for the time that the rocket is at a given height other than zero.\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nUse the formula for the height of the rocket in the previous example to find the time when the rocket is 4 feet from hitting the ground on it's way back down. \u00a0Refer to the image.\r\n\r\n[latex]h(t)=\u22122t^{2}+7t+4[\/latex]\r\n\r\n<img class=\"wp-image-4892 aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/121\/2016\/07\/20161951\/Screen-Shot-2016-06-14-at-5.39.53-PM-300x247.png\" alt=\"Parabolic motion of rocket which starts four feet up from the ground. t=0 is labeled at the starti of hte parabolic motion adn t=? is labeled at four feet from the ground on the other side of the parabola.\" width=\"413\" height=\"340\" \/>\r\n[reveal-answer q=\"198118\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"198118\"]\r\n\r\nWe are given that the height of the rocket is 4 feet from the ground on it's way back down. We want to know how long it has taken the rocket to get to that point in it's path, we are going to solve for t.\r\n\r\nSubstitute [latex]h(t) = 4[\/latex] into the formula for height, and try to get zero on one side since we know we can use the zero product principle to solve polynomials.\r\n\r\n<strong>Write and Solve:<\/strong>\r\n<p style=\"text-align: center\">[latex]\\begin{array}{l}h(t)=4=\u22122t^{2}+7t+4\\\\4=-2t^2+7t+4\\\\\\underline{-4}\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\underline{-4}\\\\0=-2t^2+7t\\end{array}[\/latex]<\/p>\r\n<p style=\"text-align: left\">Now we can factor out a t from each term:<\/p>\r\n<p style=\"text-align: center\">[latex]0=t\\left(-2t+7\\right)[\/latex]<\/p>\r\n<p style=\"text-align: left\">Solve each equation for t using the zero product principle:<\/p>\r\n<p style=\"text-align: center\">[latex]\\begin{array}{l}t=0\\text{ OR }-2t+7=0\\\\\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\underline{-7}\\,\\,\\,\\,\\,\\,\\,\\underline{-7}\\\\\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\frac{-2t}{-2}=\\frac{-7}{-2}\\\\\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,t=\\frac{7}{2}=3.5\\end{array}[\/latex]<\/p>\r\n<p style=\"text-align: left\">It doesn't make sense for us to choose t=0 because we are interested in the amount of time that has passed when the projectile is 4 feet from hitting the ground on it's way back down. We will choose t=3.5<\/p>\r\n<p style=\"text-align: left\">Check the answer on your own for practice.<\/p>\r\n\r\n<h4 style=\"text-align: left\">Answer<\/h4>\r\n[latex]t=3.5\\text{ seconds }[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nIn the following video we show another example of how to find the time when a object following a parabolic trajectory hits the ground.\r\n\r\nhttps:\/\/youtu.be\/hsWSzu3KcPU\r\n\r\nIn this section we introduced the concept of projectile motion, and showed that it can be modeled with polynomial function. \u00a0While the models used in these examples are simple, the concepts and interpretations are the same as what would happen in \"real life\".\r\n<h2>Pythagorean Theorem<\/h2>\r\nThe <b>Pythagorean theorem<\/b> or <b>Pythagoras's theorem<\/b> is a statement about the sides of a right triangle.\u00a0One of the angles of a right triangle is always equal to 90 degrees. This angle is the right angle. The two sides next to the right angle are called the legs and the other side is called the hypotenuse. The hypotenuse is the side opposite to the right angle, and it is always the longest side. The image above\u00a0shows four common kinds of triangle, including a right triangle.\r\n\r\n[caption id=\"attachment_4904\" align=\"alignleft\" width=\"205\"]<img class=\"wp-image-4904\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/121\/2016\/07\/20161955\/Screen-Shot-2016-06-14-at-8.32.56-PM-268x300.png\" alt=\"right triangle labeled with teh longest length = a, and the other two b and c.\" width=\"205\" height=\"229\" \/> Right Triangle Labeled[\/caption]\r\n<p style=\"text-align: left\">The Pythagorean theorem is often used to find unknown lengths of the sides of right triangles. If the longest leg of a right triangle is labeled c, and the other two a, and b as in the image on teh left, \u00a0The Pythagorean Theorem states that<\/p>\r\n<p style=\"text-align: center\">[latex]a^2+b^2=c^2[\/latex]<\/p>\r\n<p style=\"text-align: left\">Given enough information, we can solve for an unknown length. \u00a0This relationship has been used for many, many years for things such as celestial navigation and early civil engineering projects. We now have digital GPS and survey equipment that have been programmed to do the calculations for us.<\/p>\r\n<p style=\"text-align: left\">In the next example we will combine the power of the Pythagorean theorem and what we know about solving quadratic equations to find unknown lengths of right triangles.<\/p>\r\n\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nA\u00a0right triangle has one leg with length x, another whose length is greater by two, \u00a0and the length of the hypotenuse is greater by four. \u00a0Find the lengths of the sides of the triangle. Use the image below.\r\n\r\n<img class=\"alignnone size-medium wp-image-4907\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/121\/2016\/07\/20161957\/Screen-Shot-2016-06-14-at-8.49.02-PM-264x300.png\" alt=\"Right triangle with one leg having length = x, one with length= x+2 and the hypotenuse = x+4\" width=\"264\" height=\"300\" \/>\r\n[reveal-answer q=\"133740\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"133740\"]\r\n\r\n<strong><strong>Read and understand:\u00a0<\/strong><\/strong>We know the lengths of all the sides of a triangle in terms of one side. We also know that the Pythagorean theorem will give us a relationship between the side lengths of a right triangle.\r\n\r\n<strong>Translate:\u00a0<\/strong>\r\n<p style=\"text-align: center\">[latex]\\begin{array}{l}a^2+b^2=c^2\\\\x^2+\\left(x+2\\right)^2=\\left(x+4\\right)^2\\end{array}[\/latex]<\/p>\r\n<strong>Solve:<\/strong>\u00a0 If we can move all the terms\u00a0to one side and factor, we can use the zero product principle to solve. \u00a0Since this is the only method we know - let's hope it works!\r\n\r\nFirst, multiply the binomials and simplify so we can see what we are working with.\r\n<p style=\"text-align: center\">[latex]\\begin{array}{l}x^2+\\left(x+2\\right)^2=\\left(x+4\\right)^2\\\\x^2+x^2+4x+4=x^2+8x+16\\\\2x^2+4x+4=x^2+8x+16\\end{array}[\/latex]<\/p>\r\n<p style=\"text-align: left\">Now move all the terms to one side and see if we can factor.<\/p>\r\n<p style=\"text-align: center\">[latex]\\begin{array}{l}2x^2+4x+4=x^2+8x+16\\\\\\underline{-x^2}\\,\\,\\,\\underline{-8x}\\,\\,\\,\\underline{-16}\\,\\,\\,\\,\\,\\underline{-x^2}\\,\\,\\,\\underline{-8x}\\,\\,\\,\\underline{-16}\\\\x^2-4x-12=0\\end{array}[\/latex]<\/p>\r\n<p style=\"text-align: left\">This went from a messy looking problem to something promising. We can factor using the shortcut:<\/p>\r\n<p style=\"text-align: left\">[latex]-6\\cdot{2}=-12,\\text{ and }-6+2=-4[\/latex]<\/p>\r\n<p style=\"text-align: left\">So we can build our binomial factors with -6 and 2:<\/p>\r\n<p style=\"text-align: center\">[latex]\\left(x-6\\right)\\left(x+2\\right)=0[\/latex]<\/p>\r\n<p style=\"text-align: left\">Set each factor equal to zero:<\/p>\r\n<p style=\"text-align: left\">[latex]x-6=0, x=6[\/latex]<\/p>\r\n<p style=\"text-align: left\">[latex]x+2=0, x=-2[\/latex]<\/p>\r\n<p style=\"text-align: left\"><strong>Interpret:\u00a0<\/strong>Ok, it doesn't make sense to have a length equal to -2, so we can safely throw that solution out. \u00a0The lengths of the sides are as follows:<\/p>\r\n<p style=\"text-align: left\">[latex]x=6[\/latex]<\/p>\r\n<p style=\"text-align: left\">[latex]x+2=6+2=8[\/latex]<\/p>\r\n<p style=\"text-align: left\">[latex]x+4=6+4=10[\/latex]<\/p>\r\n<p style=\"text-align: left\"><strong>Check:\u00a0<\/strong>Since we know the relationship between the sides of a right triangle we can check that we are correct. Sometimes it helps to draw a picture<\/p>\r\n<p style=\"text-align: left\">.<img class=\"alignnone size-medium wp-image-4908\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/121\/2016\/07\/20161959\/Screen-Shot-2016-06-14-at-9.20.06-PM-300x263.png\" alt=\"Screen Shot 2016-06-14 at 9.20.06 PM\" width=\"300\" height=\"263\" \/><\/p>\r\n<p style=\"text-align: left\">We know that [latex]a^2+b^2=c^2[\/latex], so we can substitute the values we found:<\/p>\r\n<p style=\"text-align: left\">[latex]\\begin{array}{l}6^2+8^2=10^2\\\\36+64=100\\\\100=100\\end{array}[\/latex]<\/p>\r\n<p style=\"text-align: left\">Our solution checks out.<\/p>\r\n\r\n<h4 style=\"text-align: left\">Answer<\/h4>\r\nThe lengths of the sides of the right triangle are 6, 8, and 10\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nThis video example shows another way a quadratic equation can be used to find and unknown length of a right triangle.\r\n\r\nhttps:\/\/youtu.be\/xeP5pRBqsNs\r\n\r\n&nbsp;\r\n<h2>Geometric Applications<\/h2>\r\nIn this section we will explore ways that polynomials are used in applications of area problems.\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Example<\/h3>\r\nThe length of a rectangle is 3 more than the width. If the area is 40 square inches, what are the dimensions?\r\n\r\n&nbsp;\r\n\r\n[reveal-answer q=\"627193\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"627193\"]\r\n\r\n[latex]\\begin{eqnarray*}\r\nx &amp; &amp; \\text{ We} \\text{ do} \\text{ not} \\text{ know} \\text{ the}\r\n\\text{ width}, x.\\\\\r\nx + 3 &amp; &amp; \\text{ Length} \\text{ is}\\, 4 \\text{ more}, \\text{ or}\\, x + 4,\r\n\\text{ and} \\text{ area} \\text{ is}\\, 40.\\\\\r\nx (x + 3) = 40 &amp; &amp; \\text{ Multiply} \\text{ length} \\text{ by} \\text{ width}\r\n\\text{ to} \\text{ get} \\text{ area}\\\\\r\nx^2 + 3 x = 40 &amp; &amp; \\text{ Distribute}\\\\\r\n\\underline{- 40 - 40} &amp; &amp; \\text{ Make} \\text{ equation} \\text{ equal}\r\n\\text{ zero}\\\\\r\nx^2 + 3 x - 40 = 0 &amp; &amp; \\text{ Factor}\\\\\r\n(x - 5) (x + 8) = 0 &amp; &amp; \\text{ Set} \\text{ each} \\text{ factor} \\text{ equal}\r\n\\text{ to} \\text{ zero}\\\\\r\nx - 5 = 0 \\text{ or}\\, x + 8 = 0 &amp; &amp; \\text{ Solve} \\text{ each}\r\n\\text{ equation}\\\\\r\nx = 5 \\text{ or}\\, x = - 8 &amp; &amp; \\text{ Our}\\, x \\text{ is} a \\text{ width},\r\n\\text{ cannot} \\text{ be} \\text{ negative} .\\\\\r\n(5) + 3 = 8 &amp; &amp; \\text{ Length} \\text{ is}\\, x + 3, \\text{ substitute} 5\r\n\\text{ for}\\, x \\text{ to} \\text{ find} \\text{ length}\\\\\r\n5 \\text{ in} \\text{ by}\\, 8 \\text{ in} &amp; &amp; \\text{ Our} \\text{ Solution}\r\n\\end{eqnarray*}[\/latex]\r\n<h4>Answer<\/h4>\r\nThe dimensions of the rectangle are 5 inches by 8 inches.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nIn the following video you are shown how to find the dimensions of a border around a rectangle.\r\n\r\nhttps:\/\/www.youtube.com\/watch?v=AlIoxXQ-V50\r\n<div class=\"textbox tryit\">\r\n<h3>Try It<\/h3>\r\n[ohm_question]74417[\/ohm_question]\r\n\r\n<\/div>\r\n&nbsp;\r\n<h2>Cost, Revenue, and Profit Polynomials<\/h2>\r\nIn the systems of linear equations section, we discussed how a company's cost and revenue can be modeled with two linear equations. We found that the profit region for a company was the area between the two lines where the company would make money based on how much was produced. In this section, we will see that sometimes polynomials are used to describe cost and revenue.\r\n\r\nProfit is typically defined in business as the difference between the amount of money earned (revenue) by producing a certain number of items and the amount of money it takes to produce that number of items. When you are in business, you definitely want to see profit, so it is important to know what your cost and revenue is.\r\n\r\n[caption id=\"attachment_4645\" align=\"alignleft\" width=\"300\"]<img class=\"size-medium wp-image-4645\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/117\/2016\/06\/07014910\/Screen-Shot-2016-06-06-at-6.48.45-PM-300x194.png\" alt=\"Pile of cell phones\" width=\"300\" height=\"194\" \/> Cell Phones[\/caption]\r\n\r\nFor example, let's say that the cost to a manufacturer to produce a certain number of things is C and the revenue generated by selling those things is R. \u00a0The profit, P, can then be defined as\r\n<p style=\"text-align: center\">P = R-C<\/p>\r\n<p style=\"text-align: left\">The example we will work with is a hypothetical cell phone manufacturer whose\u00a0cost to manufacture x number of phones is [latex]C=2000x+750,000[\/latex], and the\u00a0Revenue generated from manufacturing x number of cell phones is [latex]R=-0.09x^2+7000x[\/latex].<\/p>\r\n\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nDefine a Profit polynomial for the hypothetical cell phone manufacturer.\r\n[reveal-answer q=\"50187\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"50187\"]\r\n\r\n<strong>Read and Understand: <\/strong>Profit is the difference between revenue and cost, so we will need to define P = R - C for the company.\r\n\r\n<strong>Define and Translate:<\/strong>\u00a0[latex]\\begin{array}{c}R=-0.09x^2+7000x\\\\C=2000x+750,000\\end{array}[\/latex]\r\n\r\n<strong>Write and Solve:\u00a0<\/strong>Substitute the expressions for R and C into the Profit equation.\r\n<p style=\"text-align: center\">[latex]\\begin{array}{c}P=R-C\\\\=-0.09x^2+7000x-\\left(2000x+750,000\\right)\\\\=-0.09x^2+7000x-2000x-750,000\\\\=-0.09x^2+5000x-750,000\\end{array}[\/latex]<\/p>\r\n<p style=\"text-align: left\">Remember that when you subtract a polynomial, you have to subtract every term of the polynomial.<\/p>\r\n\r\n<h4 style=\"text-align: left\">Answer<\/h4>\r\n[latex]P=-0.09x^2+5000x-750,000[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nMathematical models are great when you use them to learn important information. \u00a0The cell phone manufacturing company can use the profit equation to find out how much profit they will make given x number of phones are manufactured. \u00a0In the next example, we will explore some profit values for this company.\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nGiven the following numbers of cell phones manufactured, find the profit for the cell phone manufacturer:\r\n<ol>\r\n \t<li>x = 100 phones<\/li>\r\n \t<li>x = 25,000 phones<\/li>\r\n \t<li>x=60,000 phones<\/li>\r\n<\/ol>\r\nInterpret your results.\r\n[reveal-answer q=\"398706\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"398706\"]\r\n\r\n<strong>Read and Understand:\u00a0<\/strong>The profit polynomial defined in the previous example,[latex]P=-0.09x^2+5000x-750,000[\/latex], gives profit based on x number of phones manufactured. \u00a0We need to substitute the given numbers of phones manufactured into the equation, then try to understand what our answer means in terms of profit and number of phones manufactured.\r\n\r\nWe will move straight into write and solve since we already have our polynomial. It is probably easiest to use a calculator since the numbers in this problem are so large.\r\n\r\n<strong>Write and Solve:\u00a0<\/strong>\r\n\r\nSubstitute x = 100\r\n<p style=\"text-align: center\">[latex]\\begin{array}{c}P=-0.09x^2+5000x-750,000\\\\=-0.09\\left(100\\right)^2+5000\\left(100\\right)-750,000\\\\=-900+500,000-750,000\\\\=-250,900\\end{array}[\/latex]<\/p>\r\n<p style=\"text-align: left\"><strong>Interpret:\u00a0<\/strong>When the number of phones manufactured is 100, the profit for the business is $-250,000. \u00a0This is not what we want! \u00a0The company must produce more than 100 phones to make a profit.<\/p>\r\n<strong>Write and Solve:\u00a0<\/strong>\r\n\r\nSubstitute x = 25,000\r\n<p style=\"text-align: center\">[latex]\\begin{array}{c}P=-0.09x^2+5000x-750,000\\\\=-0.09\\left(25000\\right)^2+5000\\left(25000\\right)-750,000\\\\=-6120000+125,000,000-750,000\\\\=118,130,000\\end{array}[\/latex]<\/p>\r\n<p style=\"text-align: left\"><strong>Interpret:\u00a0<\/strong>When the number of phones manufactured is 25,000, the profit for the business is $118,130,000. \u00a0This is more like it! \u00a0If the company makes 25,000 phones it will make a profit after it pays all it's bills.<\/p>\r\n<p style=\"text-align: left\">\u00a0If this is true, then the company should make even more phones so it can make even more money, right? \u00a0Actually, something different happens as the number of items manufactured increases without bound.<\/p>\r\n<strong>Write and Solve:\u00a0<\/strong>\r\n\r\nSubstitute x = 60,000\r\n<p style=\"text-align: center\">[latex]\\begin{array}{c}P=-0.09x^2+5000x-750,000\\\\=-0.09\\left(60000\\right)^2+5000\\left(60000\\right)-750,000\\\\=-324,000,000+300,000,000-750,000\\\\=-24,750,000\\end{array}[\/latex]<\/p>\r\n<p style=\"text-align: left\"><strong>Interpret:\u00a0<\/strong>When the number of phones manufactured is 60,000, the profit for the business is $-24,750,000. \u00a0Wait a minute! If the company makes 60,000 phones it will\u00a0lose money, what happened? At some point, the cost to manufacture the phones will overcome the amount of profit that the business can make. \u00a0If this is interesting to you, you may enjoy reading about Economics and Business models.<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nIn the video that follows, we present another example of finding a polynomial profit equation.\r\n\r\nhttps:\/\/youtu.be\/-TWjDC4g9dU","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Objectives<\/h3>\n<ul>\n<li>Polynomial Equations\n<ul>\n<li>Use factoring methods to factor polynomial equations<\/li>\n<li>Use the principle of zero products to solve polynomial equations<\/li>\n<\/ul>\n<\/li>\n<li>Applications of Polynomial Equations\n<ul>\n<li>Projectiles<\/li>\n<li>Use the Pythagorean theorem to find the lengths of a right triangle<\/li>\n<li>Geometric Applications<\/li>\n<li>Cost, Revenue, and Profit Polynomials<\/li>\n<\/ul>\n<\/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<div id=\"attachment_4778\" style=\"width: 65px\" class=\"wp-caption alignleft\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-4778\" 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 id=\"caption-attachment-4778\" class=\"wp-caption-text\">Zero<\/p>\n<\/div>\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 2 and 5? Could they be 9 and 1? 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<p>&nbsp;<\/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 0, then at least one of the factors is 0.\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 0.<\/p>\n<\/div>\n<p>Let&#8217;s start with a simple example. \u00a0We 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 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>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 factor a trinomial using methods previously learned 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>You and I both know that it is rare to be given an equation to solve that has zero on one side, so let&#8217;s try another one.<\/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. \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}\\,\\,\\,\\,\\,\\,\\,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. (imagine that)<\/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<h4 style=\"text-align: left\">Answer<\/h4>\n<p>[latex]s=-1\\text{ OR }s=5[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>&nbsp;<\/p>\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\">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 2 to eliminate fractions:<\/p>\n<p>[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}\\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>[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>29<\/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>13<\/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>7<\/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 term of [latex]10y[\/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<h4>\u00a0Answer<\/h4>\n<p style=\"text-align: left\">[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<h2>\u00a0Projectile Motion<\/h2>\n<p>Projectile motion happens when you throw a ball into the air and it comes back down because of gravity. \u00a0A projectile will follow a curved path that behaves in a predictable way. \u00a0This predictable motion has been studied for centuries, and in simple cases it&#8217;s height from the ground\u00a0at a given time, t, can be modeled with a polynomial function of the form [latex]h(t)=at^2+bt+c[\/latex], where h(t) = height of an object at a given time, t. \u00a0Projectile motion is also called a parabolic trajectory because of the shape of the path of a projectile&#8217;s motion, as in the image of water in the fountain below.<\/p>\n<div id=\"attachment_4890\" style=\"width: 446px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-4890\" class=\"wp-image-4890\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/121\/2016\/07\/20161949\/ParabolicWaterTrajectory-225x300.jpg\" alt=\"Water from a fountain shoing classic parabolic motion.\" width=\"436\" height=\"581\" \/><\/p>\n<p id=\"caption-attachment-4890\" class=\"wp-caption-text\">Parabolic WaterTrajectory<\/p>\n<\/div>\n<p>Parabolic motion and it&#8217;s related functions\u00a0allow us to launch satellites for telecommunications, and rockets for space exploration. Recently, police departments have even begun using projectiles with GPS to track fleeing suspects in vehicles, rather than pursuing them by high-speed chase\u00a0<a class=\"footnote\" title=\"&quot;Cops' Latest Tool in High-speed Chases: GPS Projectiles.&quot; CBSNews. CBS Interactive, n.d. Web. 14 June 2016.\" id=\"return-footnote-2719-1\" href=\"#footnote-2719-1\" aria-label=\"Footnote 1\"><sup class=\"footnote\">[1]<\/sup><\/a>.<\/p>\n<p>In this section we will use\u00a0polynomial functions\u00a0to answer questions about the parabolic motion of a projectile. The real mathematical model for the path of a rocket or a police GPS projectile may have different coefficients or more variables, but the concept remains the same. We will also learn to interpret the meaning of the variables in a polynomial function that models projectile motion.<\/p>\n<div class=\"bcc-box bcc-info\">\n<h3>Example<\/h3>\n<p>A small toy rocket is launched from a 4-foot pedestal. The height (<i>h, <\/i>in feet) of the rocket <i>t<\/i> seconds after taking off is given by the function [latex]h(t)=\u22122t^{2}+7t+4[\/latex]. How long will it take the rocket to hit the ground?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q679533\">Show Solution<\/span><\/p>\n<div id=\"q679533\" class=\"hidden-answer\" style=\"display: none\">\n<p><strong><strong>\u00a0<\/strong><\/strong>The rocket will be on the ground when the [latex]h(t)=0[\/latex]. We want to know how long, t, \u00a0the rocket is in the air.<\/p>\n<p>[latex]\\begin{array}{l}h(t)=\u22122t^{2}+7t+4=0\\\\0=\u22122t^{2}+7t+4\\end{array}[\/latex]<\/p>\n<p>We can factor the polynomial [latex]\u22122t^{2}+7t+4[\/latex] more easily by first factoring out a [latex]-1[\/latex]<\/p>\n<p>[latex]\\begin{array}{c}0=-1(2t^{2}-7t-4)\\\\0=-1\\left(2t+1\\right)\\left(t-4\\right)\\end{array}[\/latex]<\/p>\n<p>Use the Zero Product Property. There is no need to set the constant factor [latex]-1[\/latex] to zero, because [latex]-1[\/latex] will never equal zero.<\/p>\n<p style=\"text-align: center\">[latex]2t+1=0\\,\\,\\,\\,\\,\\,\\text{or}\\,\\,\\,\\,\\,\\,t-4=0[\/latex]<\/p>\n<p>Solve each equation.<\/p>\n<p style=\"text-align: center\">[latex]t=-\\frac{1}{2}\\,\\,\\,\\,\\,\\,\\text{or}\\,\\,\\,\\,\\,\\,t=4[\/latex]<\/p>\n<p>Interpret the answer. Since <i>t<\/i> represents time, it cannot be a negative number; only [latex]t=4[\/latex]\u00a0makes sense in this context.<\/p>\n<p>[latex]t=4[\/latex]<\/p>\n<p>We can check our answer: [latex]\\begin{array}{c}h(4)=\u22122(4)^{2}+7(4)+4=0\\\\h(4)=-2(16)+28+4=0\\\\h(4)-32+32=0\\\\h(4)=0\\end{array}[\/latex]<\/p>\n<h4>Answer<\/h4>\n<p>The rocket will hit the ground 4 seconds after being launched.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>In the next example we will solve for the time that the rocket is at a given height other than zero.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Use the formula for the height of the rocket in the previous example to find the time when the rocket is 4 feet from hitting the ground on it&#8217;s way back down. \u00a0Refer to the image.<\/p>\n<p>[latex]h(t)=\u22122t^{2}+7t+4[\/latex]<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"wp-image-4892 aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/121\/2016\/07\/20161951\/Screen-Shot-2016-06-14-at-5.39.53-PM-300x247.png\" alt=\"Parabolic motion of rocket which starts four feet up from the ground. t=0 is labeled at the starti of hte parabolic motion adn t=? is labeled at four feet from the ground on the other side of the parabola.\" width=\"413\" height=\"340\" \/><\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q198118\">Show Solution<\/span><\/p>\n<div id=\"q198118\" class=\"hidden-answer\" style=\"display: none\">\n<p>We are given that the height of the rocket is 4 feet from the ground on it&#8217;s way back down. We want to know how long it has taken the rocket to get to that point in it&#8217;s path, we are going to solve for t.<\/p>\n<p>Substitute [latex]h(t) = 4[\/latex] into the formula for height, and try to get zero on one side since we know we can use the zero product principle to solve polynomials.<\/p>\n<p><strong>Write and Solve:<\/strong><\/p>\n<p style=\"text-align: center\">[latex]\\begin{array}{l}h(t)=4=\u22122t^{2}+7t+4\\\\4=-2t^2+7t+4\\\\\\underline{-4}\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\underline{-4}\\\\0=-2t^2+7t\\end{array}[\/latex]<\/p>\n<p style=\"text-align: left\">Now we can factor out a t from each term:<\/p>\n<p style=\"text-align: center\">[latex]0=t\\left(-2t+7\\right)[\/latex]<\/p>\n<p style=\"text-align: left\">Solve each equation for t using the zero product principle:<\/p>\n<p style=\"text-align: center\">[latex]\\begin{array}{l}t=0\\text{ OR }-2t+7=0\\\\\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\underline{-7}\\,\\,\\,\\,\\,\\,\\,\\underline{-7}\\\\\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\frac{-2t}{-2}=\\frac{-7}{-2}\\\\\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,t=\\frac{7}{2}=3.5\\end{array}[\/latex]<\/p>\n<p style=\"text-align: left\">It doesn&#8217;t make sense for us to choose t=0 because we are interested in the amount of time that has passed when the projectile is 4 feet from hitting the ground on it&#8217;s way back down. We will choose t=3.5<\/p>\n<p style=\"text-align: left\">Check the answer on your own for practice.<\/p>\n<h4 style=\"text-align: left\">Answer<\/h4>\n<p>[latex]t=3.5\\text{ seconds }[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>In the following video we show another example of how to find the time when a object following a parabolic trajectory hits the ground.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-4\" title=\"Factoring Application - Find the Time When a Projectile Hits and Ground\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/hsWSzu3KcPU?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<p>In this section we introduced the concept of projectile motion, and showed that it can be modeled with polynomial function. \u00a0While the models used in these examples are simple, the concepts and interpretations are the same as what would happen in &#8220;real life&#8221;.<\/p>\n<h2>Pythagorean Theorem<\/h2>\n<p>The <b>Pythagorean theorem<\/b> or <b>Pythagoras&#8217;s theorem<\/b> is a statement about the sides of a right triangle.\u00a0One of the angles of a right triangle is always equal to 90 degrees. This angle is the right angle. The two sides next to the right angle are called the legs and the other side is called the hypotenuse. The hypotenuse is the side opposite to the right angle, and it is always the longest side. The image above\u00a0shows four common kinds of triangle, including a right triangle.<\/p>\n<div id=\"attachment_4904\" style=\"width: 215px\" class=\"wp-caption alignleft\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-4904\" class=\"wp-image-4904\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/121\/2016\/07\/20161955\/Screen-Shot-2016-06-14-at-8.32.56-PM-268x300.png\" alt=\"right triangle labeled with teh longest length = a, and the other two b and c.\" width=\"205\" height=\"229\" \/><\/p>\n<p id=\"caption-attachment-4904\" class=\"wp-caption-text\">Right Triangle Labeled<\/p>\n<\/div>\n<p style=\"text-align: left\">The Pythagorean theorem is often used to find unknown lengths of the sides of right triangles. If the longest leg of a right triangle is labeled c, and the other two a, and b as in the image on teh left, \u00a0The Pythagorean Theorem states that<\/p>\n<p style=\"text-align: center\">[latex]a^2+b^2=c^2[\/latex]<\/p>\n<p style=\"text-align: left\">Given enough information, we can solve for an unknown length. \u00a0This relationship has been used for many, many years for things such as celestial navigation and early civil engineering projects. We now have digital GPS and survey equipment that have been programmed to do the calculations for us.<\/p>\n<p style=\"text-align: left\">In the next example we will combine the power of the Pythagorean theorem and what we know about solving quadratic equations to find unknown lengths of right triangles.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>A\u00a0right triangle has one leg with length x, another whose length is greater by two, \u00a0and the length of the hypotenuse is greater by four. \u00a0Find the lengths of the sides of the triangle. Use the image below.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-medium wp-image-4907\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/121\/2016\/07\/20161957\/Screen-Shot-2016-06-14-at-8.49.02-PM-264x300.png\" alt=\"Right triangle with one leg having length = x, one with length= x+2 and the hypotenuse = x+4\" width=\"264\" height=\"300\" \/><\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q133740\">Show Solution<\/span><\/p>\n<div id=\"q133740\" class=\"hidden-answer\" style=\"display: none\">\n<p><strong><strong>Read and understand:\u00a0<\/strong><\/strong>We know the lengths of all the sides of a triangle in terms of one side. We also know that the Pythagorean theorem will give us a relationship between the side lengths of a right triangle.<\/p>\n<p><strong>Translate:\u00a0<\/strong><\/p>\n<p style=\"text-align: center\">[latex]\\begin{array}{l}a^2+b^2=c^2\\\\x^2+\\left(x+2\\right)^2=\\left(x+4\\right)^2\\end{array}[\/latex]<\/p>\n<p><strong>Solve:<\/strong>\u00a0 If we can move all the terms\u00a0to one side and factor, we can use the zero product principle to solve. \u00a0Since this is the only method we know &#8211; let&#8217;s hope it works!<\/p>\n<p>First, multiply the binomials and simplify so we can see what we are working with.<\/p>\n<p style=\"text-align: center\">[latex]\\begin{array}{l}x^2+\\left(x+2\\right)^2=\\left(x+4\\right)^2\\\\x^2+x^2+4x+4=x^2+8x+16\\\\2x^2+4x+4=x^2+8x+16\\end{array}[\/latex]<\/p>\n<p style=\"text-align: left\">Now move all the terms to one side and see if we can factor.<\/p>\n<p style=\"text-align: center\">[latex]\\begin{array}{l}2x^2+4x+4=x^2+8x+16\\\\\\underline{-x^2}\\,\\,\\,\\underline{-8x}\\,\\,\\,\\underline{-16}\\,\\,\\,\\,\\,\\underline{-x^2}\\,\\,\\,\\underline{-8x}\\,\\,\\,\\underline{-16}\\\\x^2-4x-12=0\\end{array}[\/latex]<\/p>\n<p style=\"text-align: left\">This went from a messy looking problem to something promising. We can factor using the shortcut:<\/p>\n<p style=\"text-align: left\">[latex]-6\\cdot{2}=-12,\\text{ and }-6+2=-4[\/latex]<\/p>\n<p style=\"text-align: left\">So we can build our binomial factors with -6 and 2:<\/p>\n<p style=\"text-align: center\">[latex]\\left(x-6\\right)\\left(x+2\\right)=0[\/latex]<\/p>\n<p style=\"text-align: left\">Set each factor equal to zero:<\/p>\n<p style=\"text-align: left\">[latex]x-6=0, x=6[\/latex]<\/p>\n<p style=\"text-align: left\">[latex]x+2=0, x=-2[\/latex]<\/p>\n<p style=\"text-align: left\"><strong>Interpret:\u00a0<\/strong>Ok, it doesn&#8217;t make sense to have a length equal to -2, so we can safely throw that solution out. \u00a0The lengths of the sides are as follows:<\/p>\n<p style=\"text-align: left\">[latex]x=6[\/latex]<\/p>\n<p style=\"text-align: left\">[latex]x+2=6+2=8[\/latex]<\/p>\n<p style=\"text-align: left\">[latex]x+4=6+4=10[\/latex]<\/p>\n<p style=\"text-align: left\"><strong>Check:\u00a0<\/strong>Since we know the relationship between the sides of a right triangle we can check that we are correct. Sometimes it helps to draw a picture<\/p>\n<p style=\"text-align: left\">.<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-medium wp-image-4908\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/121\/2016\/07\/20161959\/Screen-Shot-2016-06-14-at-9.20.06-PM-300x263.png\" alt=\"Screen Shot 2016-06-14 at 9.20.06 PM\" width=\"300\" height=\"263\" \/><\/p>\n<p style=\"text-align: left\">We know that [latex]a^2+b^2=c^2[\/latex], so we can substitute the values we found:<\/p>\n<p style=\"text-align: left\">[latex]\\begin{array}{l}6^2+8^2=10^2\\\\36+64=100\\\\100=100\\end{array}[\/latex]<\/p>\n<p style=\"text-align: left\">Our solution checks out.<\/p>\n<h4 style=\"text-align: left\">Answer<\/h4>\n<p>The lengths of the sides of the right triangle are 6, 8, and 10<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>This video example shows another way a quadratic equation can be used to find and unknown length of a right triangle.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-5\" title=\"Factoring Application - Find the Lengths of Three Sides of a Right Triangle (Pythagorean Theorem)\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/xeP5pRBqsNs?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<p>&nbsp;<\/p>\n<h2>Geometric Applications<\/h2>\n<p>In this section we will explore ways that polynomials are used in applications of area problems.<\/p>\n<div class=\"bcc-box bcc-info\">\n<h3>Example<\/h3>\n<p>The length of a rectangle is 3 more than the width. If the area is 40 square inches, what are the dimensions?<\/p>\n<p>&nbsp;<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q627193\">Show Solution<\/span><\/p>\n<div id=\"q627193\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]\\begin{eqnarray*}  x & & \\text{ We} \\text{ do} \\text{ not} \\text{ know} \\text{ the}  \\text{ width}, x.\\\\  x + 3 & & \\text{ Length} \\text{ is}\\, 4 \\text{ more}, \\text{ or}\\, x + 4,  \\text{ and} \\text{ area} \\text{ is}\\, 40.\\\\  x (x + 3) = 40 & & \\text{ Multiply} \\text{ length} \\text{ by} \\text{ width}  \\text{ to} \\text{ get} \\text{ area}\\\\  x^2 + 3 x = 40 & & \\text{ Distribute}\\\\  \\underline{- 40 - 40} & & \\text{ Make} \\text{ equation} \\text{ equal}  \\text{ zero}\\\\  x^2 + 3 x - 40 = 0 & & \\text{ Factor}\\\\  (x - 5) (x + 8) = 0 & & \\text{ Set} \\text{ each} \\text{ factor} \\text{ equal}  \\text{ to} \\text{ zero}\\\\  x - 5 = 0 \\text{ or}\\, x + 8 = 0 & & \\text{ Solve} \\text{ each}  \\text{ equation}\\\\  x = 5 \\text{ or}\\, x = - 8 & & \\text{ Our}\\, x \\text{ is} a \\text{ width},  \\text{ cannot} \\text{ be} \\text{ negative} .\\\\  (5) + 3 = 8 & & \\text{ Length} \\text{ is}\\, x + 3, \\text{ substitute} 5  \\text{ for}\\, x \\text{ to} \\text{ find} \\text{ length}\\\\  5 \\text{ in} \\text{ by}\\, 8 \\text{ in} & & \\text{ Our} \\text{ Solution}  \\end{eqnarray*}[\/latex]<\/p>\n<h4>Answer<\/h4>\n<p>The dimensions of the rectangle are 5 inches by 8 inches.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>In the following video you are shown how to find the dimensions of a border around a rectangle.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-6\" title=\"Ex: Quadratic Equation App: Find the Width of a Frame (Factoring)\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/AlIoxXQ-V50?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<div class=\"textbox tryit\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm74417\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=74417&theme=oea&iframe_resize_id=ohm74417&show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<p>&nbsp;<\/p>\n<h2>Cost, Revenue, and Profit Polynomials<\/h2>\n<p>In the systems of linear equations section, we discussed how a company&#8217;s cost and revenue can be modeled with two linear equations. We found that the profit region for a company was the area between the two lines where the company would make money based on how much was produced. In this section, we will see that sometimes polynomials are used to describe cost and revenue.<\/p>\n<p>Profit is typically defined in business as the difference between the amount of money earned (revenue) by producing a certain number of items and the amount of money it takes to produce that number of items. When you are in business, you definitely want to see profit, so it is important to know what your cost and revenue is.<\/p>\n<div id=\"attachment_4645\" style=\"width: 310px\" class=\"wp-caption alignleft\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-4645\" class=\"size-medium wp-image-4645\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/117\/2016\/06\/07014910\/Screen-Shot-2016-06-06-at-6.48.45-PM-300x194.png\" alt=\"Pile of cell phones\" width=\"300\" height=\"194\" \/><\/p>\n<p id=\"caption-attachment-4645\" class=\"wp-caption-text\">Cell Phones<\/p>\n<\/div>\n<p>For example, let&#8217;s say that the cost to a manufacturer to produce a certain number of things is C and the revenue generated by selling those things is R. \u00a0The profit, P, can then be defined as<\/p>\n<p style=\"text-align: center\">P = R-C<\/p>\n<p style=\"text-align: left\">The example we will work with is a hypothetical cell phone manufacturer whose\u00a0cost to manufacture x number of phones is [latex]C=2000x+750,000[\/latex], and the\u00a0Revenue generated from manufacturing x number of cell phones is [latex]R=-0.09x^2+7000x[\/latex].<\/p>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Define a Profit polynomial for the hypothetical cell phone manufacturer.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q50187\">Show Solution<\/span><\/p>\n<div id=\"q50187\" class=\"hidden-answer\" style=\"display: none\">\n<p><strong>Read and Understand: <\/strong>Profit is the difference between revenue and cost, so we will need to define P = R &#8211; C for the company.<\/p>\n<p><strong>Define and Translate:<\/strong>\u00a0[latex]\\begin{array}{c}R=-0.09x^2+7000x\\\\C=2000x+750,000\\end{array}[\/latex]<\/p>\n<p><strong>Write and Solve:\u00a0<\/strong>Substitute the expressions for R and C into the Profit equation.<\/p>\n<p style=\"text-align: center\">[latex]\\begin{array}{c}P=R-C\\\\=-0.09x^2+7000x-\\left(2000x+750,000\\right)\\\\=-0.09x^2+7000x-2000x-750,000\\\\=-0.09x^2+5000x-750,000\\end{array}[\/latex]<\/p>\n<p style=\"text-align: left\">Remember that when you subtract a polynomial, you have to subtract every term of the polynomial.<\/p>\n<h4 style=\"text-align: left\">Answer<\/h4>\n<p>[latex]P=-0.09x^2+5000x-750,000[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>Mathematical models are great when you use them to learn important information. \u00a0The cell phone manufacturing company can use the profit equation to find out how much profit they will make given x number of phones are manufactured. \u00a0In the next example, we will explore some profit values for this company.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Given the following numbers of cell phones manufactured, find the profit for the cell phone manufacturer:<\/p>\n<ol>\n<li>x = 100 phones<\/li>\n<li>x = 25,000 phones<\/li>\n<li>x=60,000 phones<\/li>\n<\/ol>\n<p>Interpret your results.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q398706\">Show Solution<\/span><\/p>\n<div id=\"q398706\" class=\"hidden-answer\" style=\"display: none\">\n<p><strong>Read and Understand:\u00a0<\/strong>The profit polynomial defined in the previous example,[latex]P=-0.09x^2+5000x-750,000[\/latex], gives profit based on x number of phones manufactured. \u00a0We need to substitute the given numbers of phones manufactured into the equation, then try to understand what our answer means in terms of profit and number of phones manufactured.<\/p>\n<p>We will move straight into write and solve since we already have our polynomial. It is probably easiest to use a calculator since the numbers in this problem are so large.<\/p>\n<p><strong>Write and Solve:\u00a0<\/strong><\/p>\n<p>Substitute x = 100<\/p>\n<p style=\"text-align: center\">[latex]\\begin{array}{c}P=-0.09x^2+5000x-750,000\\\\=-0.09\\left(100\\right)^2+5000\\left(100\\right)-750,000\\\\=-900+500,000-750,000\\\\=-250,900\\end{array}[\/latex]<\/p>\n<p style=\"text-align: left\"><strong>Interpret:\u00a0<\/strong>When the number of phones manufactured is 100, the profit for the business is $-250,000. \u00a0This is not what we want! \u00a0The company must produce more than 100 phones to make a profit.<\/p>\n<p><strong>Write and Solve:\u00a0<\/strong><\/p>\n<p>Substitute x = 25,000<\/p>\n<p style=\"text-align: center\">[latex]\\begin{array}{c}P=-0.09x^2+5000x-750,000\\\\=-0.09\\left(25000\\right)^2+5000\\left(25000\\right)-750,000\\\\=-6120000+125,000,000-750,000\\\\=118,130,000\\end{array}[\/latex]<\/p>\n<p style=\"text-align: left\"><strong>Interpret:\u00a0<\/strong>When the number of phones manufactured is 25,000, the profit for the business is $118,130,000. \u00a0This is more like it! \u00a0If the company makes 25,000 phones it will make a profit after it pays all it&#8217;s bills.<\/p>\n<p style=\"text-align: left\">\u00a0If this is true, then the company should make even more phones so it can make even more money, right? \u00a0Actually, something different happens as the number of items manufactured increases without bound.<\/p>\n<p><strong>Write and Solve:\u00a0<\/strong><\/p>\n<p>Substitute x = 60,000<\/p>\n<p style=\"text-align: center\">[latex]\\begin{array}{c}P=-0.09x^2+5000x-750,000\\\\=-0.09\\left(60000\\right)^2+5000\\left(60000\\right)-750,000\\\\=-324,000,000+300,000,000-750,000\\\\=-24,750,000\\end{array}[\/latex]<\/p>\n<p style=\"text-align: left\"><strong>Interpret:\u00a0<\/strong>When the number of phones manufactured is 60,000, the profit for the business is $-24,750,000. \u00a0Wait a minute! If the company makes 60,000 phones it will\u00a0lose money, what happened? At some point, the cost to manufacture the phones will overcome the amount of profit that the business can make. \u00a0If this is interesting to you, you may enjoy reading about Economics and Business models.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>In the video that follows, we present another example of finding a polynomial profit equation.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-7\" title=\"Polynomial Subtraction App - Profit Equation from Revenue and Cost\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/-TWjDC4g9dU?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><li>Parabolic motion description and example. <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><li>Factoring Application - Find the Time When a Projectile Hits and Ground. <strong>Authored by<\/strong>: James Sousa (Mathispower4u.com) for Lumen Learning. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/hsWSzu3KcPU\">https:\/\/youtu.be\/hsWSzu3KcPU<\/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>Pythagorean Theorem, Description and Examples. <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><li>Screenshot: Thumbsdown. <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><li>Screenshot PI. <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><li>Polynomial Multiplication Application - Volume of a Cylinder. . <strong>Authored by<\/strong>: James Sousa (Mathispower4u.com) for Lumen Learning. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/g-g_nSsfGs4\">https:\/\/youtu.be\/g-g_nSsfGs4<\/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>Polynomial Subtracton App - Profit Equation from Revene and Cost.. <strong>Authored by<\/strong>: James Sousa (Mathispower4u.com) for Lumen Learning.. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/-TWjDC4g9dU\">https:\/\/youtu.be\/-TWjDC4g9dU<\/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>Screenshot: Cell Phones.. <strong>Authored 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><li>Profit Polynomial Examples. <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><li>Parabolic water trajectory. <strong>Authored by<\/strong>: By GuidoB. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/commons.wikimedia.org\/w\/index.php?curid=8015696\">https:\/\/commons.wikimedia.org\/w\/index.php?curid=8015696<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by-sa\/4.0\/\">CC BY-SA: Attribution-ShareAlike<\/a><\/em><\/li><li>Pythagorean Theorem. <strong>Provided by<\/strong>: Wikipedia. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/en.wikipedia.org\/wiki\/Pythagorean_theorem\">https:\/\/en.wikipedia.org\/wiki\/Pythagorean_theorem<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by-sa\/4.0\/\">CC BY-SA: Attribution-ShareAlike<\/a><\/em><\/li><li>Ex: Polynomial Addition Application - Perimeter.. <strong>Authored by<\/strong>: James Sousa (Mathispower4u.com). . <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/BhRpZv0_0jE\">https:\/\/youtu.be\/BhRpZv0_0jE<\/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: Find the Area of a Rectangle Using a Polynomial.. <strong>Authored by<\/strong>: James Sousa (Mathispower4u.com) .. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/5Q2htATOIik\">https:\/\/youtu.be\/5Q2htATOIik<\/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 11: Exponents and Polynomials, from Developmental Math: An Open Program. . <strong>Provided by<\/strong>: Monterey Institute of Technology and Education. <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><li>Applied Optimization Problems. <strong>Provided by<\/strong>: OpenStax. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"http:\/\/cnx.org\/contents\/svyieFe9@2\/Applied-Optimization-Problems\">http:\/\/cnx.org\/contents\/svyieFe9@2\/Applied-Optimization-Problems<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by-nc-sa\/4.0\/\">CC BY-NC-SA: Attribution-NonCommercial-ShareAlike<\/a><\/em>. <strong>License Terms<\/strong>: Download for free at http:\/\/cnx.org\/contents\/b2fca278-57bd-421d-aa85-21f539b4cc6f@2<\/li><li>Applications of Quadratic Equations. <strong>Authored by<\/strong>: James Sousa. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"http:\/\/www.youtube.com\/watch?v=AlIoxXQ-V50\">http:\/\/www.youtube.com\/watch?v=AlIoxXQ-V50<\/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>Quadratics - Rectangle. <strong>Authored by<\/strong>: Tyler Wallace. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"http:\/\/wallace.ccfaculty.org\/book\/book.html\">http:\/\/wallace.ccfaculty.org\/book\/book.html<\/a>. <strong>Project<\/strong>: Beginning and Intermediate Algebra. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/about\/pdm\">Public Domain: No Known Copyright<\/a><\/em><\/li><li>Question ID#74417. <strong>Authored by<\/strong>: Nearing,Daniel. <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\">All rights reserved content<\/div><ul class=\"citation-list\"><li>Celestial Navigation Math. <strong>Authored by<\/strong>: TabletClass Math. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/www.youtube.com\/watch?v=XWLZKmPU17M\">https:\/\/www.youtube.com\/watch?v=XWLZKmPU17M<\/a>. <strong>License<\/strong>: <em>All Rights Reserved<\/em>. <strong>License Terms<\/strong>: Standard YouTube License<\/li><\/ul><\/div>\n\t\t\t\t\t\t <\/div>\n\t\t\t\t\t <\/div>\n\t\t\t <\/section><hr class=\"before-footnotes clear\" \/><div class=\"footnotes\"><ol><li id=\"footnote-2719-1\">\"Cops' Latest Tool in High-speed Chases: GPS Projectiles.\" <i>CBSNews<\/i>. CBS Interactive, n.d. Web. 14 June 2016. <a href=\"#return-footnote-2719-1\" class=\"return-footnote\" aria-label=\"Return to footnote 1\">&crarr;<\/a><\/li><\/ol><\/div>","protected":false},"author":21,"menu_order":3,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc\",\"description\":\"Ex: Factor and Solve Quadratic Equation - Greatest Common Factor Only\",\"author\":\"James Sousa (Mathispower4u.com) \",\"organization\":\"\",\"url\":\"https:\/\/youtu.be\/gIwMkTAclw8\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"cc\",\"description\":\"Ex: Factor and Solve Quadratic Equations When A equals 1\",\"author\":\"James Sousa (Mathispower4u.com) \",\"organization\":\"\",\"url\":\"https:\/\/youtu.be\/bi7i_RuIGl0\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"original\",\"description\":\"Solve a Quadratic Equations with Fractions by Factoring (a not 1)\",\"author\":\"James Sousa (Mathispower4u.com) for Lumen Learning\",\"organization\":\"\",\"url\":\"https:\/\/youtu.be\/kDj_qdKW-ls\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"cc\",\"description\":\"Unit 12: Factoring, from Developmental Math: An Open Program\",\"author\":\"\",\"organization\":\"Monterey Institute of Technology\",\"url\":\"http:\/\/nrocnetwork.org\/dm-opentext\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"original\",\"description\":\"Revision and Adaptation\",\"author\":\"\",\"organization\":\"Lumen Learning\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"cc\",\"description\":\"Parabolic water trajectory\",\"author\":\"By GuidoB\",\"organization\":\"\",\"url\":\"https:\/\/commons.wikimedia.org\/w\/index.php?curid=8015696\",\"project\":\"\",\"license\":\"cc-by-sa\",\"license_terms\":\"\"},{\"type\":\"original\",\"description\":\"Parabolic motion description and example\",\"author\":\"\",\"organization\":\"Lumen Learning\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"original\",\"description\":\"Factoring Application - Find the Time When a Projectile Hits and Ground\",\"author\":\"James Sousa (Mathispower4u.com) for Lumen Learning\",\"organization\":\"\",\"url\":\"https:\/\/youtu.be\/hsWSzu3KcPU\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"copyrighted_video\",\"description\":\"Celestial Navigation Math\",\"author\":\"TabletClass Math\",\"organization\":\"\",\"url\":\"https:\/\/www.youtube.com\/watch?v=XWLZKmPU17M\",\"project\":\"\",\"license\":\"arr\",\"license_terms\":\"Standard YouTube License\"},{\"type\":\"original\",\"description\":\"Pythagorean Theorem, Description and Examples\",\"author\":\"\",\"organization\":\"Lumen Learning\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"cc\",\"description\":\"Pythagorean Theorem\",\"author\":\"\",\"organization\":\"Wikipedia\",\"url\":\"https:\/\/en.wikipedia.org\/wiki\/Pythagorean_theorem\",\"project\":\"\",\"license\":\"cc-by-sa\",\"license_terms\":\"\"},{\"type\":\"original\",\"description\":\"Screenshot: Thumbsdown\",\"author\":\"\",\"organization\":\"Lumen Learning\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"original\",\"description\":\"Screenshot PI\",\"author\":\"\",\"organization\":\"Lumen Learning\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"original\",\"description\":\"Polynomial Multiplication Application - Volume of a Cylinder. \",\"author\":\"James Sousa (Mathispower4u.com) for Lumen Learning\",\"organization\":\"\",\"url\":\"https:\/\/youtu.be\/g-g_nSsfGs4\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"original\",\"description\":\"Polynomial Subtracton App - Profit Equation from Revene and Cost.\",\"author\":\"James Sousa (Mathispower4u.com) for Lumen Learning.\",\"organization\":\"\",\"url\":\"https:\/\/youtu.be\/-TWjDC4g9dU\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"original\",\"description\":\"Screenshot: Cell Phones.\",\"author\":\"Lumen Learning\",\"organization\":\"\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"original\",\"description\":\"Profit Polynomial Examples\",\"author\":\"\",\"organization\":\"Lumen Learning\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"cc\",\"description\":\"Ex: Polynomial Addition Application - Perimeter.\",\"author\":\"James Sousa (Mathispower4u.com). \",\"organization\":\"\",\"url\":\" https:\/\/youtu.be\/BhRpZv0_0jE\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"cc\",\"description\":\"Ex: Find the Area of a Rectangle Using a Polynomial.\",\"author\":\"James Sousa (Mathispower4u.com) .\",\"organization\":\"\",\"url\":\" https:\/\/youtu.be\/5Q2htATOIik\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"cc\",\"description\":\"Unit 11: Exponents and Polynomials, from Developmental Math: An Open Program. \",\"author\":\"\",\"organization\":\"Monterey Institute of Technology and Education\",\"url\":\"http:\/\/nrocnetwork.org\/dm-opentext\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"cc\",\"description\":\"Applied Optimization Problems\",\"author\":\"\",\"organization\":\"OpenStax\",\"url\":\"http:\/\/cnx.org\/contents\/svyieFe9@2\/Applied-Optimization-Problems\",\"project\":\"\",\"license\":\"cc-by-nc-sa\",\"license_terms\":\"Download for free at http:\/\/cnx.org\/contents\/b2fca278-57bd-421d-aa85-21f539b4cc6f@2\"},{\"type\":\"cc\",\"description\":\"Applications of Quadratic Equations\",\"author\":\"James Sousa\",\"organization\":\"\",\"url\":\"www.youtube.com\/watch?v=AlIoxXQ-V50\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"cc\",\"description\":\"Quadratics - Rectangle\",\"author\":\"Tyler Wallace\",\"organization\":\"\",\"url\":\"http:\/\/wallace.ccfaculty.org\/book\/book.html\",\"project\":\"Beginning and Intermediate Algebra\",\"license\":\"pd\",\"license_terms\":\"\"},{\"type\":\"cc\",\"description\":\"Question ID#74417\",\"author\":\"Nearing,Daniel\",\"organization\":\"\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"}]","CANDELA_OUTCOMES_GUID":"","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-2719","chapter","type-chapter","status-publish","hentry"],"part":2069,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/odessa-coreq-collegealgebra\/wp-json\/pressbooks\/v2\/chapters\/2719","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/odessa-coreq-collegealgebra\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/odessa-coreq-collegealgebra\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/odessa-coreq-collegealgebra\/wp-json\/wp\/v2\/users\/21"}],"version-history":[{"count":27,"href":"https:\/\/courses.lumenlearning.com\/odessa-coreq-collegealgebra\/wp-json\/pressbooks\/v2\/chapters\/2719\/revisions"}],"predecessor-version":[{"id":5185,"href":"https:\/\/courses.lumenlearning.com\/odessa-coreq-collegealgebra\/wp-json\/pressbooks\/v2\/chapters\/2719\/revisions\/5185"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/odessa-coreq-collegealgebra\/wp-json\/pressbooks\/v2\/parts\/2069"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/odessa-coreq-collegealgebra\/wp-json\/pressbooks\/v2\/chapters\/2719\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/odessa-coreq-collegealgebra\/wp-json\/wp\/v2\/media?parent=2719"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/odessa-coreq-collegealgebra\/wp-json\/pressbooks\/v2\/chapter-type?post=2719"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/odessa-coreq-collegealgebra\/wp-json\/wp\/v2\/contributor?post=2719"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/odessa-coreq-collegealgebra\/wp-json\/wp\/v2\/license?post=2719"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}