{"id":16010,"date":"2019-09-26T16:15:54","date_gmt":"2019-09-26T16:15:54","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/wm-developmentalemporium\/chapter\/read-algebra-of-polynomial-functions\/"},"modified":"2024-05-02T15:56:54","modified_gmt":"2024-05-02T15:56:54","slug":"read-algebra-of-polynomial-functions","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/wm-developmentalemporium\/chapter\/read-algebra-of-polynomial-functions\/","title":{"raw":"Algebra of Polynomial Functions","rendered":"Algebra of Polynomial Functions"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Outcomes<\/h3>\r\n<ul>\r\n \t<li>Perform algebraic operations on polynomial functions<\/li>\r\n<\/ul>\r\n<\/div>\r\nJust as we have performed algebraic operations on polynomials, we can do the same with polynomial functions. In this section, we will show you\u00a0how to perform\u00a0algebraic operations on polynomial functions and introduce related notation.\r\n\r\nThe four basic operations on functions are adding, subtracting, multiplying, and dividing. The notation for these functions is as follows:\r\n\r\nAddition [latex](f + g)(x) = f(x)+ g(x)[\/latex]\r\n\r\nSubtraction [latex](f \u2212 g)(x)= f(x) \u2212 g(x)[\/latex]\r\n\r\nMultiplication [latex](f \u00b7 g)(x)= f(x)g(x)[\/latex]\r\n\r\nDivision [latex]\\frac{f}{ g} (x) = \\frac{f(x)}{g(x)}[\/latex]\r\n\r\nWe will focus on applying these operations to polynomial functions in this section.\r\n<h2>Add and Subtract Polynomial Functions<\/h2>\r\nAdding and subtracting polynomial functions is the same as adding and subtracting polynomials. When you evaluate a sum or difference of functions, you can either evaluate first or perform the operation on the functions first as we will see. Our next examples describe the notation used to add and subtract polynomial functions.\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nLet [latex]f(x)=2x^3-5x+3[\/latex] and [latex]h(x)=x-5[\/latex],\r\n\r\nFind the following:\r\n\r\n[latex](f+h)(x)[\/latex] and [latex](h-f)(x)[\/latex]\r\n[reveal-answer q=\"295585\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"295585\"]\r\n\r\n[latex]\\begin{array}{lll}(f+h)(x)=f(x)+ h(x) &amp; =(2x^3-5x+3)+(x-5) \\\\ &amp; =2x^3-5x+3+x-5 &amp; \\text{combine like terms} \\\\ &amp; =2x^3-4x-2 &amp; \\text{simplify}\\end{array}[\/latex]\r\n\r\n[latex]\\begin{array}{lll}(h-f)(x)=h(x)-f(x) &amp; =(x-5)-(2x^3-5x+3) \\\\ &amp; =x-5-2x^3+5x-3 &amp; \\text{combine like terms} \\\\ &amp; =-2x^3+6x-8 &amp; \\text{simplify}\\end{array}[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nIn our next example, we will evaluate a sum of functions and show that you can get to the same result in one of two ways.\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nLet [latex]f(x)=2x^3-5x+3[\/latex] and [latex]h(x)=x-5[\/latex]\r\n\r\nEvaluate:\u00a0[latex](f+h)(2)[\/latex]\r\n\r\nShow that you get the same result by\r\n\r\n1) Evaluating the functions first then performing the indicated operation on the result.\r\n\r\n2) Performing the operation on the functions first then evaluating the result.\r\n\r\n[reveal-answer q=\"754772\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"754772\"]\r\n\r\n1) First, we will evaluate the functions separately:\r\n\r\n[latex]f(2)=2(2)^3-5(2)+3=16-10+3=9[\/latex]\r\n\r\n[latex]h(2)=(2)-5=-3[\/latex]\r\n\r\nNow we will perform the indicated operation using the results:\r\n\r\n[latex](f+h)(2)=f(2)+h(2)=9+(-3)=6[\/latex]\r\n\r\n&nbsp;\r\n\r\n2) We can get the same result by adding the functions first and then evaluating the result at\u00a0[latex]x=2[\/latex].\r\n\r\n[latex](f+h)(x)=f(x)+h(x)=2x^3-4x-2[\/latex] from the previous example.\r\n\r\nNow we can evaluate this result at\u00a0[latex]x=2[\/latex]\r\n\r\n[latex](f+h)(2)=2(2)^3-4(2)-2=16-8-2=6[\/latex]\r\n\r\nBoth methods give the same result, and both require about the same amount of work.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<h2>Multiply and Divide Polynomial Functions<\/h2>\r\nWe saw that multiplying polynomials often required the use of the distributive property and that the algebra of dividing polynomials could get messy fast! \u00a0The same techniques can be used to multiply and divide polynomial functions. Additionally, the same idea applies to evaluating a product or quotient of functions as we discovered in the previous example. We can either evaluate the function and then perform the indicated operation or vice-versa. You may already be thinking it will be a lot less work to evaluate the polynomials and then divide the results!\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nLet [latex]g(t)=2t^3-t^2+7[\/latex] and [latex]f(t)=5t^2-3[\/latex]\r\n\r\nFind [latex](g \u00b7 f)(t)[\/latex], and evaluate [latex](g \u00b7 f)(-1)[\/latex]\r\n[reveal-answer q=\"48983\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"48983\"]\r\n<p style=\"text-align: left;\">[latex]\\begin{array}{lll}(g \u00b7 f)(t) &amp; =(2t^3-t^2+7)(5t^2-3) \\\\ &amp; =(2t^3\\cdot(5t^2)-t^2\\cdot(5t^2)+7\\cdot(5t^2))+(2t^3\\cdot(-3)-t^2\\cdot(-3)+7\\cdot(-3))\\,\\,\\,\\,\\,\\text{apply the distributive property} \\\\ &amp; =(10t^5-5t^4+35t^2)+(-6t^3+3t^2-21)\\,\\,\\,\\,\\text{simplify}\\\\ &amp; =10t^5-5t^4-6t^3+38t^2-21\\,\\,\\,\\,\\,\\text{combine like terms}\\end{array}[\/latex]<\/p>\r\nEvaluate\u00a0[latex](g \u00b7 f)(-1)[\/latex]\r\n\r\n[latex]\\begin{array}{lll}(g \u00b7 f)(t)=10t^5-5t^4-6t^3+38t^2-21\\\\(g \u00b7 f)(-1)=10(-1)^5-5(-1)^4-6(-1)^3+38(-1)^2-21=-10-5+6+38-21=8\\end{array}[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nIn the next example, we will divide polynomial functions and then evaluate the new function.\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nGiven [latex]p(x)=2x^2+x-15[\/latex] and [latex]r(x)=x+3[\/latex]\r\n\r\nFind [latex]\\frac{p}{r}(x)[\/latex] and evaluate\u00a0[latex]\\frac{p}{r}(2)[\/latex]\r\n[reveal-answer q=\"374371\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"374371\"]\r\n\r\nWe can use synthetic division for this polynomial division since the coefficient on [latex]r(x)=x+3[\/latex] is [latex]1[\/latex].\r\n\r\n<img class=\"size-medium wp-image-2655 aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/121\/2016\/07\/15231016\/Screen-Shot-2016-07-15-at-4.09.44-PM-300x205.png\" alt=\"Screen Shot 2016-07-15 at 4.09.44 PM\" width=\"300\" height=\"205\" \/>\r\n\r\nThis result means that\u00a0[latex]\\frac{p}{r}(x)=\\frac{2x^2+x-15}{x+3}=2x-5[\/latex]\r\n\r\nNow evaluate this quotient for\u00a0[latex]x = -3[\/latex] both ways as we did in a previous example.\r\n\r\nFirst, we will evaluate the result after polynomial division:\r\n\r\n[latex]\\begin{array}{lll}\\frac{p}{r}(x)=2x-5\\\\\\frac{p}{r}(2)=2(2)-5=4-5=-1\\end{array}[\/latex]\r\n\r\nNext, we will evaluate each function for\u00a0[latex]x = 2[\/latex], then we will divide the results.\r\n\r\n[latex]p(2)=2(2)^2+(2)-15=8+2-15=-5[\/latex]\r\n\r\n[latex]r(2)=(2)+3=5[\/latex]\r\n\r\nDivide the results:\r\n\r\n[latex]\\frac{p}{r}(2)=\\frac{-5}{5}=-1[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li>Perform algebraic operations on polynomial functions<\/li>\n<\/ul>\n<\/div>\n<p>Just as we have performed algebraic operations on polynomials, we can do the same with polynomial functions. In this section, we will show you\u00a0how to perform\u00a0algebraic operations on polynomial functions and introduce related notation.<\/p>\n<p>The four basic operations on functions are adding, subtracting, multiplying, and dividing. The notation for these functions is as follows:<\/p>\n<p>Addition [latex](f + g)(x) = f(x)+ g(x)[\/latex]<\/p>\n<p>Subtraction [latex](f \u2212 g)(x)= f(x) \u2212 g(x)[\/latex]<\/p>\n<p>Multiplication [latex](f \u00b7 g)(x)= f(x)g(x)[\/latex]<\/p>\n<p>Division [latex]\\frac{f}{ g} (x) = \\frac{f(x)}{g(x)}[\/latex]<\/p>\n<p>We will focus on applying these operations to polynomial functions in this section.<\/p>\n<h2>Add and Subtract Polynomial Functions<\/h2>\n<p>Adding and subtracting polynomial functions is the same as adding and subtracting polynomials. When you evaluate a sum or difference of functions, you can either evaluate first or perform the operation on the functions first as we will see. Our next examples describe the notation used to add and subtract polynomial functions.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Let [latex]f(x)=2x^3-5x+3[\/latex] and [latex]h(x)=x-5[\/latex],<\/p>\n<p>Find the following:<\/p>\n<p>[latex](f+h)(x)[\/latex] and [latex](h-f)(x)[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q295585\">Show Solution<\/span><\/p>\n<div id=\"q295585\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]\\begin{array}{lll}(f+h)(x)=f(x)+ h(x) & =(2x^3-5x+3)+(x-5) \\\\ & =2x^3-5x+3+x-5 & \\text{combine like terms} \\\\ & =2x^3-4x-2 & \\text{simplify}\\end{array}[\/latex]<\/p>\n<p>[latex]\\begin{array}{lll}(h-f)(x)=h(x)-f(x) & =(x-5)-(2x^3-5x+3) \\\\ & =x-5-2x^3+5x-3 & \\text{combine like terms} \\\\ & =-2x^3+6x-8 & \\text{simplify}\\end{array}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>In our next example, we will evaluate a sum of functions and show that you can get to the same result in one of two ways.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Let [latex]f(x)=2x^3-5x+3[\/latex] and [latex]h(x)=x-5[\/latex]<\/p>\n<p>Evaluate:\u00a0[latex](f+h)(2)[\/latex]<\/p>\n<p>Show that you get the same result by<\/p>\n<p>1) Evaluating the functions first then performing the indicated operation on the result.<\/p>\n<p>2) Performing the operation on the functions first then evaluating the result.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q754772\">Show Solution<\/span><\/p>\n<div id=\"q754772\" class=\"hidden-answer\" style=\"display: none\">\n<p>1) First, we will evaluate the functions separately:<\/p>\n<p>[latex]f(2)=2(2)^3-5(2)+3=16-10+3=9[\/latex]<\/p>\n<p>[latex]h(2)=(2)-5=-3[\/latex]<\/p>\n<p>Now we will perform the indicated operation using the results:<\/p>\n<p>[latex](f+h)(2)=f(2)+h(2)=9+(-3)=6[\/latex]<\/p>\n<p>&nbsp;<\/p>\n<p>2) We can get the same result by adding the functions first and then evaluating the result at\u00a0[latex]x=2[\/latex].<\/p>\n<p>[latex](f+h)(x)=f(x)+h(x)=2x^3-4x-2[\/latex] from the previous example.<\/p>\n<p>Now we can evaluate this result at\u00a0[latex]x=2[\/latex]<\/p>\n<p>[latex](f+h)(2)=2(2)^3-4(2)-2=16-8-2=6[\/latex]<\/p>\n<p>Both methods give the same result, and both require about the same amount of work.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<h2>Multiply and Divide Polynomial Functions<\/h2>\n<p>We saw that multiplying polynomials often required the use of the distributive property and that the algebra of dividing polynomials could get messy fast! \u00a0The same techniques can be used to multiply and divide polynomial functions. Additionally, the same idea applies to evaluating a product or quotient of functions as we discovered in the previous example. We can either evaluate the function and then perform the indicated operation or vice-versa. You may already be thinking it will be a lot less work to evaluate the polynomials and then divide the results!<\/p>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Let [latex]g(t)=2t^3-t^2+7[\/latex] and [latex]f(t)=5t^2-3[\/latex]<\/p>\n<p>Find [latex](g \u00b7 f)(t)[\/latex], and evaluate [latex](g \u00b7 f)(-1)[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q48983\">Show Solution<\/span><\/p>\n<div id=\"q48983\" class=\"hidden-answer\" style=\"display: none\">\n<p style=\"text-align: left;\">[latex]\\begin{array}{lll}(g \u00b7 f)(t) & =(2t^3-t^2+7)(5t^2-3) \\\\ & =(2t^3\\cdot(5t^2)-t^2\\cdot(5t^2)+7\\cdot(5t^2))+(2t^3\\cdot(-3)-t^2\\cdot(-3)+7\\cdot(-3))\\,\\,\\,\\,\\,\\text{apply the distributive property} \\\\ & =(10t^5-5t^4+35t^2)+(-6t^3+3t^2-21)\\,\\,\\,\\,\\text{simplify}\\\\ & =10t^5-5t^4-6t^3+38t^2-21\\,\\,\\,\\,\\,\\text{combine like terms}\\end{array}[\/latex]<\/p>\n<p>Evaluate\u00a0[latex](g \u00b7 f)(-1)[\/latex]<\/p>\n<p>[latex]\\begin{array}{lll}(g \u00b7 f)(t)=10t^5-5t^4-6t^3+38t^2-21\\\\(g \u00b7 f)(-1)=10(-1)^5-5(-1)^4-6(-1)^3+38(-1)^2-21=-10-5+6+38-21=8\\end{array}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>In the next example, we will divide polynomial functions and then evaluate the new function.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Given [latex]p(x)=2x^2+x-15[\/latex] and [latex]r(x)=x+3[\/latex]<\/p>\n<p>Find [latex]\\frac{p}{r}(x)[\/latex] and evaluate\u00a0[latex]\\frac{p}{r}(2)[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q374371\">Show Solution<\/span><\/p>\n<div id=\"q374371\" class=\"hidden-answer\" style=\"display: none\">\n<p>We can use synthetic division for this polynomial division since the coefficient on [latex]r(x)=x+3[\/latex] is [latex]1[\/latex].<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"size-medium wp-image-2655 aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/121\/2016\/07\/15231016\/Screen-Shot-2016-07-15-at-4.09.44-PM-300x205.png\" alt=\"Screen Shot 2016-07-15 at 4.09.44 PM\" width=\"300\" height=\"205\" \/><\/p>\n<p>This result means that\u00a0[latex]\\frac{p}{r}(x)=\\frac{2x^2+x-15}{x+3}=2x-5[\/latex]<\/p>\n<p>Now evaluate this quotient for\u00a0[latex]x = -3[\/latex] both ways as we did in a previous example.<\/p>\n<p>First, we will evaluate the result after polynomial division:<\/p>\n<p>[latex]\\begin{array}{lll}\\frac{p}{r}(x)=2x-5\\\\\\frac{p}{r}(2)=2(2)-5=4-5=-1\\end{array}[\/latex]<\/p>\n<p>Next, we will evaluate each function for\u00a0[latex]x = 2[\/latex], then we will divide the results.<\/p>\n<p>[latex]p(2)=2(2)^2+(2)-15=8+2-15=-5[\/latex]<\/p>\n<p>[latex]r(2)=(2)+3=5[\/latex]<\/p>\n<p>Divide the results:<\/p>\n<p>[latex]\\frac{p}{r}(2)=\\frac{-5}{5}=-1[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n","protected":false},"author":169554,"menu_order":11,"template":"","meta":{"_candela_citation":"[]","CANDELA_OUTCOMES_GUID":"e830134231634384b6394412db8f801b, 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