{"id":174,"date":"2023-11-08T16:10:14","date_gmt":"2023-11-08T16:10:14","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/chapter\/read-define-and-simplify-rational-expressions\/"},"modified":"2025-02-20T21:31:41","modified_gmt":"2025-02-20T21:31:41","slug":"4-1-introduction-to-rational-expressions-and-functions","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/chapter\/4-1-introduction-to-rational-expressions-and-functions\/","title":{"raw":"4.1 Introduction to Rational Expressions and Functions","rendered":"4.1 Introduction to Rational Expressions and Functions"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Outcomes<\/h3>\r\n<ul>\r\n \t<li>Evaluate rational functions.<\/li>\r\n \t<li>Identify and list restricted values of the variable in a rational expression (in particular, using factoring).<\/li>\r\n \t<li>Find the domain of a rational function and state it in set-builder or interval notation.<\/li>\r\n \t<li>Simplify rational expressions by removing a factor equal to [latex]1[\/latex].<\/li>\r\n<\/ul>\r\n<\/div>\r\n<strong>Rational expressions<\/strong> are fractions with a polynomial in the numerator and a non-zero polynomial in the denominator. Since any real number can be perceived as a polynomial, rational numbers like\u00a0[latex]\\dfrac{5}{13}[\/latex] or [latex]-3 = \\dfrac{-3}{1}[\/latex], and polynomials like\u00a0[latex]x=\\dfrac{x}{1}[\/latex] or\u00a0[latex]2x^3-5x+4=\\dfrac{2x^3-5x+4}{1}[\/latex], are examples of rational expressions. Other easier to recognize examples include\u00a0[latex]\\dfrac{1}{x}[\/latex],\u00a0[latex]\\dfrac{x+1}{x-3}[\/latex], or\u00a0[latex]\\dfrac{-2x^4 + 5x^2 - x}{x^3 - 7x^2 + 4x - 1}[\/latex].\r\n<h2>Evaluating Rational Functions<\/h2>\r\nRational expressions are quotients of real numbers or involve polynomials with a variable representing a real number. They can be evaluated at a chosen value of the variable and hence can be used to define functions. For example, to evaluate\u00a0[latex]f(x) = \\dfrac{-2x^4 + 5x^2 - x}{x^3 - 7x^2 + 4x - 1}[\/latex] at\u00a0[latex]x = -2[\/latex] means to substitute every occurrence of [latex]\\require{color}{\\color{Green}{x}}[\/latex] in [latex]f({\\color{Green}{x}}) = \\dfrac{-2{\\color{Green}{x}}^4 + 5{\\color{Green}{x}}^2 - {\\color{Green}{x}}}{{\\color{Green}{x}}^3 - 7{\\color{Green}{x}}^2 + 4{\\color{Green}{x}} - 1}[\/latex] with\u00a0[latex]{(\\color{Green}-2\\color{black})}[\/latex] (make sure to put parentheses around the input to properly evaluate)\u00a0and follow the order of operations to simplify it:\r\n<p style=\"text-align: center;\">[latex]\r\n\\begin{align}\r\nf({\\color{Green}{-2}}) &amp;= \\dfrac{-2 \\cdot ({\\color{Green}{-2}})^4 + 5 \\cdot ({\\color{Green}{-2}})^2 - ({\\color{Green}{-2}})}{({\\color{Green}{-2}})^3 - 7 \\cdot ({\\color{Green}{-2}})^2 + 4 \\cdot ({\\color{Green}{-2}}) - 1}\\\\[5pt]\r\n&amp;= \\dfrac{-2 \\cdot 16 + 5 \\cdot 4 + 2}{-8 - 7 \\cdot 4 - 8 - 1}\\\\[5pt]\r\n&amp;= \\dfrac{-32 + 20 + 2}{-8 - 28 - 9}\\\\[5pt]\r\n&amp;= \\dfrac{-10}{-45}\\\\[5pt]\r\n&amp;= \\dfrac{(-5) \\cdot 2}{(-5) \\cdot 9}\\\\[5pt]\r\n&amp;= \\dfrac{\\color{red} \\cancel{\\color{black}(-5)} \\color{black} \\cdot 2}{\\color{red} \\cancel{\\color{black}(-5)} \\color{black} \\cdot 9}\\\\[5pt]\r\n&amp;= \\dfrac{2}{9}\r\n\\end{align}[\/latex]<\/p>\r\n\r\n<h2>Restricted Values and the Domain of a Rational Function<\/h2>\r\nThere are a couple of ways to get yourself into trouble when working with rational expressions, equations, and functions. One of them is dividing by zero, and the other is trying to divide across addition or subtraction. Let's focus first on preventing division by zero.\u00a0The reason you cannot divide any nonzero number [latex]c[\/latex] by zero [latex] \\left( \\dfrac{c}{0}=? \\right)[\/latex] is that you would have to find a number such that when you multiply it by\u00a0[latex]0[\/latex], you would get back [latex]c[\/latex] [latex]\\left( ?\\cdot 0=c \\right)[\/latex]. There are no numbers that can do this, so we say \u201cdivision by zero is undefined.\u201d Consider the following rational expression evaluated at\u00a0[latex]x = 2[\/latex]:\r\n<p style=\"text-align: center;\">Evaluate \u00a0[latex]\\dfrac{x}{x-2}[\/latex] at [latex]x=2[\/latex].<\/p>\r\n<p style=\"text-align: center;\">\u00a0[latex]\r\n\\begin{align}\r\n&amp;\\quad\\; \\dfrac{{\\color{Green}{2}}}{{\\color{Green}{2}} - 2} &amp;&amp; \\color{blue}{\\textsf{substitute $x=2$}}\\\\[5pt]\r\n&amp;= \\dfrac{2}{0} &amp;&amp; \\color{blue}{\\textsf{simplify}}\r\n\\end{align}[\/latex]<\/p>\r\nThis means that for the expression [latex]\\dfrac{x}{x-2}[\/latex],\u00a0[latex]x[\/latex] cannot be\u00a0[latex]2[\/latex] because it will result in an undefined ratio. We acknowledge that [latex]2[\/latex] is the only value of the variable [latex]x[\/latex] that after simplifications results in [latex]0[\/latex] in the denominator, or as we will usually say, makes the denominator equal to zero. Therefore [latex]2[\/latex] will be referred to as a <strong>restricted <\/strong>(or <strong>excluded<\/strong>)<strong> value<\/strong> for the expression\u00a0[latex]\\dfrac{x}{x-2}[\/latex].\r\n<div class=\"textbox shaded\">\r\n<h3>restricted values in a rational expression<\/h3>\r\nThe restricted (or excluded) values of the variable in the given rational expression are the values that make the denominator equal to zero.\r\n\r\n<\/div>\r\nA function defined using a rational expression is called a <strong>rational function<\/strong>. The restricted values of the variable in the expression defining a rational function must be excluded from the set of real numbers to identify the <strong>domain<\/strong> of the rational function.\r\n<div class=\"textbox shaded\">\r\n<h3>Domain of a rational function<\/h3>\r\n<p style=\"text-align: left;\">The domain of a rational function is the set of all real numbers except for the restricted values of the variable in the expression defining the function.<\/p>\r\n\r\n<\/div>\r\nWe found above that [latex]2[\/latex] was the only restricted value of the variable in the expression [latex]\\dfrac{x}{x-2}[\/latex], so the domain of the rational function,\u00a0[latex]f(x)=\\dfrac{x}{x-2}[\/latex], consists of all real numbers [latex]x[\/latex] except [latex]2[\/latex]. The domain, as a set, can be expressed in different ways:\r\n<ul>\r\n \t<li>by simply listing all restrictions indicating that we \u201cstart\u201d with all real numbers: all real numbers [latex]x[\/latex] such that [latex]x\\ne 2[\/latex] (or all real numbers [latex]x[\/latex] except [latex]2[\/latex]),<\/li>\r\n \t<li>using set-builder notation: [latex]\\left\\{x\\, |\\, x\\ne 2\\right\\}[\/latex],<\/li>\r\n \t<li>using interval notation:\u00a0[latex](-\\infty, 2)\\cup (2,\\infty)[\/latex].<\/li>\r\n<\/ul>\r\nSince the restricted values of the variable are the values that make the denominator equal to zero, we will identify them by solving the equation [latex]\\mathsf{\u201cThe\\ Denominator\u201d} = 0[\/latex]. Recall that the only method we know so far for solving polynomial equations is factoring followed by using the Zero-Product Property. The solutions are the restricted values of the variable.\u00a0(Note that although the <i>denominator<\/i> cannot be equal to\u00a0[latex]0[\/latex], the <i>numerator<\/i> can \u2013 this is why we only look for restricted values in the <em>denominator<\/em> of a rational expression.)\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Example<\/h3>\r\nIdentify the domain of the following rational function.\r\n<p style=\"text-align: center;\">[latex]f(x)=\\dfrac{x+7}{x^2+8x-9}[\/latex]<\/p>\r\n<p style=\"text-align: left;\">[reveal-answer q=\"318517\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"318517\"]<\/p>\r\nWe find the restricted values of [latex]x[\/latex] by solving the equation\u00a0[latex]\\mathsf{\u201cThe\\ Denominator\u201d} = 0[\/latex]. We solve the equation by factoring and using the Zero-Product Property.\r\n<p style=\"text-align: center;\">[latex]\r\n\\begin{align}\r\nx^{2}+8x-9 &amp;= 0 &amp;&amp; \\color{blue}{\\textsf{set the denominator equal to $0$}}\\\\[5pt]\r\n(x+9)(x-1) &amp;= 0 &amp;&amp; \\color{blue}{\\textsf{factor}}\\\\[5pt]\r\nx+\\underset{\\large{\\color{red}{-9}}}{9} = \\underset{\\large{\\color{red}{-9}}}{0}\\ \\ \\textsf{or}\\ \\ x-\\underset{\\large{\\color{red}{+1}}}{1} &amp;= \\underset{\\large{\\color{red}{+1}}}{0} &amp;&amp; \\color{blue}{\\textsf{Zero-Product Property}}\\\\[5pt]\r\nx = -9 \\ \\ \\textsf{or}\\ \\ x &amp;= 1 &amp;&amp;\r\n\\end{align}\r\n[\/latex]<\/p>\r\nThe solutions are the restricted values that are excluded from the domain. The domain is all real numbers [latex]x[\/latex] except [latex]\u22129[\/latex] and [latex]1[\/latex]. The domain restrictions are\u00a0[latex]x\\ne \u22129[\/latex] and [latex]x\\ne 1[\/latex]. The domain in\u00a0set-builder notation is [latex]\\left\\{x\\, |\\, x\\ne -9 \\textsf{ and } x\\ne 1\\right\\}[\/latex]. The domain in interval notation is [latex](-\\infty, -9)\\cup (-9, 1)\\cup (1,\\infty)[\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<h2>Simplifying Rational Expressions<\/h2>\r\nIn the simplification of [latex]\\dfrac{-10}{-45}[\/latex] earlier in this section, we reviewed a technique of simplifying fractions by removing a factor equal to [latex]1[\/latex]. After acknowledging that the greatest common factor of [latex]10[\/latex] and [latex]45[\/latex] is [latex]5[\/latex] and that both the numerator and denominator are negative, we presented [latex]-10[\/latex] as [latex](-5) \\cdot 2[\/latex] and [latex]-45[\/latex] as [latex](-5) \\cdot 9[\/latex]. Then, we\u00a0<strong>removed a factor equal to 1<\/strong>\u00a0or <strong>divided out a common factor between the numerator and denominator <\/strong>(in our case, the common factor\u00a0of [latex]-5[\/latex]). It is justified by the following\u00a0mathematical facts:\r\n<ul>\r\n \t<li>the definition of multiplication of fractions: [latex]\\dfrac{(-5) \\cdot 2}{(-5) \\cdot 9} = \\dfrac{(-5)}{(-5)} \\cdot \\dfrac{2}{9}[\/latex],<\/li>\r\n \t<li>dividing a non-zero number by itself results in [latex]1[\/latex]: [latex]\\dfrac{(-5)}{(-5)}=1[\/latex],<\/li>\r\n \t<li>the product of a number and [latex]1[\/latex] is the number:\u00a0[latex]1\\cdot\\dfrac{2}{9}=\\dfrac{2}{9}[\/latex].<\/li>\r\n<\/ul>\r\nSince rational expressions are fractions with a polynomial in the numerator and a non-zero polynomial in the denominator, a similar technique can be used to simplify them. Consider the following three rational expressions:\u00a0[latex]\\dfrac{2x^2}{12x}[\/latex],\u00a0[latex]\\dfrac{x+1}{(x+1)(x-2)}[\/latex], and\u00a0[latex]\\dfrac{2-x}{(x+1)(x-2)}[\/latex].\r\n\r\nSince [latex]2x^2 = 2\\cdot x\\cdot x[\/latex] and [latex]12x = 2\\cdot x\\cdot 6[\/latex], the numerator and denominator of\u00a0[latex]\\dfrac{2x^2}{12x}[\/latex] have a common factor of [latex]2x[\/latex]. Hence,\r\n<p style=\"text-align: center;\">[latex]\\dfrac{2x^2}{12x}=\\dfrac{{\\color{red} \\cancel{{\\color{black}{2x}}}}\\cdot x}{{\\color{red} \\cancel{{\\color{black}{2x}}}}\\cdot 6}=\\dfrac{x}{6}[\/latex].<\/p>\r\nSince [latex]x+1 = (x+1)\\cdot 1[\/latex], the numerator and denominator of\u00a0[latex]\\dfrac{x+1}{(x+1)(x-2)}[\/latex] have a common factor of [latex]x+1[\/latex]. Hence,\r\n<p style=\"text-align: center;\">[latex]\\dfrac{x+1}{(x+1)(x-2)}\r\n=\\dfrac{{\\color{red} \\cancel{{\\color{black}{(x+1)}}}}\\cdot 1}{{\\color{red} \\cancel{{\\color{black}{(x+1)}}}}\\cdot (x-2)}\r\n=\\dfrac{1}{x-2}[\/latex].<\/p>\r\nIn the rational expression [latex]\\dfrac{2-x}{(x+1)(x-2)}[\/latex], the polynomial factors [latex]2-x[\/latex] and [latex]x-2[\/latex] look similar, though if we write the former in descending powers of the variable as [latex]-x+2[\/latex], we observe that it's actually the opposite of the latter. Thus, [latex]2-x = -(x-2)=(-1)\\cdot (x-2)[\/latex], and the numerator and denominator have a common factor of [latex]x-2[\/latex]. Hence,\r\n<p style=\"text-align: center;\">[latex]\\dfrac{2-x}{(x+1)(x-2)}\r\n=\\dfrac{(-1)\\cdot {\\color{red} \\cancel{{\\color{black}{(x-2)}}}}}{{\\color{red} \\cancel{{\\color{black}{(x-2)}}}}\\cdot (x+1)}\r\n=\\dfrac{-1}{x+1}[\/latex].<\/p>\r\nNote that\u00a0[latex]\\dfrac{-1}{x+1}[\/latex] could also be presented as\u00a0[latex]-\\dfrac{1}{x+1}[\/latex].\r\n\r\nTo simplify a general rational expression, we might have to use factoring to recognize common factors between the numerator and denominator. For example,\r\n<p style=\"text-align: center;\">[latex]\r\n\\begin{align}\r\n\\dfrac{2x^2-7x-4}{x^2-6x+8}&amp;=\\dfrac{(x-4)(2x+1)}{(x-4)(x-2)}\\\\[5pt]\r\n&amp;=\\dfrac{{\\color{red} \\cancel{{\\color{black}{(x-4)}}}}\\cdot (2x+1)}{{\\color{red} \\cancel{{\\color{black}{(x-4)}}}}\\cdot (x-2)}\\\\[5pt]\r\n&amp;=\\dfrac{2x+1}{x-2}\r\n\\end{align}\r\n[\/latex]<\/p>\r\nNote that we didn't perform any further simplifications to the rational expression\u00a0[latex]\\dfrac{2x+1}{x-2}[\/latex]. Recall that another way\u00a0to get yourself into trouble when working with rational expressions, equations, and functions is trying to divide across addition or subtraction. Removing a factor equal to [latex]1[\/latex] requires a common <strong>factor<\/strong> between the numerator and denominator. Factors are the building blocks of multiplication. In the polynomial [latex]2x+1[\/latex], we can only perceive the number [latex]2[\/latex] as a factor of the product [latex]2\\cdot x[\/latex], but not of the entire polynomial\u00a0[latex]2x+1[\/latex]. According to the terminology associated with polynomials, [latex]2x[\/latex] and [latex]1[\/latex] are terms of the polynomial [latex]2x+1[\/latex]. [latex]2[\/latex] is only a factor of the term [latex]2x[\/latex]. Similarly, in the polynomial [latex]x-2[\/latex], [latex]2[\/latex] is just one of the terms, not a factor of the entire polynomial. Therefore [latex]2[\/latex] is not a common factor\u00a0between the numerator and denominator of\u00a0[latex]\\dfrac{2x+1}{x-2}[\/latex] and <b>cannot be removed<\/b>. Also, in the polynomial [latex]2x+1[\/latex], we can only perceive the variable [latex]x[\/latex] as a factor of the product [latex]2\\cdot x[\/latex], but not of the entire polynomial, and\u00a0in the polynomial [latex]x-2[\/latex], [latex]x[\/latex] is just one of the terms, not a factor of the entire polynomial. Therefore [latex]x[\/latex] is not a common factor\u00a0between the numerator and denominator of\u00a0[latex]\\dfrac{2x+1}{x-2}[\/latex] and <b>cannot be removed<\/b>.\r\n\r\nThere is one more thing we have to acknowledge when simplifying rational expressions, the restricted values of the variable. We said that the restricted values of the variable are determined by solving the equation [latex]\\mathsf{\u201cThe\\ Denominator\u201d} = 0[\/latex]. To be exact, it's always <strong>the original denominator<\/strong>\u00a0(before we perform any simplifications to the rational expression) that we use to determine the restricted values of the variable since simplifications might make some restricted values appear to\u00a0<em>disappear<\/em>. For example, the restricted values of the variable in [latex]\\dfrac{2x^2-7x-4}{x^2-6x+8}[\/latex] are [latex]2[\/latex] and [latex]4[\/latex], the solutions of the equation [latex]x^2-6x+8=0[\/latex]. If we used the denominator of the simplified form, [latex]\\dfrac{2x+1}{x-2}[\/latex], we'd <em>lose<\/em> the restricted value [latex]4[\/latex] since the only solution of the equation [latex]x-2=0[\/latex] is [latex]2[\/latex].\r\n\r\nIn general, if we want to be mathematically precise, two algebraic expressions are equivalent if they define equivalent functions. In addition to the expression defining a function, we must also specify the domain. Only if the domains are the same and the expressions defining the functions are equivalent for all inputs in that common domain, the functions are equivalent. In particular, [latex]f(x)=\\dfrac{2x^2-7x-4}{x^2-6x+8}[\/latex] with its implied domain, [latex]\\left\\{x\\, |\\, x\\ne 2\\ \\textsf{and}\\ x\\ne 4\\right\\}[\/latex], and [latex]g(x)=\\dfrac{2x+1}{x-2}[\/latex]\u00a0with its implied domain, [latex]\\left\\{x\\, |\\, x\\ne 2\\right\\}[\/latex] are <strong>not<\/strong> equivalent, while\u00a0[latex]h(x)=\\dfrac{2x+1}{x-2}[\/latex]\u00a0with the domain [latex]\\left\\{x\\, |\\, x\\ne 2\\ \\textsf{and}\\ x\\ne 4\\right\\}[\/latex] and [latex]f(x)[\/latex] with its implied domain <strong>are<\/strong> equivalent.\r\n<div class=\"textbox shaded\">\r\n<h3>Steps for Simplifying a Rational Expression<\/h3>\r\nTo simplify a rational expression, follow these steps:\r\n<ul>\r\n \t<li>Determine the restricted values of the variable, the solutions of the equation [latex]\\mathsf{\u201cThe\\ Original\\ Denominator\u201d} = 0[\/latex].<\/li>\r\n \t<li>Factor the numerator and denominator.<\/li>\r\n \t<li>Simplify by removing all factors equal to [latex]1[\/latex]. (Divide out common factors between the numerator and denominator.)<\/li>\r\n<\/ul>\r\n<\/div>\r\nLet's practice finding the restricted values of the variable and simplifying rational expressions in additional examples.\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Example<\/h3>\r\nFor the following expressions, state the restricted values of the variable. Then simplify the expression if possible.\r\n<ol>\r\n \t<li style=\"text-align: left;\">[latex] \\dfrac{x+3}{x^2+12x+27}[\/latex]<\/li>\r\n \t<li>[latex] \\dfrac{2x-7}{x^2+1}[\/latex]<\/li>\r\n \t<li>[latex] \\dfrac{3x-1}{x}[\/latex]<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"623785\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"623785\"]\r\n\r\n1. Find the restricted values of [latex]x[\/latex] by solving the equation\u00a0[latex]\\mathsf{\u201cThe\\ Original\\ Denominator\u201d} = 0[\/latex].\r\n<p style=\"text-align: center;\">[latex]\r\n\\begin{align}\r\nx^2+12x+27 &amp;= 0 &amp;&amp; \\color{blue}{\\textsf{set the original denominator equal to $0$}}\\\\[5pt]\r\n(x+9)(x+3) &amp;= 0 &amp;&amp; \\color{blue}{\\textsf{factor}}\\\\[5pt]\r\nx+\\underset{\\large{\\color{red}{-9}}}{9} = \\underset{\\large{\\color{red}{-9}}}{0}\\ \\ \\textsf{or}\\ \\ x+\\underset{\\large{\\color{red}{-3}}}{3} &amp;= \\underset{\\large{\\color{red}{-3}}}{0} &amp;&amp; \\color{blue}{\\textsf{Zero-Product Property}}\\\\[5pt]\r\nx = -9 \\ \\ \\textsf{or}\\ \\ x &amp;= -3 &amp;&amp;\r\n\\end{align}\r\n[\/latex]<\/p>\r\nThe restricted values of the variable are [latex]-9[\/latex] and [latex]-3[\/latex].\r\n\r\nTo simplify, start with factoring, and continue by removing all factors equal to [latex]1[\/latex].\r\n<p style=\"text-align: center;\">[latex]\r\n\\begin{align}\r\n\\dfrac{x+3}{x^2+12x+27} &amp;= \\dfrac{(x+3)\\cdot 1}{(x+3)\\cdot (x+9)} &amp;&amp; {\\color{blue}{\\textsf{factor}}}\\\\[5pt]\r\n&amp;= \\dfrac{{\\color{red} \\cancel{{\\color{black}{(x+3)}}}}\\cdot 1}{{\\color{red} \\cancel{{\\color{black}{(x+3)}}}}\\cdot (x+9)} &amp;&amp; {\\color{blue}{\\textsf{remove all factors equal to $1$}}}\\\\[5pt]\r\n&amp;= \\dfrac{1}{x+9} &amp;&amp; {\\color{blue}{\\textsf{simplify}}}\r\n\\end{align}\r\n[\/latex]<\/p>\r\n\r\n\r\n<hr style=\"width: 80%;\" \/>\r\n\r\n2. Since [latex]x^2+1[\/latex] is always greater than or equal to [latex]1[\/latex] for any real number [latex]x[\/latex] (can you see, why?), the equation\u00a0[latex]x^2+1=0[\/latex] has no real solutions. Thus, there are no restricted values of [latex]x[\/latex] for [latex] \\dfrac{2x-7}{x^2+1}[\/latex]. There are also no common factors between the numerator and denominator to divide out, so it's already simplified.\r\n\r\n<hr style=\"width: 80%;\" \/>\r\n\r\n3. Since the only solution to the equation [latex]x=0[\/latex] is the number [latex]0[\/latex], it's the only restricted value of [latex]x[\/latex] for [latex] \\dfrac{3x-1}{x}[\/latex].\u00a0There are also no common factors between the numerator and denominator to divide out (don't even think about dividing out [latex]x[\/latex]!), so it's already simplified.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Example<\/h3>\r\nFor the following rational function, state the domain and write it in interval notation. Then simplify the expression defining the function.\r\n<p style=\"text-align: center;\">[latex] g(x)=\\dfrac{x^2+10x+24}{x^3-x^2-20x}[\/latex]<\/p>\r\n[reveal-answer q=\"861958\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"861958\"]\r\n\r\nFind the restricted values of [latex]x[\/latex] by solving the equation\u00a0[latex]\\mathsf{\u201cThe\\ Original\\ Denominator\u201d} = 0[\/latex].\r\n<p style=\"text-align: center;\">[latex]\r\n\\begin{align}\r\nx^3-x^2-20x &amp;= 0 &amp;&amp; \\color{blue}{\\textsf{set the original denominator equal to $0$}}\\\\[5pt]\r\nx(x^2-x-20) &amp;= 0 &amp;&amp; \\color{blue}{\\textsf{factor $x$ out}}\\\\[5pt]\r\nx(x-5)(x+4) &amp;= 0 &amp;&amp; \\color{blue}{\\textsf{factor the trinomial}}\\\\[5pt]\r\nx=0\\ \\ \\textsf{or}\\ \\ x-\\underset{\\large{\\color{red}{+5}}}{5} = \\underset{\\large{\\color{red}{+5}}}{0}\\ \\ \\textsf{or}\\ \\ x+\\underset{\\large{\\color{red}{-4}}}{4} &amp;= \\underset{\\large{\\color{red}{-4}}}{0} &amp;&amp; \\color{blue}{\\textsf{Zero-Product Property}}\\\\[5pt]\r\nx=0\\ \\ \\textsf{or}\\ \\ x = 5 \\ \\ \\textsf{or}\\ \\ x &amp;= -4 &amp;&amp;\r\n\\end{align}\r\n[\/latex]<\/p>\r\nThe domain of [latex]g(x)[\/latex] is all real numbers [latex]x[\/latex] except [latex]0[\/latex], [latex]5[\/latex], and [latex]-4[\/latex]. Written in interval notation, the domain is [latex](-\\infty, -4)\\cup (-4, 0)\\cup (0, 5)\\cup (5, \\infty)[\/latex].\r\n\r\nTo simplify, start with factoring, and continue by removing all factors equal to [latex]1[\/latex].\r\n<p style=\"text-align: center;\">[latex]\r\n\\begin{align}\r\n\\dfrac{x^2+10x+24}{x^3-x^2-20x} &amp;= \\dfrac{(x+4)\\cdot (x+6)}{x \\cdot (x-5)\\cdot (x+4)} &amp;&amp; {\\color{blue}{\\textsf{factor}}}\\\\[5pt]\r\n&amp;= \\dfrac{{\\color{red} \\cancel{{\\color{black}{(x+4)}}}}\\cdot (x+6)}{x \\cdot (x-5)\\cdot {\\color{red} \\cancel{{\\color{black}{(x+4)}}}}} &amp;&amp; {\\color{blue}{\\textsf{remove all factors equal to $1$}}}\\\\[5pt]\r\n&amp;= \\dfrac{x+6}{x(x-5)} &amp;&amp; {\\color{blue}{\\textsf{simplify}}}\r\n\\end{align}\r\n[\/latex]<\/p>\r\nIt is acceptable to either leave the denominator in factored form or to distribute\/multiply, so\u00a0[latex]\\dfrac{x+6}{x^2-5x}[\/latex] would be another acceptable simplified form.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nWe will show one last example of simplifying a rational expression. See if you can recognize the special product in the numerator.\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nFor the following expression, state the restricted values of the variable. Then simplify the expression.\r\n<p style=\"text-align: center;\">[latex] \\dfrac{x^2-9}{x^2+4x+3}[\/latex]<\/p>\r\n[reveal-answer q=\"773059\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"773059\"]\r\n\r\nFind the restricted values of [latex]x[\/latex] by solving the equation\u00a0[latex]\\mathsf{\u201cThe\\ Original\\ Denominator\u201d} = 0[\/latex].\r\n<p style=\"text-align: center;\">[latex]\r\n\\begin{align}\r\nx^2+4x+3 &amp;= 0 &amp;&amp; \\color{blue}{\\textsf{set the original denominator equal to $0$}}\\\\[5pt]\r\n(x+3)(x+1) &amp;= 0 &amp;&amp; \\color{blue}{\\textsf{factor}}\\\\[5pt]\r\nx+\\underset{\\large{\\color{red}{-3}}}{3} = \\underset{\\large{\\color{red}{-3}}}{0}\\ \\ \\textsf{or}\\ \\ x+\\underset{\\large{\\color{red}{-1}}}{1} &amp;= \\underset{\\large{\\color{red}{-1}}}{0} &amp;&amp; \\color{blue}{\\textsf{Zero-Product Property}}\\\\[5pt]\r\nx = -3 \\ \\ \\textsf{or}\\ \\ x &amp;= -1 &amp;&amp;\r\n\\end{align}\r\n[\/latex]<\/p>\r\nThe restricted values are [latex]-3[\/latex] and [latex]-1[\/latex].\r\n\r\nTo simplify, start with factoring, and continue by removing all factors equal to [latex]1[\/latex].\u00a0Notice the numerator is a difference of squares.\r\n<p style=\"text-align: center;\">[latex]\r\n\\begin{align}\r\n\\dfrac{x^2-9}{x^2+4x+3} &amp;= \\dfrac{(x+3)\\cdot (x-3)}{(x+3)\\cdot (x+1)} &amp;&amp; {\\color{blue}{\\textsf{factor}}}\\\\[5pt]\r\n&amp;= \\dfrac{{\\color{red} \\cancel{{\\color{black}{(x+3)}}}}\\cdot (x-3)}{{\\color{red} \\cancel{{\\color{black}{(x+3)}}}}\\cdot (x+1)} &amp;&amp; {\\color{blue}{\\textsf{remove all factors equal to $1$}}}\\\\[5pt]\r\n&amp;= \\dfrac{x-3}{x+1} &amp;&amp; {\\color{blue}{\\textsf{simplify}}}\r\n\\end{align}\r\n[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nIn the following video, we present additional examples of simplifying and finding the restricted values of the variable in a rational expression.\r\n\r\n[embed]https:\/\/youtu.be\/tJiz5rEktBs[\/embed]\r\n<h2>Summary<\/h2>\r\nRational expressions extend the notion of fractions to quotients of polynomials. An additional consideration for rational expressions is to determine restricted values of the variable. Since division by\u00a0[latex]0[\/latex] is undefined, any values of the variable that result in a denominator of\u00a0[latex]0[\/latex] must be restricted. Restricted values of the variable must be identified in the original expression, not in its simplified form. Rational expressions can be simplified much like fractions. To simplify a rational expression, first determine common factors between the numerator and denominator, then divide them out, removing factors equal to [latex]1[\/latex].","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li>Evaluate rational functions.<\/li>\n<li>Identify and list restricted values of the variable in a rational expression (in particular, using factoring).<\/li>\n<li>Find the domain of a rational function and state it in set-builder or interval notation.<\/li>\n<li>Simplify rational expressions by removing a factor equal to [latex]1[\/latex].<\/li>\n<\/ul>\n<\/div>\n<p><strong>Rational expressions<\/strong> are fractions with a polynomial in the numerator and a non-zero polynomial in the denominator. Since any real number can be perceived as a polynomial, rational numbers like\u00a0[latex]\\dfrac{5}{13}[\/latex] or [latex]-3 = \\dfrac{-3}{1}[\/latex], and polynomials like\u00a0[latex]x=\\dfrac{x}{1}[\/latex] or\u00a0[latex]2x^3-5x+4=\\dfrac{2x^3-5x+4}{1}[\/latex], are examples of rational expressions. Other easier to recognize examples include\u00a0[latex]\\dfrac{1}{x}[\/latex],\u00a0[latex]\\dfrac{x+1}{x-3}[\/latex], or\u00a0[latex]\\dfrac{-2x^4 + 5x^2 - x}{x^3 - 7x^2 + 4x - 1}[\/latex].<\/p>\n<h2>Evaluating Rational Functions<\/h2>\n<p>Rational expressions are quotients of real numbers or involve polynomials with a variable representing a real number. They can be evaluated at a chosen value of the variable and hence can be used to define functions. For example, to evaluate\u00a0[latex]f(x) = \\dfrac{-2x^4 + 5x^2 - x}{x^3 - 7x^2 + 4x - 1}[\/latex] at\u00a0[latex]x = -2[\/latex] means to substitute every occurrence of [latex]\\require{color}{\\color{Green}{x}}[\/latex] in [latex]f({\\color{Green}{x}}) = \\dfrac{-2{\\color{Green}{x}}^4 + 5{\\color{Green}{x}}^2 - {\\color{Green}{x}}}{{\\color{Green}{x}}^3 - 7{\\color{Green}{x}}^2 + 4{\\color{Green}{x}} - 1}[\/latex] with\u00a0[latex]{(\\color{Green}-2\\color{black})}[\/latex] (make sure to put parentheses around the input to properly evaluate)\u00a0and follow the order of operations to simplify it:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}  f({\\color{Green}{-2}}) &= \\dfrac{-2 \\cdot ({\\color{Green}{-2}})^4 + 5 \\cdot ({\\color{Green}{-2}})^2 - ({\\color{Green}{-2}})}{({\\color{Green}{-2}})^3 - 7 \\cdot ({\\color{Green}{-2}})^2 + 4 \\cdot ({\\color{Green}{-2}}) - 1}\\\\[5pt]  &= \\dfrac{-2 \\cdot 16 + 5 \\cdot 4 + 2}{-8 - 7 \\cdot 4 - 8 - 1}\\\\[5pt]  &= \\dfrac{-32 + 20 + 2}{-8 - 28 - 9}\\\\[5pt]  &= \\dfrac{-10}{-45}\\\\[5pt]  &= \\dfrac{(-5) \\cdot 2}{(-5) \\cdot 9}\\\\[5pt]  &= \\dfrac{\\color{red} \\cancel{\\color{black}(-5)} \\color{black} \\cdot 2}{\\color{red} \\cancel{\\color{black}(-5)} \\color{black} \\cdot 9}\\\\[5pt]  &= \\dfrac{2}{9}  \\end{align}[\/latex]<\/p>\n<h2>Restricted Values and the Domain of a Rational Function<\/h2>\n<p>There are a couple of ways to get yourself into trouble when working with rational expressions, equations, and functions. One of them is dividing by zero, and the other is trying to divide across addition or subtraction. Let&#8217;s focus first on preventing division by zero.\u00a0The reason you cannot divide any nonzero number [latex]c[\/latex] by zero [latex]\\left( \\dfrac{c}{0}=? \\right)[\/latex] is that you would have to find a number such that when you multiply it by\u00a0[latex]0[\/latex], you would get back [latex]c[\/latex] [latex]\\left( ?\\cdot 0=c \\right)[\/latex]. There are no numbers that can do this, so we say \u201cdivision by zero is undefined.\u201d Consider the following rational expression evaluated at\u00a0[latex]x = 2[\/latex]:<\/p>\n<p style=\"text-align: center;\">Evaluate \u00a0[latex]\\dfrac{x}{x-2}[\/latex] at [latex]x=2[\/latex].<\/p>\n<p style=\"text-align: center;\">\u00a0[latex]\\begin{align}  &\\quad\\; \\dfrac{{\\color{Green}{2}}}{{\\color{Green}{2}} - 2} && \\color{blue}{\\textsf{substitute $x=2$}}\\\\[5pt]  &= \\dfrac{2}{0} && \\color{blue}{\\textsf{simplify}}  \\end{align}[\/latex]<\/p>\n<p>This means that for the expression [latex]\\dfrac{x}{x-2}[\/latex],\u00a0[latex]x[\/latex] cannot be\u00a0[latex]2[\/latex] because it will result in an undefined ratio. We acknowledge that [latex]2[\/latex] is the only value of the variable [latex]x[\/latex] that after simplifications results in [latex]0[\/latex] in the denominator, or as we will usually say, makes the denominator equal to zero. Therefore [latex]2[\/latex] will be referred to as a <strong>restricted <\/strong>(or <strong>excluded<\/strong>)<strong> value<\/strong> for the expression\u00a0[latex]\\dfrac{x}{x-2}[\/latex].<\/p>\n<div class=\"textbox shaded\">\n<h3>restricted values in a rational expression<\/h3>\n<p>The restricted (or excluded) values of the variable in the given rational expression are the values that make the denominator equal to zero.<\/p>\n<\/div>\n<p>A function defined using a rational expression is called a <strong>rational function<\/strong>. The restricted values of the variable in the expression defining a rational function must be excluded from the set of real numbers to identify the <strong>domain<\/strong> of the rational function.<\/p>\n<div class=\"textbox shaded\">\n<h3>Domain of a rational function<\/h3>\n<p style=\"text-align: left;\">The domain of a rational function is the set of all real numbers except for the restricted values of the variable in the expression defining the function.<\/p>\n<\/div>\n<p>We found above that [latex]2[\/latex] was the only restricted value of the variable in the expression [latex]\\dfrac{x}{x-2}[\/latex], so the domain of the rational function,\u00a0[latex]f(x)=\\dfrac{x}{x-2}[\/latex], consists of all real numbers [latex]x[\/latex] except [latex]2[\/latex]. The domain, as a set, can be expressed in different ways:<\/p>\n<ul>\n<li>by simply listing all restrictions indicating that we \u201cstart\u201d with all real numbers: all real numbers [latex]x[\/latex] such that [latex]x\\ne 2[\/latex] (or all real numbers [latex]x[\/latex] except [latex]2[\/latex]),<\/li>\n<li>using set-builder notation: [latex]\\left\\{x\\, |\\, x\\ne 2\\right\\}[\/latex],<\/li>\n<li>using interval notation:\u00a0[latex](-\\infty, 2)\\cup (2,\\infty)[\/latex].<\/li>\n<\/ul>\n<p>Since the restricted values of the variable are the values that make the denominator equal to zero, we will identify them by solving the equation [latex]\\mathsf{\u201cThe\\ Denominator\u201d} = 0[\/latex]. Recall that the only method we know so far for solving polynomial equations is factoring followed by using the Zero-Product Property. The solutions are the restricted values of the variable.\u00a0(Note that although the <i>denominator<\/i> cannot be equal to\u00a0[latex]0[\/latex], the <i>numerator<\/i> can \u2013 this is why we only look for restricted values in the <em>denominator<\/em> of a rational expression.)<\/p>\n<div class=\"bcc-box bcc-info\">\n<h3>Example<\/h3>\n<p>Identify the domain of the following rational function.<\/p>\n<p style=\"text-align: center;\">[latex]f(x)=\\dfrac{x+7}{x^2+8x-9}[\/latex]<\/p>\n<p style=\"text-align: left;\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q318517\">Show Solution<\/span><\/p>\n<div id=\"q318517\" class=\"hidden-answer\" style=\"display: none\">\n<p>We find the restricted values of [latex]x[\/latex] by solving the equation\u00a0[latex]\\mathsf{\u201cThe\\ Denominator\u201d} = 0[\/latex]. We solve the equation by factoring and using the Zero-Product Property.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}  x^{2}+8x-9 &= 0 && \\color{blue}{\\textsf{set the denominator equal to $0$}}\\\\[5pt]  (x+9)(x-1) &= 0 && \\color{blue}{\\textsf{factor}}\\\\[5pt]  x+\\underset{\\large{\\color{red}{-9}}}{9} = \\underset{\\large{\\color{red}{-9}}}{0}\\ \\ \\textsf{or}\\ \\ x-\\underset{\\large{\\color{red}{+1}}}{1} &= \\underset{\\large{\\color{red}{+1}}}{0} && \\color{blue}{\\textsf{Zero-Product Property}}\\\\[5pt]  x = -9 \\ \\ \\textsf{or}\\ \\ x &= 1 &&  \\end{align}[\/latex]<\/p>\n<p>The solutions are the restricted values that are excluded from the domain. The domain is all real numbers [latex]x[\/latex] except [latex]\u22129[\/latex] and [latex]1[\/latex]. The domain restrictions are\u00a0[latex]x\\ne \u22129[\/latex] and [latex]x\\ne 1[\/latex]. The domain in\u00a0set-builder notation is [latex]\\left\\{x\\, |\\, x\\ne -9 \\textsf{ and } x\\ne 1\\right\\}[\/latex]. The domain in interval notation is [latex](-\\infty, -9)\\cup (-9, 1)\\cup (1,\\infty)[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<h2>Simplifying Rational Expressions<\/h2>\n<p>In the simplification of [latex]\\dfrac{-10}{-45}[\/latex] earlier in this section, we reviewed a technique of simplifying fractions by removing a factor equal to [latex]1[\/latex]. After acknowledging that the greatest common factor of [latex]10[\/latex] and [latex]45[\/latex] is [latex]5[\/latex] and that both the numerator and denominator are negative, we presented [latex]-10[\/latex] as [latex](-5) \\cdot 2[\/latex] and [latex]-45[\/latex] as [latex](-5) \\cdot 9[\/latex]. Then, we\u00a0<strong>removed a factor equal to 1<\/strong>\u00a0or <strong>divided out a common factor between the numerator and denominator <\/strong>(in our case, the common factor\u00a0of [latex]-5[\/latex]). It is justified by the following\u00a0mathematical facts:<\/p>\n<ul>\n<li>the definition of multiplication of fractions: [latex]\\dfrac{(-5) \\cdot 2}{(-5) \\cdot 9} = \\dfrac{(-5)}{(-5)} \\cdot \\dfrac{2}{9}[\/latex],<\/li>\n<li>dividing a non-zero number by itself results in [latex]1[\/latex]: [latex]\\dfrac{(-5)}{(-5)}=1[\/latex],<\/li>\n<li>the product of a number and [latex]1[\/latex] is the number:\u00a0[latex]1\\cdot\\dfrac{2}{9}=\\dfrac{2}{9}[\/latex].<\/li>\n<\/ul>\n<p>Since rational expressions are fractions with a polynomial in the numerator and a non-zero polynomial in the denominator, a similar technique can be used to simplify them. Consider the following three rational expressions:\u00a0[latex]\\dfrac{2x^2}{12x}[\/latex],\u00a0[latex]\\dfrac{x+1}{(x+1)(x-2)}[\/latex], and\u00a0[latex]\\dfrac{2-x}{(x+1)(x-2)}[\/latex].<\/p>\n<p>Since [latex]2x^2 = 2\\cdot x\\cdot x[\/latex] and [latex]12x = 2\\cdot x\\cdot 6[\/latex], the numerator and denominator of\u00a0[latex]\\dfrac{2x^2}{12x}[\/latex] have a common factor of [latex]2x[\/latex]. Hence,<\/p>\n<p style=\"text-align: center;\">[latex]\\dfrac{2x^2}{12x}=\\dfrac{{\\color{red} \\cancel{{\\color{black}{2x}}}}\\cdot x}{{\\color{red} \\cancel{{\\color{black}{2x}}}}\\cdot 6}=\\dfrac{x}{6}[\/latex].<\/p>\n<p>Since [latex]x+1 = (x+1)\\cdot 1[\/latex], the numerator and denominator of\u00a0[latex]\\dfrac{x+1}{(x+1)(x-2)}[\/latex] have a common factor of [latex]x+1[\/latex]. Hence,<\/p>\n<p style=\"text-align: center;\">[latex]\\dfrac{x+1}{(x+1)(x-2)}  =\\dfrac{{\\color{red} \\cancel{{\\color{black}{(x+1)}}}}\\cdot 1}{{\\color{red} \\cancel{{\\color{black}{(x+1)}}}}\\cdot (x-2)}  =\\dfrac{1}{x-2}[\/latex].<\/p>\n<p>In the rational expression [latex]\\dfrac{2-x}{(x+1)(x-2)}[\/latex], the polynomial factors [latex]2-x[\/latex] and [latex]x-2[\/latex] look similar, though if we write the former in descending powers of the variable as [latex]-x+2[\/latex], we observe that it&#8217;s actually the opposite of the latter. Thus, [latex]2-x = -(x-2)=(-1)\\cdot (x-2)[\/latex], and the numerator and denominator have a common factor of [latex]x-2[\/latex]. Hence,<\/p>\n<p style=\"text-align: center;\">[latex]\\dfrac{2-x}{(x+1)(x-2)}  =\\dfrac{(-1)\\cdot {\\color{red} \\cancel{{\\color{black}{(x-2)}}}}}{{\\color{red} \\cancel{{\\color{black}{(x-2)}}}}\\cdot (x+1)}  =\\dfrac{-1}{x+1}[\/latex].<\/p>\n<p>Note that\u00a0[latex]\\dfrac{-1}{x+1}[\/latex] could also be presented as\u00a0[latex]-\\dfrac{1}{x+1}[\/latex].<\/p>\n<p>To simplify a general rational expression, we might have to use factoring to recognize common factors between the numerator and denominator. For example,<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}  \\dfrac{2x^2-7x-4}{x^2-6x+8}&=\\dfrac{(x-4)(2x+1)}{(x-4)(x-2)}\\\\[5pt]  &=\\dfrac{{\\color{red} \\cancel{{\\color{black}{(x-4)}}}}\\cdot (2x+1)}{{\\color{red} \\cancel{{\\color{black}{(x-4)}}}}\\cdot (x-2)}\\\\[5pt]  &=\\dfrac{2x+1}{x-2}  \\end{align}[\/latex]<\/p>\n<p>Note that we didn&#8217;t perform any further simplifications to the rational expression\u00a0[latex]\\dfrac{2x+1}{x-2}[\/latex]. Recall that another way\u00a0to get yourself into trouble when working with rational expressions, equations, and functions is trying to divide across addition or subtraction. Removing a factor equal to [latex]1[\/latex] requires a common <strong>factor<\/strong> between the numerator and denominator. Factors are the building blocks of multiplication. In the polynomial [latex]2x+1[\/latex], we can only perceive the number [latex]2[\/latex] as a factor of the product [latex]2\\cdot x[\/latex], but not of the entire polynomial\u00a0[latex]2x+1[\/latex]. According to the terminology associated with polynomials, [latex]2x[\/latex] and [latex]1[\/latex] are terms of the polynomial [latex]2x+1[\/latex]. [latex]2[\/latex] is only a factor of the term [latex]2x[\/latex]. Similarly, in the polynomial [latex]x-2[\/latex], [latex]2[\/latex] is just one of the terms, not a factor of the entire polynomial. Therefore [latex]2[\/latex] is not a common factor\u00a0between the numerator and denominator of\u00a0[latex]\\dfrac{2x+1}{x-2}[\/latex] and <b>cannot be removed<\/b>. Also, in the polynomial [latex]2x+1[\/latex], we can only perceive the variable [latex]x[\/latex] as a factor of the product [latex]2\\cdot x[\/latex], but not of the entire polynomial, and\u00a0in the polynomial [latex]x-2[\/latex], [latex]x[\/latex] is just one of the terms, not a factor of the entire polynomial. Therefore [latex]x[\/latex] is not a common factor\u00a0between the numerator and denominator of\u00a0[latex]\\dfrac{2x+1}{x-2}[\/latex] and <b>cannot be removed<\/b>.<\/p>\n<p>There is one more thing we have to acknowledge when simplifying rational expressions, the restricted values of the variable. We said that the restricted values of the variable are determined by solving the equation [latex]\\mathsf{\u201cThe\\ Denominator\u201d} = 0[\/latex]. To be exact, it&#8217;s always <strong>the original denominator<\/strong>\u00a0(before we perform any simplifications to the rational expression) that we use to determine the restricted values of the variable since simplifications might make some restricted values appear to\u00a0<em>disappear<\/em>. For example, the restricted values of the variable in [latex]\\dfrac{2x^2-7x-4}{x^2-6x+8}[\/latex] are [latex]2[\/latex] and [latex]4[\/latex], the solutions of the equation [latex]x^2-6x+8=0[\/latex]. If we used the denominator of the simplified form, [latex]\\dfrac{2x+1}{x-2}[\/latex], we&#8217;d <em>lose<\/em> the restricted value [latex]4[\/latex] since the only solution of the equation [latex]x-2=0[\/latex] is [latex]2[\/latex].<\/p>\n<p>In general, if we want to be mathematically precise, two algebraic expressions are equivalent if they define equivalent functions. In addition to the expression defining a function, we must also specify the domain. Only if the domains are the same and the expressions defining the functions are equivalent for all inputs in that common domain, the functions are equivalent. In particular, [latex]f(x)=\\dfrac{2x^2-7x-4}{x^2-6x+8}[\/latex] with its implied domain, [latex]\\left\\{x\\, |\\, x\\ne 2\\ \\textsf{and}\\ x\\ne 4\\right\\}[\/latex], and [latex]g(x)=\\dfrac{2x+1}{x-2}[\/latex]\u00a0with its implied domain, [latex]\\left\\{x\\, |\\, x\\ne 2\\right\\}[\/latex] are <strong>not<\/strong> equivalent, while\u00a0[latex]h(x)=\\dfrac{2x+1}{x-2}[\/latex]\u00a0with the domain [latex]\\left\\{x\\, |\\, x\\ne 2\\ \\textsf{and}\\ x\\ne 4\\right\\}[\/latex] and [latex]f(x)[\/latex] with its implied domain <strong>are<\/strong> equivalent.<\/p>\n<div class=\"textbox shaded\">\n<h3>Steps for Simplifying a Rational Expression<\/h3>\n<p>To simplify a rational expression, follow these steps:<\/p>\n<ul>\n<li>Determine the restricted values of the variable, the solutions of the equation [latex]\\mathsf{\u201cThe\\ Original\\ Denominator\u201d} = 0[\/latex].<\/li>\n<li>Factor the numerator and denominator.<\/li>\n<li>Simplify by removing all factors equal to [latex]1[\/latex]. (Divide out common factors between the numerator and denominator.)<\/li>\n<\/ul>\n<\/div>\n<p>Let&#8217;s practice finding the restricted values of the variable and simplifying rational expressions in additional examples.<\/p>\n<div class=\"bcc-box bcc-info\">\n<h3>Example<\/h3>\n<p>For the following expressions, state the restricted values of the variable. Then simplify the expression if possible.<\/p>\n<ol>\n<li style=\"text-align: left;\">[latex]\\dfrac{x+3}{x^2+12x+27}[\/latex]<\/li>\n<li>[latex]\\dfrac{2x-7}{x^2+1}[\/latex]<\/li>\n<li>[latex]\\dfrac{3x-1}{x}[\/latex]<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q623785\">Show Solution<\/span><\/p>\n<div id=\"q623785\" class=\"hidden-answer\" style=\"display: none\">\n<p>1. Find the restricted values of [latex]x[\/latex] by solving the equation\u00a0[latex]\\mathsf{\u201cThe\\ Original\\ Denominator\u201d} = 0[\/latex].<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}  x^2+12x+27 &= 0 && \\color{blue}{\\textsf{set the original denominator equal to $0$}}\\\\[5pt]  (x+9)(x+3) &= 0 && \\color{blue}{\\textsf{factor}}\\\\[5pt]  x+\\underset{\\large{\\color{red}{-9}}}{9} = \\underset{\\large{\\color{red}{-9}}}{0}\\ \\ \\textsf{or}\\ \\ x+\\underset{\\large{\\color{red}{-3}}}{3} &= \\underset{\\large{\\color{red}{-3}}}{0} && \\color{blue}{\\textsf{Zero-Product Property}}\\\\[5pt]  x = -9 \\ \\ \\textsf{or}\\ \\ x &= -3 &&  \\end{align}[\/latex]<\/p>\n<p>The restricted values of the variable are [latex]-9[\/latex] and [latex]-3[\/latex].<\/p>\n<p>To simplify, start with factoring, and continue by removing all factors equal to [latex]1[\/latex].<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}  \\dfrac{x+3}{x^2+12x+27} &= \\dfrac{(x+3)\\cdot 1}{(x+3)\\cdot (x+9)} && {\\color{blue}{\\textsf{factor}}}\\\\[5pt]  &= \\dfrac{{\\color{red} \\cancel{{\\color{black}{(x+3)}}}}\\cdot 1}{{\\color{red} \\cancel{{\\color{black}{(x+3)}}}}\\cdot (x+9)} && {\\color{blue}{\\textsf{remove all factors equal to $1$}}}\\\\[5pt]  &= \\dfrac{1}{x+9} && {\\color{blue}{\\textsf{simplify}}}  \\end{align}[\/latex]<\/p>\n<hr style=\"width: 80%;\" \/>\n<p>2. Since [latex]x^2+1[\/latex] is always greater than or equal to [latex]1[\/latex] for any real number [latex]x[\/latex] (can you see, why?), the equation\u00a0[latex]x^2+1=0[\/latex] has no real solutions. Thus, there are no restricted values of [latex]x[\/latex] for [latex]\\dfrac{2x-7}{x^2+1}[\/latex]. There are also no common factors between the numerator and denominator to divide out, so it&#8217;s already simplified.<\/p>\n<hr style=\"width: 80%;\" \/>\n<p>3. Since the only solution to the equation [latex]x=0[\/latex] is the number [latex]0[\/latex], it&#8217;s the only restricted value of [latex]x[\/latex] for [latex]\\dfrac{3x-1}{x}[\/latex].\u00a0There are also no common factors between the numerator and denominator to divide out (don&#8217;t even think about dividing out [latex]x[\/latex]!), so it&#8217;s already simplified.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"bcc-box bcc-info\">\n<h3>Example<\/h3>\n<p>For the following rational function, state the domain and write it in interval notation. Then simplify the expression defining the function.<\/p>\n<p style=\"text-align: center;\">[latex]g(x)=\\dfrac{x^2+10x+24}{x^3-x^2-20x}[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q861958\">Show Solution<\/span><\/p>\n<div id=\"q861958\" class=\"hidden-answer\" style=\"display: none\">\n<p>Find the restricted values of [latex]x[\/latex] by solving the equation\u00a0[latex]\\mathsf{\u201cThe\\ Original\\ Denominator\u201d} = 0[\/latex].<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}  x^3-x^2-20x &= 0 && \\color{blue}{\\textsf{set the original denominator equal to $0$}}\\\\[5pt]  x(x^2-x-20) &= 0 && \\color{blue}{\\textsf{factor $x$ out}}\\\\[5pt]  x(x-5)(x+4) &= 0 && \\color{blue}{\\textsf{factor the trinomial}}\\\\[5pt]  x=0\\ \\ \\textsf{or}\\ \\ x-\\underset{\\large{\\color{red}{+5}}}{5} = \\underset{\\large{\\color{red}{+5}}}{0}\\ \\ \\textsf{or}\\ \\ x+\\underset{\\large{\\color{red}{-4}}}{4} &= \\underset{\\large{\\color{red}{-4}}}{0} && \\color{blue}{\\textsf{Zero-Product Property}}\\\\[5pt]  x=0\\ \\ \\textsf{or}\\ \\ x = 5 \\ \\ \\textsf{or}\\ \\ x &= -4 &&  \\end{align}[\/latex]<\/p>\n<p>The domain of [latex]g(x)[\/latex] is all real numbers [latex]x[\/latex] except [latex]0[\/latex], [latex]5[\/latex], and [latex]-4[\/latex]. Written in interval notation, the domain is [latex](-\\infty, -4)\\cup (-4, 0)\\cup (0, 5)\\cup (5, \\infty)[\/latex].<\/p>\n<p>To simplify, start with factoring, and continue by removing all factors equal to [latex]1[\/latex].<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}  \\dfrac{x^2+10x+24}{x^3-x^2-20x} &= \\dfrac{(x+4)\\cdot (x+6)}{x \\cdot (x-5)\\cdot (x+4)} && {\\color{blue}{\\textsf{factor}}}\\\\[5pt]  &= \\dfrac{{\\color{red} \\cancel{{\\color{black}{(x+4)}}}}\\cdot (x+6)}{x \\cdot (x-5)\\cdot {\\color{red} \\cancel{{\\color{black}{(x+4)}}}}} && {\\color{blue}{\\textsf{remove all factors equal to $1$}}}\\\\[5pt]  &= \\dfrac{x+6}{x(x-5)} && {\\color{blue}{\\textsf{simplify}}}  \\end{align}[\/latex]<\/p>\n<p>It is acceptable to either leave the denominator in factored form or to distribute\/multiply, so\u00a0[latex]\\dfrac{x+6}{x^2-5x}[\/latex] would be another acceptable simplified form.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>We will show one last example of simplifying a rational expression. See if you can recognize the special product in the numerator.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>For the following expression, state the restricted values of the variable. Then simplify the expression.<\/p>\n<p style=\"text-align: center;\">[latex]\\dfrac{x^2-9}{x^2+4x+3}[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q773059\">Show Solution<\/span><\/p>\n<div id=\"q773059\" class=\"hidden-answer\" style=\"display: none\">\n<p>Find the restricted values of [latex]x[\/latex] by solving the equation\u00a0[latex]\\mathsf{\u201cThe\\ Original\\ Denominator\u201d} = 0[\/latex].<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}  x^2+4x+3 &= 0 && \\color{blue}{\\textsf{set the original denominator equal to $0$}}\\\\[5pt]  (x+3)(x+1) &= 0 && \\color{blue}{\\textsf{factor}}\\\\[5pt]  x+\\underset{\\large{\\color{red}{-3}}}{3} = \\underset{\\large{\\color{red}{-3}}}{0}\\ \\ \\textsf{or}\\ \\ x+\\underset{\\large{\\color{red}{-1}}}{1} &= \\underset{\\large{\\color{red}{-1}}}{0} && \\color{blue}{\\textsf{Zero-Product Property}}\\\\[5pt]  x = -3 \\ \\ \\textsf{or}\\ \\ x &= -1 &&  \\end{align}[\/latex]<\/p>\n<p>The restricted values are [latex]-3[\/latex] and [latex]-1[\/latex].<\/p>\n<p>To simplify, start with factoring, and continue by removing all factors equal to [latex]1[\/latex].\u00a0Notice the numerator is a difference of squares.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}  \\dfrac{x^2-9}{x^2+4x+3} &= \\dfrac{(x+3)\\cdot (x-3)}{(x+3)\\cdot (x+1)} && {\\color{blue}{\\textsf{factor}}}\\\\[5pt]  &= \\dfrac{{\\color{red} \\cancel{{\\color{black}{(x+3)}}}}\\cdot (x-3)}{{\\color{red} \\cancel{{\\color{black}{(x+3)}}}}\\cdot (x+1)} && {\\color{blue}{\\textsf{remove all factors equal to $1$}}}\\\\[5pt]  &= \\dfrac{x-3}{x+1} && {\\color{blue}{\\textsf{simplify}}}  \\end{align}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>In the following video, we present additional examples of simplifying and finding the restricted values of the variable in a rational expression.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-1\" title=\"Simplify and Give the Domain of Rational Expressions\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/tJiz5rEktBs?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<h2>Summary<\/h2>\n<p>Rational expressions extend the notion of fractions to quotients of polynomials. An additional consideration for rational expressions is to determine restricted values of the variable. Since division by\u00a0[latex]0[\/latex] is undefined, any values of the variable that result in a denominator of\u00a0[latex]0[\/latex] must be restricted. Restricted values of the variable must be identified in the original expression, not in its simplified form. Rational expressions can be simplified much like fractions. To simplify a rational expression, first determine common factors between the numerator and denominator, then divide them out, removing factors equal to [latex]1[\/latex].<\/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-174\">\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>Screenshot: Breaking Math. <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>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>Simplify and Give the Domain of Rational Expressions. <strong>Authored by<\/strong>: James Sousa (Mathispower4u.com) for Lumen Learning. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/tJiz5rEktBs\">https:\/\/youtu.be\/tJiz5rEktBs<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><\/ul><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Shared previously<\/div><ul class=\"citation-list\"><li>Unit 15: Rational Expressions, 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>College Algebra. <strong>Authored by<\/strong>: Abramson, Jay et al.. <strong>Provided by<\/strong>: OpenStax. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"http:\/\/cnx.org\/contents\/9b08c294-057f-4201-9f48-5d6ad992740d@3.278:1\/Preface\">http:\/\/cnx.org\/contents\/9b08c294-057f-4201-9f48-5d6ad992740d@3.278:1\/Preface<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em>. <strong>License Terms<\/strong>: Download for free at : http:\/\/cnx.org\/contents\/9b08c294-057f-4201-9f48-5d6ad992740d@3.278:1\/Preface<\/li><\/ul><\/div>\n\t\t\t\t\t\t <\/div>\n\t\t\t\t\t <\/div>\n\t\t\t <\/section>","protected":false},"author":395986,"menu_order":1,"template":"","meta":{"_candela_citation":"[{\"type\":\"original\",\"description\":\"Screenshot: Breaking Math\",\"author\":\"\",\"organization\":\"Lumen Learning\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"cc\",\"description\":\"Unit 15: Rational Expressions, 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\":\"original\",\"description\":\"Revision and Adaptation\",\"author\":\"\",\"organization\":\"Lumen Learning\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"original\",\"description\":\"Simplify and Give the Domain of Rational Expressions\",\"author\":\"James Sousa (Mathispower4u.com) for Lumen Learning\",\"organization\":\"\",\"url\":\"https:\/\/youtu.be\/tJiz5rEktBs\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"cc\",\"description\":\"College Algebra\",\"author\":\"Abramson, Jay et al.\",\"organization\":\"OpenStax\",\"url\":\"http:\/\/cnx.org\/contents\/9b08c294-057f-4201-9f48-5d6ad992740d@3.278:1\/Preface\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"Download for free at : http:\/\/cnx.org\/contents\/9b08c294-057f-4201-9f48-5d6ad992740d@3.278:1\/Preface\"}]","CANDELA_OUTCOMES_GUID":"49980ff9-6f02-40d5-9481-1b5d25e00715","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-174","chapter","type-chapter","status-publish","hentry"],"part":171,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-json\/pressbooks\/v2\/chapters\/174","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-json\/wp\/v2\/users\/395986"}],"version-history":[{"count":64,"href":"https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-json\/pressbooks\/v2\/chapters\/174\/revisions"}],"predecessor-version":[{"id":2073,"href":"https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-json\/pressbooks\/v2\/chapters\/174\/revisions\/2073"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-json\/pressbooks\/v2\/parts\/171"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-json\/pressbooks\/v2\/chapters\/174\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-json\/wp\/v2\/media?parent=174"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-json\/pressbooks\/v2\/chapter-type?post=174"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-json\/wp\/v2\/contributor?post=174"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-json\/wp\/v2\/license?post=174"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}