Rational expressions 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 [latex]\dfrac{5}{13}[/latex] or [latex]-3 = \dfrac{-3}{1}[/latex], and polynomials like [latex]x=\dfrac{x}{1}[/latex] or [latex]2x^3-5x+4=\dfrac{2x^3-5x+4}{1}[/latex], are examples of rational expressions. Other easier to recognize examples include [latex]\dfrac{1}{x}[/latex], [latex]\dfrac{x+1}{x-3}[/latex], or [latex]\dfrac{-2x^4 + 5x^2 - x}{x^3 - 7x^2 + 4x - 1}[/latex].

Evaluating Rational Functions

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 [latex]f(x) = \dfrac{-2x^4 + 5x^2 - x}{x^3 - 7x^2 + 4x - 1}[/latex] at [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 [latex]{(\color{Green}-2\color{black})}[/latex] (make sure to put parentheses around the input to properly evaluate) and follow the order of operations to simplify it:

[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]

Restricted Values and the Domain of a Rational Function

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’s focus first on preventing division by zero. The reason you cannot divide any nonzero number $$c$$ 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 [latex]0[/latex], you would get back $$c$$ [latex]\left( ?\cdot 0=c \right)[/latex]. There are no numbers that can do this, so we say “division by zero is undefined.” Consider the following rational expression evaluated at [latex]x = 2[/latex]:

Evaluate  [latex]\dfrac{x}{x-2}[/latex] at [latex]x=2[/latex].

 [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]

This means that for the expression [latex]\dfrac{x}{x-2}[/latex], [latex]x[/latex] cannot be [latex]2[/latex] because it will result in an undefined ratio. We acknowledge that $$2$$ is the only value of the variable $$x$$ that after simplifications results in $$0$$ in the denominator, or as we will usually say, makes the denominator equal to zero. Therefore $$2$$ will be referred to as a restricted (or excluded) value for the expression [latex]\dfrac{x}{x-2}[/latex].

A function defined using a rational expression is called a rational function. 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 domain of the rational function.

We found above that $$2$$ was the only restricted value of the variable in the expression [latex]\dfrac{x}{x-2}[/latex], so the domain of the rational function, [latex]f(x)=\dfrac{x}{x-2}[/latex], consists of all real numbers $$x$$ except $$2$$. The domain, as a set, can be expressed in different ways:

• by simply listing all restrictions indicating that we “start” with all real numbers: all real numbers $$x$$ such that $$x\ne 2$$ (or all real numbers $$x$$ except $$2$$),
• using set-builder notation: [latex]\left\{x\, |\, x\ne 2\right\}[/latex],
• using interval notation: [latex](-\infty, 2)\cup (2,\infty)[/latex].

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{“The\ Denominator”} = 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. (Note that although the denominator cannot be equal to [latex]0[/latex], the numerator can – this is why we only look for restricted values in the denominator of a rational expression.)

Simplifying Rational Expressions

In the simplification of $$\dfrac{-10}{-45}$$ earlier in this section, we reviewed a technique of simplifying fractions by removing a factor equal to $$1$$. After acknowledging that the greatest common factor of $$10$$ and $$45$$ is $$5$$ and that both the numerator and denominator are negative, we presented $$-10$$ as $$(-5) \cdot 2$$ and $$-45$$ as $$(-5) \cdot 9$$. Then, we removed a factor equal to 1 or divided out a common factor between the numerator and denominator (in our case, the common factor of $$-5$$). It is justified by the following mathematical facts:

• the definition of multiplication of fractions: [latex]\dfrac{(-5) \cdot 2}{(-5) \cdot 9} = \dfrac{(-5)}{(-5)} \cdot \dfrac{2}{9}[/latex],
• dividing a non-zero number by itself results in $$1$$: [latex]\dfrac{(-5)}{(-5)}=1[/latex],
• the product of a number and $$1$$ is the number: [latex]1\cdot\dfrac{2}{9}=\dfrac{2}{9}[/latex].

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: [latex]\dfrac{2x^2}{12x}[/latex], [latex]\dfrac{x+1}{(x+1)(x-2)}[/latex], and [latex]\dfrac{2-x}{(x+1)(x-2)}[/latex].

Since $$2x^2 = 2\cdot x\cdot x$$ and $$12x = 2\cdot x\cdot 6$$, the numerator and denominator of [latex]\dfrac{2x^2}{12x}[/latex] have a common factor of $$2x$$. Hence,

[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].

Since $$x+1 = (x+1)\cdot 1$$, the numerator and denominator of [latex]\dfrac{x+1}{(x+1)(x-2)}[/latex] have a common factor of $$x+1$$. Hence,

[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].

In the rational expression [latex]\dfrac{2-x}{(x+1)(x-2)}[/latex], the polynomial factors $$2-x$$ and $$x-2$$ look similar, though if we write the former in descending powers of the variable as $$-x+2$$, we observe that it’s actually the opposites of the latter. Thus, [latex]2-x = -(x-2)=(-1)\cdot (x-2)[/latex], and the numerator and denominator have a common factor of $$x-2$$. Hence,

[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].

Note that [latex]\dfrac{-1}{x+1}[/latex] could also be presented as [latex]-\dfrac{1}{x+1}[/latex].

To simplify a general rational expression, we might have to use factoring to recognize common factors between the numerator and denominator. For example,

[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]

Note that we didn’t perform any further simplifications to the rational expression [latex]\dfrac{2x+1}{x-2}[/latex]. Recall that another way to 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 $$1$$ requires a common factor between the numerator and denominator. Factors are the building blocks of multiplication. In the polynomial $$2x+1$$, we can only perceive the number $$2$$ as a factor of the product $$2\cdot x$$, but not of the entire polynomial $$2x+1$$. According to the terminology associated with polynomials, $$2x$$ and $$1$$ are terms of the polynomial $$2x+1$$. $$2$$ is only a factor of the term $$2x$$. Similarly, in the polynomial $$x-2$$, $$2$$ is just one of the terms, not a factor of the entire polynomial. Therefore $$2$$ is not a common factor between the numerator and denominator of [latex]\dfrac{2x+1}{x-2}[/latex] and cannot be removed. Also, in the polynomial $$2x+1$$, we can only perceive the variable $$x$$ as a factor of the product $$2\cdot x$$, but not of the entire polynomial, and in the polynomial $$x-2$$, $$x$$ is just one of the terms, not a factor of the entire polynomial. Therefore $$x$$ is not a common factor between the numerator and denominator of [latex]\dfrac{2x+1}{x-2}[/latex] and cannot be removed.

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{“The\ Denominator”} = 0[/latex]. To be exact, it’s always the original denominator (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 disappear. For example, the restricted values of the variable in [latex]\dfrac{2x^2-7x-4}{x^2-6x+8}[/latex] are $$2$$ and $$4$$, the solutions of the equation $$x^2-6x+8=0$$. If we used the denominator of the simplified form, $$\dfrac{2x+1}{x-2}$$, we’d lose the restricted value $$4$$ since the only solution of the equation $$x-2=0$$ is $$2$$.

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, $$f(x)=\dfrac{2x^2-7x-4}{x^2-6x+8}$$ with its implied domain, [latex]\left\{x\, |\, x\ne 2\ \textsf{and}\ x\ne 4\right\}[/latex], and $$g(x)=\dfrac{2x+1}{x-2}$$ with its implied domain, [latex]\left\{x\, |\, x\ne 2\right\}[/latex] are not equivalent, while $$h(x)=\dfrac{2x+1}{x-2}$$ with the domain [latex]\left\{x\, |\, x\ne 2\ \textsf{and}\ x\ne 4\right\}[/latex] and $$f(x)$$ with its implied domain are equivalent.

Let’s practice finding the restricted values of the variable and simplifying rational expressions in additional examples.

We will show one last example of simplifying a rational expression. See if you can recognize the special product in the numerator.

In the following video, we present additional examples of simplifying and finding the restricted values of the variable in a rational expression.

Summary

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 [latex]0[/latex] is undefined, any values of the variable that result in a denominator of [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 $$1$$.