0.6: Review  Rational Expressions
 Page ID
 38227
\( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)
\( \newcommand{\vecd}[1]{\overset{\!\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)
\( \newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\)
( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\)
\( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\)
\( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\ #1 \}\)
\( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\)
\( \newcommand{\Span}{\mathrm{span}}\)
\( \newcommand{\id}{\mathrm{id}}\)
\( \newcommand{\Span}{\mathrm{span}}\)
\( \newcommand{\kernel}{\mathrm{null}\,}\)
\( \newcommand{\range}{\mathrm{range}\,}\)
\( \newcommand{\RealPart}{\mathrm{Re}}\)
\( \newcommand{\ImaginaryPart}{\mathrm{Im}}\)
\( \newcommand{\Argument}{\mathrm{Arg}}\)
\( \newcommand{\norm}[1]{\ #1 \}\)
\( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\)
\( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\AA}{\unicode[.8,0]{x212B}}\)
\( \newcommand{\vectorA}[1]{\vec{#1}} % arrow\)
\( \newcommand{\vectorAt}[1]{\vec{\text{#1}}} % arrow\)
\( \newcommand{\vectorB}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)
\( \newcommand{\vectorC}[1]{\textbf{#1}} \)
\( \newcommand{\vectorD}[1]{\overrightarrow{#1}} \)
\( \newcommand{\vectorDt}[1]{\overrightarrow{\text{#1}}} \)
\( \newcommand{\vectE}[1]{\overset{\!\!\rightharpoonup}{\vphantom{a}\smash{\mathbf {#1}}}} \)
\( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)
\( \newcommand{\vecd}[1]{\overset{\!\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)
Rational Expressions
A rational expression is the quotient of two polynomials. Some examples are
\[ \dfrac { 3 } { 6 x  5 }, \qquad \dfrac { x+1 } { 2 }, \qquad \dfrac { 8x^3 } { x^2  6x+9 }, \qquad \text{ and } \dfrac { ( x + 3 ) } { ( x + 1 ) ( x  4 ) } \nonumber \]
The set of real numbers for which an expression is defined is the domain of the expression. Because rational expressions have denominators, and division by zero is undefined, the values of the variable that make the denominator zero are not included in the domain of the expression. These values of the variable are restrictions to the domain.
How to: find the domain of a rational expression
 Set the denominator equal to zero and solve for \(x\).
 The solutions are the restrictions to the domain.
Example \(\PageIndex{1x}\): Simplifying Rational Expressions
Find the restrictions to the domain for the following rational functions
 \(\dfrac{x8}{x11}\)
 \(\dfrac{x^21}{x^2+5x+4}\)
Solution
 Answer: 11. The expression \(\dfrac{x8}{x11}\) is undefined when its denominator is zero. Solving \( x11 = 0 \) produces \( x = 11 \). Therefore the restriction to the domain of \(\dfrac{x8}{x11}\) is \( x = 11 \). Notice that the denominator CAN be zero, so \( x = 8 \) is NOT a restricted value for the variable.
 Answer: 4 and 1. The expression \(\dfrac{x^21}{x^2+5x+4}\) is undefined when its denominator is zero. Solving \( x^2+5x+4 = 0 \) produces \( (x+4)(x+1) = 0 \). Therefore the restricted values for \(x\) are \( 4 \) and \( 1 \). Again it should be pointed out that the numerator is completely ignored when determining the domain of an expression.
Try It \(\PageIndex{1x}\)
Find the restrictions to the domain for the rational function \(\dfrac{x5}{x^236}\)
 Answer

\( 6 \text{ and } 6\)
Simplifying Rational Expressions
The quotient of two polynomial expressions is called a rational expression. We can apply the properties of fractions to rational expressions, such as simplifying the expressions by canceling common factors from the numerator and the denominator. To do this, we first need to factor both the numerator and denominator. Let’s start with the rational expression shown.
\[\dfrac{x^2+8x+16}{x^2+11x+28} \nonumber \]
We can factor the numerator and denominator to rewrite the expression.
\[\dfrac{{(x+4)}^2}{(x+4)(x+7)} \nonumber \]
Then we can simplify that expression by canceling the common factor \((x+4)\) to obtain \(\dfrac{x+4}{x+7} \)
How to: Given a rational expression, simplify it
 Factor the numerator and denominator.
 Cancel any common factors. (This is permitted because any expression divided by itself is equal to \(1\).)
Example \(\PageIndex{2}\): Simplifying Rational Expressions
Simplify \(\dfrac{x^29}{x^2+4x+3}\)
Solution
\[\begin{align*} &\dfrac{(x+3)(x3)}{(x+3)(x+1)} && \text{Factor the numerator and the denominator}\\ &\dfrac{x3}{x+1} && \text{Cancel common factor } (x+3) \end{align*}\]
Can the \(x^2\) term be cancelled in the last example?
No. A factor is an expression that is multiplied by another expression. The \(x^2\) term is not a factor of the numerator or the denominator.Try It \(\PageIndex{2}\)
Simplify \(\dfrac{x6}{x^236}\)
 Answer

\(\dfrac{1}{x+6}\)
Multiplying Rational Expressions
Multiplication of rational expressions works the same way as multiplication of any other fractions. We multiply the numerators to find the numerator of the product, and then multiply the denominators to find the denominator of the product. Before multiplying, it is helpful to factor the numerators and denominators just as we did when simplifying rational expressions. We are often able to simplify the product of rational expressions.
How to: Given two rational expressions, multiply them
 Factor the numerator and denominator.
 Multiply the numerators. (Keep as a product of factors  do not multiply out).
 Multiply the denominators. (Keep as a product of factors  do not multiply out).
 Simplify. (Leave in factored form!)
Example \(\PageIndex{3}\): Multiplying Rational Expressions
Multiply the rational expressions and show the product in simplest form:
\(\dfrac{(x+5)(x1)}{3(x+6)}\times\dfrac{(2x1)}{(x+5)}\)
Solution
\[\begin{align*} &\dfrac{(x+5)(x1)}{3(x+6)}\times\dfrac{(2x1)}{(x+5)} && \text{Factor the numerator and denominator.}\\[4pt] &\dfrac{(x+5)(x1)(2x1)}{3(x+6)(x+5)} && \text{Multiply numerators and denominators}\\[4pt] &\dfrac{(x1)(2x1)}{3(x+6)} && \text{Cancel common factors to simplify} \end{align*}\]
Try It \(\PageIndex{3}\)
Multiply the rational expressions and show the product in simplest form:
\(\dfrac{x^2+11x+30}{x^2+5x+6}\times\dfrac{x^2+7x+12}{x^2+8x+16}\)
 Answer

\(\dfrac{(x+5)(x+6)}{(x+2)(x+4)}\)
Dividing Rational Expressions
Division of rational expressions works the same way as division of other fractions. To divide a rational expression by another rational expression, multiply the first expression by the reciprocal of the second. Using this approach, we would rewrite \(\dfrac{1}{x}÷\dfrac{x^2}{3}\) as the product \(\dfrac{1}{x}⋅\dfrac{3}{x^2}\). Once the division expression has been rewritten as a multiplication expression, we can multiply as we did before.
\[\dfrac{1}{x}⋅\dfrac{3}{x^2}=\dfrac{3}{x^3} \nonumber \]
How to: Given two rational expressions, divide them
 Rewrite as the first rational expression multiplied by the reciprocal of the second.
 Factor the numerators and denominators.
 Multiply the numerators. (Keep as a product of factors  do not multiply out).
 Multiply the denominators. (Keep as a product of factors  do not multiply out).
 Simplify. (Leave in factored form!)
Example \(\PageIndex{4}\): Dividing Rational Expressions
Divide the rational expressions and express the quotient in simplest form:
\(\dfrac{2x^2+x6}{x^21}÷\dfrac{x^24}{x^2+2x+1}\)
Solution
\[\begin{align*} &\dfrac{2x^2+x6}{x^21}÷\dfrac{x^24}{x^2+2x+1} \\[4pt]
&\dfrac{2x^2+x6}{x^21}\times\dfrac{x^2+2x+1}{x^24} && \text{Rewrite as a multiplication problem} \\[4pt]
&\dfrac{(2x3)(x+2)}{(x1)(x+1)}\times\dfrac{(x+1)(x+1)}{(x2)(x+2)} && \text{Factor the numerator and denominator.}\\[6pt]
&\dfrac{(2x3)(x+2)(x+1)(x+1)}{(x1)(x+1)(x2)(x+2)} && \text{Multiply numerators and denominators}\\[6pt]
&\dfrac{(2x3)(x+1)}{(x1)(x2)} && \text{Cancel common factors to simplify} \end{align*}\]
Try It \(\PageIndex{4}\)
Divide the rational expressions and express the quotient in simplest form:
\[\dfrac{9x^216}{3x^2+17x28}÷\dfrac{3x^22x8}{x^2+5x14} \nonumber \]
 Answer

\(0\)
Adding and Subtracting Rational Expressions
Adding and subtracting rational expressions works just like adding and subtracting numerical fractions. To add fractions, we need to find a common denominator. Let’s look at an example of fraction addition.
\[\begin{align*} \dfrac{5}{24}+\dfrac{1}{40} &= \dfrac{5}{24} \cdot \dfrac{5}{5}+\dfrac{1}{40}\cdot \dfrac{3}{3}\\&= \dfrac{25}{120}+\dfrac{3}{120}\\ &= \dfrac{28}{120} \qquad= \dfrac{7}{30} \end{align*}\]
We have to rewrite the fractions so they share a common denominator before we are able to add. We must do the same thing when adding or subtracting rational expressions.
The easiest common denominator to use will be the least common denominator, or LCD. The LCD is the smallest multiple that the denominators have in common. To find the LCD of two rational expressions, we factor the expressions and multiply all of the distinct factors. For instance, if the factored denominators were \((x+3)(x+4)\) and \((x+4)(x+5)\), then the LCD would be \((x+3)(x+4)(x+5)\).
Once we find the LCD, we need to multiply each expression by the form of \(1\) that will change the denominator to the LCD. We would need to multiply the expression with a denominator of \((x+3)(x+4)\) by \(\dfrac{x+5}{x+5}\) and the expression with a denominator of \((x+4)(x+5)\) by \(\dfrac{x+3}{x+3}\).
How to: Given two rational expressions, add or subtract them
 Factor the numerator and denominator.
 Find the LCD of the expressions.
 Multiply the expressions by a form of 1 that changes the denominators to the LCD. (Keep the denominator factored.)
 Add or subtract the numerators. (Multiply out numerators only, combine like terms, then factor the result.)
 Simplify. (Keep in factored form!)
Example \(\PageIndex{5}\): Adding Rational Expressions
Add the rational expressions: \[\dfrac{5}{x}+\dfrac{6}{y} \nonumber \]
Solution
First, we have to find the LCD. In this case, the LCD will be \(xy\). We then multiply each expression by the appropriate form of \(1\) to obtain \(xy\) as the denominator for each fraction.
\[\begin{align*} &\dfrac{5}{x}\times\dfrac{y}{y}+\dfrac{6}{y}\times\dfrac{x}{x}\\ &\dfrac{5y}{xy}+\dfrac{6x}{xy} \end{align*}\]
Now that the expressions have the same denominator, we simply add the numerators to find the sum.
\[\dfrac{6x+5y}{xy} \nonumber \]
Analysis
Multiplying by \(\dfrac{y}{y}\) or \(\dfrac{x}{x}\) does not change the value of the original expression because multiplying an expression by \(1\) does not change the value of the original expression.
Example \(\PageIndex{6}\): Subtracting Rational Expressions
Subtract the rational expressions: \( \dfrac{6}{x^2+4x+4}\dfrac{2}{x^24} \)
Solution
\[\begin{align*}
&\dfrac{6}{{(x+2)}^2}\dfrac{2}{(x+2)(x2)} && \text{Factor}\\
&\dfrac{6}{{(x+2)}^2}\times\dfrac{x2}{x2}\dfrac{2}{(x+2)(x2)}\times\dfrac{x+2}{x+2} && \text{Multiply each fraction to get LCD as denominator}\\
&\dfrac{6(x2)}{{(x+2)}^2(x2)}\dfrac{2(x+2)}{{(x+2)}^2(x2)} && \text{Multiply}\\
&\dfrac{6x12(2x+4)}{{(x+2)}^2(x2)} && \text{Apply distributive property (be careful here!!)}\\
&\dfrac{4x16}{{(x+2)}^2(x2)} && \text{Subtract}\\
&\dfrac{4(x4)}{{(x+2)}^2(x2)} && \text{Simplify}
\end{align*}\]
Do we have to use the LCD to add or subtract rational expressions?
No. Any common denominator will work, but it is easiest to use the LCD.Try It \(\PageIndex{6}\)
Subtract the rational expressions: \(\dfrac{3}{x+5}\dfrac{1}{x3}\)
 Answer

\(\dfrac{2(x7)}{(x+5)(x3)}\)
Simplifying Complex Rational Expressions
A complex rational expression is a rational expression that contains additional rational expressions in the numerator, the denominator, or both. We can simplify complex rational expressions by rewriting the numerator and denominator as single rational expressions and dividing. The complex rational expression \(\dfrac{a}{\dfrac{1}{b}+c}\) can be simplified by rewriting the numerator as the fraction \(\dfrac{a}{1}\) and combining the expressions in the denominator as \(\dfrac{1+bc}{b}\). We can then rewrite the expression as a multiplication problem using the reciprocal of the denominator. We get \(\dfrac{a}{1}⋅\dfrac{b}{1+bc}\), which is equal to \(\dfrac{ab}{1+bc}\).
How to: Simplify a complex rational expression (Method I)
 Combine the expressions in the numerator into a single rational expression by adding or subtracting.
 Combine the expressions in the denominator into a single rational expression by adding or subtracting.
 Rewrite as the numerator divided by the denominator.
 Rewrite as multiplication.
 Multiply.
 Simplify.
Example \(\PageIndex{7}\): Simplifying Complex Rational Expressions
Simplify: \(\dfrac{y+\dfrac{1}{x}}{\dfrac{x}{y}}\)
Solution
Begin by combining the expressions in the numerator into one expression.
\[\begin{align*} &y\times\dfrac{x}{x}+\dfrac{1}{x}\qquad \text{Multiply by } \dfrac{x}{x} \text{ to get LCD as denominator}\\ &\dfrac{xy}{x}+\dfrac{1}{x}\\ &\dfrac{xy+1}{x}\qquad \text{Add numerators} \end{align*}\]
Now the numerator is a single rational expression and the denominator is a single rational expression.
\[\begin{align*} &\dfrac{\dfrac{xy+1}{x}}{\dfrac{x}{y}}\\ \text{We can rewrite this as division, and then multiplication.}\\ &\dfrac{xy+1}{x}÷\dfrac{x}{y}\\ &\dfrac{xy+1}{x}\times\dfrac{y}{x}\qquad \text{Rewrite as multiplication}\\ &\dfrac{y(xy+1)}{x^2}\qquad \text{Multiply} \end{align*}\]
Try It \(\PageIndex{7}\)
Simplify: \(\dfrac{\dfrac{x}{y}\dfrac{y}{x}}{y}\)
 Answer

\(\dfrac{x^2y^2}{xy^2}\)
Can a complex rational expression always be simplified?
Yes. We can always rewrite a complex rational expression as a simplified rational expression.Method 2: Simplify Using the LCD
An alternative method for simplifying complex rational expressions involves clearing the fractions in the numerator and denominator by multiplying the rational expression by a special form of \(1\). In this method, multiply the numerator and denominator by the least common denominator (LCD) of all the fractions present in both the numerator and the denominator.
How to: Simplify a complex rational expression (Method II)
 Determine the LCD of all the fractions present in both the numerator and the denominator.
 Multiply the complex fraction by a form of 1 composed of the fraction \( \dfrac{\text{LCD}}{\text{ LCD }} \). This effectively results in every term in the numerator and every term in the denominator getting multiplied by this LCD.
 Simplify the resulting rational expression.
Example \(\PageIndex{8x}\):
Simplify: \(\dfrac { 4  \dfrac { 12 } { x } + \dfrac { 9 } { x ^ { 2 } } } { 2  \dfrac { 5 } { x } + \dfrac { 3 } { x ^ { 2 } } }\).
Solution
Step 1: Determine the LCD of all the fractions in the numerator and denominator. In this case, the denominators of the given fractions are \(1, x\), and \(x^{2}\). Therefore, the LCD is \(x^{2}\).
Step 2: Multiply the numerator and denominator by the LCD. This step should clear the fractions in both the numerator and denominator.
\(\begin{aligned} \dfrac { 4  \dfrac { 12 } { x } + \dfrac { 9 } { x ^ { 2 } } } { 2  \dfrac { 5 } { x } + \dfrac { 3 } { x ^ { 2 } } } & =
\dfrac { \left( 4  \dfrac { 12 } { x } + \dfrac { 9 } { x ^ { 2 } } \right) \cdot \color{Cerulean}{x ^ { 2} } } { \left( 2  \dfrac { 5 } { x } + \dfrac { 3 } { x ^ { 2 } } \right) \cdot \color{Cerulean}{x ^ { 2} } } \quad\quad\:\:\quad\quad\color{Cerulean}{Multiply\:numerator\:and\:denominator.}
\\
& = \dfrac { 4 \cdot \color{Cerulean}{x ^ { 2} }\color{black}{ } \dfrac { 12 } { x } \cdot \color{Cerulean}{x ^ { 2} }\color{black}{ +} \dfrac { 9 } { x ^ { 2 } } \cdot \color{Cerulean}{x ^ { 2} } } { 2 \cdot \color{Cerulean}{x ^ { 2} }\color{black}{ } \dfrac { 5 } { x } \cdot \color{Cerulean}{x ^ { 2} }\color{black}{ +} \dfrac { 3 } { x ^ { 2 } } \cdot \color{Cerulean}{x ^ { 2} } }\quad\quad\quad\color{Cerulean}{Distribute\:and\:then\:cancel.}
\\
& = \dfrac { 4 x ^ { 2 }  12 x + 9 } { 2 x ^ { 2 }  5 x + 3 } \end{aligned}\)
This leaves us with a single algebraic fraction with a polynomial in the numerator and in the denominator.
Step 3: Factor the numerator and denominator completely.
\(\begin{aligned} & = \dfrac { 4 x ^ { 2 }  12 x + 9 } { 2 x ^ { 2 }  5 x + 3 } \\ & = \dfrac { ( 2 x  3 ) ( 2 x  3 ) } { ( x  1 ) ( 2 x  3 ) } \end{aligned}\)
Step 4: Cancel all common factors.
\(\begin{aligned} & = \dfrac { ( 2 x  3 )\cancel{ ( 2 x  3 )} } { ( x  1 ) \cancel{( 2 x  3 )} } \\ & = \dfrac { 2 x  3 } { x  1 } \end{aligned}\)
It is important to point out that multiplying the numerator and denominator by the same nonzero factor is equivalent to multiplying by 1 and does not change the problem.
Try It \(\PageIndex{8x}\)
Simplify using the LCD: \(\dfrac { \dfrac { 1 } { y ^ { 2 } }  \dfrac { 1 } { x ^ { 2 } } } { \dfrac { 1 } { y } + \dfrac { 1 } { x } }\).
 Answer

\(\dfrac { x  y } { x y }\)