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Study Guides > Intermediate Algebra

Operations on Rational Expressions

10.1 Learning Objectives

  • Introduction to rational expressions
    • Recognize and define a rational expression
    • Determine the domain of a rational expression
    • Simplify a rational expression
  • Multiply and Divide Rational Expressions
  • Add and Subtract Rational Expressions
    • Identify the domain of a sum or difference of rational expressions
    • Identify the least dcommon denominator of two rational expressions
    • Add and subtract rational expressions using a greatest common denominator
  • Complex Rational Expressions
    • Rewrite a complex fraction as a division problem and simplify
    • Rewrite a complex rational expression as a division problem and simplify
Rational expressions are fractions that have a polynomial in the numerator, denominator, or both. Although rational expressions can seem complicated because they contain variables, they can be simplified using the techniques used to simplify expressions such as 4x312x2\frac{4x^3}{12x^2} combined with techniques for factoring polynomials. There are a couple 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.

10.1.1 Determine the domain of a rational expression

One sure way you can break math is to divide by zero. Consider the following rational expression evaluated at x = 2:

Evaluate  xx−2\frac{x}{x-2} for x=2x=2

Substitute x=2x=2

22−2=20\begin{array}{l}\frac{2}{2-2}\\\text{}\\=\frac{2}{0}\end{array}

This means that for the expression xx−2\frac{x}{x-2}, x cannot be 2 because it will result in an undefined ratio. In general, finding values for a variable that will not result in division by zero is called finding the domain. Finding the domain of a rational expression or function will help you not break math.

Domain of a rational expression or equation

The domain of a rational expression or equation is a collection of the values for the variable that will not result in an undefined mathematical operation such as division by zero.  For a = any real number, we can notate the domain in the following way:

x is all real numbers where x≠ax\neq{a}

The reason you cannot divide any number c by zero (c0  =  ?) \left( \frac{c}{0}\,\,=\,\,? \right)\\ is that you would have to find a number that when you multiply it by 0 you would get back c(?  ⋅  0  =  c)c \left( ?\,\,\cdot \,\,0\,\,=\,\,c \right). There are no numbers that can do this, so we say “division by zero is undefined”. In simplifying rational expressions you need to pay attention to what values of the variable(s) in the expression would make the denominator equal zero. These values cannot be included in the domain, so they're called excluded values. Discard them right at the start, before you go any further. (Note that although the denominator cannot be equivalent to 0, the numerator can—this is why you only look for excluded values in the denominator of a rational expression.) For rational expressions, the domain will exclude values for which the value of the denominator is 0. The following example illustrates finding the domain of an expression. Note that this is exactly the same algebra used to find the domain of a function.

Example 10.1.a

Identify the domain of the expression. x+7x2+8x−9 \frac{x+7}{{{x}^{2}}+8x-9}

Answer: Find any values for x that would make the denominator equal to 0 by setting the denominator equal to 0 and solving the equation.

x2+8x−9=0x^{2}+8x-9=0

Solve the equation by factoring. The solutions are the values that are excluded from the domain.

(x+9)(x−1)=0x=−9   or   x=1\begin{array}{c}(x+9)(x-1)=0\\x=-9\,\,\,\text{or}\,\,\,x=1\end{array}

Answer

The domain is all real numbers except −9−9 and 11.

 

10.1.2 Simplify Rational Expressions

Before we dive in to simplifying rational expressions, let's review the difference between a factor,  a term,  and an expression.  This will hopefully help you avoid another way to break math when you are simplifying rational expressions. Factors are the building blocks of multiplication. They are the numbers that you can multiply together to produce another number: 2 and 10 are factors of 20, as are 4, 5, 1, 20. Terms are single numbers, or variables and numbers connected by multiplication. -4, 6x and x2x^2 are all terms. Expressions are groups of terms connected by addition and subtraction.  2x2−52x^2-5 is an expression. This distinction is important when you are required to divide.  Let's use an example to show why this is important. Simplify: 2x212x\large\frac{2x^2}{12x} The numerator and denominator of this fraction consist of factors. To simplify it, we can divide without being impeded by addition or subtraction. 2x212x=2⋅x⋅x2⋅3⋅2⋅x=2⋅x⋅x2⋅3⋅2⋅x\begin{array}{cc}\large\frac{2x^2}{12x}\\=\large\frac{2\cdot{x}\cdot{x}}{2\cdot3\cdot2\cdot{x}}\\=\large\frac{\cancel{2}\cdot{\cancel{x}}\cdot{x}}{\cancel{2}\cdot3\cdot2\cdot{\cancel{x}}}\end{array} We can do this because 22=1 and xx=1\frac{2}{2}=1\text{ and }\frac{x}{x}=1, so our expression simplifies to x6\large\frac{x}{6} Compare that to the expression 2x2+x12−2x\large\frac{2x^2+x}{12-2x}, notice the denominator and numerator consist of two terms connected by addition and subtraction.  We have to tip-toe around the addition and subtraction.  When asked to simplify it is tempting to want to cancel out like terms as we did when we just had factors. But you can't do that, it will break math!
Shattered pottery strewn across the floor. Breaking Math
In the examples that follow, the numerator and the denominator are polynomials with more than one term, and we will show you how to properly simplify them by factoring - which turns expressions connected by addition and subtraction into terms connected by multiplication.

Example 10.1.b

Simplify and state the domain for the expression. x+3x2+12x+27 \frac{x+3}{{{x}^{2}}+12x+27}

Answer: To find the domain (and the excluded values), find the values for which the denominator is equal to 0. Factor the quadratic, and apply the zero product principle.

x+3=0      or      x+9=0x=0−3      or      x=0−9x=−3      or      x=−9x=−3      or      x=−9\begin{array}{c}x+3=0\,\,\,\,\,\,\text{or}\,\,\,\,\,\,x+9=0\\x=0-3\,\,\,\,\,\,\text{or}\,\,\,\,\,\,x=0-9\\x=-3\,\,\,\,\,\,\text{or}\,\,\,\,\,\,x=-9\\\\x=-3\,\,\,\,\,\,\text{or}\,\,\,\,\,\,x=-9\end{array}

The domain is all real numbers except x=−3x=-3 or x=−9x=-9. Factor the numerator and denominator.  Identify the factors that are the same in the numerator and denominator, and simplify.

x+3x2+12x+27=x+3(x+3)(x+9)x+3(x+3)(x+9)=1⋅1x+9\large\begin{array}{c}\frac{x+3}{x^{2}+12x+27}\\\\=\frac{x+3}{\left(x+3\right)\left(x+9\right)}\\\\\frac{\cancel{x+3}}{\cancel{\left(x+3\right)}\left(x+9\right)}\\\\\normalsize=1\cdot\large\frac{1}{x+9}\end{array}

Answer

x+3x2+12x+27=1x+9 \frac{x+3}{{{x}^{2}}+12x+27}=\frac{1}{x+9} The domain is all real numbers except −3−3 and −9−9.

Example 10.1.c

Simplify and state the domain for the expression. x2+10x+24x3−x2−20x\frac{x^{2}+10x+24}{x^{3}-x^{2}-20x}

Answer: To find the domain, determine the values for which the denominator is equal to 0.

x3−x2−20x=0x(x2−x−20)=0x(x−5)(x+4)=0\begin{array}{r}x^{3}-x^{2}-20x=0\\x\left(x^{2}-x-20\right)=0\\x\left(x-5\right)\left(x+4\right)=0\end{array}

The domain is all real numbers except 0, 5, and −4. To simplify, factor the numerator and denominator of the rational expression. Identify the factors that are the same in the numerator and denominator, and simplify.

x2+10x+24x3−x2−20x=(x+4)(x+6)x(x−5)(x+4)=(x+4)(x+6)x(x−5)(x+4) \large\begin{array}{c}\frac{x^{2}+10x+24}{x^{3}-x^{2}-20x}\\\\=\frac{\left(x+4\right)\left(x+6\right)}{x\left(x-5\right)\left(x+4\right)}\\\\=\frac{\cancel{\left(x+4\right)}\left(x+6\right)}{x\left(x-5\right)\cancel{\left(x+4\right)}}\end{array}

Simplify. It is acceptable to either leave the denominator in factored form or to distribute multiplication.

x+6x(x−5)   or   x+6x2−5x\frac{x+6}{x\left(x-5\right)}\,\,\,\text{or}\,\,\,\frac{x+6}{x^{2}-5x}

Answer

x+6x(x−5)[/latex]or[latex]x+6x2−5x \frac{x+6}{x(x-5)}[/latex] or [latex] \frac{x+6}{{{x}^{2}}-5x} The domain is all real numbers except 0, 5, and −4−4.

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

Example 10.1.d

Simplify x2−9x2+4x+3\frac{{x}^{2}-9}{{x}^{2}+4x+3}, state the domain.

Answer: The special product in the numerator is a difference of squares. \begin{array}\frac{\left(x+3\right)\left(x - 3\right)}{\left(x+3\right)\left(x+1\right)}\hfill & \hfill & \hfill & \hfill & \text{Factor the numerator and the denominator}.\hfill \\ \frac{x - 3}{x+1}\hfill & \hfill & \hfill & \hfill & \text{Cancel common factor }\left(x+3\right).\hfill \end{array} With the denominator factored it is easier to find the domain of the expression. Determine the values for which the denominator is equal to 0. (x+3)=0,(x+1)=0x≠−3, AND x≠−1\begin{array}{cc}\left(x+3\right)=0,\left(x+1\right)=0\\x\ne-3,\text{ AND }x\ne-1\end{array}

Answer

x2−9x2+4x+3=x−3x+1[/latex],Domain:[latex]x≠−3, AND x≠−1\frac{{x}^{2}-9}{{x}^{2}+4x+3}=\frac{x - 3}{x+1}[/latex], Domain: [latex]x\ne-3,\text{ AND }x\ne-1

  In the following video we present another example of finding the domain of a rational expression. https://youtu.be/tJiz5rEktBs

Steps for Simplifying a Rational Expression

To simplify a rational expression, follow these steps:
  • Determine the domain. The excluded values are those values for the variable that result in the expression having a denominator of 0.
  • Factor the numerator and denominator.
  • Find common factors for the numerator and denominator and simplify.

10.1.2 Multiply and Divide Rational Expressions

Just as you can multiply and divide fractions, you can multiply and divide rational expressions. In fact, you use the same processes for multiplying and dividing rational expressions as you use for multiplying and dividing numeric fractions. The process is the same even though the expressions look different!
Multiply and Divide Multiply and Divide

Multiply Rational Expressions

Remember that there are two ways to multiply numeric fractions. One way is to multiply the numerators and the denominators and then simplify the product, as shown here.

45⋅98=3640=3⋅3⋅2⋅25⋅2⋅2⋅2=3⋅3⋅2⋅25⋅2⋅2⋅2=3⋅35⋅2⋅1=910 \displaystyle \frac{4}{5}\cdot \frac{9}{8}=\frac{36}{40}=\frac{3\cdot 3\cdot 2\cdot 2}{5\cdot 2\cdot 2\cdot 2}=\frac{3\cdot 3\cdot \cancel{2}\cdot\cancel{2}}{5\cdot \cancel{2}\cdot\cancel{2}\cdot 2}=\frac{3\cdot 3}{5\cdot 2}\cdot 1=\frac{9}{10}

A second way is to factor and simplify the fractions before performing the multiplication.

45⋅98=2⋅25⋅3⋅32⋅2⋅2=2⋅2⋅3⋅32⋅5⋅2⋅2=1⋅3⋅35⋅2=910\frac{4}{5}\cdot\frac{9}{8}=\frac{2\cdot2}{5}\cdot\frac{3\cdot3}{2\cdot2\cdot2}=\frac{\cancel{2}\cdot\cancel{2}\cdot3\cdot3}{\cancel{2}\cdot5\cdot\cancel{2}\cdot2}=1\cdot\frac{3\cdot3}{5\cdot2}=\frac{9}{10}

Notice that both methods result in the same product. In some cases you may find it easier to multiply and then simplify, while in others it may make more sense to simplify fractions before multiplying. The same two approaches can be applied to rational expressions. Our first two examples apply both techniques to one expression. After that we will let you decide which works best for you.

Example 10.1.e

Multiply.5a214⋅710a3 \displaystyle \frac{5{{a}^{2}}}{14}\cdot \frac{7}{10{{a}^{3}}} State the product in simplest form.

Answer: Multiply the numerators, and then multiply the denominators.

5a214⋅710a3=35a2140a3\frac{5a^{2}}{14}\cdot\frac{7}{10a^{3}}=\frac{35a^{2}}{140a^{3}}

Simplify by finding common factors in the numerator and denominator. Simplify the common factors.

35a2140a3=5⋅7⋅a25⋅7⋅2⋅2⋅a2⋅a               =5⋅7⋅a25⋅7⋅2⋅2⋅a2⋅a               =1⋅14a\large\begin{array}{l}\frac{35a^{2}}{140a^{3}}=\frac{5\cdot7\cdot{a}^{2}}{5\cdot7\cdot2\cdot2\cdot{a}^{2}\cdot{a}}\\\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=\frac{\cancel{5}\cdot\cancel{7}\cdot\cancel{{a}^{2}}}{\cancel{5}\cdot\cancel{7}\cdot2\cdot2\cdot\cancel{{a}^{2}}\cdot{a}}\\\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=\normalsize1\cdot\large\frac{1}{4a}\end{array}

Simplify.

14a \displaystyle \frac{1}{4a}

Answer

5a214⋅710a3=14a[/latex][latex] \displaystyle \frac{5{{a}^{2}}}{14}\cdot \frac{7}{10{{a}^{3}}}=\frac{1}{4a}[/latex][latex] \displaystyle

Okay, that worked. But this time let’s simplify first, then multiply. When using this method, it helps to look for the greatest common factor. You can factor out any common factors, but finding the greatest one will take fewer steps.

Example 10.1.f

Multiply.  5a214⋅710a3\frac{5a^{2}}{14}\cdot\frac{7}{10a^{3}} State the product in simplest form.

Answer: Factor the numerators and denominators. Look for the greatest common factors.

5⋅a27⋅2⋅75⋅2⋅a2⋅a \displaystyle \frac{5\cdot {{a}^{2}}}{7\cdot 2}\cdot \frac{7}{5\cdot 2\cdot {{a}^{2}}\cdot a}

Simplify common factors, then multiply.

5⋅a27⋅2⋅75⋅2⋅a2⋅a=5⋅a27⋅2⋅75⋅2⋅a2⋅a=1⋅1⋅12⋅2⋅a=14a\large\begin{array}{c}\frac{5\cdot {{a}^{2}}}{7\cdot 2}\cdot \frac{7}{5\cdot 2\cdot {{a}^{2}}\cdot a}\\\\=\frac{\cancel{5}\cdot\cancel{{a}^{2}}}{\cancel{7}\cdot 2}\cdot \frac{\cancel{7}}{\cancel{5}\cdot 2\cdot\cancel{{a}^{2}}\cdot a}\\\\=\frac{1\cdot1\cdot1}{2\cdot2\cdot{a}}=\frac{1}{4a}\end{array}

Answer

5a214⋅710a3=14a\frac{5a^{2}}{14}\cdot\frac{7}{10a^{3}}=\frac{1}{4a}

Both methods produced the same answer. Also, remember that when working with rational expressions, you should get into the habit of identifying any values for the variables that would result in division by 0. These excluded values must be eliminated from the domain, the set of all possible values of the variable. In the example above, 5a214⋅710a3 \displaystyle \frac{5{{a}^{2}}}{14}\cdot \frac{7}{10{{a}^{3}}}, the domain is all real numbers where a is not equal to 0. When a=0a=0, the denominator of the fraction 710a3\frac{7}{10a^{3}} equals 0, which will make the fraction undefined. Some rational expressions contain quadratic expressions and other multi-term polynomials. To multiply these rational expressions, the best approach is to first factor the polynomials and then look for common factors. (Multiplying the terms before factoring will often create complicated polynomials…and then you will have to factor these polynomials anyway! For this reason, it is easier to factor, simplify, and then multiply.) Just take it step by step, like in the examples below.

Example 10.1.g

Multiply.  a2−a−25a⋅10aa+1  ,      a ≠  −1 ,  0 \displaystyle \frac{{{a}^{2}}-a-2}{5a}\cdot \frac{10a}{a+1}\,\,,\,\,\,\,\,\,a\,\ne \,\,-1\,,\,\,0 State the product in simplest form.

Answer:

Factor the numerators and denominators.

(a−2)(a+1)5⋅a⋅5⋅2⋅a(a+1)\frac{\left(a-2\right)\left(a+1\right)}{5\cdot{a}}\cdot\frac{5\cdot2\cdot{a}}{\left(a+1\right)}

Simplify common factors:

(a−2)(a+1)5⋅a⋅5⋅2⋅a(a+1)=a−21⋅21\large\begin{array}{c}\frac{\left(a-2\right)\cancel{\left(a+1\right)}}{\cancel{5}\cdot{\cancel{a}}}\cdot\frac{\cancel{5}\cdot2\cdot{\cancel{a}}}{\cancel{\left(a+1\right)}}\\\\=\frac{a-2}{1}\cdot\frac{2}{1}\end{array}

Multiply simplified rational expressions. This expression can be left with the numerator in factored form or multiplied out.

(a−2)1⋅21=2(a−2)\begin{array}{c}\frac{\left(a-2\right)}{1}\cdot\frac{2}{1}\\\\=2\left(a-2\right)\end{array}

Answer

a2−a−25a⋅10aa+1=2a−4 \displaystyle \frac{{{a}^{2}}-a-2}{5a}\cdot \frac{10a}{a+1}=2a-4

Example 10.1.h

Multiply.  a2+4a+42a2−a−10⋅a+5a2+2a,   a≠−2,0,52\frac{a^{2}+4a+4}{2a^{2}-a-10}\cdot\frac{a+5}{a^{2}+2a},\,\,\,a\neq-2,0,\frac{5}{2} State the product in simplest form.

Answer: Factor the numerators and denominators.

(a+2)(a+2)(2a−5)(a+2)⋅a+5a(a+2)\frac{\left(a+2\right)\left(a+2\right)}{\left(2a-5\right)\left(a+2\right)}\cdot\frac{a+5}{a\left(a+2\right)}

Simplify common factors.

(a+2)(a+2)(2a−5)(a+2)⋅a+5a(a+2)\large\frac{\cancel{\left(a+2\right)}\cancel{\left(a+2\right)}}{\left(2a-5\right)\cancel{\left(a+2\right)}}\cdot\frac{a+5}{a\cancel{\left(a+2\right)}}

Multiply simplified rational expressions. This expression can be left with the denominator in factored form or multiplied out.

1(2a−5)⋅a+5a=a+5a(2a−5)\frac{1}{\left(2a-5\right)}\cdot\frac{a+5}{a}=\frac{a+5}{a\left(2a-5\right)}

Answer

a2+4a+42a2−a−10⋅a+5a2+2a=a+5a(2a−5)\frac{a^{2}+4a+4}{2a^{2}-a-10}\cdot\frac{a+5}{a^{2}+2a}=\frac{a+5}{a\left(2a-5\right)}

Note that in the answer above, you cannot simplify the rational expression any further. It may be tempting to express the 5’s in the numerator and denominator as the fraction 55\frac{5}{5}, but these 5’s are terms because they are being added or subtracted. Remember that only common factors, not terms, can be regrouped to form factors of 1! In the following video we present another example of multiplying rational expressions. https://www.youtube.com/watch?v=Hj6gF1SNttk&feature=youtu.be

10.1.3 Divide Rational Expressions

You've seen that you multiply rational expressions as you multiply numeric fractions. It should come as no surprise that you also divide rational expressions the same way you divide numeric fractions. Specifically, to divide rational expressions, keep the first rational expression, change the division sign to multiplication, and then take the reciprocal of the second rational expression. Let’s begin by recalling division of numerical fractions.

23÷59=23⋅95=1815=65\frac{2}{3}\div\frac{5}{9}=\frac{2}{3}\cdot\frac{9}{5}=\frac{18}{15}=\frac{6}{5}

Use the same process to divide rational expressions. You can think of division as multiplication by the reciprocal, and then use what you know about multiplication to simplify.
Reciprocal Architecture Reciprocal Architecture
You do still need to think about the domain, specifically the variable values that would make either denominator equal zero. But there's a new consideration this time—because you divide by multiplying by the reciprocal of one of the rational expressions, you also need to find the values that would make the numerator of that expression equal zero. Have a look.
Knowing how to find the domain may seem unimportant here, but it will help you when you learn how to solve rational equations. To divide, multiply by the reciprocal.

Example 10.1.i

State the domain, then divide.  5x29÷15x327\frac{5x^{2}}{9}\div\frac{15x^{3}}{27}

Answer: State the Domain: Find excluded values. 9 and 27 can never equal 0. Because 15x315x^{3} becomes the denominator in the reciprocal of 15x327 \displaystyle \frac{15{{x}^{3}}}{27}, you must find the values of x that would make 15x315x^{3} equal 0.

15x3=0x=0 is an excluded value.\begin{array}{c}15x^{3}=0\\x=0\,\text{is an excluded value}.\end{array}

Divide: State the quotient in simplest form.  Rewrite division as multiplication by the reciprocal.

5x29⋅2715x3\frac{5x^{2}}{9}\cdot\frac{27}{15x^{3}}

Factor the numerators and denominators.

5⋅x⋅x3⋅3⋅3⋅3⋅35⋅3⋅x⋅x⋅x\frac{5\cdot{x}\cdot{x}}{3\cdot3}\cdot\frac{3\cdot3\cdot3}{5\cdot3\cdot{x}\cdot{x}\cdot{x}}

Simplify common factors. Simplify.

5⋅x⋅x3⋅3⋅3⋅3⋅35⋅3⋅x⋅x⋅x=1x\large\begin{array}{c}\frac{\cancel{5}\cdot{\cancel{x}}\cdot{\cancel{x}}}{\cancel{3}\cdot\cancel{3}}\cdot\frac{\cancel{3}\cdot\cancel{3}\cdot\cancel{3}}{\cancel{5}\cdot\cancel{3}\cdot{\cancel{x}}\cdot{\cancel{x}}\cdot{x}}\\\\=\frac{1}{x}\end{array}

Answer

5x29÷15x327=1x,x≠0 \displaystyle \frac{5{{x}^{2}}}{9}\div \frac{15{{x}^{3}}}{27}=\frac{1}{x},x\ne 0

Example 10.1.j

Divide.  3x2x+2÷6x4(x2+5x+6)\frac{3x^{2}}{x+2}\div\frac{6x^{4}}{\left(x^{2}+5x+6\right)} State the quotient in simplest form, and express the domain of the expression.

Answer: Determine the excluded values that make the denominators and the numerator of the divisor equal to 0.

(x+2)=0     x=−2(x2+5x+6)=0     (x+3)(x+2)=0     x=−3    or    −26x4=0     x=0     \begin{array}{r}\left(x+2\right)=0\,\,\,\,\,\\x=-2\\\left({{x}^{2}}+5x+6 \right)=0\,\,\,\,\,\\\left(x+3\right)\left(x+2\right)=0\,\,\,\,\,\\x=-3\,\,\,\,\text{or}\,\,\,\,-2\\6x^{4}=0\,\,\,\,\,\\x=0\,\,\,\,\,\end{array}

Domain is all real numbers except 00, −2−2, and −3−3. Rewrite division as multiplication by the reciprocal.

3x2x+2⋅(x2+5x+6)6x4\frac{3x^{2}}{x+2}\cdot\frac{\left(x^{2}+5x+6\right)}{6x^{4}}

Factor the numerators and denominators.

3⋅x⋅xx+2⋅(x+2)(x+3)2⋅3⋅x⋅x⋅x⋅x\frac{3\cdot{x}\cdot{x}}{x+2}\cdot\frac{\left(x+2\right)\left(x+3\right)}{2\cdot3\cdot{x}\cdot{x}\cdot{x}\cdot{x}}

Simplify common factors

3⋅x⋅xx+2⋅(x+2)(x+3)2⋅3⋅x⋅x⋅x⋅x\large\frac{\cancel{3}\cdot{\cancel{x}}\cdot{\cancel{x}}}{\cancel{x+2}}\cdot\frac{\cancel{\left(x+2\right)}\left(x+3\right)}{2\cdot\cancel{3}\cdot{\cancel{x}}\cdot{\cancel{x}}\cdot{x}\cdot{x}}

Simplify.

(x+3)2x2\frac{(x+3)}{2{{x}^{2}}}

Answer

3x2x+2÷6x4(x2+5x+6)=x+32x2 \displaystyle \frac{3{{x}^{2}}}{x+2}\div \frac{6{{x}^{4}}}{({{x}^{2}}+5x+6)}=\frac{x+3}{2{{x}^{2}}}. The domain is all real numbers except 0, −2−2, and −3−3.

Notice that once you rewrite the division as multiplication by a reciprocal, you follow the same process you used to multiply rational expressions. In the video that follows, we present another example of dividing rational expressions. https://www.youtube.com/watch?v=B1tigfgs268&feature=youtu.be

10.1.4 Add and Subtract Rational Expressions

In beginning math, students usually learn how to add and subtract whole numbers before they are taught multiplication and division. However, with fractions and rational expressions, multiplication and division are sometimes taught first because these operations are easier to perform than addition and subtraction. Addition and subtraction of rational expressions are not as easy to perform as multiplication because, as with numeric fractions, the process involves finding common denominators.
Add and Subtract Add and Subtract

10.1.5 Adding Rational Expressions

To find the LCD of two rational expressions, we factor the expressions and multiply all of the distinct factors. For instance, given the rational expressions

6(x+3)(x+4), and 9x(x+4)(x+5)\large\frac{6}{\left(x+3\right)\left(x+4\right)},\text{ and }\frac{9x}{\left(x+4\right)\left(x+5\right)}

The LCD would be (x+3)(x+4)(x+5)\left(x+3\right)\left(x+4\right)\left(x+5\right).

To find the LCD, we count the greatest number of times a factor appears  in each denominator, and make sure it is represented in the LCD that many times. For example, in 6(x+3)(x+4)\large\frac{6}{\left(x+3\right)\left(x+4\right)}, (x+3)\left(x+3\right) is represented once and  (x+4)\left(x+4\right) is represented once, so they both appear exactly once in the LCD. In 9x(x+4)(x+5)\large\frac{9x}{\left(x+4\right)\left(x+5\right)}, (x+4)\left(x+4\right) appears once, and (x+5)\left(x+5\right) appears once. We have already accounted for (x+4)\left(x+4\right), so the LCD just needs one factor of (x+5)\left(x+5\right) to be complete. Once we find the LCD, we need to multiply each expression by the form of 1 that will change the denominator to the LCD. What do we mean by " the form of 1"? x+5x+5=1\frac{x+5}{x+5}=1 so multiplying an expression by it will not change it's value. For example, we would need to multiply the expression 6(x+3)(x+4)\large\frac{6}{\left(x+3\right)\left(x+4\right)} by x+5x+5\frac{x+5}{x+5} and the expression 9x(x+4)(x+5)\frac{9x}{\left(x+4\right)\left(x+5\right)} by x+3x+3\frac{x+3}{x+3}. Hopefully this process will become clear after you practice it yourself.  As you look through the examples on this page, try to identify the LCD before you look at the answers. Also, try figuring out which "form of 1" you will need to multiply each expression by so that it has the LCD.

Example 10.1.k

Add the rational expressions: 5x+6y\frac{5}{x}+\frac{6}{y}, and define the domain. State the sum in simplest form.

Answer: First, let's define the domain of each term. Since we have x and y in the denominators, we can say x≠0, and y≠0x\ne0 ,\text{ and }y\ne0. Now we have to find the LCD. Since x appears once and y appears once,  the LCD will be xyxy.  We then multiply each expression by the appropriate form of 1 to obtain xyxy as the denominator for each fraction.

5x⋅yy+6y⋅xx5yxy+6xxy\begin{array}{l}\frac{5}{x}\cdot \frac{y}{y}+\frac{6}{y}\cdot \frac{x}{x}\\ \frac{5y}{xy}+\frac{6x}{xy}\end{array}
Now that the expressions have the same denominator, we simply add the numerators to find the sum.
6x+5yxy\frac{6x+5y}{xy}

The domain is x≠0 and y≠0x\ne0 \text{ and }y\ne0

Answer

5x+6y=6x+5yxy, x≠0 and y≠0 \displaystyle \frac{5}{x}+\frac{6}{y}=\frac{6x+5y}{xy}, \text{ }x\ne0 \text{ and }y\ne0

Analysis of the Solution

Multiplying by yy\frac{y}{y} or xx\frac{x}{x} does not change the value of the original expression because any number divided by itself is 1, and multiplying an expression by 1 gives the original expression. Here is one more example of adding rational expressions, but in this case, the expressions have denominators with multi-term polynomials. First, we will factor, then find the LCD. Note that x2−4x^2-4 is a difference of squares and can be factored using special products.

Example 10.1.L

Simplify2x2x2−4+xx−2\frac{2{{x}^{2}}}{{{x}^{2}}-4}+\frac{x}{x-2}, and give the domain. State the result in simplest form.

Answer: Find the least common multiple by factoring each denominator. Multiply each factor the maximum number of times it appears in a single factorization. Remember that x cannot be 22 or −2-2 because the denominators would be 0. (x+2)\left(x+2\right) appears a maximum of one time, as does (x–2)\left(x–2\right). This means the LCM is (x+2)(x–2)\left(x+2\right)\left(x–2\right).

x2−4=(x+2)(x−2)  x−2=x−2  x+2=x+2  LCM=(x+2)(x−2)\begin{array}{l}x^{2}-4=\left(x+2\right)\left(x-2\right)\\\,\,x-2=x-2\\\,\,x+2=x+2\\\,\,\text{LCM}=\left(x+2\right)\left(x-2\right)\end{array}

The LCM becomes the common denominator. Multiply each expression by the equivalent of 1 that will give it the common denominator.

2x2x2−4=2x2(x+2)(x−2)xx−2⋅x+2x+2=x(x+2)(x+2)(x−2)\begin{array}{r}\frac{2{{x}^{2}}}{{{x}^{2}}-4}=\frac{2{{x}^{2}}}{(x+2)(x-2)}\\\frac{x}{x-2}\cdot \frac{x+2}{x+2}=\frac{x(x+2)}{(x+2)(x-2)}\end{array}

Rewrite the original problem with the common denominator. It makes sense to keep the denominator in factored form in order to check for common factors.

2x2(x+2)(x−2)+x(x+2)(x+2)(x−2) \displaystyle \frac{2{{x}^{2}}}{(x+2)(x-2)}+\frac{x(x+2)}{(x+2)(x-2)}

Combine the numerators.

2x2+x(x+2)(x+2)(x−2)2x2+x2+2x(x+2)(x−2) \begin{array}{c}\frac{2{{x}^{2}}+x(x+2)}{(x+2)(x-2)}\\\\\frac{2{{x}^{2}}+{{x}^{2}}+2x}{(x+2)(x-2)}\end{array}

Check for simplest form. Since neither (x+2)\left(x+2\right) nor (x−2)\left(x-2\right) is a factor of 3x2+2x3{{x}^{2}}+2x, this expression is in simplest form.

3x2+2x(x+2)(x−2)\frac{3{{x}^{2}}+2x}{(x+2)(x-2)}

Answer

2x2x2−4+xx−2=3x2+2x(x+2)(x−2), x≠2,−2 \displaystyle \frac{2{{x}^{2}}}{{{x}^{2}}-4}+\frac{x}{x-2}=\frac{3{{x}^{2}}+2x}{(x+2)(x-2)}, \text{ } \displaystyle x\ne 2,-2

In the video that follows, we present an example of adding two rational expression whose denominators are binomials with no common factors. https://www.youtube.com/watch?v=CKGpiTE5vIg&feature=youtu.be

10.1.6 Subtracting Rational Expressions

To subtract rational expressions, follow the same process you use to add rational expressions. You will need to be careful with signs, though.

Example 10.1.M

Subtract2t+1−t−2t2−t−2\frac{2}{t+1}-\frac{t-2}{{{t}^{2}}-t-2}, define the domain. State the difference in simplest form.

Answer: Find the LCD of each expression. t+1t+1 cannot be factored any further, but t2−t−2{{t}^{2}}-t-2 can be. Note that t cannot be −1-1 or 22 because the denominators would be 0.

t+1=t+1t2−t−2=(t−2)(t+1)\begin{array}{c}t+1=t+1\\t^{2}-t-2=\left(t-2\right)\left(t+1\right)\end{array}

Find the least common multiple. t+1t+1 appears exactly once in both of the expressions, so it will appear once in the least common denominator. t–2t–2 also appears once. This means that (t−2)(t+1)\left(t-2\right)\left(t+1\right) is the least common multiple. In this case, it is easier to leave the common multiple in terms of the factors, so you will not multiply it out. Use the least common multiple for your new common denominator, it will be the LCD.

t+1=t+1t2−t−2=(t−2)(t+1)LCM:(t+1)(t−1)\begin{array}{c}t+1=t+1\\t^{2}-t-2=\left(t-2\right)\left(t+1\right)\\\text{LCM}:\left(t+1\right)\left(t-1\right)\end{array}

Compare each original denominator and the new common denominator. Now rewrite the rational expressions to each have the common denominator of (t+1)(t–2)\left(t+1\right)\left(t–2\right). You need to multiply t+1t+1 by t–2t–2 to get the LCD, so multiply the entire rational expression by t−2t−2 \displaystyle \frac{t-2}{t-2}. The second expression already has a denominator of (t+1)(t–2)\left(t+1\right)\left(t–2\right), so you do not need to multiply it by anything.

2t+1⋅t−2t−2=2(t−2)(t+1)(t−2)   t−2t2−t−2=t−2(t+1)(t−2) \begin{array}{c}\frac{2}{t+1}\cdot \frac{t-2}{t-2}=\frac{2(t-2)}{(t+1)(t-2)}\\\\\,\,\,\frac{t-2}{{{t}^{2}}-t-2}=\frac{t-2}{(t+1)(t-2)}\end{array}

Then rewrite the subtraction problem with the common denominator.

2(t−2)(t+1)(t−2)−t−2(t+1)(t−2) \frac{2\left(t-2\right)}{\left(t+1\right)\left(t-2\right)}-\frac{t-2}{\left(t+1\right)\left(t-2\right)}

Subtract the numerators and simplify. Remember that parentheses need to be included around the second (t–2)\left(t–2\right) in the numerator because the whole quantity is subtracted. Otherwise you would be subtracting just the t.

2(t−2)−(t−2)(t+1)(t−2)2t−4−t+2(t+1)(t−2)t−2(t+1)(t−2) \begin{array}{c}\frac{2(t-2)-(t-2)}{(t+1)(t-2)}\\\\\frac{2t-4-t+2}{(t+1)(t-2)}\\\\\frac{t-2}{(t+1)(t-2)}\end{array}

The numerator and denominator have a common factor of t–2t–2, so the rational expression can be simplified.

t−2(t+1)(t−2)=1t+1\large\begin{array}{c}\frac{\cancel{t-2}}{(t+1)\cancel{(t-2)}}\\\\=\frac{1}{t+1}\end{array}

Answer

2t+1−t−2t2−t−2=1t+1, t≠−1,2 \displaystyle \frac{2}{t+1}-\frac{t-2}{{{t}^{2}}-t-2}=\frac{1}{t+1}, \text{ }t\ne -1,2

In the next example, we will give less instruction.  See if you can find the LCD yourself before you look at the answer.

Example 10.1.N

Subtract the rational expressions: 6x2+4x+4−2x2−4\frac{6}{{x}^{2}+4x+4}-\frac{2}{{x}^{2}-4}, and define the domain. State the difference in simplest form.

Answer: Note that the denominator of the first expression is a perfect square trinomial, and the denominator of the second expression is a difference of squares so they can be factored using special products. 6(x+2)2−2(x+2)(x−2)Factor.6(x+2)2⋅x−2x−2−2(x+2)(x−2)⋅x+2x+2Multiply each fraction to get LCD as denominator.6(x−2)(x+2)2(x−2)−2(x+2)(x+2)2(x−2)Multiply.6x−12−(2x+4)(x+2)2(x−2)Apply distributive property.4x−16(x+2)2(x−2)Subtract.4(x−4)(x+2)2(x−2)Simplify.\begin{array}{cc}\frac{6}{{\left(x+2\right)}^{2}}-\frac{2}{\left(x+2\right)\left(x - 2\right)}\hfill & \text{Factor}.\hfill \\ \frac{6}{{\left(x+2\right)}^{2}}\cdot \frac{x - 2}{x - 2}-\frac{2}{\left(x+2\right)\left(x - 2\right)}\cdot \frac{x+2}{x+2}\hfill & \text{Multiply each fraction to get LCD as denominator}.\hfill \\ \frac{6\left(x - 2\right)}{{\left(x+2\right)}^{2}\left(x - 2\right)}-\frac{2\left(x+2\right)}{{\left(x+2\right)}^{2}\left(x - 2\right)}\hfill & \text{Multiply}.\hfill \\ \frac{6x - 12-\left(2x+4\right)}{{\left(x+2\right)}^{2}\left(x - 2\right)}\hfill & \text{Apply distributive property}.\hfill \\ \frac{4x - 16}{{\left(x+2\right)}^{2}\left(x - 2\right)}\hfill & \text{Subtract}.\hfill \\ \frac{4\left(x - 4\right)}{{\left(x+2\right)}^{2}\left(x - 2\right)}\hfill & \text{Simplify}.\hfill \end{array}

The domain is x≠−6x\ne-6

Answer

6(x+2)2−2(x+2)(x−2)=4(x−4)(x+2)2(x−2), x≠−6 \displaystyle \frac{6}{{\left(x+2\right)}^{2}}-\frac{2}{\left(x+2\right)\left(x - 2\right)}=\frac{4\left(x-4\right)}{{\left(x+2\right)}^{2}\left(x-2\right)}, \text{ }x\ne-6

Analysis of the solution

In the last example, the LCD was  (x+2)2(x−2)\left(x+2\right)^2\left(x-2\right).  The reason we need to include (x+2)\left(x+2\right) two times is because it appears two times in the expression 6x2+4x+4\frac{6}{{x}^{2}+4x+4}. The video that follows contains an example of subtracting rational expressions. https://www.youtube.com/watch?v=MMlNtCrkakI&feature=youtu.be

10.1.7 Complex Rational Expressions

Fractions and rational expressions can be interpreted as quotients. When both the dividend (numerator) and divisor (denominator) include fractions or rational expressions, you have something more complex than usual. Don’t fear—you have all the tools you need to simplify these quotients! A complex fraction is the quotient of two fractions. These complex fractions are never considered to be in simplest form, but they can always be simplified using division of fractions. Remember, to divide fractions, you multiply by the reciprocal. Before you multiply the numbers, it’s often helpful to factor the numbers. You can then use the factors to create a fraction equal to 1.

Example 10.1.o

Simplify.

 1235  67 \displaystyle\large\frac{\,\frac{12}{35}\,}{\,\frac{6}{7}\,}

Answer: Rewrite the complex fraction as a division problem.

 1235  67 =1235÷67\displaystyle\large \frac{\,\frac{12}{35}\,}{\,\frac{6}{7}\,}=\normalsize\frac{12}{35}\div \frac{6}{7}

Rewrite the division as multiplication, using the reciprocal of the divisor.

=1235⋅76 =\frac{12}{35}\cdot \frac{7}{6}

Factor the numerator and denominator, looking for common factors, before multiplying numbers together.

=2⋅6⋅75⋅7⋅6=25⋅6⋅76⋅7=25⋅1 \begin{array}{l}=\frac{2\cdot 6\cdot 7}{5\cdot 7\cdot 6}\\\\=\frac{2}{5}\cdot \frac{6\cdot 7}{6\cdot 7}\\\\=\frac{2}{5}\cdot 1\end{array}

Answer

 1235  67 =25\displaystyle\Large \frac{\,\frac{12}{35}\,}{\,\frac{6}{7}\,}=\normalsize\frac{2}{5}

If two fractions appear in the numerator or denominator (or both), first combine them. Then simplify the quotient as shown above.

Example 10.1.P

Simplify.

 34+12  45−110 \displaystyle\Large \frac{\,\frac{3}{4}+\frac{1}{2}\,}{\,\frac{4}{5}-\frac{1}{10}\,}

Answer: First combine the numerator and denominator by adding or subtracting, you may need to find a common denominator first. Note that we do not show the steps for finding a common denominator - so please review that if you are confused.

 34+12  45−110 = 54  710 \displaystyle\Large \frac{\,\frac{3}{4}+\frac{1}{2}\,}{\,\frac{4}{5}-\frac{1}{10}\,}=\frac{\,\frac{5}{4}\,}{\,\frac{7}{10}\,}

Rewrite the complex fraction as a division problem.

  54    710  =54÷710\displaystyle\Large \frac{\,\,\frac{5}{4}\,\,}{\,\,\frac{7}{10}\,\,}=\normalsize\frac{5}{4}\div \frac{7}{10}

Rewrite the division as multiplication, using the reciprocal of the divisor.

=54⋅107 =\Large\frac{5}{4}\cdot \frac{10}{7}

Multiply and simplify as needed.

54⋅107=5⋅5⋅22⋅2⋅7=2514\Large\frac{5}{4}\cdot \frac{10}{7}=\frac{5\cdot5\cdot2}{2\cdot2\cdot7}=\frac{25}{14}

Answer

 34+12  45−110 =2514\displaystyle\Large \frac{\,\frac{3}{4}+\frac{1}{2}\,}{\,\frac{4}{5}-\frac{1}{10}\,}=\normalsize\frac{25}{14}

In the following video we will show a couple more examples of how to simplify complex fractions. https://youtu.be/lQCwze2w7OU

10.1.8 Complex Rational Expressions

A complex rational expression is a quotient with rational expressions in the dividend, divisor, or in both. Simplify these in the exact same way as you would a complex fraction.

Example 10.1.q

Simplify.

  x+5x2−16   x2−  25x−4 \displaystyle\Large \frac{\,\,\frac{x+5}{{{x}^{2}}-16}\,}{\,\,\frac{{{x}^{2}}-\,\,25}{x-4}\,}

Answer: Rewrite the complex rational expression as a division problem.

=x+5x2−16÷x2−25x−4 =\frac{x+5}{{{x}^{2}}-16}\div \frac{{{x}^{2}}-25}{x-4}

Rewrite the division as multiplication, using the reciprocal of the divisor. Note that the excluded values for this are −4-4, 44 and 55, because those values make the denominators of one of the fractions zero.

=x+5x2−16⋅x−4x2−25 =\frac{x+5}{{{x}^{2}}-16}\cdot \frac{x-4}{{{x}^{2}}-25}

Factor the numerator and denominator, looking for common factors. In this case, x+5x+5 and x–4x–4 are common factors of the numerator and denominator. Notice that (x+5)(x−4)(x+5)(x−4) \frac{(x+5)(x-4)}{(x+5)(x-4)} is equal to 1.

=(x+5)(x−4)(x+4)(x−4)(x+5)(x−5)=(x+5)(x−4)(x+5)(x−4)⋅1(x+4)(x−5) \begin{array}{l}=\frac{(x+5)(x-4)}{(x+4)(x-4)(x+5)(x-5)}\\\\=\frac{(x+5)(x-4)}{(x+5)(x-4)}\cdot \frac{1}{(x+4)(x-5)}\end{array}

Answer

  x+5x2−16   x2−25x−4 =1(x+4)(x−5),x≠−4,4,5\displaystyle\Large \frac{\,\,\frac{x+5}{{{x}^{2}}-16}\,}{\,\,\frac{{{x}^{2}}-25}{x-4}\,}\normalsize=\frac{1}{(x+4)(x-5)},x\ne -4,4,5

In the next video example we will show that simplifying a complex fraction may require factoring first. https://youtu.be/fAaqo8gGW9Y The same ideas can be used when simplifying complex rational expressions that include more than one rational expression in the numerator or denominator. However, there is a shortcut that can be used. Compare these two examples of simplifying a complex fraction.

Example 10.1.r

Simplify.

  1−9x2    1+5x+6x2   \displaystyle\Large\frac{\,\,\normalsize1-\Large\frac{9}{{{x}^{2}}}\,\,}{\,\,\normalsize1+\Large\frac{5}{x}\normalsize+\Large\frac{6}{{{x}^{2}}}\,\,}

Answer: Combine the expressions in the numerator and denominator. To do this, rewrite the expressions using a common denominator. There is an excluded value of 0 because this makes the denominators of the fractions zero.

=x2x2−9x2x2x2+5xx2+6x2=x2−9x2x2+5x+6x2 \begin{array}{l}=\frac{\frac{{{x}^{2}}}{{{x}^{2}}}-\frac{9}{{{x}^{2}}}}{\frac{{{x}^{2}}}{{{x}^{2}}}+\frac{5x}{{{x}^{2}}}+\frac{6}{{{x}^{2}}}}\\\\=\frac{\frac{{{x}^{2}}-9}{{{x}^{2}}}}{\frac{{{x}^{2}}+5x+6}{{{x}^{2}}}}\end{array}

Rewrite the complex rational expression as a division problem. (When you are comfortable with the step of rewriting the complex rational fraction as a division problem, you might skip this step and go straight to rewriting it as multiplication.)

=x2−9x2÷x2+5x+6x2 =\frac{{{x}^{2}}-9}{{{x}^{2}}}\div \frac{{{x}^{2}}+5x+6}{{{x}^{2}}}

Rewrite the division as multiplication, using the reciprocal of the divisor.

=x2−9x2⋅x2x2+5x+6 =\frac{{{x}^{2}}-9}{{{x}^{2}}}\cdot \frac{{{x}^{2}}}{{{x}^{2}}+5x+6}

Factor the numerator and denominator, looking for common factors. In this case, x+3x+3 and x2x^{2} are common factors. We can now see there are two additional excluded values, −2-2 and −3-3.

=(x+3)(x−3)x2x2(x+3)(x+2)=(x−3)(x+2)⋅x2(x+3)x2(x+3)\begin{array}{l}=\frac{(x+3)(x-3){{x}^{2}}}{{{x}^{2}}(x+3)(x+2)}\\\\=\frac{(x-3)}{(x+2)}\cdot \frac{{{x}^{2}}(x+3)}{{{x}^{2}}(x+3)}\end{array}

Answer

1−9x21+5x+6x2=x−3x+2,x≠−3,−2,0 \frac{1-\frac{9}{{{x}^{2}}}}{1+\frac{5}{x}+\frac{6}{{{x}^{2}}}}=\frac{x-3}{x+2},x\ne -3,-2,0

Example 10.1.s

Simplify.

1−9x21+5x+6x2 \frac{1-\frac{9}{{{x}^{2}}}}{1+\frac{5}{x}+\frac{6}{{{x}^{2}}}}

Answer: Before combining the expressions, find a common denominator for all of the rational expressions. (In this case, x2x^{2} is a common denominator.) Multiply by 1 in the form of a fraction with the common denominator in both numerator and denominator. (In this case, multiply by x2x2 \frac{{{x}^{2}}}{{{x}^{2}}}.) There is an excluded value of 0 because this makes the denominators of the fractions zero.

=1−9x21+5x+6x2⋅x2x2=(1−9x2)x2(1+5x+6x2)x2=x2−9x2+5x+6 \begin{array}{l}=\frac{1-\frac{9}{{{x}^{2}}}}{1+\frac{5}{x}+\frac{6}{{{x}^{2}}}}\cdot \frac{{{x}^{2}}}{{{x}^{2}}}\\\\=\frac{\left( 1-\frac{9}{{{x}^{2}}} \right){{x}^{2}}}{\left( 1+\frac{5}{x}+\frac{6}{{{x}^{2}}} \right){{x}^{2}}}\\\\=\frac{{{x}^{2}}-9}{{{x}^{2}}+5x+6}\end{array}

Notice that the expression is no longer complex! You can simplify by factoring and identifying common factors. We can now see there are two additional excluded values, −2-2 and −3-3.

=(x+3)(x−3)(x+3)(x+2)=x+3x+3⋅x−3x+2=1⋅x−3x+2 \begin{array}{l}=\frac{(x+3)(x-3)}{(x+3)(x+2)}\\\\=\frac{x+3}{x+3}\cdot \frac{x-3}{x+2}\\\\=1\cdot \frac{x-3}{x+2}\end{array}

Answer

1−9x21+5x+6x2=x−3x+2,x≠−3,−2,0 \frac{1-\frac{9}{{{x}^{2}}}}{1+\frac{5}{x}+\frac{6}{{{x}^{2}}}}=\frac{x-3}{x+2},x\ne -3,-2,0

You may find the second method easier to use, but do try both ways to see what you prefer. In our last example, we show a similar example as the one above. https://youtu.be/P5dfmX_FNPk

Summary

An additional consideration for rational expressions is to determine what values are excluded from the domain. Since division by 0 is undefined, any values of the variables that result in a denominator of 0 must be excluded. Excluded values must be identified in the original equation, not from its factored form.Rational expressions are fractions containing polynomials. They can be simplified much like numeric fractions. To simplify a rational expression, first determine common factors of the numerator and denominator, and then remove them by rewriting them as expressions equal to 1. Rational expressions are multiplied and divided the same way as numeric fractions. To multiply, first find the greatest common factors of the numerator and denominator. Next, regroup the factors to make fractions equivalent to one. Then, multiply any remaining factors. To divide, first rewrite the division as multiplication by the reciprocal of the denominator. The steps are then the same as for multiplication. When expressing a product or quotient, it is important to state the excluded values. These are all values of a variable that would make a denominator equal zero at any step in the calculations. Complex rational expressions are quotients with rational expressions in the divisor, dividend, or both. When written in fractional form, they appear to be fractions within a fraction. These can be simplified by first treating the quotient as a division problem. Then you can rewrite the division as multiplication using the reciprocal of the divisor. Or you can simplify the complex rational expression by multiplying both the numerator and denominator by a denominator common to all rational expressions within the complex expression. This can help simplify the complex expression even faster.

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