4.4 Simplifying Complex Rational Expressions

Learning Outcomes

Simplify 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. Do not fear – you have all the tools to simplify these quotients!

A complex fraction is a fraction in which the numerator and/or denominator include a fraction, e.g.,

[latex]\displaystyle\dfrac{\,\dfrac{12}{35}\,}{\,\dfrac{6}{7}\,}[/latex].

These complex fractions are never considered to be in the simplest form, but they can always be simplified using different methods. For the simplicity of explanation, we will refer to the numerator of the complex fraction as the “big numerator” and to the denominator of the complex fraction as the “big denominator”. More complicated complex fractions might involve several fractions in the “big numerator” or “big denominator” (or both). We also introduce the notion of “small fractions”:

[latex]\require{color}\dfrac{\quad \fcolorbox{Green}{SpringGreen}{$\dfrac{3}{4} + \dfrac{1}{2}$}\quad }{\quad \fcolorbox{Blue}{SkyBlue}{$\dfrac{4}{5} - \dfrac{1}{10}$}\quad }\begin{matrix} {\color{Green}\textsf{“big numerator”}} \\ \\ {\color{Blue}\textsf{“big denominator”}} \end{matrix}\hspace{1in} \dfrac{\quad \fcolorbox{BrickRed}{Lavender}{$\dfrac{3}{4}$} + \fcolorbox{BrickRed}{Lavender}{$\dfrac{1}{2}$}\quad }{\quad \fcolorbox{BrickRed}{Lavender}{$\dfrac{4}{5}$} - \fcolorbox{BrickRed}{Lavender}{$\dfrac{1}{10}$}\quad }\ {\color{BrickRed}\textsf{“small fractions”}}[/latex]

We will describe two methods for simplifying complex fractions. The first method relies on the fact that to divide means to multiply by the reciprocal. Once you present the “big numerator” and “big denominator”, respectively, as simplified fractions (the “big numerator” should be just a single fraction and “big denominator” should be just a single fraction), rewrite the complex fraction as division and then rewrite the division as multiplication by the reciprocal.

Before you multiply the numbers, it is often helpful to factor the fractions. You can then divide out common factors.

Example

Simplify.

[latex]\displaystyle\dfrac{\,\dfrac{12}{35}\,}{\,\dfrac{6}{7}\,}[/latex]

Show Solution

The second method relies on “eliminating” all “small fractions” to reduce the complex fraction to a single fraction. You first find the LCM of the denominators of all “small fractions” and multiply the complex fraction by the form of [latex]1[/latex] consisting of that LCM in the numerator and denominator. In other words, you multiply that LCM into both, the “big numerator” and “big denominator”. Then, focusing separately on the “big numerator” and “big denominator”, you simplify each to obtain a single fraction that you simplify at the end if possible. Presenting all denominators of “small fractions” and the multiplied-in LCM in factored forms may expediate the simplification process.

ExAMPLE

Simplify.

[latex]\displaystyle\dfrac{\,\dfrac{12}{35}\,}{\,\dfrac{6}{7}\,}[/latex]

Show Solution

The second method is usually faster but we will practice both and let you decide your preference.

When using the first method for more complicated complex fractions, start with combining the fractions in the “big numerator” and “big denominator” accordingly.

Example

Simplify using both methods.

[latex]\displaystyle\frac{\,\dfrac{3}{4}+\dfrac{1}{2}\,}{\,\dfrac{4}{5}-\dfrac{1}{10}\,}[/latex]

Show Solution

For the first method, we start with combining the “big numerator” and “big denominator” to single fractions accordingly. For that purpose, we acknowledge that [latex]\operatorname{LCM}(2,4) = 4[/latex] and [latex]\operatorname{LCM}(5,10) = 10[/latex]. Now,

[latex]\begin{align} \frac{\,\dfrac{3}{4}+\dfrac{1}{2}\,}{\,\dfrac{4}{5}-\dfrac{1}{10}\,} &= \frac{\,\dfrac{3}{4}+\dfrac{1{\color{red}{} \cdot 2}}{2{\color{red}{} \cdot 2}}\,}{\,\dfrac{4{\color{red}{} \cdot 2}}{5{\color{red}{} \cdot 2}}-\dfrac{1}{10}\,}\\[5pt] &= \frac{\,\dfrac{3}{4}+\dfrac{2}{4}\,}{\,\dfrac{8}{10}-\dfrac{1}{10}\,}\\[5pt] &= \frac{\,\dfrac{5}{4}\,}{\,\dfrac{7}{10}\,}\\[5pt] &= \frac{5}{4}\div \frac{7}{10}\\[5pt] &= \frac{5}{4}\cdot \frac{10}{7}\\[5pt] &= \frac{5}{2\cdot {\color{red}\cancel{\color{black}{2}}}}\cdot \frac{{\color{red}\cancel{\color{black}{2}}} \cdot 5}{7}\\[5pt] &= \frac{25}{14} \end{align}[/latex]

For the second method, we start with the LCM of the denominators of all “small fractions”. (Please refer to this description to review finding the LCM.)

[latex]\begin{align} \operatorname{LCM}(4,2,5,10) &= \operatorname{LCM}\left(2^2,2^1,5^1,5^1\cdot 2^1\right)\\ &= 2^2\cdot 5^1\\ &= 20 \end{align}[/latex]

Now,

[latex]\begin{align} \frac{\,\dfrac{3}{4}+\dfrac{1}{2}\,}{\,\dfrac{4}{5}-\dfrac{1}{10}\,} &= \frac{\,\left(\dfrac{3}{4}+\dfrac{1}{2}\right){\color{red}{} \cdot 20}\,}{\,\left(\dfrac{4}{5}-\dfrac{1}{10}\right){\color{red}{} \cdot 20}\,}\\[5pt] &= \frac{\,\dfrac{3}{4}\cdot 20+\dfrac{1}{2}\cdot 20\,}{\,\dfrac{4}{5}\cdot 20-\dfrac{1}{10}\cdot 20\,}\\[5pt] &= \frac{\,\dfrac{3}{{\color{red}\cancel{\color{black}{4}}}\cdot 1}\cdot \dfrac{{\color{red}\cancel{\color{black}{4}}}\cdot 5}{1}+\dfrac{1}{{\color{red}\cancel{\color{black}{2}}}\cdot 1}\cdot \dfrac{{\color{red}\cancel{\color{black}{2}}}\cdot 10}{1}\,}{\,\dfrac{4}{{\color{red}\cancel{\color{black}{5}}}\cdot 1}\cdot \dfrac{{\color{red}\cancel{\color{black}{5}}}\cdot 4}{1}-\dfrac{1}{{\color{red}\cancel{\color{black}{10}}}\cdot 1}\cdot \dfrac{{\color{red}\cancel{\color{black}{10}}}\cdot 2}{1}\,}\\[5pt] &= \frac{15+10}{16-2}\\[5pt] &= \frac{25}{14} \end{align}[/latex]

In the following video, we will show a couple more examples of how to simplify complex fractions.

Complex Rational Expressions

A complex rational expression is a quotient with rational expressions in the dividend, divisor, or in both. We simplify these in the exact same way as we would a complex fraction, remembering to identify the restricted values of the variable.

Example

State the restricted values of the variable and simplify using both methods.

[latex]\displaystyle\frac{\,\dfrac{x+5}{x^2-16}\,}{\,\dfrac{x^2-25}{x-4}\,}[/latex]

Show Solution

For the restricted values of [latex]x[/latex], we need to acknowledge that we have three denominators – the denominators of the “small fractions” and the “big denominator”. The “big denominator” is itself a rational expression, so it’s equal to zero only if its numerator is equal to zero. Since [latex]x^2-16=(x+4)(x-4)[/latex], this denominator is equal to zero if [latex]x=-4[/latex] or if [latex]x=4[/latex] by the Zero-Product Property. The denominator [latex]x-4[/latex] is equal to zero if [latex]x=4[/latex], and since [latex]x^2-25 = (x+5)(x-5)[/latex], the numerator of the “big denominator” is equal to zero if [latex]x=-5[/latex] or if [latex]x=5[/latex] by the Zero-Product Property. Summarizing, the restricted values of [latex]x[/latex] are [latex]-5[/latex], [latex]-4[/latex], [latex]4[/latex], and [latex]5[/latex].

For the first method, we start with rewriting the complex rational expression as division.

[latex]\begin{align} \frac{\,\dfrac{x+5}{x^2-16}\,}{\,\dfrac{x^2-25}{x-4}\,} &= \frac{x+5}{x^2-16}\div\frac{x^2-25}{x-4}\\[5pt] &= \frac{x+5}{x^2-16}\cdot \frac{x-4}{x^2-25}\\[5pt] &= \frac{(x+5)}{(x+4)(x-4)}\cdot \frac{(x-4)}{(x+5)(x-5)}\\[5pt] &= \frac{{\color{red}\cancel{\color{black}{(x+5)}}}\cdot 1}{(x+4)\cdot {\color{red}\cancel{\color{black}{(x-4)}}}}\cdot \frac{{\color{red}\cancel{\color{black}{(x-4)}}}\cdot 1}{{\color{red}\cancel{\color{black}{(x+5)}}}\cdot (x-5)}\\[5pt] &= \frac{1}{(x+4)(x-5)} \end{align}[/latex]

For the second method, we start with the LCM of the denominators of all “small fractions”.

[latex]\begin{align} \operatorname{LCM}\left(x^2-16, x-4\right) &= \operatorname{LCM}\left((x+4)^1\cdot (x-4)^1, (x-4)^1\right)\\ &= (x+4)^1\cdot (x-4)^1\\ &= (x+4)(x-4) \end{align}[/latex]

Now,

[latex]\begin{align} \frac{\,\dfrac{x+5}{x^2-16}\,}{\,\dfrac{x^2-25}{x-4}\,} &= \frac{\, \dfrac{x+5}{x^2-16}{\color{red}{} \cdot (x+4)\cdot (x-4)}\,}{\,\dfrac{x^2-25}{x-4}\,{\color{red}{} \cdot (x+4)\cdot (x-4)}}\\[5pt] &= \frac{\, \dfrac{(x+5)}{{\color{red}\cancel{\color{black}{(x+4)\cdot (x-4)}}}\cdot 1}\cdot \dfrac{{\color{red}\cancel{\color{black}{(x+4)\cdot (x-4)}}}\cdot 1}{1}\,}{\,\dfrac{(x+5)(x-5)}{{\color{red}\cancel{\color{black}{(x-4)}}} \cdot 1} \cdot \dfrac{(x+4)\cdot {\color{red}\cancel{\color{black}{(x-4)}}}}{1}}\\[5pt] &= \frac{{\color{red}\cancel{\color{black}{(x+5)}}} \cdot 1}{{\color{red}\cancel{\color{black}{(x+5)}}}\cdot (x-5)\cdot (x+4)}\\[5pt] &= \frac{1}{(x-5)(x+4)} \end{align}[/latex]

This time, in the final answer we may conveniently multiply the denominator out and simplify to obtain an equivalent form [latex]\dfrac{1}{x^2-x-20}[/latex]. It’s not expected or necessary though.

The next video contains another example of simplifying a complex rational expression with just two “small fractions”.

The same ideas can be used when simplifying complex rational expressions that include more than one rational expression in the “big numerator” or “big denominator” (or both).

Example

State the restricted values of the variable and simplify using both methods.

[latex]\displaystyle\frac{\,1-\dfrac{9}{x^2}\,}{\,1+\dfrac{5}{x}+\dfrac{6}{x^2}\,}[/latex]

Show Solution

For the restricted values of [latex]x[/latex], we need to acknowledge that we have four denominators – the denominators of the “small fractions” and the “big denominator”. While the denominators of all “small fractions” are just powers of [latex]x[/latex], meaning only [latex]x=0[/latex] would make any of them equal to zero, the “big denominator” is not yet a single rational expression, so we first combine the “big denominator” into a single rational expression to acknowledge its numerator for restricted values of the variable purposes. The LCM of [latex]x[/latex] and [latex]x^2[/latex] is [latex]x^2[/latex], and [latex]1=\dfrac{1}{1}[/latex], thus

[latex]\begin{align} \frac{1}{1}+\frac{5}{x}+\frac{6}{x^2} &= \frac{1{\color{red}{} \cdot x^2}}{1{\color{red}{} \cdot x^2}}+\frac{5{\color{red}{} \cdot x}}{x{\color{red}{} \cdot x}}+\frac{6}{x^2}\\[5pt] &= \frac{x^2+5x+6}{x^2}\\[5pt] &= \frac{(x+2)(x+3)}{x^2} \end{align}[/latex]

Hence, the numerator of the “big denominator” rewritten as a single rational expression is equal to zero if [latex]x=-2[/latex] or if [latex]x=-3[/latex] by the Zero-Product Property. Summarizing, the restricted values of [latex]x[/latex] are [latex]0[/latex], [latex]-2[/latex], and [latex]-3[/latex].

[latex]\begin{align} \frac{\,1-\dfrac{9}{x^2}\,}{\,1+\dfrac{5}{x}+\dfrac{6}{x^2}\,} &= \frac{\,\dfrac{1}{1}-\dfrac{9}{x^2}\,}{\,\dfrac{(x+2)(x+3)}{x^2}\,}\\[5pt] &= \frac{\,\dfrac{1{\color{red}{} \cdot x^2}}{1{\color{red}{} \cdot x^2}}-\dfrac{9}{x^2}\,}{\,\dfrac{(x+2)(x+3)}{x^2}\,}\\[5pt] &= \frac{\,\dfrac{x^2-9}{x^2}\,}{\,\dfrac{(x+2)(x+3)}{x^2}\,}\\[5pt] &= \frac{\,\dfrac{(x+3)(x-3)}{x^2}\,}{\,\dfrac{(x+2)(x+3)}{x^2}\,}\\[5pt] &= \dfrac{(x+3)(x-3)}{x^2}\div\dfrac{(x+2)(x+3)}{x^2}\\[5pt] &= \dfrac{(x+3)(x-3)}{x^2}\cdot\dfrac{x^2}{(x+2)(x+3)}\\[5pt] &= \dfrac{{\color{red}\cancel{\color{black}{(x+3)}}}\cdot (x-3)}{{\color{red}\cancel{\color{black}{x^2}}} \cdot 1}\cdot\dfrac{{\color{red}\cancel{\color{black}{x^2}}} \cdot 1}{(x+2)\cdot {\color{red}\cancel{\color{black}{(x+3)}}}} \end{align}[/latex]

For the second method, we acknowledge that the LCM of the denominators of all “small fractions” is [latex]x^2[/latex], and proceed accordingly.

[latex]\begin{align} \frac{\,1-\dfrac{9}{x^2}\,}{\,1+\dfrac{5}{x}+\dfrac{6}{x^2}\,} &= \frac{\,\left(1-\dfrac{9}{x^2}\right){\color{red}\cdot\: x^2}\,}{\,\left(1+\dfrac{5}{x}+\dfrac{6}{x^2}\right){\color{red} \cdot\: x^2}\,}\\[5pt] &= \frac{\,1\cdot x^2-\dfrac{9}{{\color{red}\cancel{\color{black}{x^2}}} \cdot 1}\cdot \dfrac{{\color{red}\cancel{\color{black}{x^2}}} \cdot 1}{1}\,}{\,1\cdot x^2+\dfrac{5}{{\color{red}\cancel{\color{black}{x}}} \cdot 1}\cdot \dfrac{{\color{red}\cancel{\color{black}{x}}}\cdot x}{1}+\dfrac{6}{{\color{red}\cancel{\color{black}{x^2}}} \cdot 1}\cdot \dfrac{{\color{red}\cancel{\color{black}{x^2}}} \cdot 1}{1}\,}\\[5pt] &= \frac{x^2-9}{x^2+5x+6}\\[5pt] &= \frac{{\color{red}\cancel{\color{black}{(x+3)}}}\cdot (x-3)}{{\color{red}\cancel{\color{black}{(x+3)}}}\cdot(x+2)}\\[5pt] &= \frac{x-3}{x+2} \end{align}[/latex]

The next video contains another example of simplifying a complex rational expression similar to the one in the example above.

We conclude with an example in which the LCM of the denominators of all “small fractions” will be more complicated.

Example

State the restricted values of the variable and simplify using both methods.

[latex]\displaystyle\frac{\,\dfrac{x}{x^2-1}-\dfrac{x+1}{x-3}\,}{\,\dfrac{2}{x+1}-\dfrac{3x}{x^2-2x-3}\,}[/latex]

Show Solution

For the restricted values of [latex]x[/latex], we need to acknowledge that we have five denominators – the denominators of the “small fractions” and the “big denominator”. For the denominators of all “small fractions”, we solve the following equations:

[latex]x^2-1=0[/latex], equivalent to [latex](x-1)(x+1)=0[/latex], resulting in [latex]x=1[/latex] or [latex]x=-1[/latex] by the Zero-Factor Property,
[latex]x-3=0[/latex], resulting in [latex]x=3[/latex],
[latex]x+1=0[/latex], resulting in [latex]x=-1[/latex], and
[latex]x^2-2x-3=0[/latex], equivalent to [latex](x-3)(x+1)=0[/latex], resulting in [latex]x=3[/latex] or [latex]x=-1[/latex] by the Zero-Factor Property.

The “big denominator” is not yet a single rational expression, so we first combine the “big denominator” into a single rational expression to acknowledge its numerator for restricted values of the variable purposes. Based on the factorization [latex](x-3)(x+1)[/latex] of [latex]x^2-2x-3[/latex], the LCM of [latex]x+1[/latex] and [latex]x^2-2x-3[/latex] is [latex](x-3)(x+1)[/latex] and thus

[latex]\begin{align} \frac{2}{x+1}-\frac{3x}{x^2-2x-3} &= \frac{2{\color{red}\: \cdot\: (x-3)}}{(x+1){\color{red}\: \cdot\: (x-3)}}-\frac{3x}{(x-3)(x+1)}\\[5pt] &= \frac{2x-6-3x}{(x-3)(x+1)}\\[5pt] &= \frac{-x-6}{(x-3)(x+1)} \end{align}[/latex]

Hence, the numerator of the “big denominator” rewritten as a single rational expression, [latex]-x-6[/latex] is equal to zero if [latex]x=-6[/latex]. Summarizing, the restricted values of [latex]x[/latex] are [latex]1[/latex], [latex]-1[/latex], [latex]3[/latex], and [latex]-6[/latex].

Since we already have the “big denominator” rewritten as a single rational expression, for the first method we only need to rewrite the “big numerator” as a single rational expression and proceed accordingly. Notice that [latex]\operatorname{LCM}\left(x^2-1, x-3\right)=\operatorname{LCM}\left((x+1)(x-1), x-3\right)=(x+1)(x-1)(x-3)[/latex]. Now,

[latex]\begin{align} \frac{\,\dfrac{x}{x^2-1}-\dfrac{x+1}{x-3}\,}{\,\dfrac{2}{x+1}-\dfrac{3x}{x^2-2x-3}\,} &= \frac{\,\dfrac{x{\color{red}\: \cdot\: (x-3)}}{(x+1)\cdot(x-1){\color{red}\: \cdot\: (x-3)}}-\dfrac{(x+1){\color{red}\: \cdot\: (x-1)\cdot(x+1)}}{(x-3){\color{red}\: \cdot\: (x-1)\cdot(x+1)}}\,}{\,\dfrac{-x-6}{(x-3)\cdot(x+1)}\,}\\[5pt] &= \frac{\,\dfrac{x^2-3x-\left(\left(x^2-1\right)\cdot(x+1)\right)}{(x-3)\cdot(x-1)\cdot(x+1)}}{\,\dfrac{-x-6}{(x-3)\cdot(x+1)}\,}\\[5pt] &= \frac{\,\dfrac{x^2-3x-\left(x^3+x^2-x-1\right)}{(x-3)\cdot(x-1)\cdot(x+1)}}{\,\dfrac{-x-6}{(x-3)\cdot(x+1)}\,}\\[5pt] &= \frac{\,\dfrac{x^2-3x-x^3-x^2+x+1}{(x-3)\cdot(x-1)\cdot(x+1)}}{\,\dfrac{-x-6}{(x-3)\cdot(x+1)}\,}\\[5pt] &= \frac{\,\dfrac{-x^3-2x+1}{(x-3)\cdot(x-1)\cdot(x+1)}}{\,\dfrac{-x-6}{(x-3)\cdot(x+1)}\,}\\[5pt] &= \frac{-x^3-2x+1}{(x-3)\cdot(x-1)\cdot(x+1)}\div\dfrac{-x-6}{(x-3)\cdot(x+1)}\\[5pt] &= \frac{{\color{red}\cancel{\color{black}{(-1)}}}\cdot\left(x^3+2x-1\right)}{{\color{red}\cancel{\color{black}{(x-3)}}}\cdot(x-1)\cdot{\color{red}\cancel{\color{black}{(x+1)}}}}\cdot\dfrac{{\color{red}\cancel{\color{black}{(x-3)}}}\cdot{\color{red}\cancel{\color{black}{(x+1)}}}}{{\color{red}\cancel{\color{black}{(-1)}}}\cdot(x+6)}\\[5pt] &= \frac{x^3+2x-1}{(x-1)(x+6)} \end{align}[/latex]

For the second method, let’s begin with the LCM of the denominators of all “small fractions”:

[latex]\begin{align} \operatorname{LCM}\left(x^2-1, x-3, x+1, x^2-2x-3\right)&=\operatorname{LCM}\left((x+1)^1\cdot(x-1)^1, (x-3)^1, (x+1)^1, (x-3)^1\cdot(x+1)^1\right)\\[5pt] &=(x+1)(x-1)(x-3) \end{align}[/latex]

Now,

[latex]\begin{align} \frac{\,\dfrac{x}{x^2-1}-\dfrac{x+1}{x-3}\,}{\,\dfrac{2}{x+1}-\dfrac{3x}{x^2-2x-3}\,} &= \frac{\,\left(\dfrac{x}{x^2-1}-\dfrac{x+1}{x-3}\right){\color{red}\: \cdot\: (x+1)\cdot(x-1)\cdot(x-3)}\,}{\,\left(\dfrac{2}{x+1}-\dfrac{3x}{x^2-2x-3}\right){\color{red}\: \cdot\: (x+1)\cdot(x-1)\cdot(x-3)}\,}\\[5pt] &=\frac{\,\dfrac{x}{{\color{red}\cancel{\color{black}{\left(x^2-1\right)}}}\cdot 1}\cdot\dfrac{{\color{red}\cancel{\color{black}{\left(x^2-1\right)}}}\cdot (x-3)}{1}-\dfrac{(x+1)}{{\color{red}\cancel{\color{black}{(x-3)}}}\cdot 1}\cdot\dfrac{\left(x^2-1\right)\cdot {\color{red}\cancel{\color{black}{(x-3)}}}}{1}\,}{\,\dfrac{2}{{\color{red}\cancel{\color{black}{(x+1)}}}\cdot 1}\cdot\dfrac{{\color{red}\cancel{\color{black}{(x+1)}}}\cdot\left(x^2-4x+3\right)}{1}-\dfrac{3x}{{\color{red}\cancel{\color{black}{\left(x^2-2x-3\right)}}}\cdot 1}\cdot\dfrac{{\color{red}\cancel{\color{black}{\left(x^2-2x-3\right)}}}\cdot(x-1)}{1}\,}\\[5pt] &= \frac{x^2-3x-\left(x^3-x+x^2-1\right)}{2x^2-8x+6-\left(3x^2-3x\right)}\\[5pt] &= \frac{x^2-3x-x^3+x-x^2+1}{2x^2-8x+6-3x^2+3x}\\[5pt] &= \frac{-x^3-2x+1}{-x^2-5x+6}\\[5pt] &= \frac{{\color{red}\cancel{\color{black}{(-1)}}}\cdot\left(x^3+2x-1\right)}{{\color{red}\cancel{\color{black}{(-1)}}}\cdot\left(x^2+5x-6\right)}\\[5pt] &= \frac{x^3+2x-1}{(x-1)(x+6)} \end{align}[/latex]

Summary

Complex rational expressions are quotients with rational expressions in the divisor, dividend, or both. When written in fraction form, they appear to be fractions within a fraction. These can be simplified by first treating the quotient as a division problem, after you present the “big numerator” and “big denominator”, respectively, as simplified fractions. Then you can rewrite the division as multiplication by the reciprocal of the divisor. Or you can simplify the complex rational expression by multiplying both the “big numerator” and “big denominator” by the LCM of the denominators of all “small fractions”. This can help simplify the complex expression even faster.

Module 4: Rational Expressions, Functions, and Equations