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Rational Expressions

Simplify rational expressions.

Rational Expressions

  • Identify the skills leading to successful preparation for a college level mathematics exam.
  • Create a plan for success when taking mathematics exams.
  1. It’s important to take the opportunity to reflect on your past experiences in taking math exams as you begin a new term. We can learn a lot from these reflections and thus work toward developing a strategy for improvement.
    In the table below list 5 challenges you have had in past math courses when taking an exam and list a possible solution that you could try this semester.
    Challenge:Possible Solution:
    1. ________
    2. ________
    3. ________
    4. ________
    5. ________
  2. Develop your plan for success. Keep in mind the idea of mindsets and try to approach your test taking strategies with a growth mindset. Now is the time for growth as you begin a new term. Share your plan with your study group members.

Condensed — the full section is in OpenStax College Algebra 2e.

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.

\[\frac{{x}^{2}+8x+16}{{x}^{2}+11x+28}\]

We can factor the numerator and denominator to rewrite the expression.

\[\frac{{(x+4)}^{2}}{(x+4)(x+7)}\]

Then we can simplify that expression by canceling the common factor \((x+4).\)

\[\frac{x+4}{x+7}\]
Example

Try it.

Simplify \(\frac{{x}^{2}-9}{{x}^{2}+4x+3}.\)

Solution\[\begin{array}{lllll}\frac{(x+3)(x-3)}{(x+3)(x+1)} & & & & \text{Factor the numerator and the denominator}. \\ \frac{x-3}{x+1} & & & & \text{Cancel common factor }(x+3).\end{array}\]

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.

Example

Try it.

Multiply the rational expressions and show the product in simplest form:

\[\frac{{x}^{2}+4x-5}{3x+18}⋅\frac{2x-1}{x+5}\]
Solution\[\begin{array}{lllll}\frac{(x+5)(x-1)}{3(x+6)}⋅\frac{(2x-1)}{(x+5)} & & & & \text{Factor the numerator and denominator}. \\ \frac{(x+5)(x-1)(2x-1)}{3(x+6)(x+5)} & & & & \text{Multiply numerators and denominators}. \\ \frac{(x+5)(x-1)(2x-1)}{3(x+6)(x+5)} & & & & \text{Cancel common factors to simplify}. \\ \frac{(x-1)(2x-1)}{3(x+6)} & & & & \end{array}\]

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 \(\frac{1}{x}\div \frac{{x}^{2}}{3}\) as the product \(\frac{1}{x}⋅\frac{3}{{x}^{2}}.\) Once the division expression has been rewritten as a multiplication expression, we can multiply as we did before.

\[\frac{1}{x}⋅\frac{3}{{x}^{2}}=\frac{3}{{x}^{3}}\]
Example

Try it.

Divide the rational expressions and express the quotient in simplest form:

\[\frac{2{x}^{2}+x-6}{{x}^{2}-1}\div \frac{{x}^{2}-4}{{x}^{2}+2x+1}\]
Solution\[\begin{array}{ll}\frac{2{x}^{2}+x-6}{{x}^{2}-1}⋅\frac{{x}^{2}+2x+1}{{x}^{2}-4} & \ \text{Rewrite as multiplication.} \\ \frac{(2x-3)(x+2)}{(x+1)(x-1)}⋅\frac{{(x+1)}^{2}}{(x+2)(x-2)} & \ \text{Factor.} \\ \frac{(2x-3)(x+2){(x+1)}^{2}}{(x+1)(x-1)(x+2)(x-2)} & \ \text{Multiply.} \\ \frac{(2x-3)(x+1)}{(x-1)(x-2)} & \ \text{Cancel common factors to simplify.}\end{array}\]

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.

\[\begin{array}{lll}\frac{5}{24}+\frac{1}{40} & = & \frac{25}{120}+\frac{3}{120} \\ & = & \frac{28}{120} \\ & = & \frac{7}{30}\end{array}\]

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 \(\frac{x+5}{x+5}\) and the expression with a denominator of \((x+4)(x+5)\) by \(\frac{x+3}{x+3}.\)

Example

Try it.

Add the rational expressions:

\[\frac{5}{x}+\frac{6}{y}\]
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.

\[\begin{array}{l}\frac{5}{x}⋅\frac{y}{y}+\frac{6}{y}⋅\frac{x}{x} \\ \frac{5y}{xy}+\frac{6x}{xy}\end{array}\]

Now that the expressions have the same denominator, we simply add the numerators to find the sum.

\[\frac{6x+5y}{xy}\]
Example

Try it.

Subtract the rational expressions:

\[\frac{6}{{x}^{2}+4x+4}-\frac{2}{{x}^{2}-4}\]
Solution\[\begin{array}{ll}\frac{6}{{(x+2)}^{2}}-\frac{2}{(x+2)(x-2)} & \ \text{Factor}. \\ \frac{6}{{(x+2)}^{2}}⋅\frac{x-2}{x-2}-\frac{2}{(x+2)(x-2)}⋅\frac{x+2}{x+2} & \ \text{Multiply each fraction to get LCD as denominator}. \\ \frac{6(x-2)}{{(x+2)}^{2}(x-2)}-\frac{2(x+2)}{{(x+2)}^{2}(x-2)} & \ \text{Multiply}. \\ \frac{6x-12-(2x+4)}{{(x+2)}^{2}(x-2)} & \ \text{Apply distributive property}. \\ \frac{4x-16}{{(x+2)}^{2}(x-2)} & \ \text{Subtract}. \\ \frac{4(x-4)}{{(x+2)}^{2}(x-2)} & \ \text{Simplify}.\end{array}\]

Condensed — the full section is in OpenStax College Algebra 2e.

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 \(\frac{a}{\frac{1}{b}+c}\) can be simplified by rewriting the numerator as the fraction \(\frac{a}{1}\) and combining the expressions in the denominator as \(\frac{1+bc}{b}.\) We can then rewrite the expression as a multiplication problem using the reciprocal of the denominator. We get \(\frac{a}{1}⋅\frac{b}{1+bc},\) which is equal to \(\frac{ab}{1+bc}.\)

Example

Try it.

Simplify: \(\frac{y+\frac{1}{x}}{\frac{x}{y}}\) .

Solution

Begin by combining the expressions in the numerator into one expression.

\[\begin{array}{ll}y⋅\frac{x}{x}+\frac{1}{x} & \ \text{Multiply by }\frac{x}{x}\ \text{to get LCD as denominator}. \\ \frac{xy}{x}+\frac{1}{x} & \\ \frac{xy+1}{x} & \ \text{Add numerators}.\end{array}\]

Now the numerator is a single rational expression and the denominator is a single rational expression.

\[\frac{\frac{xy+1}{x}}{\frac{x}{y}}\]

We can rewrite this as division, and then multiplication.

\[\begin{array}{ll}\frac{xy+1}{x}\div \frac{x}{y} & \\ \frac{xy+1}{x}⋅\frac{y}{x} & \ \text{Rewrite as multiplication}\text{.} \\ \frac{y(xy+1)}{{x}^{2}} & \ \text{Multiply}\text{.}\end{array}\]

Key Concepts

  • Rational expressions can be simplified by cancelling common factors in the numerator and denominator. See .
  • We can multiply rational expressions by multiplying the numerators and multiplying the denominators. See .
  • To divide rational expressions, multiply by the reciprocal of the second expression. See .
  • Adding or subtracting rational expressions requires finding a common denominator. See and .
  • Complex rational expressions have fractions in the numerator or the denominator. These expressions can be simplified. See .

Chapter Practice Test

For the following exercises, identify the number as rational, irrational, whole, or natural. Choose the most descriptive answer.

For the following exercises, evaluate the expression.

For the following exercises, simplify the expression.

For the following exercises, factor the polynomial.

For the following exercises, simplify the expression.

Practice (40)

Try each one on paper first. Reveal the answer to check; verified ones can be opened in the solver for every step.

  1. Each of the behaviors or attitudes listed in the table above are associated with successful college mathematics exam preparation. This means that students who use these strategies or are open to these beliefs pass their college math courses. Compute your total score and share your score with your study group in class. Be supportive of your fellow students and offer encouragement!

    Total score =__________

  2. Based on this survey, create a list of the top 5 test preparation and taking strategies that you currently utilize, and feel are most helpful to you.

  3. Based on this survey, create a list of the top 5 test preparation and taking strategies that interest you, and that you feel could be most helpful to you this term. Plan on implementing these strategies.

  4. Simplify \(\frac{{x}^{2}-9}{{x}^{2}+4x+3}.\)

    Asehoy ny valinteny
    \[\begin{array}{lllll}\frac{(x+3)(x-3)}{(x+3)(x+1)} & & & & \text{Factor the numerator and the denominator}. \\ \frac{x-3}{x+1} & & & & \text{Cancel common factor }(x+3).\end{array}\]

  5. Simplify \(\frac{x-6}{{x}^{2}-36}.\)

    Asehoy ny valinteny

    \(\frac{1}{x+6}\)

  6. Multiply the rational expressions and show the product in simplest form:

    \[\frac{{x}^{2}+4x-5}{3x+18}⋅\frac{2x-1}{x+5}\]
    Asehoy ny valinteny
    \[\begin{array}{lllll}\frac{(x+5)(x-1)}{3(x+6)}⋅\frac{(2x-1)}{(x+5)} & & & & \text{Factor the numerator and denominator}. \\ \frac{(x+5)(x-1)(2x-1)}{3(x+6)(x+5)} & & & & \text{Multiply numerators and denominators}. \\ \frac{(x+5)(x-1)(2x-1)}{3(x+6)(x+5)} & & & & \text{Cancel common factors to simplify}. \\ \frac{(x-1)(2x-1)}{3(x+6)} & & & & \end{array}\]

  7. Multiply the rational expressions and show the product in simplest form:

    \[\frac{{x}^{2}+11x+30}{{x}^{2}+5x+6}⋅\frac{{x}^{2}+7x+12}{{x}^{2}+8x+16}\]
    Asehoy ny valinteny

    \(\frac{(x+5)(x+6)}{(x+2)(x+4)}\)

  8. Divide the rational expressions and express the quotient in simplest form:

    \[\frac{2{x}^{2}+x-6}{{x}^{2}-1}\div \frac{{x}^{2}-4}{{x}^{2}+2x+1}\]
    Asehoy ny valinteny
    \[\begin{array}{ll}\frac{2{x}^{2}+x-6}{{x}^{2}-1}⋅\frac{{x}^{2}+2x+1}{{x}^{2}-4} & \ \text{Rewrite as multiplication.} \\ \frac{(2x-3)(x+2)}{(x+1)(x-1)}⋅\frac{{(x+1)}^{2}}{(x+2)(x-2)} & \ \text{Factor.} \\ \frac{(2x-3)(x+2){(x+1)}^{2}}{(x+1)(x-1)(x+2)(x-2)} & \ \text{Multiply.} \\ \frac{(2x-3)(x+1)}{(x-1)(x-2)} & \ \text{Cancel common factors to simplify.}\end{array}\]

  9. Divide the rational expressions and express the quotient in simplest form:

    \[\frac{9{x}^{2}-16}{3{x}^{2}+17x-28}\div \frac{3{x}^{2}-2x-8}{{x}^{2}+5x-14}\]
    Asehoy ny valinteny

    \(1\)

  10. Add the rational expressions:

    \[\frac{5}{x}+\frac{6}{y}\]
    Asehoy ny valinteny

    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.

    \[\begin{array}{l}\frac{5}{x}⋅\frac{y}{y}+\frac{6}{y}⋅\frac{x}{x} \\ \frac{5y}{xy}+\frac{6x}{xy}\end{array}\]

    Now that the expressions have the same denominator, we simply add the numerators to find the sum.

    \[\frac{6x+5y}{xy}\]

  11. Subtract the rational expressions:

    \[\frac{6}{{x}^{2}+4x+4}-\frac{2}{{x}^{2}-4}\]
    Asehoy ny valinteny
    \[\begin{array}{ll}\frac{6}{{(x+2)}^{2}}-\frac{2}{(x+2)(x-2)} & \ \text{Factor}. \\ \frac{6}{{(x+2)}^{2}}⋅\frac{x-2}{x-2}-\frac{2}{(x+2)(x-2)}⋅\frac{x+2}{x+2} & \ \text{Multiply each fraction to get LCD as denominator}. \\ \frac{6(x-2)}{{(x+2)}^{2}(x-2)}-\frac{2(x+2)}{{(x+2)}^{2}(x-2)} & \ \text{Multiply}. \\ \frac{6x-12-(2x+4)}{{(x+2)}^{2}(x-2)} & \ \text{Apply distributive property}. \\ \frac{4x-16}{{(x+2)}^{2}(x-2)} & \ \text{Subtract}. \\ \frac{4(x-4)}{{(x+2)}^{2}(x-2)} & \ \text{Simplify}.\end{array}\]

  12. Subtract the rational expressions: \(\frac{3}{x+5}-\frac{1}{x-3}.\)

    Asehoy ny valinteny

    \(\frac{2(x-7)}{(x+5)(x-3)}\)

  13. Simplify: \(\frac{y+\frac{1}{x}}{\frac{x}{y}}\) .

    Asehoy ny valinteny

    Begin by combining the expressions in the numerator into one expression.

    \[\begin{array}{ll}y⋅\frac{x}{x}+\frac{1}{x} & \ \text{Multiply by }\frac{x}{x}\ \text{to get LCD as denominator}. \\ \frac{xy}{x}+\frac{1}{x} & \\ \frac{xy+1}{x} & \ \text{Add numerators}.\end{array}\]

    Now the numerator is a single rational expression and the denominator is a single rational expression.

    \[\frac{\frac{xy+1}{x}}{\frac{x}{y}}\]

    We can rewrite this as division, and then multiplication.

    \[\begin{array}{ll}\frac{xy+1}{x}\div \frac{x}{y} & \\ \frac{xy+1}{x}⋅\frac{y}{x} & \ \text{Rewrite as multiplication}\text{.} \\ \frac{y(xy+1)}{{x}^{2}} & \ \text{Multiply}\text{.}\end{array}\]

  14. Simplify: \(\frac{\frac{x}{y}-\frac{y}{x}}{y}\)

    Asehoy ny valinteny

    \(\frac{{x}^{2}-{y}^{2}}{x{y}^{2}}\)

  15. How can you use factoring to simplify rational expressions?

    Asehoy ny valinteny

    You can factor the numerator and denominator to see if any of the terms can cancel one another out.

  16. How do you use the LCD to combine two rational expressions?

  17. Tell whether the following statement is true or false and explain why: You only need to find the LCD when adding or subtracting rational expressions.

    Asehoy ny valinteny

    True. Multiplication and division do not require finding the LCD because the denominators can be combined through those operations, whereas addition and subtraction require like terms.

  18. \(\frac{{x}^{2}-16}{{x}^{2}-5x+4}\)

  19. \(\frac{{y}^{2}+10y+25}{{y}^{2}+11y+30}\)

    Asehoy ny valinteny

    \(\frac{y+5}{y+6}\)

  20. \(\frac{6{a}^{2}-24a+24}{6{a}^{2}-24}\)

  21. \(\frac{9{b}^{2}+18b+9}{3b+3}\)

    Asehoy ny valinteny

    \(3b+3\)

  22. \(\frac{m-12}{{m}^{2}-144}\)

  23. \(\frac{2{x}^{2}+7x-4}{4{x}^{2}+2x-2}\)

    Asehoy ny valinteny

    \(\frac{x+4}{2x+2}\)

  24. \(\frac{6{x}^{2}+5x-4}{3{x}^{2}+19x+20}\)

  25. \(\frac{{a}^{2}+9a+18}{{a}^{2}+3a-18}\)

    Asehoy ny valinteny

    \(\frac{a+3}{a-3}\)

  26. \(\frac{3{c}^{2}+25c-18}{3{c}^{2}-23c+14}\)

  27. \(\frac{12{n}^{2}-29n-8}{28{n}^{2}-5n-3}\)

    Asehoy ny valinteny

    \(\frac{3n-8}{7n-3}\)

  28. \(\frac{{x}^{2}-x-6}{2{x}^{2}+x-6}⋅\frac{2{x}^{2}+7x-15}{{x}^{2}-9}\)

  29. \(\frac{{c}^{2}+2c-24}{{c}^{2}+12c+36}⋅\frac{{c}^{2}-10c+24}{{c}^{2}-8c+16}\)

    Asehoy ny valinteny

    \(\frac{c-6}{c+6}\)

  30. \(\frac{2{d}^{2}+9d-35}{{d}^{2}+10d+21}⋅\frac{3{d}^{2}+2d-21}{3{d}^{2}+14d-49}\)

  31. \(\frac{10{h}^{2}-9h-9}{2{h}^{2}-19h+24}⋅\frac{{h}^{2}-16h+64}{5{h}^{2}-37h-24}\)

    Asehoy ny valinteny

    \(1\)

  32. \(\frac{6{b}^{2}+13b+6}{4{b}^{2}-9}⋅\frac{6{b}^{2}+31b-30}{18{b}^{2}-3b-10}\)

  33. \(\frac{2{d}^{2}+15d+25}{4{d}^{2}-25}⋅\frac{2{d}^{2}-15d+25}{25{d}^{2}-1}\)

    Asehoy ny valinteny

    \(\frac{{d}^{2}-25}{25{d}^{2}-1}\)

  34. \(\frac{6{x}^{2}-5x-50}{15{x}^{2}-44x-20}⋅\frac{20{x}^{2}-7x-6}{2{x}^{2}+9x+10}\)

  35. \(\frac{{t}^{2}-1}{{t}^{2}+4t+3}⋅\frac{{t}^{2}+2t-15}{{t}^{2}-4t+3}\)

    Asehoy ny valinteny

    \(\frac{t+5}{t+3}\)

  36. \(\frac{2{n}^{2}-n-15}{6{n}^{2}+13n-5}⋅\frac{12{n}^{2}-13n+3}{4{n}^{2}-15n+9}\)

  37. \(\frac{36{x}^{2}-25}{6{x}^{2}+65x+50}⋅\frac{3{x}^{2}+32x+20}{18{x}^{2}+27x+10}\)

    Asehoy ny valinteny

    \(\frac{6x-5}{6x+5}\)

  38. \(\frac{3{y}^{2}-7y-6}{2{y}^{2}-3y-9}\div \frac{{y}^{2}+y-2}{2{y}^{2}+y-3}\)

  39. \(\frac{6{p}^{2}+p-12}{8{p}^{2}+18p+9}\div \frac{6{p}^{2}-11p+4}{2{p}^{2}+11p-6}\)

    Asehoy ny valinteny

    \(\frac{p+6}{4p+3}\)

  40. \(\frac{{q}^{2}-9}{{q}^{2}+6q+9}\div \frac{{q}^{2}-2q-3}{{q}^{2}+2q-3}\)

Symbols used here

\pm
plus or minus
Both signs at once: x = 3 ± 2 means 5 and 1.
\neq
not equal
The two sides are different.
\leq,\ \geq
less/greater than or equal
Inequalities that allow equality; < and > exclude it.
\sqrt{x},\ \sqrt[n]{x}
square root, n-th root
The non-negative number whose square (n-th power) is x.
|x|
absolute value / modulus
Distance from zero: |−3| = 3. For a complex number, distance from the origin.
i
imaginary unit
i² = −1.
\log_b x,\ \ln x
logarithm, natural log
The exponent b must be raised to for x; ln uses base e.
\mathbb{N},\ \mathbb{Z},\ \mathbb{Q},\ \mathbb{R},\ \mathbb{C}
number sets
Naturals, integers, rationals, reals, complex numbers.

How to: Rational Expressions

  1. Simplify rational expressions.
  2. Multiply rational expressions.
  3. Divide rational expressions.
  4. Add and subtract rational expressions.
  5. Simplify complex rational expressions.
  6. Identify the skills leading to successful preparation for a college level mathematics exam.
  7. Create a plan for success when taking mathematics exams.
  8. It’s important to take the opportunity to reflect on your past experiences in taking math exams as you begin a new term. We can learn a lot from these reflections and thus work toward developing a strategy for improvement.

Questions people ask

What does it mean to solve an equation?

To find every value of the unknown that makes both sides equal. Each step is an operation applied to both sides that keeps the solution set the same, until the unknown stands alone.

Why do I sometimes get two answers?

A quadratic can cross the axis twice, so it can have two solutions. A degree-n polynomial has up to n. The graph shows where each one comes from.

How do I know whether to factor or use the quadratic formula?

Try factoring for a few seconds: look for two numbers that multiply to a·c and add to b. If nothing obvious appears, the discriminant b² − 4ac tells you how many real roots there are, and the formula finds them without guessing.

Andramo ny anao manokana

Parts of this page are adapted from OpenStax College Algebra 2e (CC BY-NC-SA 4.0). Condensed and re-explained here; errors are ours.

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