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2.13: Simplifying, Multiplying, and Dividing Rational Expressions

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    You have learned about the properties of fractions and their operations. Rational numbers are just fractions where the numerators and denominators are integers. We will now work with fractions whose numerators and denominators are polynomials. We call this kind of expression a rational expression.

    Rational Expression

    A rational expression is an expression of the form \(\dfrac{p}{q}\), where p and q are polynomials and \(q\neq 0\).

    Here are some examples of rational expressions:

    \[−\dfrac{24}{56} \qquad \dfrac{5x}{12y} \qquad \dfrac{4x+1}{x^2−9} \qquad \dfrac{4x^2+3x−1}{2x−8}\nonumber\]

    Notice that the first rational expression listed above, \(−\dfrac{24}{56}\), is just a fraction. Since a constant is a polynomial with degree zero, the ratio of two constants is a rational expression, provided the denominator is not zero.

    We will do the same operations with rational expressions that we did with fractions. We will simplify, add, subtract, multiply, divide and use them in applications.

    Determining the Values for Which a Rational Expression is Undefined

    If the denominator is zero, the rational expression is undefined. The numerator of a rational expression may be 0—but not the denominator.

    When we work with a numerical fraction, it is easy to avoid dividing by zero because we can see the number in the denominator. In order to avoid dividing by zero in a rational expression, we must not allow values of the variable that will make the denominator be zero.

    So before we begin any operation with a rational expression, we examine it first to find the values that would make the denominator zero. That way, when we solve a rational equation for example, we will know whether the algebraic solutions we find are allowed or not.

    Example \(\PageIndex{1}\)

    Determine the value for which each rational expression is undefined.

    1. \(\dfrac{8a^2b}{3c}\)
    2. \(\dfrac{4b−3}{2b+5}\)
    3. \(\dfrac{x+4}{x^2+5x+6}\)
    Solution

    The expression will be undefined when the denominator is zero.

    1. \(\begin{array} {ll} &\dfrac{8a^2b}{3c} \\ \begin{array} {l} \text{Set the denominator equal to zero and solve} \\ \text{for the variable.} \end{array} &3c=0 \\ &c=0 \\ &\dfrac{8a^2b}{3c}\text{ is undefined for }c=0 \end{array} \)
    2. \(\begin{array} {ll} &\dfrac{4b-3}{2b+5} \\ \begin{array} {l} \text{Set the denominator equal to zero and solve} \\ \text{for the variable.} \end{array} &2b+5=0 \\ &2b=-5 \\ &b=-\dfrac{5}{2} \\ & \\ &\dfrac{4b-3}{2b+5} \text{ is undefined for }b=-\dfrac{5}{2} \end{array} \)
    3. \(\begin{array} {ll} &\dfrac{x+4}{x^2 + 5x + 6} \\ \begin{array} {l} \text{Set the denominator equal to zero and solve } \\ \text{for the variable.} \end{array} &x^2+5x+6=0 \\ &(x+2)(x+3)=0 \\ &x+2=0\text{ or }x+3=0 \\ &x=-2\text{ or }x=-3 \\ & \\ &\dfrac{x+4}{x^2+5x+6}\text{ is undefined for }x=-2\text{ or }x=-3 \end{array} \)
    You Try \(\PageIndex{1}\)

    Determine the value for which each rational expression is undefined.

    1. \(\dfrac{4p}{5q}\)
    2. \(\dfrac{y−1}{3y+2}\)
    3. \(\dfrac{m−5}{m^2+m−6}\)
    Answer
    1. \(q=0\)
    2. \(y=−\dfrac{2}{3}\)
    3. \(m=2,m=−3\)

    Simplifying Rational Expressions

    A fraction is considered simplified if there are no common factors, other than 1, in its numerator and denominator. Similarly, a simplified rational expression has no common factors, other than 1, in its numerator and denominator.

    Simplified Rational Expression

    A rational expression is considered simplified if there are no common factors in its numerator and denominator.

    For example,

    \[ \begin{array} {l} \dfrac{x+2}{x+3} \text{ is simplified because there are no common factors of } x+2 \text{ and }x+3. \\ \dfrac{2x}{3x} \text{ is not simplified because x is a common factor of }2x\text{ and }3x. \\ \end{array} \nonumber\]

    We use the Equivalent Fractions Property to simplify numerical fractions. We restate it here as we will also use it to simplify rational expressions.

    Equivalent Fractions Property

    If a, b, and c are numbers where \(b\neq 0,c\neq 0,\)

    \[\text {then } \dfrac{a}{b}=\dfrac{a·c}{b·c} \text{ and } \dfrac{a·c}{b·c}=\dfrac{a}{b}\nonumber\]

    Notice that in the Equivalent Fractions Property, the values that would make the denominators zero are specifically disallowed. We see \(b\neq 0,c\neq 0\) clearly stated.

    To simplify rational expressions, we first write the numerator and denominator in factored form. Then we remove the common factors using the Equivalent Fractions Property.

    Be very careful as you remove common factors. Factors are multiplied to make a product. You can remove a factor from a product. You cannot remove a term from a sum.

    The rational expression is the quantity 2 times 3 times 7 divided by the quantity 3 times 5 times 7 are 3 and 7. Its common factors are 3 and 7, which are factors of the product. When they are removed, the result is two-fifths. The rational expression is the product of 3 x and the quantity x minus 9 divided by the product of 5 and the quantity x minus 9. The common factor is x minus 9, which is a factor of the product. When it is removed, the result is 3 x divided by 5. The rational expression is the quantity x plus 5 divided by 5. There is an x both the numerator and denominator. However, it is a term of the sum in the numerator. The rational expression has no common factors.

    Removing the x’s from \(\dfrac{x+5}{x}\) would be like cancelling the 2’s in the fraction \(\dfrac{2+5}{2}!\)

    How to Simplify a Rational Expression

    Example \(\PageIndex{2}\)

    Simplify: \(\dfrac{x^2+5x+6}{x^2+8x+12}\)

    Solution

    Step 1 is to factor the numerator and denominator completely in the rational expression, the quantity x squared plus 5 x plus six divided by the quantity x squared 8 x plus 12. The numerator, x squared plus 5 x plus six, factors into the quantity x plus 2 times the quantity x plus 3. The denominator, x squared 8 x plus 12, factors into the quantity x plus 2 times the quantity x plus 6.Step 2 is to simplify the rational expression, the quantity x plus 2 times the quantity x plus 3 all divided by the quantity x plus 2 times the quantity x plus 6, by dividing out the common factor, x plus 6. The result of removing the common factor is the quantity x plus 3 divided by the quantity x plus 6, where x is not equal to 2 and x is not equal to -6.

    You Try \(\PageIndex{2}\)

    Simplify: \(\dfrac{a^2−3a−10}{a^2+a−2}\)

    Answer

    \(\dfrac{a−5}{a−1},a\neq −2,a\neq 1\)

    We now summarize the steps you should follow to simplify rational expressions.

    Simplifying a Rational Expression
    1. Factor the numerator and denominator completely.
    2. Simplify by dividing out common factors.

    Usually, we leave the simplified rational expression in factored form. This way, it is easy to check that we have removed all the common factors.

    We’ll use the methods we have learned to factor the polynomials in the numerators and denominators in the following examples.

    Every time we write a rational expression, we should make a statement disallowing values that would make a denominator zero. However, to let us focus on the work at hand, we will omit writing it in the examples.

    Example \(\PageIndex{3}\)

    Simplify: \(\dfrac{3a^2−12ab+12b^2}{6a^2−24b^2}\)

    Solution

    \(\begin{array} {ll} &\dfrac{3a^2−12ab+12b^2}{6a^2−24b^2} \\ & \\ & \\ \begin{array} {l} \text{Factor the numerator and denominator,} \\ \text{first factoring out the GCF.} \end{array} &\dfrac{3(a^2−4ab+4b^2)}{6(a^2−4b^2)} \\ & \\ &\dfrac{3(a−2b)(a−2b)}{6(a+2b)(a−2b)} \\ & \\ \text{Remove the common factors of }a−2b\text{ and }3. &\dfrac{\cancel{3}(a−2b)\cancel{(a−2b)}}{\cancel{3}·2(a+2b)\cancel{(a−2b)}} \\ &\dfrac{a−2b}{2(a+2b)} \end{array} \)

    You Try \(\PageIndex{3}\)

    Simplify: \(\dfrac{2x^2−12xy+18y^2}{3x^2−27y^2}\)

    Answer

    \(\dfrac{2(x−3y)}{3(x+3y)}\)

    Now we will see how to simplify a rational expression whose numerator and denominator have opposite factors. We previously introduced opposite notation: the opposite of a is \(−a\) and \(−a=−1·a\).

    The numerical fraction, say \(\dfrac{7}{−7}\) simplifies to \(−1\). We also recognize that the numerator and denominator are opposites.

    The fraction \(\dfrac{a}{−a}\), whose numerator and denominator are opposites also simplifies to \(−1\).

    \[\begin{array} {ll} \text{Let’s look at the expression }b−a. &b−a \\ \text{Rewrite.} &−a+b \\ \text{Factor out }–1. &−1(a−b) \nonumber\end{array} \]

    This tells us that \(b−a\) is the opposite of \(a−b\).

    In general, we could write the opposite of \(a−b\) as \(b−a\). So the rational expression \(\dfrac{a−b}{b−a}\) simplifies to \(−1\).

    Opposites in a Rational Expression

    The opposite of \(a−b\) is \(b−a\).

    \[\dfrac{a−b}{b−a}=−1 \quad a\neq b\nonumber\]

    An expression and its opposite divide to \(−1\).

    We will use this property to simplify rational expressions that contain opposites in their numerators and denominators. Be careful not to treat \(a+b\) and \(b+a\) as opposites. Recall that in addition, order doesn’t matter so \(a+b=b+a\). So if \(a\neq −b\), then \(\dfrac{a+b}{b+a}=1\).

    Example \(\PageIndex{4}\)

    Simplify: \(\dfrac{x^2−4x−32}{64−x^2}\)

    Solution
      .
    Factor the numerator and the denominator. .
    Recognize the factors that are opposites. .
    Simplify. .
    You Try \(\PageIndex{4}\)

    Simplify: \(\dfrac{b^2+b−2}{1−b^2}\)

    Answer

    \(−\dfrac{b+2}{b+1}\)

    Multiplying Rational Expressions

    To multiply rational expressions, we do just what we did with numerical fractions. We multiply the numerators and multiply the denominators. Then, if there are any common factors, we remove them to simplify the result.

    Multiplication of Rational Expressions

    If p, q, r, and s are polynomials where \(q\neq 0\), \(s\neq 0\), then

    \[\dfrac{p}{q}·\dfrac{r}{s}=\dfrac{pr}{qs}\nonumber\]

    To multiply rational expressions, multiply the numerators and multiply the denominators.

    Remember, throughout this chapter, we will assume that all numerical values that would make the denominator be zero are excluded. We will not write the restrictions for each rational expression, but keep in mind that the denominator can never be zero. So in this next example, x\neq 0,x\neq 0,x\neq 3,x\neq 3, and x\neq 4.x\neq 4.

    Example \(\PageIndex{5}\)

    Simplify: \(\dfrac{2x}{x^2−7x+12}·\dfrac{x^2−9}{6x^2}\)

    Solution

    Step 1 is to factor each numerator and the denominator completely in 2 x divided by the quantity x squared minus 7 x plus 12 times the rational expression the quantity x squared minus 9 divided by 6 x squared. The denominator, x squared minus 7 x plus 12, factors into the quantity x minus 3 times the quantity x minus 4. The numerator x squared minus 9 factors into the quantity x minus 3 times the quantity x plus 3.Step 2 is to multiply the numerators 2 x and the quantity x minus 3 times the quantity x plus 3, and the denominators the quantity x minus 3 times the quantity x minus 4 and 6 x squared. It is helpful to write the monomials in the numerator and in the denominator. first.Step 3 is to simplify 2 x times the quantity x minus 3 times the quantity x plus 3 all divided by 2 times 3 times x times x times the quantity x minus 3 times the quantity x plus 4 by dividing out the common factor, x minus 3. Leaving the denominator in factored form, the result is the quantity x plus 3 divided by 3 x times the quantity x minus 4.

    You Try \(\PageIndex{5}\)

    Simplify: \(\dfrac{5x}{x^2+5x+6}·\dfrac{x^2−4}{10x}\)

    Answer

    \(\dfrac{x−2}{2(x+3)}\)

    Multiplying Rational Expressions
    1. Factor each numerator and denominator completely.
    2. Multiply the numerators and denominators.
    3. Simplify by dividing out common factors.
    Example \(\PageIndex{6}\)

    Multiply: \(\dfrac{3a^2−8a−3}{a^2−25}·\dfrac{a^2+10a+25}{3a^2−14a−5}\)

    Solution

    \(\begin{array} {ll} &\dfrac{3a^2−8a−3}{a^2−25}·\dfrac{a^2+10a+25}{3a^2−14a−5} \\ & \\ \begin{array} {ll} \text{Factor the numerators and denominators} \\ \text{and then multiply.} \end{array} &\dfrac{(3a+1)(a−3)(a+5)(a+5)}{(a−5)(a+5)(3a+1)(a−5)} \\ & \\ \begin{array} {l} \text{Simplify by dividing out} \\ \text{common factors.} \end{array} &\dfrac{\cancel{(3a+1)}(a−3)\cancel{(a+5)}(a+5)}{(a−5)\cancel{(a+5)}\cancel{(3a+1)}(a−5)} \\ & \\ \text{Simplify.} &\dfrac{(a−3)(a+5)}{(a−5)(a−5)} \\ & \\ \text{We can also rewrite }(a−5)(a−5)\text{ using an exponent.} &\dfrac{(a−3)(a+5)}{(a−5)^2} \end{array}\)

    You Try \(\PageIndex{6}\)

    Simplify: \(\dfrac{4b^2+7b−2}{1−b^2}·\dfrac{b^2−2b+1}{4b^2+15b−4}\)

    Answer

    \(−\dfrac{(b+2)(b−1)}{(1+b)(b+4)}\)

    Dividing Rational Expressions

    Just like we did for numerical fractions, to divide rational expressions, we multiply the first fraction by the reciprocal of the second.

    Division of Rational Expressions

    If p, q, r, and s are polynomials where \(q\neq 0\), \(r\neq 0\), \(s\neq 0\), then

    \[\dfrac{p}{q}÷\dfrac{r}{s}=\dfrac{p}{q}·\dfrac{s}{r}\nonumber\]

    To divide rational expressions, multiply the first fraction by the reciprocal of the second.

    Once we rewrite the division as multiplication of the first expression by the reciprocal of the second, we then factor everything and look for common factors.

    Example \(\PageIndex{7}\)

    Divide: \(\dfrac{p^3+q^3}{2p^2+2pq+2q^2}÷\dfrac{p^2−q^2}{6}\)

    Solution

    Step 1 is to rewrite the division of the rational expression, the quantity p cubed plus q cubes divided by the quantity 2 p squared plus 2 p q plus 2 q squared divided by the rational expression, the quantity p squared minus q squared all divided by 6. Do this by flipping the rational expression, the quantity p squared minus q squared all divided by 6, and changing division to multiplication. The result is the quantity p cubed plus q cubes divided by the quantity 2 p squared plus 2 p q plus 2 q squared times the quantity 6 divided by the quantity p squared minus q squared.Step 2 is to factor the numerators, the quantity p cubed plus q cubed and 6, and the denominators, the quantity 2 p squared plus 2 p q plus 2 squared and the quantity p squared minus q squared, completely. The result is the quantity p plus q times the quantity p squared minus p q plus q squared all times the quantity 2 times 3 divided by the quantity p minus q times the quantity p plus q.Step 3 is to multiply the numerators and denominators. The result is the quantity p plus q times the quantity p squared minus p q plus q squared times 2 times 3 all divided by the 2 times the quantity p squared plus p q plus q squared times the quantity p minus q times the quantity p plus q.Step 4 is to simplify the expression by dividing out the common factors, the quantity p plus q and 2. The result is 3 times the quantity p squared minus p q plus q squared all divided by the quantity p minus q times the quantity p squared plus p q plus q squared.

    You Try \(\PageIndex{7}\)

    Simplify: \(\dfrac{2z^2}{z^2−1}÷\dfrac{z^3−z^2+z}{z^3+1}\)

    Answer

    \(\dfrac{2z}{z−1}\)

    Dividing Rational Expressions
    1. Rewrite the division as the product of the first rational expression and the reciprocal of the second.
    2. Factor the numerators and denominators completely.
    3. Multiply the numerators and denominators together.
    4. Simplify by dividing out common factors.

    A complex fraction is a fraction that contains a fraction in the numerator, the denominator or both. Also, remember a fraction bar means division. A complex fraction is another way of writing division of two fractions.

    Example \(\PageIndex{8}\)

    Divide: \(\dfrac{\dfrac{6x^2−7x+2}{4x−8}}{\dfrac{2x^2−7x+3}{x^2−5x+6}}\)

    Solution

    \(\begin{array} {ll} &\dfrac{\dfrac{6x^2−7x+2}{4x−8}}{\dfrac{2x^2−7x+3}{x^2−5x+6}} \\ & \\ \text{Rewrite with a division sign.} &\dfrac{6x^2−7x+2}{4x−8}÷\dfrac{2x^2−7x+3}{x^2−5x+6} \\ & \\ \begin{array} {l} \text{Rewrite as product of first times reciprocal} \\ \text{of second.} \end{array} &\dfrac{6x^2−7x+2}{4x−8}·\dfrac{x^2−5x+6}{2x^2−7x+3} \\ & \\ \begin{array} {l} \text{Factor the numerators and the} \\ \text{denominators, and then multiply.} \end{array} &\dfrac{(2x−1)(3x−2)(x−2)(x−3)}{4(x−2)(2x−1)(x−3)} \\ & \\ \text{Simplify by dividing out common factors.} &\dfrac{\cancel{(2x−1)}(3x−2)\cancel{(x−2)}\cancel{(x−3)}}{4\cancel{(x−2)}\cancel{(2x−1)}\cancel{(x−3)}} \\ \text{Simplify.} &\dfrac{3x−2}{4} \end{array}\)

    You Try \(\PageIndex{8}\)

    Simplify: \(\dfrac{\dfrac{y^2−36}{2y^2+11y−6}}{\dfrac{2y^2−2y−60}{8y−4}}\)

    Answer

    \(\dfrac{2}{y+5}\)

    If we have more than two rational expressions to work with, we still follow the same procedure. The first step will be to rewrite any division as multiplication by the reciprocal. Then, we factor and multiply.

    Example \(\PageIndex{9}\)

    Perform the indicated operations: \(\dfrac{3x−6}{4x−4}·\dfrac{x^2+2x−3}{x^2−3x−10}÷\dfrac{2x+12}{8x+16}\)

    Solution
      .
    Rewrite the division as multiplication
    by the reciprocal.    		 .
    Factor the numerators and the denominators. .
    Multiply the fractions. Bringing the constants to
    the front will help when removing common factors.    		  
    Simplify by dividing out common factors. .
    Simplify. .
    You Try \(\PageIndex{9}\)

    Perform the indicated operations: \(\dfrac{2n^2+10n}{n−1}÷\dfrac{n^2+10n+24}{n^2+8n−9}·\dfrac{n+4}{8n^2+12n}\)

    Answer

    \(\dfrac{(n+5)(n+9)}{2(n+6)(2n+3)}\)

    This page titled 2.13: Simplifying, Multiplying, and Dividing Rational Expressions is shared under a CC BY 4.0 license and was authored, remixed, and/or curated by Katherine Skelton.