Skip to main content
Mathematics LibreTexts

4.5: Simplifying Rational Expressions

  • Page ID
    92751
  • \( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)

    \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)

    \( \newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\)

    ( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\)

    \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\)

    \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\)

    \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\)

    \( \newcommand{\Span}{\mathrm{span}}\)

    \( \newcommand{\id}{\mathrm{id}}\)

    \( \newcommand{\Span}{\mathrm{span}}\)

    \( \newcommand{\kernel}{\mathrm{null}\,}\)

    \( \newcommand{\range}{\mathrm{range}\,}\)

    \( \newcommand{\RealPart}{\mathrm{Re}}\)

    \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\)

    \( \newcommand{\Argument}{\mathrm{Arg}}\)

    \( \newcommand{\norm}[1]{\| #1 \|}\)

    \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\)

    \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\AA}{\unicode[.8,0]{x212B}}\)

    \( \newcommand{\vectorA}[1]{\vec{#1}}      % arrow\)

    \( \newcommand{\vectorAt}[1]{\vec{\text{#1}}}      % arrow\)

    \( \newcommand{\vectorB}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)

    \( \newcommand{\vectorC}[1]{\textbf{#1}} \)

    \( \newcommand{\vectorD}[1]{\overrightarrow{#1}} \)

    \( \newcommand{\vectorDt}[1]{\overrightarrow{\text{#1}}} \)

    \( \newcommand{\vectE}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash{\mathbf {#1}}}} \)

    \( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)

    \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)

    \(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)

    Reducing Rational Expressions in x

    Now that we’ve discussed some fundamental ideas and techniques, let’s apply what we’ve learned to rational expressions that are functions of an independent variable (usually x). Let’s start with a simple example.

    Example \(\PageIndex{4}\)

    Reduce the rational expression \[\frac{2 x-6}{x^{2}-7 x+12} \nonumber \] to lowest terms. For what values of x is your result valid?

    Solution

    In the numerator, factor out a 2, as in 2x − 6 = 2(x − 3).

    The denominator is a quadratic trinomial with ac = (1)(12) = 12. The integer pair −3 and −4 has product 12 and sum −7, so the denominator factors as shown.

    \[\frac{2 x-6}{x^{2}-7 x+12}=\frac{2(x-3)}{(x-3)(x-4)} \nonumber \]

    Now that both numerator and denominator are factored, we can cancel common factors.

    \[\frac{2 x-6}{x^{2}-7 x+12}=\frac{2(x-3)}{(x-3)(x-4)}=\frac{2}{x-4} \nonumber \]

    Thus, we have shown that \[\frac{2 x-6}{x^{2}-7 x+12}=\frac{2}{x-4} \nonumber \]

    In equation (15), we are stating that the expression on the left (the original expression) is identical to the expression on the right for all values of x.

    Actually, there are two notable exceptions, the first of which is x = 3. If we substitute x = 3 into the left-hand side of equation (15), we get

    \[\frac{2 x-6}{x^{2}-7 x+12}=\frac{2(3)-6}{(3)^{2}-7(3)+12}=\frac{0}{0} \nonumber \]

    We cannot divide by zero, so the left-hand side of equation (15) is undefined if x = 3. Therefore, the result in equation (15) is not valid if x = 3.

    Similarly, if we insert x = 4 in the left-hand side of equation (15),

    \[\frac{2 x-6}{x^{2}-7 x+12}=\frac{2(4)-6}{(4)^{2}-7(4)+12}=\frac{2}{0} \nonumber \]

    Again, division by zero is undefined. The left-hand side of equation (15) is undefined if x = 4, so the result in equation (15) is not valid if x = 4. Note that the right-hand side of equation (15) is also undefined at x = 4.

    However, the algebraic work we did above guarantees that the left-hand side of equation (15) will be identical to the right-hand side of equation (15) for all other values of x. For example, if we substitute x = 5 into the left-hand side of equation (15),

    \[\frac{2 x-6}{x^{2}-7 x+12}=\frac{2(5)-6}{(5)^{2}-7(5)+12}=\frac{4}{2}=2 \nonumber \]

    On the other hand, if we substitute x = 5 into the right-hand side of equation (15),

    \[\frac{2}{x-4}=\frac{2}{5-4}=2 \nonumber \]

    Hence, both sides of equation (15) are identical when x = 5. In a similar manner, we could check the validity of the identity in equation (15) for all other values of x

    You can use the graphing calculator to verify the identity in equation (15). Load the left- and right-hand sides of equation (15) in Y= menu, as shown in Figure \(\PageIndex{1}\)(a). Press 2nd TBLSET and adjust settings as shown in Figure \(\PageIndex{1}\)(b). Be sure that you highlight AUTO for both independent and dependent variables and press ENTER on each to make the selection permanent. In Figure \(\PageIndex{1}\)(b), note that we’ve set TblStart = 0 and ∆Tbl = 1. Press 2nd TABLE to produce the tabular results shown in Figure \(\PageIndex{1}\)(c).

    Screen Shot 2019-07-12 at 3.54.33 PM.png
    Figure \(\PageIndex{1}\). Using the graphing calculator to check that the left- and right-hand sides of equation (15) are identical.

    Remember that we placed the left- and right-hand sides of equation (15) in Y1 and Y2, respectively.

    • In the tabular results of Figure \(\PageIndex{1}\)(c), note the ERR (error) message in Y1 when x = 3 and x = 4. This agrees with our findings above, where the left-hand side of equation (15) was undefined because of the presence of zero in the denominator when x = 3 or x = 4.
    • In the tabular results of Figure \(\PageIndex{1}\)(c), note that the value of Y1 and Y2 agree for all other values of x.

    We are led to the following key result.

    Restrictions

    In general, when you reduce a rational expression to lowest terms, the expression obtained should be identical to the original expression for all values of the variables in each expression, save those values of the variables that make any denominator equal to zero. This applies to the denominator in the original expression, all intermediate expressions in your work, and the final result. We will refer to any values of the variable that make any denominator equal to zero as restrictions.

    Let’s look at another example.

    Example \(\PageIndex{5}\)

    Reduce the expression \[\frac{2 x^{2}+5 x-12}{4 x^{3}+16 x^{2}-9 x-36} \nonumber \] to lowest terms. State all restrictions.

    Solution

    The numerator is a quadratic trinomial with ac = (2)(−12) = −24. The integer pair −3 and 8 have product −24 and sum 5. Break the middle term of the polynomial in the numerator into a sum using this integer pair, then factor by grouping.

    \[\begin{aligned} 2 x^{2}+5 x-12 &=2 x^{2}-3 x+8 x-12 \\ &=x(2 x-3)+4(2 x-3) \\ &=(x+4)(2 x-3) \end{aligned} \nonumber \]

    Factor the denominator by grouping.

    \[\begin{aligned} 4 x^{3}+16 x^{2}-9 x-36 &=4 x^{2}(x+4)-9(x+4) \\ &=\left(4 x^{2}-9\right)(x+4) \\ &=(2 x+3)(2 x-3)(x+4) \end{aligned} \nonumber \]

    Note how the difference of two squares pattern was used to factor \(4x^2 − 9 = (2x + 3)(2x − 3)\) in the last step.

    Now that we’ve factored both numerator and denominator, we cancel common factors.

    \[\begin{aligned} \frac{2 x^{2}+5 x-12}{4 x^{3}+16 x^{2}-9 x-36} &=\frac{(x+4)(2 x-3)}{(2 x+3)(2 x-3)(x+4)} \\ &=\frac{(x+4)(2 x-3)}{(2 x+3)(2 x-3)(x+4)} \\ &=\frac{1}{2 x+3} \end{aligned} \nonumber \]

    We must now determine the restrictions. This means that we must find those values of x that make any denominator equal to zero.

    • In the body of our work, we have the denominator (2x + 3)(2x − 3)(x + 4). If we set this equal to zero, the zero product property implies that \[2 x+3=0 \quad \text { or } \quad 2 x-3=0 \quad \text { or } \quad x+4=0 \nonumber \]

    Each of these linear factors can be solved independently \[x=-3 / 2 \quad \text { or } \quad x=3 / 2 \quad \text { or } \quad x=-4 \nonumber \]

    Each of these x-values is a restriction.

    • In the final rational expression, the denominator is 2x + 3. This expression equals zero when x = −3/2 and provides no new restrictions.
    • Because the denominator of the original expression, namely \(4 x^{3}+16 x^{2}-9 x-36\), is identical to its factored form in the body our work, this denominator will produce no new restrictions.

    Thus, for all values of x \[\frac{2 x^{2}+5 x-12}{4 x^{3}+16 x^{2}-9 x-36}=\frac{1}{2 x+3} \nonumber \]

    provided \(x \neq −3/2, 3/2, or −4\). These are the restrictions. The two expressions are identical for all other values of x.

    Finally, let’s check this result with our graphing calculator. Load each side of equation (18) into the Y= menu, as shown in Figure \(\PageIndex{2}\)(a). We know that we have a restriction at x = −3/2, so let’s set TblStart = −2 and \(\Delta \mathrm{Tbl}=0.5\), as shown in Figure \(\PageIndex{2}\)(b). Be sure that you have AUTO set for both independent and dependent variables. Push the TABLE button to produce the tabular display shown in Figure \(\PageIndex{2}\)(c).

    Screen Shot 2019-07-12 at 4.06.53 PM.png
    Figure \(\PageIndex{2}\). Using the graphing calculator to check that the left- and right-hand sides of equation (18) are identical.

    Remember that we placed the left- and right-hand sides of equation (18) in Y1 and Y2, respectively.

    • In Figure \(\PageIndex{2}\)(c), note that the expressions Y1 and Y2 agree at all values of x except x = −1.5. This is the restriction −3/2 we found above.
    • Use the down arrow key to scroll down in the table shown in Figure \(\PageIndex{2}\)(c) to produce the tabular view shown in Figure \(\PageIndex{2}\)(d). Note that Y1 and Y2 agree for all values of x except x = 1.5. This is the restriction 3/2 we found above.
    • We leave it to our readers to uncover the restriction at x = −4 by using the uparrow to scroll up in the table until you reach an x-value of −4. You should uncover another ERR (error) message at this x-value because it is a restriction. You get the ERR message due to the fact that the denominator of the left-hand side of equation (18) is zero at x = −4.

    Sign Changes

    It is not uncommon that you will have to manipulate the signs in a fraction in order to obtain common factors that can be then cancelled. Consider, for example, the rational expression

    \[\frac{3-x}{x-3} \nonumber \]

    One possible approach is to factor −1 out of the numerator to obtain

    \[\frac{3-x}{x-3}=\frac{-(x-3)}{x-3} \nonumber \]

    You can now cancel common factors.

    \[\frac{3-x}{x-3}=\frac{-(x-3)}{x-3}=\frac{-(x-3)}{x-3}=-1 \nonumber \]

    This result is valid for all values of x, provided \(x \neq 3\).

    Let’s look at another example.

    Example \(\PageIndex{6}\)

    Reduce the rational expression \[\frac{2 x-2 x^{3}}{3 x^{3}+4 x^{2}-3 x-4} \nonumber \] to lowest terms. State all restrictions.

    Solution

    In the numerator, factor out 2x, then complete the factorization using the difference of two squares pattern.

    \[2 x-2 x^{3}=2 x\left(1-x^{2}\right)=2 x(1+x)(1-x) \nonumber \]

    The denominator can be factored by grouping.

    \[\begin{aligned} 3 x^{3}+4 x^{2}-3 x-4 &=x^{2}(3 x+4)-1(3 x+4) \\ &=\left(x^{2}-1\right)(3 x+4) \\ &=(x+1)(x-1)(3 x+4) \end{aligned} \nonumber \]

    Note how the difference of two squares pattern was applied in the last step.

    At this point, \[\frac{2 x-2 x^{3}}{3 x^{3}+4 x^{2}-3 x-4}=\frac{2 x(1+x)(1-x)}{(x+1)(x-1)(3 x+4)} \nonumber \]

    Because we have 1 − x in the numerator and x − 1 in the denominator, we will factor out a −1 from 1 − x, and because the order of factors does not affect their product, we will move the −1 out to the front of the numerator.

    \[\frac{2 x-2 x^{3}}{3 x^{3}+4 x^{2}-3 x-4}=\frac{2 x(1+x)(-1)(x-1)}{(x+1)(x-1)(3 x+4)}=\frac{-2 x(1+x)(x-1)}{(x+1)(x-1)(3 x+4)} \nonumber \]

    We can now cancel common factors.

    \[\begin{aligned} \frac{2 x-2 x^{3}}{3 x^{3}+4 x^{2}-3 x-4} &=\frac{-2 x(1+x)(x-1)}{(x+1)(x-1)(3 x+4)} \\ &=\frac{-2 x(1+x)(x-1)}{(x+1)(x-1)(3 x+4)} \\ &=\frac{-2 x}{3 x+4} \end{aligned} \nonumber \]

    Note that x + 1 is identical to 1 + x and cancels. Thus,

    \[\frac{2 x-2 x^{3}}{3 x^{3}+4 x^{2}-3 x-4}=\frac{-2 x}{3 x+4} \nonumber \]

    for all values of x, provided \(x \neq-1,1,\) or \(-4 / 3\). These are the restrictions, values of x that make denominators equal to zero.

    The Sign Change Rule for Fractions

    Let’s look at an alternative approach to the last example. First, let’s share the precept that every fraction has three signs, one on the numerator, one on the denominator, and a third on the fraction bar. Thus,

    \[\frac{-2}{3} \quad \text { has understood signs } \quad+\frac{-2}{+3} \nonumber \]

    Let’s state the sign change rule for fractions.

    The Sign Change Rule for Fractions

    Every fraction has three signs, one on the numerator, one on the denominator, and one on the fraction bar. If you don’t see an explicit sign, then a plus sign is understood. If you negate any two of these parts,

    • numerator and denominator, or
    • numerator and fraction bar, or
    • fraction bar and denominator,

    then the fraction remains unchanged.

    For example, let’s start with −2/3, then do two negations: numerator and fraction bar. Then,

    \[+\frac{-2}{+3}=-\frac{+2}{+3}, \quad \text { or with understood plus signs, } \quad \frac{-2}{3}=-\frac{2}{3} \nonumber \]

    This is a familiar result, as negative two divided by a positive three equals a negative two-thirds.

    On another note, we might decide to negate numerator and denominator. Then −2/3 becomes

    \[+\frac{-2}{+3}=\frac{+2}{-3}, \quad \text { or with understood plus signs, } \quad \frac{-2}{3}=\frac{2}{-3} \nonumber \]

    Again, a familiar result. Certainly, negative two divided by positive three is the same as positive two divided by negative three. They both equal minus two-thirds.

    So there you have it. Negate any two parts of a fraction and it remains unchanged. On the surface, this seems a trivial remark, but it can be put to good use when reducing rational expressions. Suppose, for example, that we take the original rational expression from Example \(\PageIndex{6}\) and negate the numerator and fraction bar.

    \[\frac{2 x-2 x^{3}}{3 x^{3}+4 x^{2}-3 x-4}=-\frac{2 x^{3}-2 x}{3 x^{3}+4 x^{2}-3 x-4} \nonumber \]

    Note how we’ve made two sign changes. We’ve negated the fraction bar, we’ve negated the numerator \(\left(-\left(2 x-2 x^{3}\right)=2 x^{3}-2 x\right)\), and left the denominator alone. Therefore, the fraction is unchanged according to our sign change rule.

    Now, factor and cancel common factors (we leave the steps for our readers — they’re similar to those we took in Example \(\PageIndex{6}\)).

    \[\begin{aligned} \frac{2 x-2 x^{3}}{3 x^{3}+4 x^{2}-3 x-4} &=-\frac{2 x^{3}-2 x}{3 x^{3}+4 x^{2}-3 x-4} \\ &=-\frac{2 x(x+1)(x-1)}{(x+1)(x-1)(3 x+4)} \\ &=-\frac{2 x(x+1)(x-1)}{(x+1)(x-1)(3 x+4)} \\ &=-\frac{2 x}{3 x+4} \end{aligned} \nonumber \]

    But does this answer match the answer in equation (22)? It does, as can be seen by making two negations, fraction bar and numerator.

    \[-\frac{2 x}{3 x+4}=\frac{-2 x}{3 x+4} \nonumber \]


    This page titled 4.5: Simplifying Rational Expressions is shared under a CC BY-NC-SA license and was authored, remixed, and/or curated by David Arnold.

    • Was this article helpful?