8.1: Reduce Rational Expressions
- Page ID
- 45075
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)A rational expression is a ratio of two polynomials, i.e., a fraction where the numerator and denominator are polynomials.
Evaluate Rational Expressions
Evaluate \(\dfrac{x^2-4}{x^2+6x+8}\) when \(x=-6\).
Solution
\[\begin{array}{rl}\dfrac{x^2-4}{x^2+6x+8}&\text{Plug-n-chug }x=-6 \\ \dfrac{\color{blue}{(-6)}\color{black}{^2}-4}{\color{blue}{(-6)}\color{black}{^2}+6\color{blue}{(-6)}\color{black}{}+8}&\text{Simplify each numerator and denominator} \\ \dfrac{36-4}{36-36+8}&\text{Simplify} \\ \dfrac{32}{8}&\text{Reduce} \\ 4&\text{Evaluated value}\end{array}\nonumber\]
Find Excluded Values of Rational Expressions
Rational expressions are special types of fractions, but still hold the same arithmetic properties. One property of fractions we recall is that the fraction is undefined when the denominator is zero.
A rational expression is undefined where the denominator is zero.
Step 1. Set the denominator of the rational expression equal to zero.
Step 2. Solve the equation for the given variable.
Step 3. The values found in the previous step are the values excluded from the expression.
Find the excluded value(s) of the expression: \(\dfrac{-3z}{z+5}\)
Solution
Step 1. Set the denominator of the rational expression equal to zero: \[z+5=0\nonumber\]
Step 2. Solve the equation for \(z\): \[\begin{aligned}z+5&=0 \\ z&=-5\end{aligned}\]
Step 3. The values found in the previous step are the values excluded from the expression. Hence, the excluded value is \(z = −5\).
Find the excluded value(s) of the expression: \(\dfrac{x^2-1}{3x^2+5x}\)
Solution
Step 1. Set the denominator of the rational expression equal to zero: \[3x^2+5x=0\nonumber\]
Step 2. Solve the equation for \(x\): \[\begin{aligned} 3x^2+5x&=0 \\ x(3x+5)&=0 \\ x=0\quad &\text{or}\quad 3x+5=0 \\ x=0\quad &\text{or}\quad 3x=-5 \\ x=0\quad &\text{or} \quad x=-\dfrac{5}{3}\end{aligned}\]
Step 3. The values found in the previous step are the values excluded from the expression. Hence, the excluded values are \(x = 0\) and \(x = −5\).
Recall, the excluded values are values in which make the expression undefined. Hence, when evaluating rational expressions, we can evaluate the expressions for any values except the excluded values.
The number zero was not widely accepted in mathematical thought around the world for many years. It was the Mayans of Central America who first used zero to aid in the use of their base-20 system as a place holder.
Reduce Rational Expressions with Monomials
Rational expressions are reduced, just as in arithmetic, even without knowing the value of the variable. When we reduce, we divide out common factors as we discussed with polynomial division with monomials. Now, we use factoring techniques and exponent properties to reduce rational expressions.
If \(P,\: Q,\: K\) are non-zero polynomials and \(\dfrac{PK}{QK}\) is a rational expression, then \[\dfrac{P\cdot\cancel{K}}{Q\cdot\cancel{K}}=\dfrac{P}{Q}\nonumber\] We call a rational expression irreducible if there are no more common factors among the numerator and denominator.
Simplify: \(\dfrac{15x^4y^2}{25x^2y^6}\)
Solution
Since the denominator is a monomial, then we reduce as usual and apply exponent rules:
\[\begin{array}{rl}\dfrac{15x^4y^2}{25x^2y^6}&\text{Reduce by applying exponent rules} \\ \dfrac{3x^2}{5y^4}&\text{Reduced expression}\end{array}\nonumber\]
Reduce Rational Expressions with Polynomials
However, if there is a sum or difference in either the numerator or denominator, we first factor the numerator and denominator to obtain a product of factors, then reduce.
Simplify: \(\dfrac{28}{8x^2-16}\)
Solution
Since we have a difference in the denominator, we factor the denominator and then reduce.
\[\begin{array}{rl}\dfrac{28}{8x^2-16}&\text{Factor a GCF }8\text{ from the denominator} \\ \dfrac{\cancel{4}\cdot 7}{2\cdot\cancel{4}(x^2-2)}&\text{Reduce by a factor of }4 \\ \dfrac{7}{2(x^2-2)}&\text{Reduced expression}\end{array}\nonumber\]
Simplify: \(\dfrac{9x-3}{18x-6}\)
Solution
Since we have a difference in the denominator and numerator, we factor the denominator and numerator, and then reduce.
\[\begin{array}{rl}\dfrac{9x-3}{18x-6}&\text{Factor the GCF from numerator and denominator} \\ \dfrac{3\color{blue}{(3x-1)}}{6\color{blue}{(3x-1)}}&\color{black}{\text{Reduce by a factor of }}3(3x-1) \\ \dfrac{\cancel{3}\color{blue}{\cancel{(3x-1)}}}{2\cdot\cancel{3}\cdot\color{blue}{\cancel{(3x-1)}}}&\color{black}{\text{Rewrite the expression}} \\ \dfrac{1}{2}&\text{Reduced expression}\end{array}\nonumber\]
Simplify: \(\dfrac{x^2-25}{x^2+8x+15}\)
Solution
Since we have a sum and difference of terms in the denominator and numerator, we factor the denominator and numerator, and then reduce.
\[\begin{array}{rl}\dfrac{x^2-25}{x^2+8x+15}&\text{Factor using factoring techniques} \\ \dfrac{\color{blue}{(x+5)}\color{black}{}(x-5)}{(x+3)\color{blue}{(x+5)}}&\color{black}{\text{Reduce by a factor of }}(x+5) \\ \dfrac{\color{blue}{\cancel{(x+5)}}\color{black}{}(x-5)}{(x+3)\color{blue}{\cancel{(x+5)}}}&\color{black}{\text{Rewrite the expression}} \\ \dfrac{x-5}{x+3}&\text{Reduced expression}\end{array}\nonumber\]
We cannot reduce terms, only factors. This means we cannot reduce anything with a \(+\) or \(−\) between the parts. In Example 8.1.7 , we obtained the reduced expression \(\dfrac{x−5}{x+3}\). Note, we are not allowed to divide out the \(x\)’s because they are terms (separated by \(+\) or \(−\)) not factors (separated by multiplication).
Reduce Rational Expressions Homework
Evaluate the expression for the given value.
\(\dfrac{4v+2}{6}\) when \(v=4\)
\(\dfrac{x-3}{x^2-4x+3}\) when \(x=-4\)
\(\dfrac{b+2}{b^2+4b+4}\) when \(b=0\)
\(\dfrac{b-3}{3b-9}\) when \(b=-2\)
\(\dfrac{a+2}{a^2+3a+2}\) when \(a=-1\)
\(\dfrac{n^2-n-6}{n-3}\) when \(n=4\)
Find the excluded value(s).
\(\dfrac{3k^2+30k}{k+10}\)
\(\dfrac{15n^2}{10n+25}\)
\(\dfrac{10m^2+8m}{10m}\)
\(\dfrac{r^2+3r+2}{5r+10}\)
\(\dfrac{b^2+12b+32}{b^2+4b-32}\)
\(\dfrac{27p}{18p^2-36p}\)
\(\dfrac{x+10}{8x^2+80x}\)
\(\dfrac{10x+16}{6x+20}\)
\(\dfrac{6n^2-21n}{6n^2+3n}\)
Simplify each expression.
\(\dfrac{21x^2}{18x}\)
\(\dfrac{24a}{40a^2}\)
\(\dfrac{32x^3}{8x^4}\)
\(\dfrac{18m-24}{60}\)
\(\dfrac{20}{4p+2}\)
\(\dfrac{x+1}{x^2+8x+7}\)
\(\dfrac{32x^2}{28x^2+28x}\)
\(\dfrac{n^2+4n-12}{n^2-7n+10}\)
\(\dfrac{9v+54}{v^2-4v-60}\)
\(\dfrac{12x^2-42x}{30x^2-42x}\)
\(\dfrac{6a-10}{10a+4}\)
\(\dfrac{2n^2+19n-10}{9n+90}\)
\(\dfrac{21k}{24k^2}\)
\(\dfrac{90x^2}{20x}\)
\(\dfrac{10}{81n^3+36n^2}\)
\(\dfrac{n-9}{9n-81}\)
\(\dfrac{28m+12}{36}\)
\(\dfrac{49r+56}{56r}\)
\(\dfrac{b^2+14b+48}{b^2+15b+56}\)
\(\dfrac{30x-90}{50x+40}\)
\(\dfrac{k^2-12k+32}{k^2-64}\)
\(\dfrac{9p+18}{p^2+4p+4}\)
\(\dfrac{3x^2-29x+40}{5x^2-30x-80}\)
\(\dfrac{8m+16}{20m-12}\)
\(\dfrac{2x^2-10x+8}{3x^2-7x+4}\)
\(\dfrac{7n^2-32n+16}{4n-16}\)
\(\dfrac{n^2-2n+1}{6n+6}\)
\(\dfrac{7a^2-26a-45}{6a^2-34a+20}\)
\(\dfrac{56x-48}{24x^2+56x+32}\)
\(\dfrac{50b-80}{50b+20}\)
\(\dfrac{35v+35}{21v+7}\)
\(\dfrac{56x-48}{24x^2+56x+32}\)
\(\dfrac{4k^3-2k^2-2k}{9k^3-18k^2+9k}\)