8.3: Reducing Rational Expressions
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The Logic Behind The Process
When working with rational expressions, it is often best to write them in the simplest possible form. For example, the rational expression
\[\dfrac{x^2 - 4}{x^2 - 6x + 8} \nonumber\]
can be reduced to the simpler expresion \(\dfrac{x+2}{x-4}\) for all \(x\) except \(x = 2, 4\).
From our discussion of equality of fractions in Section 8.2, we know that \(\dfrac{a}{b} = \dfrac{c}{d}\) when \(ad=bc\). This fact allows us to deduce that, if \(k≠0, \dfrac{ak}{bk} = \dfrac{a}{b}\), since \(akb = abk\) (recall the commutative property of multiplication). But this fact means that if a factor (in this case, \(k\)) is common to both the numerator and denominator of a fraction, we may remove it without changing the value of the fraction.
\[\dfrac{ak}{bk} = \dfrac{a \not k}{b \not k} = \dfrac{a}{b} \nonumber\]
Cancelling
The process of removing common factors is commonly called canceling.
\(\dfrac{16}{40}\) can be reduced to \(\dfrac{2}{5}\).
Process:
\(\dfrac{16}{40} = \dfrac{2 \cdot 2 \cdot 2 \cdot 2}{2 \cdot 2 \cdot 2 \cdot 5}\)
Remove the three factors of \(1\); \(\dfrac{2}{2} \cdot \dfrac{2}{2} \cdot \dfrac{2}{2}\).
\(\dfrac{\not 2 \cdot \not 2 \cdot \not 2 \cdot 2}{\not 2 \cdot \not 2 \cdot \not 2 \cdot 5} = \dfrac{2}{5}\)
Notice that in \(\dfrac{2}{5}\), there is no factor common to the numerator and denominator.
\(\dfrac{111}{148}\) can be reduced to \(\dfrac{3}{4}\). Process:
\(\dfrac{111}{148} = \dfrac{3 \cdot 37}{4 \cdot 37}\)
Remove the factor of \(1\); \(\dfrac{37}{37}\).
\(\dfrac{3 \cdot \cancel{37}}{4 \cdot \cancel{37}}\).
\(\dfrac{3}{4}\)
Notice that in \(\dfrac{3}{4}\), there is no other factor common to the numerator and denominator.
\(\dfrac{3}{9}\) can be reduced to \(\dfrac{1}{3}\). Process:
\(\dfrac{3}{9} = \dfrac{3 \cdot 1}{3 \cdot 3}\).
Remove the factor of \(1\); \(\dfrac{3}{3}\).
\(\dfrac{\not 3 \cdot 1}{\not 3 \cdot 3} = \dfrac{1}{3}\)
Notice that in \(\dfrac{1}{3}\) there is no factor common to the numerator and denominator.
\(\dfrac{5}{7}\) cannot be reduced since there are no factors common to the numerator and denominator.
Problems 1, 2, and 3 shown above could all be reduced. The process in each reduction included the following steps:
- Both the numerator and denominator were factored.
- Factors that were common to both the numerator and denominator were noted and removed by dividing them out.
We know that we can divide both sides of an equation by the same nonzero number, but why should we be able to divide both the numerator and denominator of a fraction by the same nonzero number? The reason is that any nonzero number divided by itself is 1, and that if a number is multiplied by 1, it is left unchanged.
Consider the fraction \(\dfrac{6}{24}\). Multiply this fraction by \(1\). This is written \(\dfrac{6}{24} \cdot 1\). But \(1\) can be rewritten as \(\dfrac{\dfrac{1}{6}}{\dfrac{1}{6}}\).
\(\dfrac{6}{24} \cdot \dfrac{\dfrac{1}{6}}{\dfrac{1}{6}} = \dfrac{6 \cdot \dfrac{1}{6}}{24 \cdot \dfrac{1}{6}} = \dfrac{1}{4}\).
The answer, \(\dfrac{1}{4}\), is the reduced form. Notice that in \(\dfrac{1}{4}\) there is no factor common to both the numerator and the denominator. This reasoning provides justification for the following rule.
Multiplying or dividing the numerator and denominator by the same nonzero number does not change the value of a fraction.
The Process
We can now state a process for reducing a rational expression.
- Factor the numerator and denominator completely.
- Divide the numerator and denominator by all factors they have in common, that is, remove all factors of 1.
A rational expression is said to be reduced to lowest terms when the numerator and denominator have no factors in common.
Sample Set A
Reduce the following rational expressions.
\(\dfrac{15x}{20x}\) Factor.
\(\dfrac{15x}{20x} = \dfrac{5 \cdot 3 \cdot x}{5 \cdot 2 \cdot 2 \cdot x}\). The factors that are common to both the numerator and denominator are \(5\) and \(x\). Divide each by \(5x\).
\(\dfrac{\cancel{5} \cdot 3 \cdot \cancel{x}}{\cancel{5} \cdot 2 \cdot 2 \cdot \cancel{x}} = \dfrac{3}{4}, x \not = 0\).
It is helpful to draw a line through the divided out factors.
\(\dfrac{x^2 - 4}{x^2 - 6x + 8}\). Factor.
\(\dfrac{(x+2)(x-2)}{(x-2)(x-4)}\). The factor that is common to both the numerator and denominator is \(x-2\). Divide each by \(x-2\).
\(\dfrac{(x+2)\cancel{(x-2)}}{\cancel{(x-2)}(x-4)} = \dfrac{x+2}{x-4}, x \not = 2, 4\).
The expression \(\dfrac{x-2}{x-4}\) is the reduced form since there are no factors common to both the numerator and denominator. Although there is an \(x\) in both, it is a common term, not a common factor, and therefore cannot be divided out.
CAUTION - This is a common error: \(\dfrac{x-2}{x-4} = \dfrac{\cancel{x}-2}{\cancel{x}-4} = \dfrac{2}{3}\) is incorrect!
\(\dfrac{a+2b}{6a+12b}\). Factor.
\(\dfrac{a+2b}{6(a+2b)} = \dfrac{\cancel{a+2b}}{6 \cancel{(a+2b)}} = \dfrac{1}{6}, a \not = -2b\).
Since \(a+2b\) is a common factor to both the numerator and denominator, we divide both by \(a+2b\). Since \(\dfrac{(a+2b)}{(a+2b)} = 1\), we get \(1\) in the numerator.
Sometimes we may reduce a rational expression by using the division rule of exponents.
\(\dfrac{8x^2y^5}{4xy^2}\). Factor and use the rule \(\dfrac{a^n}{a^m} = a^{n-m}\).
\(\dfrac{8x^2y^5}{4xy^2} = \dfrac{2 \cdot 2 \cdot 2}{2 \cdot 2}x^{2-1}y^{5-2}\)
\( = 2xy^3, x \not = 0, y \not = 0\)
\(\dfrac{-10x^3a(x^2-36)}{2x^3-10x^2-12x}\). Factor.
\(\begin{aligned}
\dfrac{-10x^3a(x^2-36)}{2x^3-10x^2-12x} &= \dfrac{-5 \cdot 2x^3a(x+6)(x-6)}{2x(x^2 - 5x - 6)}\\
&= \dfrac{-5 \cdot 2x^3a(x+6)(x-6)}{2x(x-6)(x+1)}\\
&= \dfrac{-5 \cdot \cancel{2}x^{\cancel{3}}a(x+6)(x-6)}{\cancel{2} \cancel{x} \cancel{(x-6)} (x+1)}\\
&= \dfrac{-5x^2a(x+6)}{x-1}, x \not = -1, 6
\end{aligned}\)
\(\dfrac{x^2 - x - 12}{-x^2 + 2x + 8}\). Since it is most convenient to have the leading terms of a polynomial positive, factor out \(-1\) from the denominator.
\(\dfrac{x^2 - x - 12}{-(x^2 - 2x - 8)}\). Rewrite this.
\(-\dfrac{x^2 - x - 12}{x^2 - 2x - 8}\). Factor
\(-\dfrac{\cancel{(x-4)}(x+3)}{\cancel{(x-4)}(x+2)}\)
\(-\dfrac{x+3}{x+2} = \dfrac{-(x+3)}{x+2} = \dfrac{-x-3}{x+2}, x \not = -2, 4\)
\(\dfrac{a-b}{b-a}\). The numerator and denominator have the same terms but they occur with opposite signs. Factor \(-1\) from the denominator.
\(\dfrac{a-b}{-(-b+a)} = \dfrac{a-b}{-(a-b)} = -\dfrac{\cancel{a-b}}{\cancel{a-b}} = -1, a \not = b\)
Practice Set A
Reduce each of the following fractions to lowest terms.
\(\dfrac{30y}{35y}\)
- Answer
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\(\dfrac{6}{7}\)
\(\dfrac{x^2-9}{x^2+5x+6}\)
- Answer
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\(\dfrac{x-3}{x+2}\)
\(\dfrac{x+2b}{4x+8b}\)
- Answer
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\(\dfrac{1}{4}\)
\(\dfrac{18a^3b^5c^7}{3ab^3c^5}\)
- Answer
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\(6a^2b^2c^2\)
\(\dfrac{-3a^4+75a^2}{2a^3-16a^2+30a}\)
- Answer
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\(\dfrac{-3a(a+5)}{2(a-3)}\)
\(\dfrac{x^2-5x+4}{-x^2+12x-32}\)
- Answer
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\(\dfrac{-x+1}{x-8}\)
\(\dfrac{2x-y}{y-2x}\)
- Answer
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\(-1\)
Exercises
For the following problems, reduce each rational expression to the lowest terms.
\(\dfrac{6}{3x-12}\)
- Answer
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\(\dfrac{2}{(x-4)}\)
\(\dfrac{8}{4a-16}\)
\(\dfrac{9}{3y-21}\)
- Answer
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\(\dfrac{3}{(y-7)}\)
\(\dfrac{10}{5x-5}\)
\(\dfrac{7}{7x-14}\)
- Answer
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\(\dfrac{1}{(x-2)}\)
\(\dfrac{6}{6x - 18}\)
\(\dfrac{2y^2}{8y}\)
- Answer
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\(\dfrac{1}{4}y\)
\(\dfrac{4x^3}{2x}\)
\(\dfrac{16a^2b^3}{2ab^2}\)
- Answer
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\(8ab\)
\(\dfrac{20a^4b^4}{4ab^2}\)
\(\dfrac{(x+3)(x-2)}{(x+3)(x+5)}\)
- Answer
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\(\dfrac{x-2}{x+5}\)
\(\dfrac{(y-1)(y-7)}{(y-1)(y+6)}\)
\(\dfrac{(a+6)(a-5)}{(a-5)(a+2)}\)
- Answer
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\(\dfrac{a+6}{a+2}\)
\(\dfrac{(m-3)(m-1)}{(m-1)(m+4)}\)
\(\dfrac{(y-2)(y-3)}{(y-3)(y-2)}\)
- Answer
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\(1\)
\(\dfrac{(x+7)(x+8)}{(x+8)(x+7)}\)
\(\dfrac{-12x^2(x+4)}{4x}\)
- Answer
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\(−3x(x+4)\)
\(\dfrac{-3a^4(a-1)(a+5)}{-2a^3(a-1)(a+9)}\)
\(\dfrac{6x^2y^5(x-1)(x+4)}{-2xy(x+4)}\)
- Answer
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\(-3xy^4(x-1)\)
\(\dfrac{22a^4b^6c^7(a+2)(a-7)}{4c(a+2)(a-5)}\)
\(\dfrac{(x+10)^3}{x+10}\)
- Answer
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\((x+10)^2\)
\(\dfrac{(y-6)^7}{y-6}\)
\(\dfrac{(x-8)^2(x+6)^4}{(x-8)(x+6)}\)
- Answer
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\((x-8)(x+6)^3\)
\(\dfrac{(y-2)^6(y-1)^4}{(y-2)^3(y-1)^2}\)
- Answer
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\((y-2)^3(y-1)^2\)
\(\dfrac{(x+10)^5(x-6)^3}{(x-6)(x+10)^2}\)
\(\dfrac{(a+6)^2(a-7)^6}{(a+6)^5(a-7)^2}\)
- Answer
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\(\dfrac{(a-7)^4}{(a+6)^3}\)
\(\dfrac{(m+7)^4(m-8)^5}{(m+7)^7(m-8)^2}\)
\(\dfrac{(a+2)(a-1)^3}{(a+1)(a-1)}\)
- Answer
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\(\dfrac{(a+2)(a-1)^2}{(a+1)}\)
\(\dfrac{(b+6)(b-2)^4}{(b-1)(b-2)}\)
\(\dfrac{8(x+2)^3(x-5)^6}{2(x+2)(x-5)^2}\)
- Answer
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\(4(x+2)^2(x-5)^4\)
\(\dfrac{14(x-4)^3(x-10)^6}{-7(x-4)^2(x-10)^2}\)
\(\dfrac{x^2+3x-10}{x^2+2x-15}\)
\(\dfrac{x^2 - 10x + 21}{x^2 - 6x - 7}\)
- Answer
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\(\dfrac{(x-3)}{(x+1)}\)
\(\dfrac{x^2 + 10x + 24}{x^2 + 6x}\)
\(\dfrac{x^2 + 9x + 14}{x^2 + 7x}\)
- Answer
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\(\dfrac{(x+2)}{x}\)
\(\dfrac{6b^2-b}{6b^2+11b-2}\)
\(\dfrac{3b^2 + 10b + 3}{3b^2 + 7b + 2}\)
- Answer
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\(\dfrac{b+3}{b+2}\)
\(\dfrac{4b^2-1}{2b^2 + 5b - 3}\)
\(\dfrac{16a^2 - 9}{4a^2 - a - 3}\)
- Answer
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\(\dfrac{(4a-3)}{(a-1)}\)
\(\dfrac{20x^2 + 28xy + 9y^2}{4x^2 + 4xy + y^2}\)
For the following problems, reduce each rational expression if possible. If not possible, state the answer in lowest terms.
\(\dfrac{x+3}{x+4}\)
- Answer
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\(\dfrac{(x+3)}{(x+4)}\)
\(\dfrac{a+7}{a-1}\)
\(\dfrac{3a+6}{3}\)
- Answer
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\(a+2\)
\(\dfrac{4x + 12}{4}\)
\(\dfrac{5a-5}{-5}\)
- Answer
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\(-(a - 1)\) or \(-a + 1\)
\(\dfrac{6b - 6}{-3}\)
\(\dfrac{8x - 16}{-4}\)
- Answer
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\(−2(x−2)\)
\(\dfrac{4x - 7}{-7}\)
\(\dfrac{-3x + 10}{10}\)
- Answer
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\(\dfrac{-3x + 10}{10}\)
\(\dfrac{x - 2}{2 - x}\)
\(\dfrac{a - 3}{3 - a}\)
- Answer
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\(-1\)
\(\dfrac{x^3 - x}{x}\)
\(\dfrac{y^4 - y}{y}\)
- Answer
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\(y^3 - 1\)
\(\dfrac{a^5 -a^2}{a}\)
\(\dfrac{a^6 - a^4}{a^3}\)
- Answer
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\(a(a+1)(a−1)\)
\(\dfrac{4b^2 + 3b}{b}\)
\(\dfrac{2a^3 + 5a}{a}\)
- Answer
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\(2a^2 + 5\)
\(\dfrac{a}{a^3 + a}\)
\(\dfrac{x^4}{x^5 - 3x}\)
- Answer
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\(\dfrac{x^3}{x^4 - 3}\)
\(\dfrac{-a}{-a^2-a}\)
Exercises For Review
Write \((\dfrac{4^4a^8b^{10}}{4^2a^6b^2})^{-1}\) so that only positive exponenets appear.
- Answer
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\(\dfrac{1}{16a^2b^8}\)
Factor \(y^4 - 16\)
Factor \(10x^2 - 17x + 3\)
- Answer
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\((5x−1)(2x−3)\)
Supply the missing word. An equation expressed in the form \(ax+by=c\) is said to be expressed in ____ form.
Find the domain of the rational expression: \(\dfrac{2}{x^2 - 3x - 18}\)
- Answer
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\(x≠−3, 6\)