Rational expressions: reducing rational expressions (Page 2/2)

Algebra ii for the community Page 2 / 2

Consider the fraction $\frac{6}{24}$ . Multiply this fraction by 1. This is written $\frac{6}{24} \cdot 1$ . But 1 can be rewritten as $\frac{\frac{1}{6}}{\frac{1}{6}}$ .

$\frac{6}{24} \cdot \frac{\frac{1}{6}}{\frac{1}{6}} = \frac{6 \cdot \frac{1}{6}}{24 \cdot \frac{1}{6}} = \frac{1}{4}$

The answer, $\frac{1}{4}$ , is the reduced form. Notice that in $\frac{1}{4}$ there is no factor common to both the numerator and denominator. This reasoning provides justification for the following rule.

Cancelling

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.

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.

Reduced to lowest terms

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.

$\begin{array}{l} \begin{array}{l} \frac{15 x}{20 x} . & Factor . \\ \frac{15 x}{20 x} = \frac{5 \cdot 3 \cdot x}{5 \cdot 2 \cdot 2 \cdot x} & \begin{array}{l} The factors that are common to both the numerator and \\ denominator are 5 and x . Divide each by 5 x . \end{array} \end{array} \\ \frac{5 \cdot 3 \cdot x}{5 \cdot 2 \cdot 2 \cdot x} = \frac{3}{4}, x \neq 0 \\ It is helpful to draw a line through the divided-out factors . \end{array}$

$\begin{array}{l} \begin{array}{l} \frac{x^{2} - 4}{x^{2} - 6 x + 8} . & Factor. \\ \frac{(x + 2) (x - 2)}{(x - 2) (x - 4)} & \begin{array}{l} The factor that is common to both the numerator \\ and denominator is x - 2. Divide each by x - 2. \end{array} \end{array} \\ \frac{(x + 2) (x - 2)}{(x - 2) (x - 4)} = \frac{x + 2}{x - 4}, x \neq 2, 4 \end{array}$

The expression $\frac{x - 2}{x - 4}$ 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: $\frac{x - 2}{x - 4} = \frac{x - 2}{x - 4} = \frac{2}{4}$ is incorrect!

$\begin{array}{l} \begin{array}{l} \frac{a + 2 b}{6 a + 12 b} . & Factor . \end{array} \\ \frac{a + 2 b}{6 (a + 2 b)} = \frac{a + 2 b}{6 (a + 2 b)} = \frac{1}{6}, a \neq - 2 b \end{array}$
Since $a + 2 b$ is a common factor to both the numerator and denominator, we divide both by $a + 2 b$ . Since $\frac{(a + 2 b)}{(a + 2 b)} = 1$ , we get 1 in the numerator.

Sometimes we may reduce a rational expression by using the division rule of exponents.

$\begin{array}{l} \frac{8 x^{2} y^{5}}{4 x y^{2}} . & Factor and use the rule \frac{a^{n}}{a^{m}} = a^{n - m} . \\ \frac{8 x^{2} y^{5}}{4 x y^{2}} & = & \frac{2 \cdot 2 \cdot 2}{2 \cdot 2} x^{2 - 1} y^{5 - 2} \\ = & 2 x y^{3}, x \neq 0, y \neq 0 \end{array}$

$\begin{array}{l} \frac{- 10 x^{3} a (x^{2} - 36)}{2 x^{3} - 10 x^{2} - 12 x} . & Factor . \\ \frac{- 10 x^{3} a (x^{2} - 36)}{2 x^{3} - 10 x^{2} - 12 x} & = & \frac{- 5 \cdot 2 x^{3} a (x + 6) (x - 6)}{2 x (x^{2} - 5 x - 6)} \\ = & \frac{- 5 \cdot 2 x^{3} a (x + 6) (x - 6)}{2 x (x - 6) (x + 1)} \\ = & \frac{- 5 \cdot 2 x^{\begin{array}{l} 2 \\ 3 \end{array}} a (x + 6) (x - 6)}{2 x (x - 6) (x + 1)} \\ = & \frac{- 5 x^{2} a (x + 6)}{x - 1}, x \neq - 1, 6 \end{array}$

$\begin{array}{l} \begin{array}{l} \frac{x^{2} - x - 12}{- x^{2} + 2 x + 8} . & \begin{array}{l} Since it is most convenient to have the leading terms of a \\ polynomial positive, factor out - 1 from the denominator . \end{array} \\ \frac{x^{2} - x - 12}{- (x^{2} - 2 x - 8)} & Rewrite this . \\ - \frac{x^{2} - x - 12}{x^{2} - 2 x - 8} & Factor . \\ - \frac{(x - 4) (x + 3)}{(x - 4) (x + 2)} \end{array} \\ - \frac{x + 3}{x + 2} = \frac{- (x + 3)}{x + 2} = \frac{- x - 3}{x + 2}, x \neq - 2, 4 \end{array}$

$\begin{array}{l} \begin{array}{l} \frac{a - b}{b - a} . & \begin{array}{l} The numerator and denominator have the same terms but they \\ occur with opposite signs . Factor - 1 from the denominator . \end{array} \end{array} \\ \frac{a - b}{- (- b + a)} = \frac{a - b}{- (a - b)} = - \frac{a - b}{a - b} = - 1, a \neq b \end{array}$

Practice set a

Reduce each of the following fractions to lowest terms.

$\frac{30 y}{35 y}$

$\frac{6}{7}$

$\frac{x^{2} - 9}{x^{2} + 5 x + 6}$

$\frac{x - 3}{x + 2}$

$\frac{x + 2 b}{4 x + 8 b}$

$\frac{1}{4}$

$\frac{18 a^{3} b^{5} c^{7}}{3 a b^{3} c^{5}}$

$6 a^{2} b^{2} c^{2}$

$\frac{- 3 a^{4} + 75 a^{2}}{2 a^{3} - 16 a^{2} + 30 a}$

$\frac{- 3 a (a + 5)}{2 (a - 3)}$

$\frac{x^{2} - 5 x + 4}{- x^{2} + 12 x - 32}$

$\frac{- x + 1}{x - 8}$

$\frac{2 x - y}{y - 2 x}$

−1

Excercises

For the following problems, reduce each rational expression to lowest terms.

$\frac{6}{3 x - 12}$

$\frac{2}{(x - 4)}$

$\frac{8}{4 a - 16}$

$\frac{9}{3 y - 21}$

$\frac{3}{(y - 7)}$

$\frac{10}{5 x - 5}$

$\frac{7}{7 x - 14}$

$\frac{1}{(x - 2)}$

$\frac{6}{6 x - 18}$

$\frac{2 y^{2}}{8 y}$

$\frac{1}{4} y$

$\frac{4 x^{3}}{2 x}$

$\frac{16 a^{2} b^{3}}{2 a b^{2}}$

$8 a b$

$\frac{20 a^{4} b^{4}}{4 a b^{2}}$

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

$\frac{x - 2}{x + 5}$

$\frac{(y - 1) (y - 7)}{(y - 1) (y + 6)}$

$\frac{(a + 6) (a - 5)}{(a - 5) (a + 2)}$

$\frac{a + 6}{a + 2}$

$\frac{(m - 3) (m - 1)}{(m - 1) (m + 4)}$

$\frac{(y - 2) (y - 3)}{(y - 3) (y - 2)}$

$\frac{(x + 7) (x + 8)}{(x + 8) (x + 7)}$

$\frac{- 12 x^{2} (x + 4)}{4 x}$

$- 3 x (x + 4)$

$\frac{- 3 a^{4} (a - 1) (a + 5)}{- 2 a^{3} (a - 1) (a + 9)}$

$\frac{6 x^{2} y^{5} (x - 1) (x + 4)}{- 2 x y (x + 4)}$

$- 3 x y^{4} (x - 1)$

$\frac{22 a^{4} b^{6} c^{7} (a + 2) (a - 7)}{4 c (a + 2) (a - 5)}$

$\frac{{(x + 10)}^{3}}{x + 10}$

${(x + 10)}^{2}$

$\frac{{(y - 6)}^{7}}{y - 6}$

$\frac{{(x - 8)}^{2} {(x + 6)}^{4}}{(x - 8) (x + 6)}$

$(x - 8) {(x + 6)}^{3}$

$\frac{{(a + 1)}^{5} {(a - 1)}^{7}}{{(a + 1)}^{3} {(a - 1)}^{4}}$

$\frac{{(y - 2)}^{6} {(y - 1)}^{4}}{{(y - 2)}^{3} {(y - 1)}^{2}}$

${(y - 2)}^{3} {(y - 1)}^{2}$

$\frac{{(x + 10)}^{5} {(x - 6)}^{3}}{(x - 6) {(x + 10)}^{2}}$

$\frac{{(a + 6)}^{2} {(a - 7)}^{6}}{{(a + 6)}^{5} {(a - 7)}^{2}}$

$\frac{{(a - 7)}^{4}}{{(a + 6)}^{3}}$

$\frac{{(m + 7)}^{4} {(m - 8)}^{5}}{{(m + 7)}^{7} {(m - 8)}^{2}}$

$\frac{(a + 2) {(a - 1)}^{3}}{(a + 1) (a - 1)}$

$\frac{(a + 2) {(a - 1)}^{2}}{(a + 1)}$

$\frac{(b + 6) {(b - 2)}^{4}}{(b - 1) (b - 2)}$

$\frac{8 {(x + 2)}^{3} {(x - 5)}^{6}}{2 (x + 2) {(x - 5)}^{2}}$

$4 {(x + 2)}^{2} {(x - 5)}^{4}$

$\frac{14 {(x - 4)}^{3} {(x - 10)}^{6}}{- 7 {(x - 4)}^{2} {(x - 10)}^{2}}$

$\frac{x^{2} + x - 12}{x^{2} - 4 x + 3}$

$\frac{(x + 4)}{(x - 1)}$

$\frac{x^{2} + 3 x - 10}{x^{2} + 2 x - 15}$

$\frac{x^{2} - 10 x + 21}{x^{2} - 6 x - 7}$

$\frac{(x - 3)}{(x + 1)}$

$\frac{x^{2} + 10 x + 24}{x^{2} + 6 x}$

$\frac{x^{2} + 9 x + 14}{x^{2} + 7 x}$

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

$\frac{6 b^{2} - b}{6 b^{2} + 11 b - 2}$

$\frac{3 b^{2} + 10 b + 3}{3 b^{2} + 7 b + 2}$

$\frac{b + 3}{b + 2}$

$\frac{4 b^{2} - 1}{2 b^{2} + 5 b - 3}$

$\frac{16 a^{2} - 9}{4 a^{2} - a - 3}$

$\frac{(4 a - 3)}{(a - 1)}$

$\frac{20 x^{2} + 28 x y + 9 y^{2}}{4 x^{2} + 4 x y + y^{2}}$

For the following problems, reduce each rational expression if possible. If not possible, state the answer in lowest terms.

$\frac{x + 3}{x + 4}$

$\frac{(x + 3)}{(x + 4)}$

$\frac{a + 7}{a - 1}$

$\frac{3 a + 6}{3}$

$a + 2$

$\frac{4 x + 12}{4}$

$\frac{5 a - 5}{- 5}$

$\begin{array}{l} - (a - 1) & or & - a + 1 \end{array}$

$\frac{6 b - 6}{- 3}$

$\frac{8 x - 16}{- 4}$

$- 2 (x - 2)$

$\frac{4 x - 7}{- 7}$

$\frac{- 3 x + 10}{10}$

$\frac{x - 2}{2 - x}$

$\frac{a - 3}{3 - a}$

$- 1$

$\frac{x^{3} - x}{x}$

$\frac{y^{4} - y}{y}$

$y^{3} - 1$

$\frac{a^{5} - a^{2}}{a}$

$\frac{a^{6} - a^{4}}{a^{3}}$

$a (a + 1) (a - 1)$

$\frac{4 b^{2} + 3 b}{b}$

$\frac{2 a^{3} + 5 a}{a}$

$2 a^{2} + 5$

$\frac{a}{a^{3} + a}$

$\frac{x^{4}}{x^{5} - 3 x}$

$\frac{x^{3}}{x^{4} - 3}$

$\frac{- a}{- a^{2} - a}$

Excercises for review

( [link] ) Write ${(\frac{4^{4} a^{8} b^{10}}{4^{2} a^{6} b^{2}})}^{- 1}$ so that only positive exponents appear.

$\frac{1}{16 a^{2} b^{8}}$

( [link] ) Factor $y^{4} - 16$ .

( [link] ) Factor $10 x^{2} - 17 x + 3$ .

$(5 x - 1) (2 x - 3)$

( [link] ) Supply the missing word. An equation expressed in the form $a x + b y = c$ is said to be expressed in form.

( [link] ) Find the domain of the rational expression $\frac{2}{x^{2} - 3 x - 18}$ .

$x \neq - 3, 6$

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Source: OpenStax, Algebra ii for the community college. OpenStax CNX. Jul 03, 2014 Download for free at http://cnx.org/content/col11671/1.1

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