# 8.1 Rational expressions

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## Zero-factor property

Sometimes to find the domain of a rational expression, it is necessary to factor the denominator and use the zero-factor property of real numbers.

## Zero-factor property

If two real numbers $a$ and $b$ are multiplied together and the resulting product is 0, then at least one of the factors must be zero, that is, either $a=0$ , $b=0$ , or both $a=0$ and $b=0$ .

The following examples illustrate the use of the zero-factor property.

What value will produce zero in the expression $4x$ ? By the zero-factor property, if $4x=0$ , then $x=0$ .

What value will produce zero in the expression $8\left(x-6\right)$ ? By the zero-factor property, if $8\left(x-6\right)=0$ , then

$\begin{array}{lll}x-6\hfill & =\hfill & 0\hfill \\ \hfill x& =\hfill & 6\hfill \end{array}$
Thus, $8\left(x-6\right)=0$ when $x=6$ .

What value(s) will produce zero in the expression $\left(x-3\right)\left(x+5\right)$ ? By the zero-factor property, if $\left(x-3\right)\left(x+5\right)=0$ , then

$\begin{array}{lllllllll}x-3\hfill & =\hfill & 0\hfill & \hfill & \text{or}\hfill & \hfill & x+5\hfill & =\hfill & 0\hfill \\ \hfill x& =\hfill & 3\hfill & \hfill & \hfill & \hfill & \hfill x& =\hfill & -5\hfill \end{array}$
Thus, $\left(x-3\right)\left(x+5\right)=0$ when $x=3$ or $x=-5$ .

What value(s) will produce zero in the expression ${x}^{2}+6x+8$ ? We must factor ${x}^{2}+6x+8$ to put it into the zero-factor property form.

${x}^{2}+6x+8=\left(x+2\right)\left(x+4\right)$

Now, $\left(x+2\right)\left(x+4\right)=0$ when
$\begin{array}{lllllllll}x+2\hfill & =\hfill & 0\hfill & \hfill & \text{or}\hfill & \hfill & x+4\hfill & =\hfill & 0\hfill \\ \hfill x& =\hfill & -2\hfill & \hfill & \hfill & \hfill & \hfill x& =\hfill & -4\hfill \end{array}$
Thus, ${x}^{2}+6x+8=0$ when $x=-2$ or $x=-4$ .

What value(s) will produce zero in the expression $6{x}^{2}-19x-7$ ? We must factor $6{x}^{2}-19x-7$ to put it into the zero-factor property form.

$6{x}^{2}-19x-7=\left(3x+1\right)\left(2x-7\right)$

Now, $\left(3x+1\right)\left(2x-7\right)=0$ when
$\begin{array}{lllllllll}3x+1\hfill & =\hfill & 0\hfill & \hfill & \text{or}\hfill & \hfill & 2x-7\hfill & =\hfill & 0\hfill \\ \hfill 3x& =\hfill & -1\hfill & \hfill & \hfill & \hfill & \hfill 2x& =\hfill & 7\hfill \\ \hfill x& =\hfill & \frac{-1}{3}\hfill & \hfill & \hfill & \hfill & \hfill x& =\hfill & \frac{7}{2}\hfill \end{array}$
Thus, $6{x}^{2}-19x-7=0$ when $x=\frac{-1}{3}$ or $\frac{7}{2}$ .

## Sample set a

Find the domain of the following expressions.

$\frac{5}{x-1}$ .

The domain is the collection of all real numbers except 1. One is not included, for if $x=1$ , division by zero results.

$\frac{3a}{2a-8}$ .

If we set $2a-8$ equal to zero, we find that $a=4$ .

$\begin{array}{lll}2a-8\hfill & =\hfill & 0\hfill \\ \hfill 2a& =\hfill & 8\hfill \\ \hfill a& =\hfill & 4\hfill \end{array}$
Thus 4 must be excluded from the domain since it will produce division by zero. The domain is the collection of all real numbers except 4.

$\frac{5x-1}{\left(x+2\right)\left(x-6\right)}$ .

Setting $\left(x+2\right)\left(x-6\right)=0$ , we find that $x=-2$ and $x=6$ . Both these values produce division by zero and must be excluded from the domain. The domain is the collection of all real numbers except –2 and 6.

$\frac{9}{{x}^{2}-2x-\text{15}}$ .

Setting ${x}^{2}-2x-15=0$ , we get

$\begin{array}{rrr}\hfill \left(x+3\right)\left(x-5\right)& \hfill =& 0\hfill \\ \hfill x& \hfill =& \hfill -3,5\end{array}$
Thus, $x=-3$ and $x=5$ produce division by zero and must be excluded from the domain. The domain is the collection of all real numbers except –3 and 5.

$\frac{2{x}^{2}+x-7}{x\left(x-1\right)\left(x-3\right)\left(x+10\right)}$ .

Setting $x\left(x-1\right)\left(x-3\right)\left(x+10\right)=0$ , we get $x=0,1,3,-10$ . These numbers must be excluded from the domain. The domain is the collection of all real numbers except 0, 1, 3, –10.

$\frac{8b+7}{\left(2b+1\right)\left(3b-2\right)}$ .

Setting $\left(2b+1\right)\left(3b-2\right)=0$ , we get $b$ = $-\frac{1}{2}$ , $\frac{2}{3}$ . The domain is the collection of all real numbers except $-\frac{1}{2}$ and $\frac{2}{3}$ .

$\frac{4x-5}{{x}^{2}+1}$ .

No value of $x$ is excluded since for any choice of $x$ , the denominator is never zero. The domain is the collection of all real numbers.

$\frac{x-9}{6}$ .

No value of $x$ is excluded since for any choice of $x$ , the denominator is never zero. The domain is the collection of all real numbers.

## Practice set a

Find the domain of each of the following rational expressions.

$\frac{2}{x-7}$

7

$\frac{5x}{x\left(x+4\right)}$

$0,\text{\hspace{0.17em}}-4$

$\frac{2x+1}{\left(x+2\right)\left(1-x\right)}$

$-2,\text{​}\text{\hspace{0.17em}}1$

$\frac{5a+2}{{a}^{2}+6a+8}$

$-2,\text{​}\text{\hspace{0.17em}}-4$

$\frac{12y}{3{y}^{2}-2y-8}$

$-\frac{4}{3},2$

$\frac{2m-5}{{m}^{2}+3}$

All real numbers comprise the domain.

$\frac{{k}^{2}-4}{5}$

All real numbers comprise the domain.

## The equality property of fractions

From our experience with arithmetic we may recall the equality property of fractions. Let $a$ , $b$ , $c$ , $d$ be real numbers such that $b\ne 0$ and $d\ne 0$ .

## Equality property of fractions

1. If $\frac{a}{b}=\frac{c}{d}$ , then $ad=bc$ .
2. If $ad=bc$ , then $\frac{a}{b}=\frac{c}{d}$ .

Two fractions are equal when their cross-products are equal.

We see this property in the following examples:

$\frac{2}{3}=\frac{8}{\text{12}}$ , since $2·\text{12}=3·8$ .

$\frac{5y}{2}=\frac{15{y}^{2}}{6y}$ , since $5y·6y=2·\text{15}{y}^{2}$ and $30{y}^{2}=30{y}^{2}$ .

Since $9a·4=\text{18}a·2$ , $\frac{9a}{18a}=\frac{2}{4}$ .

## The negative property of fractions

A useful property of fractions is the negative property of fractions .

## Negative property of fractions

The negative sign of a fraction may be placed

1. in front of the fraction, $-\frac{a}{b}$ ,
2. in the numerator of the fraction, $\frac{-a}{b}$ ,
3. in the denominator of the fraction, $\frac{a}{-b}$ .

All three fractions will have the same value, that is,

$-\frac{a}{b}=\frac{-a}{b}=\frac{a}{-b}$

• The negative property of fractions is illustrated by the fractions
• $-\frac{3}{4}=\frac{-3}{4}=\frac{3}{-4}$

To see this, consider $-\frac{3}{4}=\frac{-3}{4}$ . Is this correct?

By the equality property of fractions, $-\left(3·4\right)=-\text{12}$ and $-3·4=-\text{12}$ . Thus, $-\frac{3}{4}=\frac{-3}{4}$ . Convince yourself that the other two fractions are equal as well.

This same property holds for rational expressions and negative signs. This property is often quite helpful in simplifying a rational expression (as we shall need to do in subsequent sections).

If either the numerator or denominator of a fraction or a fraction itself is immediately preceded by a negative sign, it is usually most convenient to place the negative sign in the numerator for later operations.

## Sample set b

$\frac{x}{-4}$ is best written as $\frac{-x}{4}$ .

$-\frac{y}{9}$ is best written as $\frac{-y}{9}$ .

$-\frac{x-4}{2x-5}$ could be written as $\frac{-\left(x-4\right)}{2x-5}$ , which would then yield $\frac{-x+4}{2x-5}$ .

$\begin{array}{lll}\frac{-5}{-10-x}.\hfill & \hfill & \text{Factor out}-\text{1 from the denominator}\text{.}\hfill \\ \frac{-5}{-\left(10+x\right)}\hfill & \hfill & \text{A negative divided by a negative is a positive}\text{.}\hfill \\ \frac{5}{10+x}\hfill & \hfill & \hfill \end{array}$

$\begin{array}{lll}-\frac{3}{7-x}.\hfill & \hfill & \text{Rewrite this}\text{.}\hfill \\ \frac{-3}{7-x}\hfill & \hfill & \text{Factor out}-\text{1 from the deno}\mathrm{min}\text{ator}\text{.}\hfill \\ \frac{-3}{-\left(-7+x\right)}\hfill & \hfill & \text{A negative divided by a negative is positive}.\hfill \\ \frac{3}{-7+x}\hfill & \hfill & \text{Rewrite}.\hfill \\ \frac{3}{x-7}\hfill & \hfill & \hfill \end{array}$

This expression seems less cumbersome than does the original (fewer minus signs).

## Practice set b

Fill in the missing term.

$-\frac{5}{y-2}=\frac{}{y-2}$

$-5$

$-\frac{a+2}{-a+3}=\frac{}{a-3}$

$a+2$

$-\frac{8}{5-y}=\frac{}{y-5}$

8

## Exercises

For the following problems, find the domain of each of the rational expressions.

$\frac{6}{x-4}$

$x\ne 4$

$\frac{-3}{x-8}$

$\frac{-\text{11}x}{x+1}$

$x\ne -1$

$\frac{x+\text{10}}{x+4}$

$\frac{x-1}{{x}^{2}-4}$

$x\ne -2,\text{\hspace{0.17em}}2$

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

$\frac{-x+4}{{x}^{2}-\text{36}}$

$x\ne -6,\text{\hspace{0.17em}}6$

$\frac{-a+5}{a\left(a-5\right)}$

$\frac{2b}{b\left(b+6\right)}$

$b\ne 0,\text{\hspace{0.17em}}-6$

$\frac{3b+1}{b\left(b-4\right)\left(b+5\right)}$

$\frac{3x+4}{x\left(x-\text{10}\right)\left(x+1\right)}$

$x\ne 0,\text{\hspace{0.17em}}10,\text{\hspace{0.17em}}-1$

$\frac{-2x}{{x}^{2}\left(4-x\right)}$

$\frac{6a}{{a}^{3}\left(a-5\right)\left(7-a\right)}$

$x\ne 0,\text{\hspace{0.17em}}5,\text{\hspace{0.17em}}7$

$\frac{-5}{{a}^{2}+6a+8}$

$\frac{-8}{{b}^{2}-4b+3}$

$b\ne 1,\text{\hspace{0.17em}}3$

$\frac{x-1}{{x}^{2}-9x+2}$

$\frac{y-9}{{y}^{2}-y-\text{20}}$

$y\ne 5,\text{\hspace{0.17em}}-4$

$\frac{y-6}{2{y}^{2}-3y-2}$

$\frac{2x+7}{6{x}^{3}+{x}^{2}-2x}$

$x\ne 0,\text{\hspace{0.17em}}\frac{1}{2},\text{\hspace{0.17em}}-\frac{2}{3}$

$\frac{-x+4}{{x}^{3}-8{x}^{2}+12x}$

For the following problems, show that the fractions are equivalent.

$\frac{-3}{5}$ and $-\frac{3}{5}$

$\left(-3\right)5=-15,\text{\hspace{0.17em}}-\left(3\text{\hspace{0.17em}}·\text{\hspace{0.17em}}5\right)=-15$

$\frac{-2}{7}$ and $-\frac{2}{7}$

$-\frac{1}{4}\text{\hspace{0.17em}}\text{and}\text{\hspace{0.17em}}\frac{-1}{4}$

$-\left(1\text{\hspace{0.17em}}·\text{\hspace{0.17em}}4\right)=-4,\text{\hspace{0.17em}}4\left(-1\right)=-4$

$\frac{-2}{3}$ and $-\frac{2}{3}$

$\frac{-9}{\text{10}}$ and $\frac{9}{-\text{10}}$

$\left(-9\right)\left(-10\right)=90\text{\hspace{0.17em}}\text{and}\text{\hspace{0.17em}}\left(9\right)\left(10\right)=90$

For the following problems, fill in the missing term.

$-\frac{4}{x-1}=\frac{}{x-1}$

$-\frac{2}{x+7}=\frac{}{x+7}$

$-2$

$-\frac{3x+4}{2x-1}=\frac{}{2x-1}$

$-\frac{2x+7}{5x-1}=\frac{}{5x-1}$

$-2x-7$

$-\frac{x-2}{6x-1}=\frac{}{6x-1}$

$-\frac{x-4}{2x-3}=\frac{}{2x-3}$

$-x+4$

$-\frac{x+5}{-x-3}=\frac{}{x+3}$

$-\frac{a+1}{-a-6}=\frac{}{a+6}$

$a+1$

$\frac{x-7}{-x+2}=\frac{}{x-2}$

$\frac{y+\text{10}}{-y-6}=\frac{}{y+6}$

$-y-10$

## Exercises for review

( [link] ) Write ${\left(\frac{15{x}^{-3}{y}^{4}}{5{x}^{2}{y}^{-7}}\right)}^{-2}$ so that only positive exponents appear.

( [link] ) Solve the compound inequality $1\le 6x-5<13$ .

$1\le x<3$

( [link] ) Factor $8{x}^{2}-18x-5$ .

( [link] ) Factor ${x}^{2}-12x+36$ .

${\left(x-6\right)}^{2}$

( [link] ) Supply the missing word. The phrase "graphing an equation" is interpreted as meaning "geometrically locate the to an equation."

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