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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.
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
$\frac{7}{2}$ .
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.
Find the domain of each of the following rational expressions.
$\frac{2x+1}{\left(x+2\right)\left(1-x\right)}$
$-2,\text{}\text{\hspace{0.17em}}1$
$\frac{5a+2}{{a}^{2}+6a+8}$
$-2,\text{}\text{\hspace{0.17em}}-4$
$\frac{2m-5}{{m}^{2}+3}$
All real numbers comprise the domain.
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 $d\ne 0$ .
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 $2\xb7\text{12}=3\xb78$ .
$\frac{5y}{2}=\frac{15{y}^{2}}{6y}$ , since $5y\xb76y=2\xb7\text{15}{y}^{2}$ and $30{y}^{2}=30{y}^{2}$ .
Since $9a\xb74=\text{18}a\xb72$ , $\frac{9a}{18a}=\frac{2}{4}$ .
A useful property of fractions is the negative property of fractions .
To see this, consider
$-\frac{3}{4}=\frac{-3}{4}$ . Is this correct?
By the equality property of fractions,
$-\left(3\xb74\right)=-\text{12}$ and
$-3\xb74=-\text{12}$ . 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.
$\frac{x}{-4}$ 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).
Fill in the missing term.
For the following problems, find the domain of each of the rational expressions.
$\frac{-3}{x-8}$
$\frac{x+\text{10}}{x+4}$
$\frac{x+7}{{x}^{2}-9}$
$\frac{-x+4}{{x}^{2}-\text{36}}$
$x\ne -6,\text{\hspace{0.17em}}6$
$\frac{-a+5}{a\left(a-5\right)}$
$\frac{2b}{b\left(b+6\right)}$
$b\ne 0,\text{\hspace{0.17em}}-6$
$\frac{3b+1}{b\left(b-4\right)\left(b+5\right)}$
$\frac{3x+4}{x\left(x-\text{10}\right)\left(x+1\right)}$
$x\ne 0,\text{\hspace{0.17em}}10,\text{\hspace{0.17em}}-1$
$\frac{-2x}{{x}^{2}\left(4-x\right)}$
$\frac{6a}{{a}^{3}\left(a-5\right)\left(7-a\right)}$
$x\ne 0,\text{\hspace{0.17em}}5,\text{\hspace{0.17em}}7$
$\frac{-5}{{a}^{2}+6a+8}$
$\frac{x-1}{{x}^{2}-9x+2}$
$\frac{y-9}{{y}^{2}-y-\text{20}}$
$y\ne 5,\text{\hspace{0.17em}}-4$
$\frac{y-6}{2{y}^{2}-3y-2}$
$\frac{2x+7}{6{x}^{3}+{x}^{2}-2x}$
$x\ne 0,\text{\hspace{0.17em}}\frac{1}{2},\text{\hspace{0.17em}}-\frac{2}{3}$
$\frac{-x+4}{{x}^{3}-8{x}^{2}+12x}$
For the following problems, show that the fractions are equivalent.
$\frac{-3}{5}$ and $-\frac{3}{5}$
$\left(-3\right)5=-15,\text{\hspace{0.17em}}-\left(3\text{\hspace{0.17em}}\xb7\text{\hspace{0.17em}}5\right)=-15$
$\frac{-2}{7}$ and $-\frac{2}{7}$
$-\frac{1}{4}\text{\hspace{0.17em}}\text{and}\text{\hspace{0.17em}}\frac{-1}{4}$
$-\left(1\text{\hspace{0.17em}}\xb7\text{\hspace{0.17em}}4\right)=-4,\text{\hspace{0.17em}}4\left(-1\right)=-4$
$\frac{-2}{3}$ 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.
$-\frac{4}{x-1}=\frac{}{x-1}$
$-\frac{3x+4}{2x-1}=\frac{}{2x-1}$
$-\frac{x-2}{6x-1}=\frac{}{6x-1}$
$-\frac{x+5}{-x-3}=\frac{}{x+3}$
$\frac{x-7}{-x+2}=\frac{}{x-2}$
( [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
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