<< Chapter < Page | Chapter >> Page > |
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.
$\frac{2}{x-7}$
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.
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.
$-\frac{5}{y-2}=\frac{}{y-2}$
$-5$
$-\frac{a+2}{-a+3}=\frac{}{a-3}$
$a+2$
$-\frac{8}{5-y}=\frac{}{y-5}$
8
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 $-\frac{3}{5}$
$\left(-3\right)5=-15,\text{\hspace{0.17em}}-\left(3\text{\hspace{0.17em}}\xb7\text{\hspace{0.17em}}5\right)=-15$
$\frac{-2}{7}$ 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.
$-\frac{4}{x-1}=\frac{}{x-1}$
$-\frac{2}{x+7}=\frac{}{x+7}$
$-2$
$-\frac{3x+4}{2x-1}=\frac{}{2x-1}$
$-\frac{2x+7}{5x-1}=\frac{}{5x-1}$
$-2x-7$
$-\frac{x-2}{6x-1}=\frac{}{6x-1}$
$-\frac{x-4}{2x-3}=\frac{}{2x-3}$
$-x+4$
$-\frac{x+5}{-x-3}=\frac{}{x+3}$
$-\frac{a+1}{-a-6}=\frac{}{a+6}$
$a+1$
$\frac{x-7}{-x+2}=\frac{}{x-2}$
$\frac{y+\text{10}}{-y-6}=\frac{}{y+6}$
$-y-10$
( [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
Notification Switch
Would you like to follow the 'Algebra ii for the community college' conversation and receive update notifications?