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

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 $6x^2-19x-7$ ? We must factor $6x^2-19x-7$ to put it into the zero-factor property form.

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

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, $6x^2-19x-7=0$ when $x=\frac{-1}{3}$ or $\frac{7}{2}$ .

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

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.

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

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.

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

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

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

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

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

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

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