# 8.4 Building rational expressions and the lcd  (Page 3/4)

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To add or subtract two or more rational expressions they must have the same denominator .

Building rational expressions allows us to transform fractions into fractions with the same denominators (which we can then add or subtract). The most convenient new denominator is the least common denominator (LCD) of the given fractions.

## The least common denominator (lcd)

In arithmetic, the least common denominator is the smallest (least) quantity that each of the given denominators will divide into without a remainder. For algebraic expressions, the LCD is the polynomial of least degree divisible by each denominator. Some examples are shown below.

$\frac{3}{4},\frac{1}{6},\frac{5}{12}.$

The LCD is 12 since 12 is the smallest number that 4, 6, and 12 will divide into without a remainder.

$\frac{1}{3},\frac{5}{6},\frac{5}{8},\frac{7}{12}.$

The LCD is 24 since 24 is the smallest number that 3, 6, 8, and 12 will divide into without a remainder.

$\frac{2}{x},\frac{3}{{x}^{2}}.$

The LCD is ${x}^{2}$ since ${x}^{2}$ is the smallest quantity that $x$ and ${x}^{2}$ will divide into without a remainder.

$\frac{5a}{6{a}^{2}b},\text{\hspace{0.17em}}\text{\hspace{0.17em}}\frac{3a}{8a{b}^{3}}.$

The LCD is $24{a}^{2}{b}^{3}$ since $24{a}^{2}{b}^{3}$ is the smallest quantity that $6{a}^{2}b$ and $8a{b}^{3}$ will divide into without a remainder.

$\frac{2y}{y-6},\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\frac{4{y}^{2}}{{\left(y-6\right)}^{3}},\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\frac{y}{y-1}.$

The LCD is ${\left(y-6\right)}^{3}\left(y-1\right)$ since ${\left(y-6\right)}^{3}\text{\hspace{0.17em}}·\text{\hspace{0.17em}}\left(y-1\right)$ is the smallest quantity that $y-6,{\left(y-6\right)}^{3}$ and $y-1$ will divide into without a remainder.

We’ll now propose and demonstrate a method for obtaining the LCD.

Method for Obtaining the LCD

1. Factor each denominator. Use exponents for repeated factors. It is usually not necessary to factor numerical quantities.
2. Write down each different factor that appears. If a factor appears more than once, use only the factor with the highest exponent.
3. The LCD is the product of the factors written in step 2.

## Sample set b

Find the LCD.

• $\frac{1}{x},\text{\hspace{0.17em}}\frac{3}{{x}^{3}},\text{\hspace{0.17em}}\frac{2}{4y}$
1. The denominators are already factored.
2. Note that $x$ appears as $x$ and ${x}^{3}$ . Use only the $x$ with the higher exponent, ${x}^{3}$ . The term $4y$ appears, so we must also use $4y$ .
3. The LCD is $4{x}^{3}y$ .
• $\frac{5}{{\left(x-1\right)}^{2}},\text{\hspace{0.17em}}\frac{2x}{\left(x-1\right)\left(x-4\right)},\text{\hspace{0.17em}}\frac{-5x}{{x}^{2}-3x+2}$
1. Only the third denominator needs to be factored.

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

Now the three denominators are ${\left(x-1\right)}^{2},\left(x-1\right)\left(x-4\right)$ , and $\left(x-2\right)\left(x-1\right).$
2. Note that $x-1$ appears as ${\left(x-1\right)}^{2},x-1$ , and $x-1.$ Use only the $x-1$ with the highest exponent, ${\left(x-1\right)}^{2}$ . Also appearing are $x-4$ and $x-2.$
3. The LCD is ${\left(x-1\right)}^{2}\left(x-4\right)\left(x-2\right).$
• $\frac{-1}{6{a}^{4}},\text{\hspace{0.17em}}\frac{3}{4{a}^{3}b},\text{\hspace{0.17em}}\frac{1}{3{a}^{3}\left(b+5\right)}$
1. The denominators are already factored.
2. We can see that the LCD of the numbers 6, 4, and 3 is 12. We also need ${a}^{4}$ , $b$ , and $b+5$ .
3. The LCD is $12{a}^{4}b\left(b+5\right).$
• $\frac{9}{x},\text{\hspace{0.17em}}\frac{4}{8y}$
1. The denominators are already factored.
2. $x,\text{\hspace{0.17em}}8y.$
3. The LCD is $8xy$ .

## Practice set b

Find the LCD.

$\frac{3}{{x}^{2}},\text{\hspace{0.17em}}\frac{4}{{x}^{5}},\frac{-6}{xy}$

${x}^{5}y$

$\frac{x+1}{x-4},\frac{x-7}{{\left(x-4\right)}^{2}},\frac{-6}{x+1}$

${\left(x-4\right)}^{2}\left(x+1\right)$

$\frac{2}{m-6},\frac{-5m}{{\left(m+1\right)}^{2}\left(m-2\right)},\frac{12{m}^{2}}{{\left(m-2\right)}^{3}\left(m-6\right)}$

$\left(m-6\right){\left(m+1\right)}^{2}{\left(m-2\right)}^{3}$

$\frac{1}{{x}^{2}-1},\text{\hspace{0.17em}}\frac{2}{{x}^{2}-2x-3},\text{\hspace{0.17em}}\frac{-3x}{{x}^{2}-6x+9}$

$\left(x+1\right)\left(x-1\right){\left(x-3\right)}^{2}$

$\frac{3}{4{y}^{2}-8y},\frac{8}{{y}^{2}-4y+4},\frac{10y-1}{3{y}^{3}-6{y}^{2}}$

$12{y}^{2}{\left(y-2\right)}^{2}$

## Sample set c

Change the given rational expressions into rational expressions having the same denominator.

$\begin{array}{lll}\frac{3}{{x}^{2}},\frac{4}{x}.\hfill & \hfill & \begin{array}{l}\text{The\hspace{0.17em}LCD,\hspace{0.17em}by\hspace{0.17em}inspection,\hspace{0.17em}is\hspace{0.17em}}{x}^{2}\text{.\hspace{0.17em}Rewrite\hspace{0.17em}each\hspace{0.17em}expression\hspace{0.17em}}\\ \text{with\hspace{0.17em}}{x}^{2}\text{\hspace{0.17em}as\hspace{0.17em}the\hspace{0.17em}new\hspace{0.17em}denominator}\text{.\hspace{0.17em}}\end{array}\hfill \\ \frac{}{{x}^{2}},\frac{}{{x}^{2}}\hfill & \hfill & \begin{array}{l}\text{Determine\hspace{0.17em}the\hspace{0.17em}numerators}\text{.\hspace{0.17em}In\hspace{0.17em}}\frac{3}{{x}^{2}}\text{,\hspace{0.17em}the\hspace{0.17em}denominator\hspace{0.17em}was\hspace{0.17em}not\hspace{0.17em}}\\ \text{changed\hspace{0.17em}so\hspace{0.17em}we\hspace{0.17em}need\hspace{0.17em}not\hspace{0.17em}change\hspace{0.17em}the\hspace{0.17em}numerator}\text{.\hspace{0.17em}}\end{array}\hfill \\ \frac{3}{{x}^{2}},\frac{}{{x}^{2}}\hfill & \hfill & \begin{array}{l}\text{In\hspace{0.17em}the\hspace{0.17em}second\hspace{0.17em}fraction,\hspace{0.17em}the\hspace{0.17em}original\hspace{0.17em}denominator\hspace{0.17em}was\hspace{0.17em}}x\text{.\hspace{0.17em}}\\ \text{We\hspace{0.17em}can\hspace{0.17em}see\hspace{0.17em}that\hspace{0.17em}}x\text{\hspace{0.17em}must\hspace{0.17em}be\hspace{0.17em}multiplied\hspace{0.17em}by\hspace{0.17em}}x\text{\hspace{0.17em}to\hspace{0.17em}build\hspace{0.17em}it\hspace{0.17em}to\hspace{0.17em}}{x}^{2}\text{.\hspace{0.17em}}\\ \text{So\hspace{0.17em}we\hspace{0.17em}must\hspace{0.17em}also\hspace{0.17em}multiply\hspace{0.17em}the\hspace{0.17em}numerator\hspace{0.17em}4\hspace{0.17em}by\hspace{0.17em}}x\text{.\hspace{0.17em}Thus,\hspace{0.17em}4}\text{\hspace{0.17em}}·\text{\hspace{0.17em}}x=4x\text{.\hspace{0.17em}}\end{array}\hfill \\ \frac{3}{{x}^{2}},\frac{4x}{{x}^{2}}\hfill & \hfill & \hfill \end{array}$

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