# 8.2: Rational Expressions

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## Rational Expressions

In arithmetic, it is noted that a fraction is a quotient of two whole numbers. The expression \(\dfrac{a}{b}\), where \(a\) and \(b\) are any two whole numbers and \(b≠0\), is called a fraction. The top number, \(a\), is called the numerator, and the bottom number, \(b\), is called the denominator.

**Simple Algebraic ****Fraction**

We define a simple algebraic fraction in a similar manner. Rather than restricting ourselves only to numbers, we use polynomials for the numerator and denominator. Another term for a simple algebraic fraction is a **rational expression**. A rational expression is an expression of the form \(\dfrac{P}{Q}\), where \(P\) and \(Q\) are both polynomials and \(Q\) never represents the zero polynomial.

A **rational expression** is an algebraic expression that can be written as the quotient of two polynomials.

**Examples 1–4 are rational expressions:**

\(\dfrac{x+9}{x-7}\) is a rational expression: \(P\) is \(x + 9\) and \(Q\) is \(x-7\).

\(\dfrac{x^3 + 5x^2 - 12x + 1}{x^4 - 10}\) is a rational expression. \(P\) is \(x^3 + 5x^2 - 12x + 1\) and \(Q\) is \(x^4 - 10\)

\(\dfrac{3}{8}\) is a rational expression: \(P\) is \(3\) and \(Q\) is \(8\).

\(4x - 5\) is a rational expression since \(4x - 5\) can be written as \(\dfrac{4x-5}{1}\): \(P\) is \(4x - 5\) and \(Q\) is \(1\).

\(\dfrac{\sqrt{5x^2-8}}{2x-1}\) is **not **a rational expression since \(\sqrt{5x^2-8}\) is not a polynomial.

In the rational expression \(\dfrac{P}{Q}\), \(P\) is called the numerator and \(Q\) is called the denominator.

**Domain of a Rational ****Expression**

Since division by zero is not defined, we must be careful to note the values for which the rational expression is valid. The collection of values for which the rational expression is defined is called the **domain** of the rational expression. (Recall our study of the domain of an equation in Section __4.8__.)

**Finding the Domain of a Rational ****Expression**

To find the domain of a rational expression we must ask, "What values, if any, of the variable will make the denominator zero?" To find these values, we set the denominator equal to zero and solve. If any zero-producing values are obtained, they are not included in the domain. All other real numbers are included in the domain (unless some have been excluded for particular situational reasons).

## 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.

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 \(8(x-6)\)? By the zero-factor property, if \(8(x-6) = 0\), then:

\(\begin{aligned}

x-6&=0\\

x&=0

\end{aligned}\)

Thus, \(8(x-6) = 0\) when \(x = 6\).

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

\(\begin{aligned}

x-3&=0&\text{ or }&x+5&=0\\

x&=3&&x&=-5

\end{aligned}\)

Thus, \((x-3)(x+5) = 0\) 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.

\(x^2 + 6x + 8 = (x+2)(x+4)\)

Now, \((x+2)(x+4) = 0\) when

\(\begin{aligned}

x+2&=0&\text{ or }&x+4&=0\\

x&=-2&&x&=-4

\end{aligned}\)

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.

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

Now, \((3x+1)(2x-7) = 0\) when

\(\begin{aligned}

3x+1&=0&\text{ or }&2x-7&=0\\

3x&=-1&&2x&=7

\end{aligned}\)

Thus, \(6x^2 - 19x - 7 = 0\) when \(x = \dfrac{-1}{3}\) or \(\dfrac{7}{2}\)

## Sample Set A

Find the domain of the following expressions.

\(\dfrac{5}{x-1}\)

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

\(\dfrac{3a}{2a-8}\)

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

\(\begin{aligned}

2a - 8 &=0\\

2a &= 8\\

a &=4

\end{aligned}\)

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.

\(\dfrac{5x-1}{(x+2)(x-6)}\).

Setting \((x+2)(x−6)=0\), 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 \(6\).

\(\dfrac{9}{(x^2-2x-15}\).

Setting \(x^2 - 2x - 15 = 0\), we get:

\(\begin{aligned}

(x+3)(x-5)&=0\\

x&=-3, 5

\end{aligned}\)

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 \(5\).

\(\dfrac{2x^2 + x - 7}{x(x-1)(x-3)(x+10)}\)

Setting \(x(x−1)(x−3)(x+10)=0\), 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 \(0, 1, 3, –10\).

\(\dfrac{8b+7}{(2b+1)(3b-2)}\).

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

\(\dfrac{4x-5}{x^2+1}\).

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.

\(\dfrac{x-9}{6}\)

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.

\(\dfrac{2}{x-7}\)

**Answer**-
\(7\)

\(\dfrac{5x}{x(x+4)}\)

**Answer**-
\(0, −4\)

\(\dfrac{2x+1}{(x+2)(1-x)}\)

**Answer**-
\(−2, 1\)

\(\dfrac{5a+2}{a^2+6a+8}\)

**Answer**-
\(−2, −4\)

\(\dfrac{12y}{3y^2-2y-8}\)

**Answer**-
\((-\dfrac{4}{3}, 2)\)

\(\dfrac{2m-5}{m^2 + 3}\)

**Answer**-
All real numbers comprise the domain.

\(\dfrac{k^2 - 4}{5}\)

**Answer**-
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 \(b≠0\) and \(d≠0\).

If \(\dfrac{a}{b} = \dfrac{c}{d}\), then \(ad = bc\).

If \(ad = bc\), then \(\dfrac{a}{b} = \dfrac{c}{d}\)

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

**We see this property in the following examples:**

\(\dfrac{2}{3} = \dfrac{8}{12}\), since \(2 \cdot 12\ = 3 \cdot 8\).

\(\dfrac{5y}{2} = \dfrac{15y^2}{6y}\), since \(5y \cdot 6y = 2 \cdot 15y^2\) and \(30y^2 = 30y^2\).

Since \(9a \cdot 4 = 18a \cdot 2\), \(\dfrac{9a}{18a} = \dfrac{2}{4}\)

## The Negative Property of Fractions

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

The negative sign of a fraction may be placed:

- in front of the fraction, \(-\dfrac{a}{b}\),

- in the numerator of the fraction, \(\dfrac{-a}{b}\),

- in the denominator of the fraction, \(\dfrac{a}{-b}\),

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

\(-\dfrac{a}{b} = \dfrac{-a}{b} = \dfrac{a}{-b}\)

The negative property of fractions is illustrated by the fractions

\(-\dfrac{3}{4} = \dfrac{-3}{4} = \dfrac{3}{-4}\)

To see this, consider \(-\dfrac{3}{4} = \dfrac{-3}{4}\). Is this correct?

By the equality property of fractions, \(-(3 \cdot 4) = -13\) and \(-3 \cdot 4 = -12\). Thus, \(-\dfrac{3}{4} = \dfrac{-3}{4}\). 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

\(\dfrac{x}{-4}\) is best written as \(\dfrac{-x}{4}\)

\(-\dfrac{y}{9}\) is best written as \(\dfrac{-y}{9}\)

\(-\dfrac{x-4}{2x-5}\) could be written as \(\dfrac{-(x-4)}{2x-5}\), which would then yield \(\dfrac{-x+4}{2x-5}\)

\(\dfrac{-5}{-10-x}\). Factor our \(-1\) from the denominator.

\(\dfrac{-5}{-(10+x)}\) A negative divided by a negative is a positive

\(\dfrac{5}{10+x}\)

\(-\dfrac{3}{7-x}\). Rewrite this.

\(\dfrac{-3}{7-x}\) Factor out \(-1\) from the denominator.

\(\dfrac{-3}{-(-7+x)}\) A negative divided by a negative is positive.

\(\dfrac{3}{-7+x}\) Rewrite.

\(\dfrac{3}{x-7}\)

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

## Practice Set B

Fill in the missing term.

\(-\dfrac{5}{y-2} = \dfrac{?}{y-2}\)

**Answer**-
\(−5\)

\(-\dfrac{a+2}{-a+3} = \dfrac{?}{a-3}\)

**Answer**-
\(a+2\)

\(-\dfrac{8}{5-y} = \dfrac{?}{y-5}\)

**Answer**-
\(8\)

## Exercises

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

\(\dfrac{6}{x-4}\)

**Answer**-
\(x \not = 4\)

\(\dfrac{-3}{x-8}\)

\(\dfrac{-11x}{x+1}\)

**Answer**-
\(x≠−1\)

\(\dfrac{x+10}{x+4}\)

\(\dfrac{x-1}{x^2-4}\)

**Answer**-
\(x≠−2, 2\)

\(\dfrac{x+7}{x^2-9}\)

\(\dfrac{-x+4}{x^2-36}\)

**Answer**-
\(x≠−6, 6\)

\(\dfrac{-a+5}{a(a-5)}\)

\(\dfrac{2b}{b(b+6)}\)

**Answer**-
\(b≠0, −6\)

\(\dfrac{3b+1}{b(b-4)(b+5)}\)

\(\dfrac{3x+4}{x(x-10)(x+1)}\)

**Answer**-
\(x≠0, 10, −1\)

\(\dfrac{-2x}{x^2(4-x)}\)

\(\dfrac{6a}{a^3(a-5)(7-a)}\)

**Answer**-
\(x≠0, 5, 7\)

\(\dfrac{-5}{a^2 + 6a + 8}\)

\(\dfrac{-8}{b^2 - 4b + 3}\)

**Answer**-
\(b≠1, 3\)

\(\dfrac{x-1}{x^2 - 9x + 2}\)

\(\dfrac{y-9}{y^2-y-20}\)

**Answer**-
\(y≠5, −4\)

\(\dfrac{y-6}{2y^2 - 3y - 2}\)

\(\dfrac{2x + 7}{6x^3 + x^2 - 2x}\)

**Answer**-
\(x \not = 0, \dfrac{1}{2}, -\dfrac{2}{3}\)

\(\dfrac{-x+4}{x^3 - 8x^2 + 12x}\)

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

\(\dfrac{-3}{5}\) and \(-\dfrac{3}{5}\)

**Answer**-
\((−3)5=−15, −(3 ⋅ 5)=−15\)

\(\dfrac{-2}{7}\) and \(-\dfrac{2}{7}\)

\(-\dfrac{1}{4}\) and \(\dfrac{-1}{4}\)

**Answer**-
\(−(1 ⋅ 4)=−4, 4(−1)=−4\)

\(\dfrac{-2}{3}\) and \(-\dfrac{2}{3}\)

\(\dfrac{-9}{10}\) and \(\dfrac{9}{-10}\)

**Answer**-
\((−9)(−10)=90\) and \((9)(10)=90\)

For the following problems, fill in the missing term.

\(-\dfrac{4}{x-1} = \dfrac{?}{x-1}\)

\(-\dfrac{2}{x+7} = \dfrac{?}{x+7}\)

**Answer**-
\(−2\)

\(-\dfrac{3x+4}{2x-1} = \dfrac{?}{2x-1}\)

\(-\dfrac{2x+7}{5x-1} = \dfrac{?}{5x-1}\)

**Answer**-
\(−2x−7\)

\(-\dfrac{x-2}{6x-1} = \dfrac{?}{6x-1}\)

\(-\dfrac{x-4}{2x-3} = \dfrac{?}{2x-3}\)

**Answer**-
\(−x+4\)

\(-\dfrac{x+5}{-x-3} = \dfrac{?}{x+3}\)

\(-\dfrac{a+1}{-a-6} = \dfrac{?}{a+6}\)

**Answer**-
\(a+1\)

\(\dfrac{x-7}{-x+2} = \dfrac{?}{x-2}\)

\(\dfrac{y+10}{-y-6} = \dfrac{?}{y+6}\)

**Answer**-
\(−y−10\)

## Exercises For Review

Write \((\dfrac{15x^{-3}y^4}{5x^2y^{-7}})^2\) so that only positive exponents appear.

Solve the compound inequality \(1≤6x−5<13\)

**Answer**-
\(1≤x<3\)

Factor \(8x^2 - 18x - 5\).

Factor \(x^2 - 12x + 36\)

**Answer**-
\((x-6)^2\)

Supply the missing word. The phrase "graphing an equation" is interpreted as meaning "geometrically locate the ____ to an equation."