2.4 Rational Expressions & Complex Fractions

2.4 Rational Expressions & Complex Fractions

The official CLEP College Algebra outline places operations with algebraic expressions inside the Algebraic Operations content area, which carries 25% of the exam. Polynomial arithmetic is only half of that skill: the other half is manipulating rational expressions — quotients of polynomials. These appear directly as simplification items and indirectly inside rational equations, function domains, composite functions, and difference quotients. Getting fluent here pays off across the whole test.

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Definition and Domain Restrictions

A rational expression is a quotient of two polynomials:

$R(x) = \frac{P(x)}{Q(x)}, \quad Q(x) \neq 0$

Because division by zero is undefined, every value that makes the denominator zero must be excluded. Two rules govern this and both are tested:

1. Read restrictions from the original denominators, before you simplify. If $\frac{x^2 - 9}{x^2 - 3x}$ is simplified to $\frac{x + 3}{x}$, the restriction $x \neq 3$ still applies even though the factor $(x - 3)$ has vanished from the written form.
2. In a product or quotient chain, collect restrictions from every denominator that appears at any stage — including the divisor's numerator, which becomes a denominator after you flip it.

Exam Trap: A very common distractor is the simplified expression paired with only the restrictions still visible in it. The cancelled restriction is the one CLEP wants you to remember.

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Simplifying: Cancel Factors, Never Terms

To reduce a rational expression to lowest terms:

1. Factor the numerator completely.
2. Factor the denominator completely.
3. Divide out factors common to both.

Worked Example 1: Full Simplification

Problem: Simplify $\frac{2x^2 - 7x - 15}{x^2 - 9}$ and state all restrictions.

1. Factor the numerator ($ac$-method: $ac = 2(-15) = -30$, and $-10 + 3 = -7$): $2x^2 - 10x + 3x - 15 = 2x(x - 5) + 3(x - 5) = (2x + 3)(x - 5)$
2. Factor the denominator (difference of squares): $x^2 - 9 = (x - 3)(x + 3)$.
3. Restrictions from the original denominator: $x \neq 3$ and $x \neq -3$.
4. Cancel: nothing is common, so the expression is already in lowest terms: $\frac{(2x + 3)(x - 5)}{(x - 3)(x + 3)}, \quad x \neq \pm 3$

The single most common algebra error on this topic: cancelling across a sum. In $\frac{x + 5}{x}$ the $x$ values do not cancel, because $x + 5$ is a sum, not a product. Cancelling is division, and division distributes over a product, not over a sum. Only a factor that multiplies the entire numerator may be divided out.

Opposite Factors

Factors that differ only by sign cancel to $-1$, because $b - a = -(a - b)$:

$\frac{x - 4}{4 - x} = \frac{x - 4}{-(x - 4)} = -1, \quad x \neq 4$

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Multiplication and Division

No common denominator is needed for either operation. Never expand the products before cancelling — that destroys the factored form you need.

Worked Example 2: Division of Rational Expressions

Problem: Simplify $\frac{x^2 - 4}{x^2 + 6x + 8} \div \frac{3x - 6}{x + 4}$.

1. Multiply by the reciprocal: $\frac{x^2 - 4}{x^2 + 6x + 8} \cdot \frac{x + 4}{3x - 6}$
2. Factor every polynomial: $\frac{(x - 2)(x + 2)}{(x + 2)(x + 4)} \cdot \frac{x + 4}{3(x - 2)}$
3. Restrictions (from every denominator that appears, plus the original divisor): $x \neq -2$, $x \neq -4$, $x \neq 2$.
4. Cancel $(x - 2)$, $(x + 2)$, and $(x + 4)$: $\frac{1}{3}$

The answer is the constant $\frac{1}{3}$ for every $x$ other than $2$, $-2$, and $-4$.

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Addition and Subtraction: Building the LCD

Unlike multiplication, adding and subtracting requires a common denominator.

How to build the LCD: factor every denominator, then take each distinct factor to the highest power it reaches in any single denominator.

Worked Example 3: Subtraction with Unlike Denominators

Problem: Simplify $\frac{5}{x^2 - x - 6} - \frac{2}{x^2 - 4}$.

1. Factor both denominators: $x^2 - x - 6 = (x - 3)(x + 2)$ and $x^2 - 4 = (x - 2)(x + 2)$.
2. LCD $= (x - 3)(x + 2)(x - 2)$. Restrictions: $x \neq 3, -2, 2$.
3. Rewrite each fraction over the LCD: $\frac{5(x - 2)}{(x - 3)(x + 2)(x - 2)} - \frac{2(x - 3)}{(x - 3)(x + 2)(x - 2)}$
4. Subtract the numerators — distribute the minus sign across the whole second numerator: $\frac{5x - 10 - (2x - 6)}{(x - 3)(x + 2)(x - 2)} = \frac{5x - 10 - 2x + 6}{(x - 3)(x + 2)(x - 2)} = \frac{3x - 4}{(x - 3)(x + 2)(x - 2)}$
5. Check for further factoring: $3x - 4$ shares no factor with the denominator, so this is final.

Exam Trap: Forgetting to distribute the subtraction across every term of the second numerator produces $5x - 10 - 2x - 6 = 3x - 16$. That wrong numerator is almost always one of the answer choices.

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Complex Fractions

A complex fraction contains a fraction inside its numerator, its denominator, or both. Two reliable methods exist; pick whichever is faster for the structure in front of you.

Method 1 (combine, then divide): simplify the numerator into a single fraction, simplify the denominator into a single fraction, then multiply by the reciprocal.

Method 2 (multiply by the LCD): multiply every term in both the numerator and the denominator by the LCD of all the small inner denominators. This clears every small fraction in one step and is usually faster when several small fractions are present.

Worked Example 4: Complex Fraction by the LCD Method

Problem: Simplify $\dfrac{\dfrac{1}{x} - \dfrac{1}{3}}{\dfrac{1}{x^2} - \dfrac{1}{9}}$.

1. Identify the inner denominators: $x$, $3$, $x^2$, $9$. Their LCD is $9x^2$.
2. Multiply every term above and below by $9x^2$: $\frac{9x^2 \cdot \frac{1}{x} - 9x^2 \cdot \frac{1}{3}}{9x^2 \cdot \frac{1}{x^2} - 9x^2 \cdot \frac{1}{9}} = \frac{9x - 3x^2}{9 - x^2}$
3. Factor both parts: $9x - 3x^2 = 3x(3 - x)$ and $9 - x^2 = (3 - x)(3 + x)$.
4. Cancel the common factor $(3 - x)$: $\frac{3x}{3 + x}, \quad x \neq 0,\ x \neq 3,\ x \neq -3$

Notice that the restriction $x \neq 3$ survives from the original expression even though $(3 - x)$ no longer appears — exactly the situation described at the start of this section.

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Where This Shows Up Elsewhere on the Exam

• Function domains (Chapter 4): the excluded values of a rational expression are precisely the values missing from the domain.
• Rational equations (Section 3.1): the LCD you build here is the multiplier that clears the equation, and the restrictions are the extraneous-root screen.
• Rational functions (Section 4.7): a factor that cancels produces a hole; a factor that survives in the denominator produces a vertical asymptote.