14.2 Rational Expressions
Key Takeaways
- A rational expression is a ratio of polynomials; simplify by factoring numerator and denominator, then cancel only common factors—not individual terms
- Excluded values are any $x$ that make the original denominator zero; list them from the unsimplified denominator
- To multiply rationals, factor all pieces, cancel across, then multiply remaining factors; to divide, multiply by the reciprocal of the second expression
- Common PERT error: canceling terms that are added or subtracted (for example, canceling $x$ in $\frac{x + 3}{x}$ is wrong)
- After simplifying, state the simplified form and, when asked, the values of $x$ that are not allowed
14.2 Rational Expressions
Quick Answer: A rational expression is a fraction whose numerator and denominator are polynomials, such as $\frac{x^2 - 9}{x^2 - x - 6}$. Simplify by factoring completely, then cancel only common factors. Multiply by factoring all numerators and denominators and canceling across; divide by multiplying by the reciprocal. Always list excluded values—any $x$ that makes a denominator zero—from the original expression before simplifying.
Rational expressions show up on PERT whenever algebra moves past plain polynomials into fractions of expressions. The arithmetic is the same as numerical fractions, but the algebra requires careful factoring and a strict rule: only factors cancel.
What Is a Rational Expression?
| Object | Form | Example |
|---|---|---|
| Rational expression | $\frac{P(x)}{Q(x)}$ with $Q(x) \neq 0$ | $\frac{x + 2}{x - 3}$ |
| Polynomial (special case) | Denominator is a nonzero constant | $\frac{x^2 + 1}{1}$ |
| Undefined at | Roots of the denominator | $\frac{x + 2}{x - 3}$ undefined at $x = 3$ |
Domain note: On PERT you usually do not write formal domain sets, but multiple-choice items frequently ask for values that make the expression undefined or for the simplified form for all allowed $x$.
Excluded Values (Critical First Step)
Excluded values (also called restrictions) are values of the variable that make any denominator equal to zero in the original expression—before you cancel factors.
Worked Example 1: Find excluded values
Problem: For $\frac{x^2 - 9}{x^2 - x - 6}$, which values of $x$ are excluded?
Step 1 — Factor the denominator only (for restrictions).
Step 2 — Set each factor to zero. $x - 3 = 0 \Rightarrow x = 3$; $x + 2 = 0 \Rightarrow x = -2$.
Step 3 — State exclusions. $x \neq 3$ and $x \neq -2$.
Even if a factor later cancels with the numerator, the value that made the original denominator zero remains excluded from the domain of the original expression. After canceling, you get an equivalent simplified expression on the common domain.
| Expression | Denominator zeros | Excluded values |
|---|---|---|
| $\frac{x + 1}{x}$ | $x = 0$ | $x \neq 0$ |
| $\frac{x - 2}{x^2 - 4}$ | $x = \pm 2$ | $x \neq 2$, $x \neq -2$ |
| $\frac{x + 5}{x^2 + 1}$ | $x^2 + 1 = 0$ has no real root | No real exclusions |
| $\frac{(x - 1)(x + 2)}{(x - 1)(x + 3)}$ | $x = 1$ or $x = -3$ | Both excluded originally |
Simplifying: Factor, Then Cancel Factors Only
The golden rule
You may cancel a factor that multiplies the entire numerator with the same factor that multiplies the entire denominator. You may not cancel a term that is only part of a sum or difference.
| Legal cancel | Illegal “cancel” |
|---|---|
| $\frac{(x - 1)(x + 2)}{(x - 1)(x + 3)} = \frac{x + 2}{x + 3}$ (for $x \neq 1$) | $\frac{x + 2}{x} \neq 2$ |
| $\frac{6x}{3x} = 2$ (for $x \neq 0$) | $\frac{x + 6}{x + 3} \neq 2$ |
| $\frac{x^2 - 9}{x - 3} = x + 3$ (for $x \neq 3$) | Crossing out the 9 and the 3 only |
Worked Example 2: Full simplify
Problem: Simplify $\frac{x^2 - 9}{x^2 - x - 6}$.
Step 1 — Factor numerator and denominator completely.
Step 2 — Cancel the common factor $(x - 3)$, noting $x \neq 3$ and also $x \neq -2$ from the other denominator factor.
Step 3 — Check with a number. Let $x = 4$: original $\frac{16 - 9}{16 - 4 - 6} = \frac{7}{6}$; simplified $\frac{4 + 3}{4 + 2} = \frac{7}{6}$. Match.
Worked Example 3: Common factor with coefficients
Problem: Simplify $\frac{6x^2 + 3x}{9x}$.
Factor: $\frac{3x(2x + 1)}{9x} = \frac{3x(2x + 1)}{3 \cdot 3 \cdot x} = \frac{2x + 1}{3}$ for $x \neq 0$.
Trap: Writing $\frac{6x^2 + 3x}{9x} = 6x^2 + \frac{1}{3}$ by “splitting” incorrectly. Always factor first; do not cancel the $x$ in only one term of the numerator.
Multiplying Rational Expressions
Process:
- Write the product as one big fraction: $\frac{A}{B} \cdot \frac{C}{D} = \frac{AC}{BD}$.
- Factor every polynomial in the numerator and denominator.
- Cancel common factors across top and bottom.
- Multiply remaining factors; state exclusions from all original denominators (and any denominator that appeared after a reciprocal in a division problem).
Worked Example 4: Multiply
Problem: Simplify $\frac{x^2 - 4}{x^2 + 5x + 6} \cdot \frac{x + 3}{x - 2}$.
Step 1 — Factor.
Step 2 — Cancel $(x - 2)$, $(x + 2)$, and $(x + 3)$ where allowed.
(Every factor canceled; the product simplifies to the constant 1 on the common domain.)
Worked Example 5: Multiply with leftover factors
Problem: $\frac{2x}{x - 1} \cdot \frac{x^2 - 1}{4x}$.
Dividing Rational Expressions
Division flips the second fraction:
Then proceed exactly as in multiplication. Exclusions include zeros of $B$, zeros of $C$ (the reciprocal’s denominator), and zeros of $D$ if it was originally a denominator in a more nested form—on standard PERT items, track zeros of every denominator that appears before canceling.
Worked Example 6: Divide
Problem: Simplify $\frac{x^2 - 9}{x + 1} \div \frac{x - 3}{x^2 - 1}$.
Step 1 — Reciprocal multiply.
Step 2 — Factor.
Step 3 — Cancel $(x - 3)$ and $(x + 1)$.
Exclusions (list before canceling): The first denominator requires $x \neq -1$. The second expression $\frac{x - 3}{x^2 - 1}$ must be defined, so $x^2 - 1 \neq 0$ ⇒ $x \neq \pm 1$, and it must be nonzero for division, so $x \neq 3$. Combined: $x \neq -1$, $x \neq 1$, and $x \neq 3$.
Check with $x = 2$: Original $\frac{4 - 9}{2 + 1} \div \frac{2 - 3}{4 - 1} = \frac{-5}{3} \div \frac{-1}{3} = \frac{-5}{3} \cdot (-3) = 5$. Simplified $2^2 + 2(2) - 3 = 5$. Match.
On multiple choice, match the simplified polynomial (and the stated restrictions if both appear).
Common Errors Catalog
| Error | Wrong move | Correct move |
|---|---|---|
| Canceling terms | $\frac{x + 5}{x} \to 5$ | Leave as $1 + \frac{5}{x}$ or do not “cancel $x$” |
| Canceling only one term | $\frac{x + 3}{x + 6} \to \frac{1}{2}$ | Factors differ; expression does not simplify that way |
| Forgetting exclusions after cancel | Cancel $(x - 2)$ and claim $x = 2$ is fine | Still exclude $x = 2$ for the original |
| Not factoring first | Cancel $x^2$ in $\frac{x^2 + 1}{x^2}$ wrongly | Factor? Cannot cancel $x^2$ from a sum; $\frac{x^2 + 1}{x^2} = 1 + \frac{1}{x^2}$ |
| Sign errors in factors | $x^2 - 9 = (x - 3)(x - 3)$ | Difference of squares is $(x - 3)(x + 3)$ |
Worked Example 7: Spot the trap
Problem: Which is equivalent to $\frac{x^2 + 5x}{x}$ for $x \neq 0$?
A. $x + 5$
B. $x + 5x$
C. $5$
D. $x^2 + 5$
Correct: Factor $\frac{x(x + 5)}{x} = x + 5$ for $x \neq 0$. Choice A.
Why C is tempting: Students cancel the $x$ in $5x$ only and drop $x^2$, which is illegal.
Mini Decision Checklist
- Identify every denominator in the original problem → excluded values.
- Factor all numerators and denominators completely.
- Cancel only identical factors (not terms).
- Multiply or take reciprocal for products/quotients, then cancel again.
- Write the simplified expression; attach exclusions if the question asks.
- Spot-check with a safe number not equal to an excluded value.
PERT-Style Comparison Table
| Operation | First rewrite | Then |
|---|---|---|
| Simplify one fraction | Factor top and bottom | Cancel common factors |
| Multiply two fractions | One fraction of products | Factor all; cancel; multiply |
| Divide two fractions | Multiply by reciprocal of second | Same as multiply |
| Find undefined values | Original denominators = 0 | Solve those equations |
Rational expression items reward methodical factoring more than speed tricks. If you can factor linear and simple quadratic polynomials confidently, these problems become mechanical—and the main way to lose points is canceling something that is not a factor.
Simplify $\frac{x^2 - 9}{x^2 - x - 6}$ for all $x$ where the expression is defined. Which is equivalent?
For the expression $\frac{x + 2}{x^2 - 4}$, which values of $x$ are excluded?
Which of the following is a correct simplification for $x \neq 0$?
Compute $\frac{2x}{x - 1} \cdot \frac{x^2 - 1}{4x}$ and simplify for all allowed $x$.