Simplifying, multiplying, and combining rational expressions leans directly on Factoring Polynomials — every technique in this lesson starts with factoring the numerator and denominator, then applies the same fraction rules you already know from arithmetic.
What Is a Rational Expression?
A rational expression is a fraction whose numerator and denominator are both polynomials, such as \(\dfrac{x^{2}-4}{x+2}\). Since division by zero is undefined, a rational expression’s domain excludes any value of \(x\) that makes the denominator zero — for \(\dfrac{x^{2}-4}{x+2}\), that’s \(x=-2\).
A simplifying convention. As in earlier lessons, every technique below still requires stating excluded values explicitly — factoring and simplifying never removes them, even when a factor cancels.
Simplifying Rational Expressions
To simplify a rational expression, factor the numerator and denominator completely, then cancel any factor common to both:
$$ \frac{x^{2}-4}{x+2} = \frac{(x+2)(x-2)}{x+2} = x-2, \qquad x \neq -2 $$
The excluded value \(x=-2\) is carried over from the original denominator — it doesn’t disappear just because the factor canceled.
Properties of Rational Expressions
| Rule | Statement |
|---|---|
| Simplifying | \(\dfrac{ac}{bc} = \dfrac{a}{b}\), for \(c \neq 0\) |
| Multiplying | \(\dfrac{a}{b} \cdot \dfrac{c}{d} = \dfrac{ac}{bd}\) |
| Dividing | \(\dfrac{a}{b} \div \dfrac{c}{d} = \dfrac{a}{b} \cdot \dfrac{d}{c}\) |
| Adding (same denominator) | \(\dfrac{a}{c} + \dfrac{b}{c} = \dfrac{a+b}{c}\) |
| Subtracting (same denominator) | \(\dfrac{a}{c} - \dfrac{b}{c} = \dfrac{a-b}{c}\) |
Every one of these is the exact rule you already use for numeric fractions — nothing new to learn, only polynomials to factor first.
Example 1: Simplifying Rational Expressions
| Expression | Factored | Simplified |
|---|---|---|
| \(\dfrac{x^{2}-9}{x+3}\) | \(\dfrac{(x+3)(x-3)}{x+3}\) | \(x-3\), \(x \neq -3\) |
| \(\dfrac{x^{2}+5x+6}{x+2}\) | \(\dfrac{(x+2)(x+3)}{x+2}\) | \(x+3\), \(x \neq -2\) |
| \(\dfrac{3x^{2}-12x}{x-4}\) | \(\dfrac{3x(x-4)}{x-4}\) | \(3x\), \(x \neq 4\) |
| \(\dfrac{x^{2}-16}{x^{2}-4x}\) | \(\dfrac{(x+4)(x-4)}{x(x-4)}\) | \(\dfrac{x+4}{x}\), \(x \neq 0, 4\) |
Multiplying and Dividing Rational Expressions
Multiplying works exactly like numeric fractions: multiply numerators together, denominators together, then simplify by canceling common factors. Dividing means multiplying by the reciprocal of the second expression — flip it, then multiply.
Example 2: Multiplying and Dividing
| Expression | Result |
|---|---|
| \(\dfrac{x}{3} \cdot \dfrac{9}{x^{2}}\) | \(\dfrac{3}{x}\), \(x \neq 0\) |
| \(\dfrac{x+2}{x-1} \cdot \dfrac{x-1}{x+5}\) | \(\dfrac{x+2}{x+5}\), \(x \neq 1, -5\) |
| \(\dfrac{x^{2}-1}{x+3} \div \dfrac{x-1}{x+3}\) | \(x+1\), \(x \neq -3, 1\) |
| \(\dfrac{4}{x+2} \div \dfrac{8}{x^{2}-4}\) | \(\dfrac{x-2}{2}\), \(x \neq -2\) |
Adding and Subtracting Rational Expressions
Same denominator combines directly, just like numeric fractions. Different denominators need a least common denominator (LCD) first — factor each denominator, then build the LCD from every factor that appears, before rewriting and combining.
$$ \frac{2}{x} + \frac{3}{x+1} = \frac{2(x+1)}{x(x+1)} + \frac{3x}{x(x+1)} = \frac{5x+2}{x(x+1)}, \qquad x \neq 0, -1 $$
Example 3: Adding and Subtracting
| Expression | Result |
|---|---|
| \(\dfrac{3}{x} + \dfrac{2}{x}\) | \(\dfrac{5}{x}\), \(x \neq 0\) |
| \(\dfrac{1}{x+1} + \dfrac{2}{x+1}\) | \(\dfrac{3}{x+1}\), \(x \neq -1\) |
| \(\dfrac{2}{x} + \dfrac{3}{x+1}\) | \(\dfrac{5x+2}{x(x+1)}\), \(x \neq 0, -1\) |
| \(\dfrac{1}{x-2} - \dfrac{1}{x+2}\) | \(\dfrac{4}{x^{2}-4}\), \(x \neq 2, -2\) |
Worked Example A: Simplify by Factoring Both Parts
Simplify \(\dfrac{x^{2}-x-6}{x^{2}-9}\).
Step 1 — factor the numerator (numbers multiplying to \(-6\), adding to \(-1\)):
$$ x^{2}-x-6 = (x-3)(x+2) $$
Step 2 — factor the denominator (difference of squares):
$$ x^{2}-9 = (x+3)(x-3) $$
Step 3 — cancel the common factor \((x-3)\):
$$ \frac{(x-3)(x+2)}{(x+3)(x-3)} = \frac{x+2}{x+3}, \qquad x \neq 3, -3 $$
Worked Example B: Multiply, Then Watch Everything Cancel
Simplify \(\dfrac{x^{2}-4}{x^{2}+5x+6} \cdot \dfrac{x+3}{x-2}\).
Step 1 — factor every part:
$$ \frac{(x-2)(x+2)}{(x+2)(x+3)} \cdot \frac{x+3}{x-2} $$
Step 2 — cancel every matching factor:
$$ \frac{(x-2)(x+2)(x+3)}{(x+2)(x+3)(x-2)} = 1, \qquad x \neq 2, -2, -3 $$
Every factor in the numerator matches one in the denominator — the entire expression collapses to \(1\), a good reminder to fully factor before deciding an expression can’t be simplified.
Worked Example C: Adding with Different Denominators
Simplify \(\dfrac{3}{x-1} + \dfrac{2}{x+4}\).
Step 1 — the LCD is the product of both denominators, since they share no common factor:
$$ \frac{3(x+4)}{(x-1)(x+4)} + \frac{2(x-1)}{(x-1)(x+4)} $$
Step 2 — combine the numerators:
$$ 3(x+4) + 2(x-1) = 3x+12+2x-2 = 5x+10 $$
Final answer: \(\dfrac{5x+10}{(x-1)(x+4)}\), \(x \neq 1, -4\). The numerator factors further to \(5(x+2)\), but since neither denominator factor matches, nothing more cancels.
Common Mistakes to Avoid
- Canceling terms that are added, not multiplied. \(\dfrac{x+3}{x} \neq 3\) — the \(x\)‘s don’t cancel, because the numerator’s \(x\) isn’t a factor of the whole numerator.
- Forgetting to state excluded values. They come from the original denominator, before any simplifying.
- Adding or subtracting without a common denominator. Just like numeric fractions, this is never valid.
- Forgetting to distribute a negative sign. Subtracting a rational expression with a multi-term numerator means flipping every term’s sign, not just the first.
- Canceling only part of a factor. \((x+2)\) in the numerator only cancels with an identical \((x+2)\) in the denominator — never with just the \(x\) or just the \(2\).
Practice Problems
Work each problem yourself before opening the answer.
Problem 1. Simplify \(\dfrac{x^{2}-25}{x+5}\).
Show answer
$$ \frac{x^{2}-25}{x+5} = \frac{(x+5)(x-5)}{x+5} = x-5 $$
Answer: \(x-5\), \(x \neq -5\)
Problem 2. Simplify \(\dfrac{x^{2}+6x+8}{x+2}\).
Show answer
$$ \frac{x^{2}+6x+8}{x+2} = \frac{(x+2)(x+4)}{x+2} = x+4 $$
Answer: \(x+4\), \(x \neq -2\)
Problem 3. Simplify \(\dfrac{4x^{2}-16x}{x-4}\).
Show answer
$$ \frac{4x^{2}-16x}{x-4} = \frac{4x(x-4)}{x-4} = 4x $$
Answer: \(4x\), \(x \neq 4\)
Problem 4. Multiply \(\dfrac{x}{4} \cdot \dfrac{8}{x^{2}}\).
Show answer
$$ \frac{x}{4} \cdot \frac{8}{x^{2}} = \frac{8x}{4x^{2}} = \frac{2}{x} $$
Answer: \(\dfrac{2}{x}\), \(x \neq 0\)
Problem 5. Multiply \(\dfrac{x+1}{x-3} \cdot \dfrac{x-3}{x+6}\).
Show answer
$$ \frac{x+1}{x-3} \cdot \frac{x-3}{x+6} = \frac{x+1}{x+6} $$
Answer: \(\dfrac{x+1}{x+6}\), \(x \neq 3, -6\)
Problem 6. Divide \(\dfrac{x^{2}-1}{x+4} \div \dfrac{x+1}{x+4}\).
Show answer
$$ \frac{x^{2}-1}{x+4} \cdot \frac{x+4}{x+1} = \frac{(x-1)(x+1)}{x+1} = x-1 $$
Answer: \(x-1\), \(x \neq -4, -1\)
Problem 7. Add \(\dfrac{4}{x} + \dfrac{3}{x}\).
Show answer
$$ \frac{4}{x} + \frac{3}{x} = \frac{7}{x} $$
Answer: \(\dfrac{7}{x}\), \(x \neq 0\)
Problem 8. Add \(\dfrac{1}{x} + \dfrac{1}{x+3}\).
Show answer
$$ \frac{x+3}{x(x+3)} + \frac{x}{x(x+3)} = \frac{2x+3}{x(x+3)} $$
Answer: \(\dfrac{2x+3}{x(x+3)}\), \(x \neq 0, -3\)
Problem 9. Subtract \(\dfrac{1}{x+1} - \dfrac{1}{x-1}\).
Show answer
$$ \frac{x-1}{(x+1)(x-1)} - \frac{x+1}{(x+1)(x-1)} = \frac{(x-1)-(x+1)}{x^{2}-1} = \frac{-2}{x^{2}-1} $$
Answer: \(\dfrac{-2}{x^{2}-1}\), \(x \neq -1, 1\)
Problem 10. Simplify \(\dfrac{x^{2}-x-6}{x^{2}-4}\).
Show answer
$$ \frac{(x-3)(x+2)}{(x-2)(x+2)} = \frac{x-3}{x-2} $$
Answer: \(\dfrac{x-3}{x-2}\), \(x \neq -2, 2\)
Quick Reference
| Situation | Move |
|---|---|
| Simplifying | Factor numerator and denominator, cancel common factors only |
| Multiplying | Multiply numerators, multiply denominators, then simplify |
| Dividing | Multiply by the reciprocal of the second expression |
| Same denominator | Add or subtract numerators directly |
| Different denominators | Find the LCD first, rewrite each fraction, then combine |
| Stating the domain | List every value that makes the original denominator zero |
Once these techniques feel automatic, they carry forward directly into solving rational equations and into Complex Numbers, which reuses the exact same conjugate trick from Radicals to clear an imaginary denominator instead of a radical one. Build speed with the Rational Expression Generator, browse the rest of the Algebra lessons as they publish, revisit Factoring Polynomials if any step above felt unfamiliar, or explore the site’s calculators to check your own arithmetic against a worked result.