Simplifying a rational expression is the same idea as reducing an arithmetic fraction like \(\dfrac{6}{8}\) to \(\dfrac{3}{4}\) — cancel whatever factor the numerator and denominator share. The difference is that a rational expression’s shared factor is almost never sitting in plain sight; it has to be found by factoring first. This lesson covers that two-step process in full, plus the one detail that’s easy to lose along the way: what cancelling a factor does, and doesn’t, do to the expression’s domain.
The Two-Step Process
- Factor the numerator and denominator completely, using whichever techniques from Factoring Polynomials apply.
- Cancel any factor that appears in both, leaving behind whatever factors don’t match.
$$ \frac{x^{2}-9}{x^{2}+5x+6} = \frac{(x+3)(x-3)}{(x+2)(x+3)} = \frac{x-3}{x+2} $$
The \((x+3)\) factor appears in both the numerator and denominator and cancels; \((x-3)\) and \((x+2)\) don’t match anything on the other side, so they stay exactly where they are.
Excluded Values: What Cancelling Doesn’t Erase
Every rational expression is undefined wherever its original, unsimplified denominator equals zero. Cancelling a shared factor can make that restriction disappear from the written expression without making it disappear from the actual domain.
$$ \frac{(x+3)(x-3)}{(x+2)(x+3)} = \frac{x-3}{x+2}, \qquad x \neq -2, \; x \neq -3 $$
The simplified form \(\dfrac{x-3}{x+2}\) only shows the restriction \(x \neq -2\) on its own — the \(x\neq -3\) restriction came from the factor that got cancelled, and it still applies, because the original expression was genuinely undefined there before any simplifying happened. Stating both restrictions alongside the simplified answer is part of a complete solution.
Cancelling Factors, Not Terms
Only a factor — something multiplying the entire numerator or denominator — can be cancelled. A term that’s added or subtracted can never be cancelled on its own, no matter how similar it looks to something in the denominator.
$$ \frac{x+3}{x} \neq \frac{3}{1} \qquad \text{(the } x \text{‘s do not cancel — } x \text{ is not a factor of the whole numerator)} $$
The numerator \(x+3\) has no common factor with the denominator \(x\) at all, so this expression is already in lowest terms — a very common wrong instinct is to “cancel the \(x\)‘s,” which is not a legal operation here.
The Opposite-Factors Pattern
A numerator and denominator that are exact opposites of each other — like \((x-5)\) and \((5-x)\) — don’t look like they share a factor, but factoring a \(-1\) out of one of them reveals that they do.
$$ 5-x = -1(x-5) \qquad \frac{x-5}{5-x} = \frac{x-5}{-1(x-5)} = \frac{1}{-1} = -1 $$
Recognizing this pattern on sight — two binomials that are identical except every sign is flipped — saves rewriting the whole expression from scratch every time it appears.
Worked Example A: Factoring Both the Numerator and Denominator
Simplify \(\dfrac{x^{2}+7x+10}{x^{2}+2x}\), and state the excluded values.
Step 1 — factor the numerator and denominator:
$$ \frac{(x+2)(x+5)}{x(x+2)} $$
Step 2 — cancel the shared factor \((x+2)\):
$$ \frac{x+5}{x} $$
Answer: \(\dfrac{x+5}{x}\), for \(x \neq 0\) and \(x \neq -2\)
Worked Example B: A GCF Hiding in Both Parts
Simplify \(\dfrac{4x^{2}-16}{2x^{2}+8x+8}\), and state the excluded values.
Step 1 — factor a GCF out of each part first:
$$ \frac{4(x^{2}-4)}{2(x^{2}+4x+4)} $$
Step 2 — factor each remaining polynomial (a difference of squares, and a perfect square trinomial):
$$ \frac{4(x+2)(x-2)}{2(x+2)^{2}} $$
Step 3 — cancel the shared factor \((x+2)\) and simplify the numeric coefficient:
$$ \frac{4(x-2)}{2(x+2)} = \frac{2(x-2)}{x+2} $$
Answer: \(\dfrac{2(x-2)}{x+2}\), for \(x \neq -2\)
Worked Example C: The Opposite-Factors Pattern
Simplify \(\dfrac{2x-8}{4-x}\), and state the excluded values.
Step 1 — factor the numerator, and factor a \(-1\) out of the denominator:
$$ \frac{2(x-4)}{-1(x-4)} $$
Step 2 — cancel the shared factor \((x-4)\):
$$ \frac{2}{-1} = -2 $$
Answer: \(-2\), for \(x \neq 4\). Even though the simplified answer is a constant, the original expression was still undefined at \(x=4\), so that restriction is still part of a complete answer.
Worked Example D: A Rational Expression Already in Lowest Terms
Simplify \(\dfrac{x^{2}+4}{x+1}\), if possible.
Step 1 — attempt to factor the numerator:
\(x^{2}+4\) is a sum of squares, which does not factor over the integers, and the denominator \(x+1\) shares no factor with it.
Answer: \(\dfrac{x^{2}+4}{x+1}\) is already in lowest terms — there is nothing to cancel.
Common Mistakes to Avoid
- Cancelling a term instead of a factor. In \(\dfrac{x+3}{x}\), nothing cancels — \(x\) is not a factor of the entire numerator, only part of a sum.
- Losing the excluded value that came from a cancelled factor. \(x \neq -3\) is still a restriction after \(\dfrac{(x+3)(x-3)}{(x+2)(x+3)}\) simplifies to \(\dfrac{x-3}{x+2}\), even though \(-3\) no longer makes the simplified expression’s denominator zero.
- Forgetting to factor a GCF out before looking for a matching binomial. A shared factor can be hidden by unfactored numeric coefficients, exactly as in Worked Example B.
- Missing the opposite-factors pattern. \((x-5)\) and \((5-x)\) are negatives of each other and do share a factor of \(-1\), even though they don’t look identical at first glance.
- Assuming every rational expression simplifies. Many genuinely don’t, once fully factored — “already in lowest terms” is a valid final answer.
- Cancelling before factoring is complete. Every factor has to be fully broken down first; cancelling based on a partially factored expression risks missing a shared factor or cancelling something that isn’t actually shared.
Where This Shows Up Later
- Multiplying and dividing rational expressions. Both operations depend on this exact factor-then-cancel process, applied across two rational expressions at once instead of one.
- Adding and subtracting rational expressions. Finding a least common denominator starts with factoring every denominator involved, using the same technique as this lesson.
- Solving rational equations. Every excluded value found while simplifying becomes a value that must be checked against (and often rejected from) the equation’s final solution set.
- Graphing rational functions. A cancelled factor corresponds to a hole in the function’s graph, while an uncancelled denominator factor corresponds to a vertical asymptote — the distinction this lesson draws has a direct graphical meaning.
Practice Problems
Work each problem yourself before opening the answer.
Problem 1. Simplify \(\dfrac{x^{2}-16}{x^{2}+4x}\), and state the excluded values.
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$$ \frac{(x+4)(x-4)}{x(x+4)} = \frac{x-4}{x} $$
Answer: \(\dfrac{x-4}{x}\), for \(x \neq 0\) and \(x \neq -4\)
Problem 2. Simplify \(\dfrac{x^{2}+5x+6}{x^{2}+3x}\), and state the excluded values.
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$$ \frac{(x+2)(x+3)}{x(x+3)} = \frac{x+2}{x} $$
Answer: \(\dfrac{x+2}{x}\), for \(x \neq 0\) and \(x \neq -3\)
Problem 3. Simplify \(\dfrac{3x-9}{x^{2}-9}\), and state the excluded values.
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$$ \frac{3(x-3)}{(x+3)(x-3)} = \frac{3}{x+3} $$
Answer: \(\dfrac{3}{x+3}\), for \(x \neq 3\) and \(x \neq -3\)
Problem 4. Simplify \(\dfrac{6-2x}{x-3}\), and state the excluded values.
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$$ \frac{-2(x-3)}{x-3} = -2 $$
Answer: \(-2\), for \(x \neq 3\)
Problem 5. Simplify \(\dfrac{x^{2}-x-6}{x^{2}-4}\), and state the excluded values.
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$$ \frac{(x-3)(x+2)}{(x+2)(x-2)} = \frac{x-3}{x-2} $$
Answer: \(\dfrac{x-3}{x-2}\), for \(x \neq 2\) and \(x \neq -2\)
Problem 6. Simplify \(\dfrac{x+7}{x^{2}+9}\), if possible.
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\(x^{2}+9\) is a sum of squares and does not factor over the integers; the numerator and denominator share no common factor.
Answer: already in lowest terms, \(\dfrac{x+7}{x^{2}+9}\)
Problem 7. Simplify \(\dfrac{2x^{2}-8}{4x^{2}+8x}\), and state the excluded values.
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Step 1 — factor out a GCF from each part:
$$ \frac{2(x^{2}-4)}{4x(x+2)} $$
Step 2 — factor the remaining difference of squares:
$$ \frac{2(x+2)(x-2)}{4x(x+2)} $$
Step 3 — cancel the shared factor and simplify the coefficient:
$$ \frac{2(x-2)}{4x} = \frac{x-2}{2x} $$
Answer: \(\dfrac{x-2}{2x}\), for \(x \neq 0\) and \(x \neq -2\)
Problem 8. Simplify \(\dfrac{x^{2}+6x+9}{x^{2}-9}\), and state the excluded values.
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$$ \frac{(x+3)^{2}}{(x+3)(x-3)} = \frac{x+3}{x-3} $$
Answer: \(\dfrac{x+3}{x-3}\), for \(x \neq -3\) and \(x \neq 3\)
Problem 9. Simplify \(\dfrac{5x-10}{2x^{2}-8}\), and state the excluded values.
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Step 1 — factor the numerator and denominator:
$$ \frac{5(x-2)}{2(x^{2}-4)} = \frac{5(x-2)}{2(x+2)(x-2)} $$
Step 2 — cancel the shared factor \((x-2)\):
$$ \frac{5}{2(x+2)} $$
Answer: \(\dfrac{5}{2(x+2)}\), for \(x \neq 2\) and \(x \neq -2\)
Problem 10. Simplify \(\dfrac{x^{2}-49}{7-x}\), and state the excluded values.
Show answer
Step 1 — factor the numerator and factor a \(-1\) out of the denominator:
$$ \frac{(x+7)(x-7)}{-1(x-7)} $$
Step 2 — cancel the shared factor \((x-7)\):
$$ \frac{x+7}{-1} = -(x+7) $$
Answer: \(-(x+7)\), for \(x \neq 7\)
Quick Reference
| Situation | Move |
|---|---|
| Any rational expression | Factor the numerator and denominator completely before doing anything else |
| Matching factor on top and bottom | Cancel it, leaving whatever doesn’t match |
| A term (added or subtracted) that looks similar | Cannot be cancelled — only multiplied factors cancel |
| Opposite binomials, like \((x-a)\) and \((a-x)\) | Factor a \(-1\) out of one to reveal the shared factor |
| Excluded values | Always come from the original denominator, and stay in force even after a factor cancels |
| No shared factor exists | The expression is already in lowest terms |
Once factor-then-cancel feels automatic, this is exactly the first step in multiplying, dividing, adding, and subtracting rational expressions. Revisit Rational Expressions for the full toolkit this lesson is drawn from, or Factoring Polynomials for the factoring techniques it depends on. Build speed with the Simplifying Rational Expressions Generator, or browse the rest of the Algebra lessons as new ones publish.