Operations on Rational Expressions

Multiplying and dividing rational expressions is factor-and-cancel, just like numerical fractions. Adding and subtracting takes one more step first: rewriting each piece over a shared denominator before the numerators can combine.

By the end you'll be able to multiply, divide, add, and subtract rational expressions — factoring and cancelling for ×/÷, building a common denominator for +/− — and track which values stay excluded from the domain along the way.

Predict: does adding two rational expressions require a common denominator? Click + below and check the steps.

Same two expressions, four operations. For × and ÷ you factor and cancel; division first flips the second expression to its reciprocal — try ÷ to see the flip. Watch how (x − 2) and (x + 3) track through each operation.

\(\dfrac{x+1}{x-2}\) and \(\dfrac{x-2}{x+3}\) — denominator factors (x − 2) and (x + 3).

× → \(\dfrac{x+1}{x+3}\), domain \(x \neq 2,\ x \neq -3\)

Every operation on rational expressions ends the same place: a single simplified quotient of polynomials, with the excluded values tracked along the way (see simplifying rational expressions).

Procedural

Multiply: \(\frac{A}{B}\cdot\frac{C}{D} = \frac{AC}{BD}\) — factor everything first and cancel any factor shared by a numerator and a denominator before multiplying out. Divide: \(\frac{A}{B}\div\frac{C}{D} = \frac{A}{B}\cdot\frac{D}{C}\); flip the divisor to its reciprocal, then proceed exactly like multiplication. Add or subtract: rewrite each expression over the least common denominator — the product of the distinct denominator factors, each raised to the highest power that appears — then combine numerators over that one denominator and simplify.

Formal

Only factors may cancel, never terms: in \(\frac{(x-1)(x+4)}{x-1}\) the (x−1) cancels because it's a factor of the whole numerator, but in \(\frac{x+4}{x}\) you cannot cancel the x's, since x is only a term of the numerator. When subtracting, distribute the minus sign across the entire second numerator, not just its first term — a frequent source of sign errors. Any value that zeroes an original denominator (or a numerator/denominator you cancelled) stays excluded from the final domain, even after it disappears from the simplified result.

Applied

Combined resistance in a parallel circuit, mixing rates, and probability odds all add rational expressions with different denominators — you can't just add the numbers on top; you first need everything measured against the same denominator before the totals mean anything.

Worked example — multiply

Compute \(\dfrac{x^2-1}{x+2}\cdot\dfrac{x+2}{x+1}\). Factor: \(\dfrac{(x-1)(x+1)}{x+2} \cdot \dfrac{x+2}{x+1}\). Cancel \((x+2)\) and \((x+1)\): the result is \(x - 1\), with \(x \neq -2\) and \(x \neq -1\) excluded — both values come from factors that appeared in a denominator before cancelling.

Your turn — divide

Compute \(\dfrac{x^2-36}{x+8} \div \dfrac{x-6}{x+8}\). Flip the divisor to its reciprocal: \(\dfrac{x^2-36}{x+8}\cdot\dfrac{x+8}{x-6}\). Factor the first numerator: \(\dfrac{(x-6)(x+6)}{x+8}\cdot\dfrac{x+8}{x-6}\). Now finish it: cancel \((x+8)\) and \((x-6)\), giving ____, with domain restrictions \(x \neq -8\) and ____.

Reveal the answer

Cancelling \((x+8)\) and \((x-6)\) leaves \(x + 6\). The domain excludes \(x \neq -8\) (original denominator) and \(x \neq 6\) — that's the value that made the divisor's numerator \((x-6)\) zero, which would make the division itself undefined, so it stays excluded even though \((x-6)\) cancelled out of the final answer.

More info — why the common denominator has to be the LCD

Any shared denominator works arithmetically, but the LCD (least common denominator) keeps the numbers as small as possible: it's the product of the distinct factors across both denominators, each raised to the highest power it appears with. Using a bigger common denominator (say, the plain product of two denominators that already share a factor) still gives a correct sum, but leaves extra cancelling to do afterward — the LCD does that cancelling up front. This is the exact same idea as simplifying a result after combining it (see simplifying rational expressions): factor, then cancel what you can.

Check your understanding

Question 1 of 4

Multiply \(\dfrac{x+5}{x-1} \times \dfrac{x-1}{x+2}\) and state the domain.

Question 2 of 4

Add \(\dfrac{2}{x+3} + \dfrac{1}{x-4}\) and state the domain.

Question 3 of 4

For \(\dfrac{A}{B} \div \dfrac{C}{D}\), what's the correct next step before multiplying?

Question 4 of 4

Adding two rational expressions gave \(\dfrac{x^2-9}{(x-3)(x+7)}\) before final simplification. Simplify it and state the domain.

Recap

  • Multiply: \(\frac{A}{B}\cdot\frac{C}{D}=\frac{AC}{BD}\) — factor first, then cancel shared factors (never terms).
  • Divide: \(\frac{A}{B}\div\frac{C}{D}=\frac{A}{B}\cdot\frac{D}{C}\) — flip the divisor to its reciprocal, then multiply.
  • Add or subtract: rewrite each expression over the least common denominator, combine numerators (distribute a minus sign across the whole second numerator), then simplify.
  • A value that zeroed any original denominator — or a factor you cancelled away — stays excluded from the domain, even if it no longer appears in the simplified result.

Dive deeper

Sources

  • Operations on Rational Expressions