Simplifying Rational Expressions

A rational expression is a quotient of polynomials — the algebraic cousin of a numeric fraction. You simplify it the same way you'd reduce 6/8: factor top and bottom completely, then cancel what they share.

By the end you'll be able to factor a rational expression's numerator and denominator, cancel their common factor, and still name the domain restriction the cancelled factor leaves behind.

Predict: after factoring \(\dfrac{x^2-4}{x-2}\) and cancelling the shared factor \((x-2)\), what happens right at \(x=2\)? Does it become a normal, defined point — or does something stay excluded? Factor and cancel below to check.

Pick an expression, then step through factoring and cancelling. Watch the shared factor light up in both the numerator and denominator, and watch for the hole that appears on the graph at that same x-value, color-linked to the cancelled factor.

Original: \(\dfrac{x^2-4}{x-2}\)

Not yet factored — the shared factor and the hole it leaves behind aren't visible yet.

Graph — the true function, hole included
simplified function hole (excluded x)

Simplifying a rational expression is exactly like reducing a numeric fraction — factor, then cancel what's shared — except the "value it's undefined at" has to be tracked by hand, because the algebra no longer shows it.

Procedural

Three steps, in order: (1) factor the numerator completely, (2) factor the denominator completely, (3) cancel any factor appearing in both. Before you cancel anything, set the ORIGINAL denominator equal to zero and solve — every one of those x-values is excluded from the domain forever, whether or not its factor survives the cancellation.

Formal

A rational expression \(P(x)/Q(x)\) is in lowest terms when \(P\) and \(Q\) share no common polynomial factor besides 1. Cancelling a shared factor \((x-c)\) produces an expression that agrees with the original everywhere \(Q(x) \ne 0\) — but at \(x=c\) itself, the original is undefined (0/0) while the reduced form is perfectly defined. That single mismatched point is the hole: the function's domain is still missing \(x=c\), even though the simplified formula no longer objects to it.

Applied

An average-speed formula like \(\dfrac{\text{distance}^2 - 4}{\text{distance} - 2}\) might simplify to something as clean as \(\text{distance} + 2\) — but if distance = 2 was the one input that made the original setup meaningless (say, a divide-by-zero in how the formula was derived), that input is still off-limits, no matter how clean the reduced formula looks.

Worked example

Simplify \(\dfrac{x^2-1}{x^2+3x+2}\). Factor the numerator (difference of squares): \(x^2-1=(x-1)(x+1)\). Factor the denominator: \(x^2+3x+2=(x+1)(x+2)\). Rewrite as \(\dfrac{(x-1)(x+1)}{(x+1)(x+2)}\); the denominator is zero at \(x=-1\) and \(x=-2\), so exclude both before cancelling. Cancel \((x+1)\): \(\dfrac{x^2-1}{x^2+3x+2} = \) \(\dfrac{x-1}{x+2}\), \(x \ne -1,\ x \ne -2\).

Your turn

Simplify \(\dfrac{x^2-16}{x^2+x-12}\). Factor the numerator: \(x^2-16=(x-4)(x+4)\). Factor the denominator: \(x^2+x-12=(x+4)(x-3)\). The original denominator is zero at \(x=-4\) and \(x=3\) — exclude both. Now cancel the shared factor \((x+4)\) and finish: \(\dfrac{x^2-16}{x^2+x-12} = \) ____, \(x \ne\) ____.

Reveal the answer

\(\dfrac{x^2-16}{x^2+x-12} = \dfrac{(x-4)(x+4)}{(x+4)(x-3)} = \) \(\dfrac{x-4}{x-3}\), \(x \ne -4,\ x \ne 3\). Select the first example above, factor, and cancel to see the same cancel-and-track pattern play out on the graph.

More info — why the ORIGINAL denominator decides the exclusions

Plug \(x=2\) into \(\dfrac{x^2-4}{x-2}\) directly and you get \(0/0\) — undefined, full stop. Plug \(x=2\) into the simplified form \(x+2\) and you get a perfectly ordinary 4. Those two functions agree everywhere else, but they are not identical functions, because one has a bigger domain than the other. Reporting the simplified form alone, without the exclusion, silently claims a domain the original expression never had. See the Paul's Online Notes link in Dive deeper for more worked reductions.

Check your understanding

Question 1 of 4

Simplify \(\dfrac{x^2-25}{x^2+8x+15}\) and state the excluded value(s).

Question 2 of 4

Simplify \(\dfrac{x+3}{x}\), if possible.

Question 3 of 4

To simplify \(\dfrac{x^3-8}{x^2-4}\), first factor the numerator as a difference of cubes and the denominator as a difference of squares. What is the simplified form, and which value(s) must be excluded?

Question 4 of 4

\(\dfrac{x^2-9}{x^2-x-6}\) simplifies to \(\dfrac{x+3}{x+2}\). Which value(s) must still be excluded from the domain?

Recap

  • A rational expression is a quotient of polynomials \(P(x)/Q(x)\), with \(Q(x)\) not the zero polynomial.
  • Simplify by factoring the numerator and denominator completely, then cancelling only shared FACTORS — never terms that are added.
  • Excluded values come from the ORIGINAL denominator's zeros; they survive simplification even after their factor cancels.
  • A cancelled factor's zero shows up on the graph as a hole — a single missing point, still outside the domain.

Dive deeper

Sources

  • Simplifying Rational Expressions