A rational expression is one polynomial over another, and it means nothing at all where the bottom is zero. Those excluded values come from the denominator you were handed — never from the tidier one you end up with.
Core ideas
Everything in the first half of this course is a rule about factors, and every one of those rules is false at a value where the expression does not exist. So the domain comes first, on its own line.
(3x + 1)/(x² − 5) is one; so is x + 2, with denominator 1. What is not one is anything with a variable under a root or in an exponent — those arrive in “Simplifying Radical Expressions” and in Exponential and Logarithmic Functions.
The numerator may be anything, zero included. (x − 4)/(x + 1) is perfectly well defined at x = 4, where its value is 0. It is x = −1 that is excluded.
0/0 is not 0, not 1, and not "infinity". Division is defined as the inverse of multiplication, and no number times 0 gives 4, while every number times 0 gives 0. Neither case yields an answer.
The material
The definition, then the question you ask before any algebra.
A rational expression is a quotient P(x)/Q(x) of two polynomials in which Q is not the zero polynomial. It is the algebraic version of a fraction, and it obeys the same rules for the same reasons.
The condition on Q is about the polynomial, not about x. Q(x) = x − 2 is a perfectly good denominator even though it takes the value zero somewhere; what is forbidden is Q(x) = 0 for every x, which would leave nothing to divide by anywhere.
The domain of a rational expression is the set of real numbers x for which Q(x) ≠ 0. A number c with Q(c) = 0 is an excluded value: the expression assigns nothing to it.
For the denominators this course asks you to solve, finding the domain starts as a factoring problem. Factor the denominator, set each factor to zero, and collect the real solutions. This is the first of many places in this course where an unfactored denominator hides the answer.
For (x + 1)/(x² − x − 6), factor the bottom: x² − x − 6 = (x − 3)(x + 2). It is zero when x = 3 and when x = −2, so the domain is every real number except those two. The numerator's zero at x = −1 excludes nothing — the expression is simply 0 there.
Now the case that gives this lesson its point. (x² − 4)/(x − 2) has denominator zero at x = 2, so 2 is excluded. Factor the top and the expression becomes ((x − 2)(x + 2))/(x − 2), which for every x ≠ 2 equals x + 2. At x = 2 it equals nothing: the substitution gives 0/0.
The values approach 4 from both sides and the expression still has no value at 2. That gap is called a hole, and “Graphs and Asymptotes” puts it on a graph. What matters now is that x + 2 — the answer to the simplification — carries no record of it. The exclusion is a fact about where you started.
Two rational expressions are equal when they agree at every value in the domain of both. So (x² − 4)/(x − 2) = x + 2 is a true statement for x ≠ 2, and writing it without that condition is writing something false at one point.
Interactive
Every denominator, set to zero, while it is still there to set
Method
Four steps, and the first one is not optional.
An unfactored quadratic hides its zeros. x² − x − 6 looks harmless; (x − 3)(x + 2) says exactly which two numbers are forbidden.
One linear equation per factor. A repeated factor such as (x + 2)² contributes the single value x = −2, not two of them.
Each solution leaves the domain. Note that you never test the numerator: it has no say in whether the expression exists.
Put x ≠ 3, x ≠ −2 beside the expression now. After you cancel, the surviving form may be defined at those values, and you will have no way to recover them.
Factor, solve, exclude. The check at the end costs one line and catches a mis-factored denominator immediately.
The last line is the one students skip. x = −1 makes the numerator zero, and a beginner who has learned "zero is a problem" excludes it. It is not a problem: 0/(−4) = 0, an ordinary value at an ordinary point of the domain.
For a faded domain pass, take (2x − 1)/(x² + x − 12). The supplied factorisation is x² + x − 12 = (x + 4)(x − 3). Write the domain, then decide whether the numerator's zero at x = 1/2 is allowed. Check against x ≠ −4, 3, with 1/2 retained. If you excluded 1/2, you tested both levels instead of asking only where division fails.
Given any rational expression you should produce its domain in one line, from the factored denominator, without simplifying first. If your habit is to simplify and then look for trouble, you will lose an exclusion in “Simplifying Rational Expressions” and an extraneous root in “Solving Rational Equations”.
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