What is a rational expression?

A rational expression is a quotient of two polynomials, such as (x + 2) / (x² − 9). It is the algebraic counterpart of an ordinary fraction: polynomials stand where the numbers used to, and the denominator must not be zero.

How do I find the domain of a rational expression?

Factor the denominator and set each factor to zero. The numbers that come out are excluded from the domain. For (x + 2) / (x² − 9) the denominator is (x − 3)(x + 3), so the domain is the reals without −3 and 3. Note that a repeated factor (x − 3)² excludes one number, not two.

What may I not cancel in an algebraic fraction?

Only a common FACTOR cancels — something multiplied by the rest. In (x + 3) / (x + 5) the x does not cancel, because it is a term of a sum, not a factor. That is why cancelling is always preceded by factoring: until then you cannot see what the factors are.

How do I add two algebraic fractions?

Exactly as numerical ones: factor the denominators, take their least common multiple, widen both fractions to it and add the numerators. The common denominator need not be the product — for 1/(x − 2) and 3/((x − 2)(x + 1)) it is just (x − 2)(x + 1).

How do I solve a rational equation?

State the domain first, then bring the equation to the form fraction = 0 and set the numerator to zero. Finally discard any candidate that fell outside the domain. That last step is not optional — without it you can quote as an answer a number for which the equation has no meaning at all.

Advanced level

Rational expressions and equations

A fraction with polynomials where the numbers used to be — and the same four operations as on ordinary fractions, only preceded by factoring. The domain and the excluded numbers, cancelling, multiplying and dividing, adding and subtracting, and rational equations in product form.

Before you start

This topic builds on earlier ideas. Before you start, it's worth working through the lessons below — they'll make everything click:

PolynomialsA polynomial is a sum of monomials — and everything you do to one is an extension of ordinary arithmetic. Degree and coefficients, adding, subtracting and multiplying, dividing by a binomial with Horner’s scheme, the remainder and factor theorems, integer roots, factoring and biquadratic equations.
FractionsA fraction writes part of a whole as a numerator and a denominator. Learn equivalent fractions, all four operations on fractions, mixed numbers, comparing fractions and reducing to lowest terms.

Where this is used

Real situations where you count exactly the way this lesson teaches:

A round trip there and back
The same route driven out at 60 km/h and back at 40 km/h averages 2 / (1/60 + 1/40) = 48 km/h, not 50 km/h. The slower leg takes longer and therefore weighs more in the average — and that only becomes visible once the two fractions are put over a common denominator.
See the lesson: Speed, distance and time
Decorating with two people
One painter does a room in 6 hours, the other in 4. Together they work at 1/6 + 1/4 = 5/12 of a room per hour, so they finish in 12/5 = 2.4 hours — 2 hours 24 minutes. The answer "5 hours, the average of 6 and 4" is more than two hours too long.
Electronics and resistors in parallel
Two resistors on parallel branches give a combined resistance R = R₁R₂ / (R₁ + R₂) — which is exactly the sum 1/R₁ + 1/R₂ inverted. For 100 Ω and 150 Ω that is 15,000 / 250 = 60 Ω, less than either of them, which looks like a mistake the first time you meet it.
Optics and pulling focus
The lens equation 1/f = 1/x + 1/y solved for the image distance gives y = xf / (x − f). A 50 mm lens focused on a subject 200 mm away throws its image 200 · 50 / 150 ≈ 66.7 mm behind the lens — that is the extension the barrel has to travel.

All formulas

Domain
$D = \{ x \in \mathbb{R} : Q(x) \neq 0 \}$
the domain first, the arithmetic second
Cancelling
$\frac{P(x) \cdot S(x)}{Q(x) \cdot S(x)} = \frac{P(x)}{Q(x)}$
only a common FACTOR cancels, never a term
Multiplying
$\frac{A}{B} \cdot \frac{C}{D} = \frac{AC}{BD}$
numerator by numerator, denominator by denominator
Dividing
$\frac{A}{B} : \frac{C}{D} = \frac{A}{B} \cdot \frac{D}{C}$
multiply by the reciprocal; the condition C ≠ 0 joins the domain
Adding
$\frac{A}{B} + \frac{C}{D} = \frac{AD + CB}{BD}$
the common denominator is the least common multiple, not always the product
Rational equation
$\frac{P(x)}{Q(x)} = 0 \iff P(x) = 0 \ \text{and} \ Q(x) \neq 0$
the numerator vanishes — but only inside the domain

The expression $\dfrac{x + 2}{x^2 - 9}$ is an algebraic fraction: a quotient of two polynomials. Every rule of the arithmetic is the one you already know from ordinary fractions — with one difference that changes the order of the work. The denominator is not a number but an expression, so for some values of the variable it can vanish. Which is why any problem with an algebraic fraction begins with the domain, not with the calculation.

This lesson is about the expression and the equation. The graph of a rational function and its asymptotes are a separate topic, in the branch on functions.

The domain: what has to be excluded

D = \{ x \in \mathbb{R} : Q(x) \neq 0 \}

To find it, factor the denominator and set each factor to zero:

x^2 - 9 = (x - 3)(x + 3) = 0 \iff x = 3 \ \text{or} \ x = -3

D = \mathbb{R} \setminus \{-3, 3\}

Two things are worth fixing straight away. A repeated factor excludes one number, not two: the denominator $(x - 5)^2$ forbids only $x = 5$ . And a quadratic factor with a negative discriminant excludes nothing: $x^2 + 4$ is never zero, so it takes not a single number out of the domain.

Cancelling

\frac{P(x) \cdot S(x)}{Q(x) \cdot S(x)} = \frac{P(x)}{Q(x)}

What cancels is a factor — something multiplied by everything else. A term of a sum does not cancel, and this is the most common mistake in the whole topic:

\frac{x + 3}{x + 5} \neq \frac{3}{5}

Which is why cancelling is always preceded by factoring: while the numerator and the denominator are sums, there is no way to see what the factors are. This is where the special products and the factor theorem from the lesson on polynomials earn their keep.

Cancel the fraction (x² − 4) / (x² + 4x + 4) and state its domain.

Multiplying and dividing

\frac{A}{B} \cdot \frac{C}{D} = \frac{AC}{BD} \qquad \frac{A}{B} : \frac{C}{D} = \frac{A}{B} \cdot \frac{D}{C}

The rule is the numerical one, but it pays to apply it in reverse order: factor everything first, cancel crosswise, and only then multiply. Otherwise you produce a quartic that has to be factored back apart a moment later.

\frac{x^2 - 1}{x + 3} \cdot \frac{x + 3}{x - 1} = \frac{(x - 1)(x + 1)}{x + 3} \cdot \frac{x + 3}{x - 1} = x + 1

Division brings one extra condition: the divisor may not be zero, so the numerator of the second fraction joins the domain as well.

Adding and subtracting

\frac{A}{B} + \frac{C}{D} = \frac{AD + CB}{BD}

In practice you do not multiply the denominators blindly — you take their least common multiple, exactly as with numerical fractions. For $\dfrac{1}{x - 2}$ and $\dfrac{3}{(x - 2)(x + 1)}$ the common denominator is $(x - 2)(x + 1)$ , not the product of both.

Write as a single fraction: 1/(x − 2) + 3/(x + 1).

Rational equations

\frac{P(x)}{Q(x)} = 0 \iff P(x) = 0 \ \text{and} \ Q(x) \neq 0

A fraction is zero exactly when its numerator is zero — the denominator has no say beyond the fact that it must not be zero. Solving a rational equation in product form therefore has three steps: the domain, the zeros of the numerator, and discarding whichever of them fell outside the domain.

The filled dots at −3 and 1 are where the expression is zero. The hollow dot at 4 is a hole in the domain — there the expression has no value at all.

The drawing shows the difference between two kinds of point on the axis. The zeros of the numerator are filled: they lie in the domain and they solve the equation. The zero of the denominator is hollow — and it stays hollow however the equation is rewritten.

Solve (x − 1)(x − 4) / (x − 4) = 0.

An equation with the unknown in a denominator that is not yet written as "a fraction equal to zero" is brought to that form: move everything to one side and put it over a common denominator. Multiplying both sides by the denominator works too — provided the domain was written down first, because the multiplication itself loses it.

Practice

Work through a set of exercises — they get harder as you go. At the end you'll see your score and the mistakes worth reviewing.

Exercise 1 of 8Score: 0

simplify (x² − 6x + 8)/(x² − 9x + 20)

Common mistakes

Cancelling a term instead of a factor — nothing cancels in $\dfrac{x + 3}{x + 5}$ . The numerical check: at $x = 1$ it is $\tfrac{4}{6}$ , not $\tfrac{3}{5}$ .
The domain after cancelling — the excluded numbers are read off the original denominator. The shortened fraction does not show them, but they still apply.
Counting a repeated factor twice — the denominator $(x - 5)^2$ excludes one number, not two.
Forgetting the divisor's domain — when dividing fractions, the numerator of the second one may not be zero either, since inverting sends it into a denominator.
The minus in front of a fraction — subtracting $\dfrac{3x - 2}{x + 1}$ flips the sign of the whole numerator, both terms, not only the first.
Multiplying the equation through without stating the domain — after that multiplication nothing in the arithmetic remembers the excluded number, and it can be handed in as a solution.

Formula card

Topic: Rational expressions and equations

Domain
$D = \{ x \in \mathbb{R} : Q(x) \neq 0 \}$
the domain first, the arithmetic second
Cancelling
$\frac{P(x) \cdot S(x)}{Q(x) \cdot S(x)} = \frac{P(x)}{Q(x)}$
only a common FACTOR cancels, never a term
Multiplying
$\frac{A}{B} \cdot \frac{C}{D} = \frac{AC}{BD}$
numerator by numerator, denominator by denominator
Dividing
$\frac{A}{B} : \frac{C}{D} = \frac{A}{B} \cdot \frac{D}{C}$
multiply by the reciprocal; the condition C ≠ 0 joins the domain
Adding
$\frac{A}{B} + \frac{C}{D} = \frac{AD + CB}{BD}$
the common denominator is the least common multiple, not always the product
Rational equation
$\frac{P(x)}{Q(x)} = 0 \iff P(x) = 0 \ \text{and} \ Q(x) \neq 0$
the numerator vanishes — but only inside the domain

The equation (x − 1)(x + 3) / (x − 4) = 0. The filled dots at −3 and 1 are solutions, the hollow dot at 4 is a number outside the domain.

Frequently asked questions

A balance scale in perfect equilibrium rendered as a geometric form, with a straight line crossing an axis like an orbital path, against a cosmic nebula in violet and magenta

Algebra

What x really means — unknown, variable and parameter

The letter x does not always mean the same thing. Sometimes it is a number you are hunting for, sometimes a quantity that varies continuously, sometimes a blank waiting for any number at all. Here is where algebraic notation came from, why it took two thousand years to arrive, what exactly separates an unknown from a variable and a parameter, and why an inequality flips when you multiply by a negative number.

11 min