XMocks · SAT Math · Rational expressions

Simplify a rational expression without losing its domain

Factoring can make a rational expression shorter, but cancellation does not make an excluded input valid again. Find every zero of the original denominator first, keep those restrictions, and only then simplify.

The restriction comes from the original denominator

A rational expression is a quotient of polynomials. Division by zero is undefined, so every input that makes the denominator zero must be excluded. Factoring can reveal those values. A common factor may cancel from the written formula, but its zero remains outside the domain of the original expression.

That distinction matters when describing a graph too: a cancelled factor creates a missing point, called a hole, while an uncancelled denominator zero generally produces a vertical asymptote.

Worked example: simplify and describe the restrictions

Consider (x² − 9) / (x² − 4x + 3).

  1. Factor the numerator and denominator: [(x − 3)(x + 3)] / [(x − 1)(x − 3)].
  2. Set the original denominator equal to zero. Its factors give x = 1 and x = 3, so the original expression’s domain excludes both values.
  3. Cancel the common factor only while preserving the restrictions: (x + 3)/(x − 1), where x ≠ 1 and x ≠ 3.

Understand the cancelled factor

The remaining denominator is zero at x = 1, so the simplified graph has a vertical asymptote there. At x = 3 the factor was cancelled, but the original expression was undefined. The simplified formula would give 3, so the graph has a hole at (3, 3). As a numerical check, x = 0 is allowed and both forms equal −3.

A reliable three-step process

  1. Factor everything. This exposes numerator and denominator zeros.
  2. Record the excluded inputs from the original denominator. Keep this list beside your work.
  3. Cancel common factors and state the restrictions with the final expression. Use a test input from the allowed domain to check the algebra.

When solving an equation containing rational expressions, restrictions still apply after clearing denominators. Check any proposed solution in the original equation because multiplying both sides by a denominator can introduce a value that was never in the domain.

Practice: identify the domain and the hole

Simplify (x² − 16) / (x² − 7x + 12). State the excluded inputs and the coordinates of any hole.

Show the worked answer

Factor to get [(x − 4)(x + 4)] / [(x − 3)(x − 4)]. The original denominator excludes x = 3 and x = 4. After cancelling x − 4, the expression is (x + 4)/(x − 3) with both exclusions retained. At x = 4 the simplified formula equals 8/1 = 8, so the hole is (4, 8). At x = 3 the remaining denominator is zero, so there is a vertical asymptote.

Practice and next steps

As you work, write the domain before reducing the expression. That one habit prevents a cancelled factor from being mistaken for an allowed input.

This independently prepared XMocks lesson is not affiliated with or endorsed by the College Board.