Rational Expressions and Domain Restrictions

5.6: Rational Expressions and Domain Restrictions. Original editorial SAT preparation with three worked examples, ten flashcards, and thirty practice questions. Educator-review draft.

Lesson 6 of 63 parts Practice Flashcards

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Canceled factors and holes

Objective: Simplify a rational expression while retaining its original restrictions.

Prerequisite skills: factoring and fraction operations. Factor numerator and denominator and cancel only common nonzero factors. The simplified formula agrees wherever the original is defined, but canceled denominator zeros still leave holes. Common error: restoring a canceled denominator zero to the domain. Restrictions and interpretation: Original denominators remain excluded even if a common factor cancels.

Formulas

$(x² − a²)/(x − a) = x + a for x ≠ a$

Worked examples

Try itFor x ≠ 23, simplify (x² − 529)/(x − 23), then find its value at x = 26.Show answer

Factor the numerator: (x − 23)(x + 23). Cancel x − 23, retaining x ≠ 23. The value at x = 26 is 26+23 = 49. Therefore the requested answer is 49.

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Canceled factors and holes

Objective: Simplify a rational expression while retaining its original restrictions.

Formulas

$(x² − a²)/(x − a) = x + a for x ≠ a$

Worked examples

Try itFor x ≠ 23, simplify (x² − 529)/(x − 23), then find its value at x = 26.Show answer

Factor the numerator: (x − 23)(x + 23). Cancel x − 23, retaining x ≠ 23. The value at x = 26 is 26+23 = 49. Therefore the requested answer is 49.

Domain exclusions

Objective: Find all zeros of an original factored denominator.

A rational expression is defined only when its denominator is nonzero. Factor that denominator and solve each factor equal to zero to list excluded inputs; numerator zeros do not by themselves cause exclusions. Common error: excluding numerator zeros while ignoring denominator factors. Restrictions and interpretation: Original denominators remain excluded even if a common factor cancels.

Formulas

$(x − a)(x + b) ≠ 0 ⇒ x ≠ a and x ≠ −b$

Worked examples

Try itWhich lists all excluded real x for (x − 1)/[(x − 24)(x + 26)]?Show answer

The denominator is zero if x − 24 = 0 or x + 26 = 0. Thus exclude x = 24 and x = -26. The numerator zero x = 1 is allowed because neither denominator factor vanishes there. Therefore the requested answer is 24 and -26.

Combining rational expressions

Objective: Use common denominators and preserve restrictions.

Addition and subtraction require a common denominator. Combine the complete numerators before simplifying, then carry forward every original denominator restriction. You cannot cancel terms that are only part of a sum. Common error: canceling a term inside an addition rather than a full factor. Restrictions and interpretation: Original denominators remain excluded even if a common factor cancels. Desmos and method choice: Use Desmos for arithmetic or a graph/table check when that reduces work. Enter parentheses explicitly and verify the requested quantity and units. Use the algebraic steps shown here for an exact result; a rounded graph display is not proof of exact equality.

Formulas

$a/x + b/x = (a + b)/x, x ≠ 0$

Worked examples

Try itFor x = 26, evaluate 25/x + 5/x².Show answer

The common denominator is x². Rewrite 25/x as 25x/x². The sum is (25x+5)/x²; at x = 26, this is 655/676 = 655/676. The input is nonzero. Therefore the requested answer is 655/676.

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