Excluded Values Practice Test

Find every input that makes an original denominator zero or a rational divisor invalid.

Excluded Values Practice Test

20 domain-restriction questions with worked explanations.

Instant feedback · Worked explanations

Excluded-values domain border control

Inspect every original denominator. Block every forbidden input.

Excluded values are inputs for which the original rational expression is undefined. They come from denominator zeros, not numerator zeros, and they remain excluded even when simplification later removes the factor that caused the restriction. When a rational expression is used as a divisor, one more rule applies: the divisor itself must also be nonzero.

Anchor 01Start with the original unsimplified denominators.

Anchor 02Factor polynomial denominators completely.

Anchor 03Set each denominator factor equal to zero.

Anchor 04If a rational expression is a divisor, also exclude its zeros.

1. Domain restrictions are collected before simplification

The safest workflow treats excluded values as a permanent record attached to the original expression.

InspectFind every denominator in the original expression.

FactorRewrite polynomial denominators as products whenever possible.

SolveSet each denominator factor equal to zero.

CombineMerge all forbidden inputs into one restriction set.

Denominator rule

Only denominator zeros matter

Numerator zeros may make the rational expression equal zero, but they do not make it undefined.

Original-form rule

Restrictions are recorded before cancellation

A canceled factor can hide a restriction but cannot erase it.

Multiple-source rule

Check every denominator

A sum or difference of several rational expressions may contribute several distinct excluded values.

2. Linear factors: solve the denominator equation carefully

For a linear denominator factor, set it equal to zero and solve. The sign in the factor is opposite the sign of the resulting excluded value.

Linear restriction checkpoint

Do not copy the visible sign from the denominator. Solve the denominator equation.

Minus form

x - 5 = 0 \Rightarrow x = 5

Plus form

x + 4 = 0 \Rightarrow x = - 4

Final languageThe solved values are excluded from the domain.

Denominator with subtraction

x - 5

excludes

5

Denominator with addition

x + 4

excludes

- 4

Quick verificationSubstitute the candidate into the denominator factor; it should produce

0

Numerator zeros are differentA zero numerator may produce an output of

0

, but it does not create an excluded value by itself.

3. Quadratic denominators must be factored or solved for real zeros

A quadratic denominator can contribute two distinct excluded values, one repeated excluded value, or no real excluded value at all.

Difference of squares

x^{2} - 9 = (x - 3) (x + 3)

The denominator vanishes at both factor zeros.

Restriction set

x \neq 3, x \neq - 3

Both real zeros must be excluded.

Factorable trinomial

x^{2} + x - 6 = (x + 3) (x - 2)

Each linear factor contributes one denominator zero.

4. Repeated denominator factors create one excluded value, not several copies

Multiplicity matters for algebraic structure, but the domain question asks which input values are forbidden. Repeating the same factor does not create a new value.

Repeated denominator factor

{(x - 2)}^{3}

The denominator is zero at $x = 2$ , regardless of whether the factor appears once or several times.

Domain record

List the excluded input once: $x \neq 2$ . The exponent describes multiplicity, not multiple different excluded values.

5. Several rational terms require a complete denominator sweep

A sum or difference can contain several unrelated denominators. Missing even one denominator produces an incomplete domain restriction set.

\frac{1}{x - 1} + \frac{2}{x + 3} - \frac{4}{(x - 5) (x + 2)}

Scan each fraction separatelyEvery original denominator contributes possible forbidden inputs.

Factor composite denominatorsThe product denominator in the third term contributes two separate zeros.

Merge the set

x \neq 1, x \neq - 3, x \neq 5, x \neq - 2

6. A canceled denominator factor still leaves its original exclusion

Simplification changes the visible formula, not the original domain.

Original expression

Restriction source is visible

\frac{(x - 4) (x + 1)}{(x - 4) (x + 7)}

The original denominator excludes both factor zeros.

Simplified expression

One factor disappears

\frac{x + 1}{x + 7}

The final restriction set is still $x \neq 4, x \neq - 7$ .

7. Not every quadratic denominator has real zeros

Do not assume that every quadratic denominator creates excluded real values. Check whether the denominator can actually equal zero over the real numbers.

Illustrative denominator

x^{2} + 4

This quadratic is always positive for real inputs.

Real-zero check

x^{2} + 4 > 0

Because the expression never reaches zero over the reals, it contributes no real excluded values.

Do not invent restrictions

Factorization or a discriminant check should support the conclusion; do not list complex zeros as real-domain exclusions unless the problem explicitly uses a complex domain.

8. A rational divisor creates two restriction checks

When a rational expression is used as a divisor, it must be defined and it must not equal zero. That means checking both its denominator and its numerator.

Divisor itself

\frac{x - 6}{x + 2}

The divisor is undefined when its denominator is zero and unusable as a divisor when its numerator is zero.

Combined divisor restrictions

x \neq - 2, x \neq 6

The first restriction keeps the rational expression defined; the second prevents division by a zero-valued rational expression.

9. Fully factored denominators make the restriction set easier to read

Factoring converts one polynomial denominator into a list of simpler zero-factor equations.

Step 1

Factor completely

x^{2} + x - 6 = (x + 3) (x - 2)

Step 2

Solve each factor

$x + 3 = 0$ and $x - 2 = 0$ are separate zero equations.

Step 3

Write the restriction set

$x \neq - 3, x \neq 2$

10. Worked mini-set: identify the source of each exclusion

These examples are illustrative teaching examples, not questions copied from the test.

Example A

Linear denominator

$x - 5 = 0 \Rightarrow x = 5$

Example B

Plus-sign factor

$x + 4 = 0 \Rightarrow x = - 4$

Example C

Quadratic denominator

$x^{2} - 9 = (x - 3) (x + 3)$

Example D

Repeated factor

${(x - 2)}^{3}$ still contributes only one distinct forbidden input.

Example E

Canceled restriction

$\frac{x + 1}{x + 7}$ still carries the exclusions from the original expression.

Example F

Rational divisor

$x \neq - 2, x \neq 6$

11. Error analysis: domain mistakes come from checking the wrong parts

The most common wrong answers use numerator zeros, miss one denominator, or forget that a rational divisor can be invalid even when it is defined.

Numerator zeros used as exclusions

A zero numerator normally makes the rational expression equal zero; it does not make the expression undefined.

Linear-factor sign reversed

Solve the denominator factor equal to zero instead of copying the visible sign.

One denominator in a sum forgotten

Every original rational term must be scanned for denominator zeros.

Canceled restriction dropped

Simplification never restores a value excluded by the original denominator.

Every quadratic assumed to have real zeros

Some quadratic denominators remain nonzero for all real inputs.

Rational divisor checked only for undefined points

A divisor must also be nonzero, so zeros of its numerator create additional restrictions.

Final excluded-values checklist

Before accepting a domain restriction set, make sure every original denominator and every rational divisor has been screened.

Did I inspect every original denominator?Work from the unsimplified expression.

Did I factor polynomial denominators completely?Linear factors make zero equations easier to see.

Did I solve every denominator factor equal to zero?Use the solved values as exclusions.

Did I combine restrictions from all fractions?Do not stop after finding the first denominator zero.

Did I preserve restrictions after cancellation?A removed factor can still represent a hole in the original domain.

If dividing by a rational expression, did I also exclude its zeros?The divisor must be defined and nonzero.

This block supports the Excluded Values Practice Test. Its examples are illustrative rather than copies of the test questions. The central habit is complete domain screening: inspect every original denominator, factor when needed, solve every zero-factor condition, preserve canceled exclusions, recognize denominators with no real zeros, and add numerator-zero restrictions when a rational expression is used as a divisor.