Factor a denominator and keep every original exclusion

Explain it like I am five

Picture taking a tiled rectangle apart so you can see the lengths that made its area. Factoring does a similar job with a formula. It rewrites one expression as multiplied pieces called factors, without changing the result. This helps you find a denominator's forbidden inputs. A multiplied result is zero when a factor is zero, so smaller equations replace one larger equation. Sometimes a matching factor appears above and below a fraction bar. You may cancel it where it is nonzero. Keep the original forbidden inputs written down. Changing the way you write the recipe cannot make its original division by zero work.

A factor pair must explain the middle coefficient and the constant term together.

Reminder

Distribution. Multiply each term in one factor by each term in the other: (x + 5)(x − 4) = $x^{2}$ − 4x + 5x − 20.
Zero product property. AB = 0 requires A = 0 or B = 0. For a denominator, those inputs are exclusions.
Negative denominators. A negative denominator is allowed: $11 over −20$ = − $11 over 20$ . The forbidden denominator is zero.
Keep all restrictions. A root and denominator must both work: x ≥ −11 and x ≠ −5 and x ≠ 4 keeps [−11, −5) ∪ (−5, 4) ∪ (4, ∞).

Why it works. Multiplying each term in x + p by each term in x + q gives + px + qx + pq. The middle terms combine into (p + q)x, so the two factor numbers must have the required sum and product. A product of nonzero real numbers cannot be zero: if AB = 0 and A ≠ 0, division by A gives B = 0. This explains why each zero factor finds a denominator exclusion. Cancellation divides the top and bottom by the same factor. That operation requires a nonzero factor, so it cannot restore an input already excluded by the original denominator.

Rule(x + p)(x + q) = + (p + q)x + pq. To factor + Bx + C, find p + q = B and pq = C.
Exclude every input making the original denominator zero, and retain those exclusions after cancellation.

The same idea, five ways

Say it

Say factor the original denominator, exclude its zeros, and retain those exclusions after cancellation.

Write it

A shorter fraction describes the original function only when you keep the original domain conditions.

In math

(x + p)(x + q) = $x^{2}$ + (p + q)x + pq
For $x^{2}$ + x − 20: p + q = 1 and pq = −20
$x^{2}$ + x − 20 = (x + 5)(x − 4)
Denominator restrictions: x ≠ −5 and x ≠ 4
(−∞, −5) ∪ (−5, 4) ∪ (4, ∞)
Graph words: hollow dots at both original denominator-zero inputs

Taking a rectangle apart reveals its multiplied lengths; shortening a recipe leaves its accepted choices unchanged.

See it

The product and sum checks rebuild the quadratic before you solve its factor equations.

The same idea, other ways

As four tile products

A rectangle cut into two rows and two columns gives four products. The two middle products are px and qx, so their combined size is (p + q)x. The last product is pq. This is why matching only the constant term is not enough.

Both the sum 5 + 7 and the product 5 × 7 are visible in the expansion.

As two locks

A factor pair must open two locks. For $x^{2}$ + 12x + 35, the product lock needs 35 and the sum lock needs 12. The pair 1 and 35 opens only the product lock. The pair 5 and 7 opens both.

As an unchanged guest list

Simplifying a fraction changes the recipe's writing, not the original guest list of permitted inputs. For $x squared − 8 x over x squared − 64$ , input 8 was already forbidden. Cancellation may shorten the recipe for other inputs, but it cannot admit 8.

.1Find a pair by its product and sum

Think of choosing two tile lengths. They must create both the right corner area and the right combined middle strip. A monic quadratic has $x^{2}$ with coefficient 1, meaning one copy of $x^{2}$ . For $x^{2}$ + Bx + C, the factor numbers multiply to C and add to B. Keep the signs attached to the numbers.

Coefficient means the number multiplying a variable or power. In 12x, the coefficient is 12.
For $x^{2}$ + Bx + C, seek two numbers whose product is C and whose sum is B.
To generate whole-number factor pairs, try positive whole-number divisors in order. Keep a divisor and quotient only when the quotient is whole. For 20, 20 ÷ 1 = 20, 20 ÷ 2 = 10, and 20 ÷ 4 = 5 give the pairs; 20 ÷ 3 is not whole. At candidate 5 the quotient is 4, so stop: larger candidates would repeat pairs in reverse. Then choose signs and check sums.
If C is negative, the factor numbers have opposite signs. If B is positive, the positive number has the larger size; if B is negative, the negative number has the larger size.
If C is positive, the factor numbers have the same sign. A positive sum uses two positives, and a negative sum uses two negatives.
This whole-number pair search works for examples with whole-number factors. If no pair works, do not invent a factorization or conclude that the domain is empty. Another method may be needed.
If C = 0, x is a common factor: $x^{2}$ + Bx = x(x + B), because distributing x gives both original terms.

The signed products explain why the constant's sign determines the factor signs.

Reminder

Signed multiplication. 5 × (−4) = −20. One positive and one negative factor give a negative product.

The same idea, five ways

Say it

Say find two numbers that multiply to negative twenty and add to one.

Write it

The factor numbers must match the constant by multiplication and the middle coefficient by addition.

In math

p + q = 1
pq = −20
p = 5, q = −4
$x^{2}$ + x − 20 = (x + 5)(x − 4)

Two tile lengths must fit both the corner area and the combined middle strip.

See it

The four products reconstruct the entire quadratic.

Worked exampleA negative constant needs opposite signs

Factor $x^{2}$ + x − 20. You need multiplied pieces that reproduce the coefficient 1 of x and the constant −20.

Treat these as signed products: 5x and −4x combine to x, and 5 × (−4) is −20.

Every pictured pair has product −20; the highlighted sum 1 selects 5 and −4.

We need p + q = 1 and pq = −20.The expansion has middle coefficient p + q and constant term pq; the unwritten coefficient of x is 1.
Generate the factor-pair sizes of 20 by division: 20 ÷ 1 = 20 gives 1 and 20; 20 ÷ 2 = 10 gives 2 and 10; 20 ÷ 3 is not whole, so skip it; 20 ÷ 4 = 5 gives 4 and 5. Stop at candidate 5 because its quotient is 4, so that pair repeats in reverse. The picture gives each larger number a positive sign and the smaller one a negative sign.A divisor and its whole-number quotient multiply back to 20. After the candidate exceeds its quotient, the smaller member of every possible pair has already been tried.
Read the sums below those pairs: 19, 8, and 1. The column for 5 and −4 is the only one with sum 1.A negative product needs one positive and one negative factor number, so their sum is a difference of sizes.
Choose +5 and −4. Their product is −20 and their sum is 1.The positive number must have the larger size to make the required positive sum.
Write $x^{2}$ + x − 20 = (x + 5)(x − 4).Both the sum and product match, so these factors rebuild the original quadratic.

Answer

+ x − 20 = (x + 5)(x − 4)

Check Multiply back: (x + 5)(x − 4) = − 4x + 5x − 20 = + x − 20. At x = 0, both forms equal −20.

Counterexamples: what it is not, and the tempting wrong moves

✗ Not this: Use (x − 2)(x + 10) to factor + x − 20.

The product −20 matches, but the factor numbers add to 8, so the middle term would be 8x.

✓ Instead: Use (x + 5)(x − 4), since 5 + (−4) = 1 and 5 × (−4) = −20.

Tips and tricks

Write sum and product on separate lines, and check both before accepting a pair.

.2Zero factors locate forbidden inputs

Imagine a product counting boxes times items per box. The total is zero if either count is zero. The zero product property says the same thing for real numbers: a product is zero exactly when at least one factor is zero. For a denominator, these zero inputs are the choices to remove, rather than the choices to keep.

The zero product property says AB = 0 exactly when A = 0 or B = 0.
Set the denominator equal to zero to locate the inputs where division fails.
Solve each factor's zero equation, then substitute its result into the original denominator.
A product equaling a nonzero number does not let you set its factors equal to zero.

(x + 5)(x − 4) = 0

x + 5 = 0 or x − 4 = 0

x = −5 or x = 4

These inputs are excluded

Zero equations find forbidden inputs for a fraction's domain.

Reminder

Solving a linear equation. x + 5 = 0 gives x = −5 by subtracting 5 from both sides; −5 + 5 = 0 checks the result.

The same idea, five ways

Say it

Say keep every real input except negative five and four.

Write it

The original denominator vanishes at −5 and 4, so both inputs are excluded.

In math

x ≠ −5 and x ≠ 4
x < −5 or −5 < x < 4 or x > 4
{x | x ≠ −5 and x ≠ 4}
(−∞, −5) ∪ (−5, 4) ∪ (4, ∞)
Graph words: hollow dots at −5 and 4, with every other position shaded

Keep the whole guest list except the two names that fail the entry check.

See it

The three intervals describe exactly the same exclusions as the two not-equal conditions.

Worked exampleA quadratic denominator excludes two inputs

Find the domain of f(x) = $7 x + 11 over x squared + x − 20$ . You need every input for which the original fraction is defined.

Both roots of the original denominator are omitted, leaving three allowed intervals.

Set $x^{2}$ + x − 20 = 0 to locate the inputs that would make division fail.The domain excludes zero denominators; this equation finds the values to throw out.
Factor the denominator as (x + 5)(x − 4).The factor numbers 5 and −4 have sum 1 and product −20.
Set x + 5 = 0 or x − 4 = 0. Subtract 5 in the first equation to get x = −5; add 4 in the second to get x = 4.A zero product requires a zero factor. Each smaller equation finds one forbidden input.
Check the original denominator: at −5 it is 25 − 5 − 20 = 0; at 4 it is 16 + 4 − 20 = 0.Substitution confirms that each solved value really makes the original fraction undefined.
Exclude −5 and 4, and keep all other real inputs.The numerator is a polynomial, and the factored denominator has no other zeros.
Write (−∞, −5) ∪ (−5, 4) ∪ (4, ∞).Three open intervals keep every real input except the two forbidden values.

Answer

Domain: (−∞, −5) ∪ (−5, 4) ∪ (4, ∞)

Check At x = 0, the denominator is −20 and the output is −, which is defined. At x = 5, the denominator is 25 + 5 − 20 = 10 and the output is = . Valid inputs on different remaining intervals confirm that negative denominators and inputs beyond 4 can work.

Counterexamples: what it is not, and the tempting wrong moves

✗ Not this: The domain is {−5, 4}, because those are the solutions of the zero equation.

That equation found the inputs making the denominator zero, which are precisely the forbidden inputs.

✓ Instead: Remove −5 and 4 from the real number line; the domain is (−∞, −5) ∪ (−5, 4) ∪ (4, ∞).

Tips and tricks

Write forbidden inputs beside the answers to a denominator's zero equation.

.3Cancel a common factor without changing the domain

Think of shortening a written recipe after checking which ingredients it can accept. You may remove a matching multiplication above and below a fraction bar, but only where that common factor is nonzero. This is cancellation. It divides the numerator and denominator by the same nonzero factor. A choice that failed the original recipe still fails it, even if the shorter recipe could calculate a number there.

Cancel factors that multiply the entire numerator and denominator. A term inside a sum is not a factor of that entire sum.
Cancellation requires the common factor to be nonzero.
The simplified expression agrees with the original expression only on the original domain.
Keep every original denominator exclusion, including an exclusion hidden by a canceled factor.

= only where x ≠ −8 and x ≠ 8

The original exclusions stay

Write the domain conditions beside the shorter formula so cancellation cannot hide them.

Reminder

Difference of squares. $x^{2}$ − 64 = (x − 8)(x + 8), because the middle terms cancel when multiplied back.
Division by zero. $0 over 0$ is undefined too. Every number multiplied by 0 gives 0, so the multiplication question cannot select one unique quotient.

The same idea, five ways

Say it

Say the shorter fraction works on the original domain, with both exclusions retained.

Write it

Canceling x − 8 does not allow input 8 into the original function.

In math

$x squared − 8 x over x squared − 64$ = $x over x + 8$ for x ≠ −8 and x ≠ 8
{x | x ≠ −8 and x ≠ 8}
(−∞, −8) ∪ (−8, 8) ∪ (8, ∞)
Graph words: omit both original denominator-zero inputs

Shortening the recipe's writing does not change which choices the original recipe accepted.

See it

Both original exclusions remain part of the simplified description.

Worked exampleThe shorter fraction keeps two original exclusions

Find the domain of f(x) = $x squared − 8 x over x squared − 64$ , then simplify the formula where it is defined. You must identify permitted inputs from the original fraction before canceling anything.

The original function excludes both −8 and 8, even after its common factor is canceled.

Factor the numerator: $x^{2}$ − 8x = x(x − 8).Both numerator terms contain a factor x; multiplying x into x − 8 gives $x^{2}$ − 8x back.
Factor the original denominator: $x^{2}$ − 64 = (x − 8)(x + 8).64 = $8 squared$ , so the difference-of-squares pattern applies.
Set the denominator factors equal to zero to find forbidden inputs. x − 8 = 0 gives x = 8, and x + 8 = 0 gives x = −8.Either zero factor makes the original denominator zero. These equations locate inputs to exclude.
Check the original denominator: $8 squared$ − 64 = 0 and (−8 $) squared$ − 64 = 0. Write x ≠ −8 and x ≠ 8 before canceling.Substitution confirms both original exclusions, including the one a shorter formula might hide.
For those permitted inputs, cancel the common factor x − 8: $x(x − 8) over (x − 8)(x + 8)$ = $x over x + 8$ .x ≠ 8 makes the common factor nonzero, so dividing the top and bottom by it is legal.
Write the shorter formula $x over x + 8$ with x ≠ −8 and x ≠ 8, and retain (−∞, −8) ∪ (−8, 8) ∪ (8, ∞) as the original domain.Cancellation changes the expression's appearance while preserving its value only where the original fraction was defined.

Answer

Original domain: (−∞, −8) ∪ (−8, 8) ∪ (8, ∞)
Simplified formula: $x over x + 8$ , with x ≠ −8 and x ≠ 8

Check At x = 16, the original fraction gives = = , and the shorter fraction gives = . At x = 8, the original gives , which has no unique quotient; the shorter formula's value does not restore that excluded input.

Counterexamples: what it is not, and the tempting wrong moves

✗ Not this: After cancellation, the only exclusion is x = −8.

The shorter denominator hides the original zero at x = 8. Canceling that factor required x ≠ 8 in the first place.

✓ Instead: Keep x ≠ −8 and x ≠ 8.

✗ Not this: Cancel x from and leave .

The x terms are added, not factors multiplying the entire top and bottom. At x = 0 the original equals , but at x = 1 it equals = , a different value.

✓ Instead: Factor first and cancel only a common factor multiplying the entire numerator and denominator.

Tips and tricks

Copy every original exclusion above the line before simplifying.
Ask whether the proposed canceled object multiplies the entire numerator and the entire denominator.

Strategy: step by step

1. Copy the original denominator before changing the fraction. Its zero inputs are the values you must exclude.
2. For $x^{2}$ + Bx + C, identify B and C, including their signs. B is the coefficient of x, meaning the number multiplying x; C is the constant term. The constant term is the fixed number added or subtracted without being multiplied by x; here it is C.
3. Generate the whole-number factor-pair sizes of the size of C. Try positive whole-number candidates in order and divide the size of C by each one. Keep a pair only when the quotient is a whole number. Stop when the candidate is larger than its quotient, because further pairs would repeat earlier pairs in reverse. Choose signs so the product is C, then find the pair whose sum is B. If C = 0, factor out x directly instead of making a nonzero-constant pair list.
4. Write (x + p)(x + q) and multiply it out to check both middle and constant terms.
5. Set each denominator factor equal to zero to find forbidden inputs. Solve each smaller equation, then substitute each result into the original denominator to confirm that it produces zero.
6. If a common factor cancels, write the simplified expression together with every original exclusion. Cancel only at inputs where the common factor is nonzero.
7. Keep any other root or context restrictions as well, then write the remaining input intervals.

Strategy

Find all exclusions before simplifying

Is the denominator a quadratic of the form + Bx + C?

YesIdentify signed B and C. If C is nonzero, generate whole-number size pairs by dividing by positive whole-number candidates in order, stopping when the candidate exceeds its quotient; then choose signs and test the sum and product.

NoUse its applicable earlier method, such as a linear zero equation or the difference-of-squares identity.

↓

Is C negative?

YesUse opposite signs; the larger size determines the sign of their sum.

NoFor positive C use matching signs; for C = 0 factor out x directly.

↓

Does your proposed pair match both the sum and product?

YesWrite the factors, multiply back, then solve the zero factor equations to locate exclusions.

NoTry another pair. If no whole-number pair works, stop this pair-search method and use another established method.

↓

Does a common factor cancel?

YesCancel only on the original domain, and retain that factor's zero input as an exclusion.

NoKeep the original denominator exclusions.

↓

Does a root or context impose another restriction?

YesKeep inputs passing that restriction and every original denominator check.

NoExclude the original denominator zeros from all real inputs.

Copy the original denominator and identify any other arithmetic or context restrictions.
For $x^{2}$ + Bx + C, try positive whole-number divisors of the size of C in order, keeping only whole quotients. Stop when the candidate exceeds its quotient; then choose signs and test product C and sum B. If C = 0, factor out x instead. Check by multiplying back.
Set each denominator factor equal to zero to locate forbidden inputs. Solve and plug each value into the original denominator.
Record every exclusion before canceling any common factor.
Cancel only a nonzero common factor multiplying the entire top and bottom.
Keep every original exclusion and any root or context condition when writing the domain.

Worked exampleFind two numbers that rebuild a quadratic

Factor $x^{2}$ + 12x + 35. This asks you to rewrite the expression as multiplied pieces with the same value.

The two middle products add to 12x, while the corner product is 35.

The column for 5 and 7 is the pair whose sum is 12; both pairs have product 35.

The coefficient of x is 12 and the constant term is 35. We need p + q = 12 and pq = 35.Expanding (x + p)(x + q) gives middle coefficient p + q and constant term pq.
Generate the positive factor-pair sizes of 35 by division. 35 ÷ 1 = 35 gives 1 and 35. Dividing by 2, 3, or 4 does not give a whole-number quotient, so skip those candidates. 35 ÷ 5 = 7 gives 5 and 7. Stop before trying larger candidates: 6 is already greater than its quotient 35 ÷ 6, because 6 × 6 = 36 > 35.A whole-number divisor and its quotient multiply back to 35. Once the candidate is larger than its quotient, any further whole-number pair reverses a pair with a smaller member that was already tried.
Read the factor-pair picture. Under 1 and 35, the sum is 36. Under 5 and 7, the sum is 12. Both pairs have product 35.A positive product with a positive sum needs two positive numbers here. Checking the pairs finds the required sum.
Choose p = 5 and q = 7, then write (x + 5)(x + 7).5 + 7 = 12 and 5 × 7 = 35, so both coefficients are reproduced.
Multiply back: (x + 5)(x + 7) = $x^{2}$ + 7x + 5x + 35 = $x^{2}$ + 12x + 35.Each term in the first factor multiplies each term in the second, and the two middle terms combine.

Answer

+ 12x + 35 = (x + 5)(x + 7)

Check At x = 2, the original expression is 4 + 24 + 35 = 63. The product is (2 + 5)(2 + 7) = 7 × 9 = 63. Multiplication back checks the identity for every x; substitution checks this numerical case.

Ladder: from easy to exam-hard. Press Try it first on any rung to hide its steps and use them as hints.

Rung 1Rung 1: a positive product and a positive sum

Factor $x^{2}$ + 12x + 35. You are rewriting the expression as multiplied factors with exactly the same value.

The four products combine into the original expression.

The column for 5 and 7 is the pair whose sum is 12; both pairs have product 35.

Seek p + q = 12 and pq = 35.Expanding two factors gives $x^{2}$ + (p + q)x + pq.
Generate the positive factor-pair sizes of 35 by division. 35 ÷ 1 = 35 gives 1 and 35. Dividing by 2, 3, or 4 does not give a whole-number quotient, so skip those candidates. 35 ÷ 5 = 7 gives 5 and 7. Stop before trying larger candidates: 6 is already greater than its quotient 35 ÷ 6, because 6 × 6 = 36 > 35.A whole-number divisor and its quotient multiply back to 35. Once the candidate is larger than its quotient, any further whole-number pair reverses a pair with a smaller member that was already tried.
Read the factor-pair picture. Under 1 and 35, the sum is 36. Under 5 and 7, the sum is 12. Both pairs have product 35.A positive product and positive sum use two positive factor numbers here.
Write (x + 5)(x + 7).5 + 7 = 12 and 5 × 7 = 35 satisfy both requirements.
Multiply back to obtain $x^{2}$ + 7x + 5x + 35 = $x^{2}$ + 12x + 35.Distribution checks the identity for every input.

Answer

+ 12x + 35 = (x + 5)(x + 7)

Check At x = 0 both forms give 35, and at x = 2 both give 63. The complete expansion, rather than these two samples alone, proves the factorization.

Rung 2Rung 2: a negative product and a small positive sum

Factor $x^{2}$ + x − 20. You need factors whose multiplication leaves middle coefficient 1 and constant −20.

Use the signs in both the product check and the sum check.

Every pictured pair has product −20; the highlighted sum 1 selects 5 and −4.

Seek p + q = 1 and pq = −20.The coefficient of x is an unwritten 1, and the constant term includes its negative sign.
Generate the factor-pair sizes of 20 by division: 20 ÷ 1 = 20 gives 1 and 20; 20 ÷ 2 = 10 gives 2 and 10; 20 ÷ 3 is not whole, so skip it; 20 ÷ 4 = 5 gives 4 and 5. Stop at candidate 5 because its quotient is 4, so that pair repeats in reverse. The picture gives each larger number a positive sign and the smaller one a negative sign.A divisor and its whole-number quotient multiply back to 20. After the candidate exceeds its quotient, the smaller member of every possible pair has already been tried.
Choose +5 and −4, and write (x + 5)(x − 4).Their product is −20 and their sum is 1.
Multiply back: $x^{2}$ − 4x + 5x − 20 = $x^{2}$ + x − 20.The opposite middle products leave exactly one copy of x.

Answer

+ x − 20 = (x + 5)(x − 4)

Check At x = 1, the original is 1 + 1 − 20 = −18; the product is (1 + 5)(1 − 4) = 6 × (−3) = −18. The expansion checks all inputs.

Rung 3Rung 3: factor to find a fraction's domain

Find the domain of f(x) = $7 x + 11 over x squared + x − 20$ . You are finding every input that keeps the original fraction defined.

A quadratic denominator can require three intervals because it has two zero inputs.

Set $x^{2}$ + x − 20 = 0 to find denominator-zero inputs, then factor it as (x + 5)(x − 4).These inputs would make division fail. The factor numbers add to 1 and multiply to −20.
Set x + 5 = 0 or x − 4 = 0. Solve to get x = −5 or x = 4.The zero product property locates every zero through the smaller factor equations.
Substitute −5 and 4 into the original denominator: 25 − 5 − 20 = 0 and 16 + 4 − 20 = 0.Each substitution confirms an input that must be excluded.
Remove both values and write (−∞, −5) ∪ (−5, 4) ∪ (4, ∞).Every remaining denominator is nonzero, and the polynomial numerator imposes no additional restriction.

Answer

Domain: (−∞, −5) ∪ (−5, 4) ∪ (4, ∞)

Check At x = 0, the denominator is −20 and the output is −. At x = 5 the output is . Both are defined, while the two excluded inputs give zero denominators.

Rung 4Rung 4: cancellation keeps the original exclusions

Find the domain of f(x) = $x squared − 8 x over x squared − 64$ , then simplify it on that domain. You must preserve inputs excluded by the original denominator.

Both zero inputs remain omitted even though one factor no longer appears in the shorter formula.

Rewrite the fraction as $x(x − 8) over (x − 8)(x + 8)$ .Factoring out x rebuilds the numerator, and the difference-of-squares identity rebuilds the denominator.
Set x − 8 = 0 or x + 8 = 0 to find forbidden inputs. Solving gives x = 8 or x = −8.A zero factor makes the original denominator zero, so these equations identify exclusions.
Check $8 squared$ − 64 = 0 and (−8 $) squared$ − 64 = 0, then record x ≠ −8 and x ≠ 8.The exclusions must be confirmed in the original denominator before any factor disappears.
Cancel x − 8 at the remaining inputs to obtain $x over x + 8$ .The recorded condition x ≠ 8 permits division by this common factor.
Keep (−∞, −8) ∪ (−8, 8) ∪ (8, ∞) as the original domain, and write the shorter formula with both exclusions.A canceled factor does not assign a value where the original fraction had a zero denominator.

Answer

Domain: (−∞, −8) ∪ (−8, 8) ∪ (8, ∞)
Simplified formula: $x over x + 8$ , for x ≠ −8 and x ≠ 8

Check At input 16, the original gives = and the shorter expression gives = . At input 8 the original gives , so the shorter formula cannot restore that input.

Rung 5Rung 5: factor and keep a root restriction too

Find the domain of f(x) = $square root of x + 11 over x squared + x − 20$ . You need inputs that make the root real and the original denominator nonzero at the same time.

The root determines the start, and the factored denominator removes two inputs from that allowed ray.

Require x + 11 ≥ 0 and subtract 11 to obtain x ≥ −11.The root's radicand must be nonnegative; this inequality identifies the inputs permitted by the numerator.
Set $x^{2}$ + x − 20 = 0 to locate forbidden denominator inputs. Factor to obtain (x + 5)(x − 4) = 0.Division fails at those zero inputs, and the factor numbers 5 and −4 add to 1 and multiply to −20.
Solve x + 5 = 0 or x − 4 = 0, giving x = −5 or x = 4. Check 25 − 5 − 20 = 0 and 16 + 4 − 20 = 0 in the original denominator.A zero product has a zero factor; substitution confirms both exclusions.
Keep x ≥ −11 and exclude −5 and 4 from that ray.The root and division must work together, and both forbidden denominator inputs lie inside the root's allowed ray.
At x = −11, the numerator is $square root of 0$ = 0 and the denominator is 121 − 11 − 20 = 90, so keep this endpoint.A zero numerator over a nonzero denominator is a valid zero output.
Write [−11, −5) ∪ (−5, 4) ∪ (4, ∞).The starting root endpoint is included, while both original denominator zeros are omitted.

Answer

Domain: [−11, −5) ∪ (−5, 4) ∪ (4, ∞)

Check At x = −12 the radicand is −1, so it fails the root check. At x = −6 the radicand is 5 and the denominator is 36 − 6 − 20 = 10, so the output is defined. At x = 5 the radicand is 16 and the denominator is 10, giving = .

Counterexamples: what it is not, and the tempting wrong moves

✗ Not this: Choose any pair multiplying to the constant term.

The pair's sum also determines the middle coefficient. A right product alone can create a different quadratic.

✓ Instead: Match product and sum, then multiply the factors back.

✗ Not this: Keep the solutions of a denominator-zero equation as the domain.

The zero equation deliberately locates where division fails.

✓ Instead: Label those solutions forbidden inputs and exclude them.

✗ Not this: Cancellation restores an input that made the original denominator zero.

Cancellation divides by the common factor and requires that factor to be nonzero.

✓ Instead: Keep every original exclusion beside the shorter formula.

Tips and tricks

For factoring, write product = C and sum = B before trying pairs.
For a domain, mark denominator zeros as forbidden before writing any intervals.
Copy original restrictions before cancellation, then compare the final domain with that list.

Trap. Canceling a factor and then forgetting its zero input. The shorter expression equals the original only where the original denominator was nonzero. Record every exclusion before canceling.