Rational expressions — simplification, operations: Question bank

Related toolkits you may want open while drilling: Polynomial factoring, Domain and range, Complex fractions, Rational equations, Polynomial long division, Partial fraction decomposition, Limits and continuity.

True or false — justify

The claim is either true or false — say which, then give the one reason that settles it.

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

after "cancelling the

x

's".

False. You may only cancel factors (things multiplied), and here

x

is a term being added; test

x = 1

gives

\frac{4}{6} = \frac{2}{3} \neq = \frac{3}{5}

\frac{x - 3}{( x - 3 ) ( x + 2 )}

simplifies to

\frac{1}{x + 2}

with the only restriction

x \neq = - 2

False. The original denominator was zero at

x = 3

too, so the simplified form must still carry

x \neq = 3

— simplifying never enlarges the domain.

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

False. The common denominator is the product

(x - 2) (x + 3)

, not the sum; adding denominators is not a rule for fractions.

Multiplying a fraction by

\frac{x - 2}{x - 2}

changes its value.

False.

\frac{x - 2}{x - 2} = 1

for every

x \neq = 2

, so this only rewrites the expression — it never changes what it equals.

\frac{x ^{2} - 4}{x ^{2} + 5 x + 6}

and

\frac{x - 2}{x + 3}

are the same function.

False. They agree everywhere except

x = - 2

, where the first has a hole and the second is defined — same formula, different domain.

If two rational expressions simplify to the same reduced form, they have the same domain.

False.

\frac{x}{x}

and

\frac{1}{1}

both reduce to

1

, but the first excludes

x = 0

; reduced forms can lose restrictions.

Dividing by

\frac{c}{d}

is the same as multiplying by

\frac{d}{c}

True. Multiplying top and bottom of the complex fraction by

\frac{d}{c}

turns the divisor into

1

, leaving

\frac{a}{b} \cdot \frac{d}{c}

— see Complex fractions.

A rational expression can equal zero even though it is a fraction.

True. A fraction is zero exactly when its numerator is zero (and denominator is nonzero), e.g.

\frac{x - 4}{x + 1} = 0

x = 4

\frac{- x + 4}{( x - 2 ) ( x + 2 )}

and

\frac{4 - x}{( x - 2 ) ( x + 2 )}

are equal expressions.

True.

- x + 4

and

4 - x

are the identical polynomial written in a different order, so nothing about the fraction changed.

The LCD of

\frac{1}{x ^{2} - 4}

and

\frac{1}{x + 2}

(x^{2} - 4) (x + 2)

False. Since

x^{2} - 4 = (x - 2) (x + 2)

already contains

x + 2

, the LCD is just

(x - 2) (x + 2)

— never multiply in a factor you already have.

Spot the error

Each line contains one flawed move. Name the exact wrong step.

\frac{2 x ^{2}}{2 x} = \frac{x ^{2}}{x} = x

— but the student writes "domain: all reals."

The cancelled factor

x

(and the

2

) required

x \neq = 0

; the true domain excludes

0

even though the final

x

hides no visible denominator.

\frac{x ^{2} + 9}{x + 3} = x + 3

"because

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

(x + 3)^{2} = x^{2} + 6 x + 9

, not

x^{2} + 9

; a sum of squares

x^{2} + 9

does not factor over the reals, so nothing cancels.

\frac{5 x + 1}{x - 5}

from adding

\frac{3 x}{x - 5} + \frac{2 x + 1}{x - 5}

"so restriction is none."

With a common denominator you keep

x - 5

, so

x \neq = 5

still applies — the denominator did not disappear.

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

, then "cancel the

x

's to get

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

You cannot cancel

x

from the sum

x^{2} - x - 2

; those are terms, and the numerator is also a sum — no common factor exists to cancel.

\frac{x ^{2} + 2 x}{x + 3} \div \frac{x}{x ^{2} - 9} = \frac{x ^{2} + 2 x}{x + 3} \cdot \frac{x}{x ^{2} - 9}

Division flips the second fraction; it should become

\cdot \frac{x ^{2} - 9}{x}

, the reciprocal of the divisor.

Simplifying

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

, the student cancels

x + 1

with

x + 3

x + 1

and

x + 3

are different factors; only identical factors cancel, so this is invalid.

\frac{x}{x ^{2} - 4} - \frac{2}{x + 2} = \frac{x - 2}{x ^{2} - 4}

"just subtract the numerators."

The second fraction wasn't rewritten over the LCD; you must multiply it by

\frac{x - 2}{x - 2}

first, giving

\frac{x - 2 ( x - 2 )}{( x - 2 ) ( x + 2 )}

\frac{x + 3}{x ^{2} - 9}

is "already simplest since top and bottom look different."

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

shares the factor

x + 3

, so it reduces to

\frac{1}{x - 3}

with

x \neq = - 3

retained.

Why questions

Explain the reason, not just the rule.

Why must we factor before cancelling, never after guessing?

Cancelling is only legal on shared multiplied factors; factoring exposes those factors so we don't accidentally strike out terms inside a sum.

Why do we find restrictions from the original denominator, not the simplified one?

The excluded values describe where the starting expression fails; simplifying is just rewriting, so it cannot rescue points that were undefined to begin with.

Why does dividing by a fraction equal multiplying by its reciprocal?

Multiply numerator and denominator of the stacked fraction by the divisor's reciprocal — the divisor becomes

1

, leaving the reciprocal multiplied on top.

Why does multiplying a fraction by

\frac{x - 2}{x - 2}

let us add unlike denominators?

It equals

1

, so value is untouched, but it re-expresses each fraction over a shared denominator — and only equal denominators can have numerators combined.

Why is the LCD built from the highest power of each factor, not the product of denominators?

Using the product wastes shared factors and bloats the numerator; the least common denominator includes each distinct factor just enough times to cover every denominator (see Partial fraction decomposition).

Why can

\frac{x ^{2} - 4}{x ^{2} + 5 x + 6}

leave a "hole" rather than a vertical asymptote at

x = - 2

The factor

(x + 2)

cancels from both top and bottom, so the discontinuity is removable — the graph is missing a single point, not exploding to infinity (Limits and continuity).

Why do restrictions matter even when the simplified expression looks perfectly defined there?

Two functions are equal only if their domains match; keeping restrictions preserves the true domain and hence the genuine identity of the expression (Domain and range).

Edge cases

The scenarios the rules quietly assume away.

What is

\frac{0}{x - 3}

and where is it undefined?

It equals

0

for every

x \neq = 3

, and is undefined at

x = 3

— a zero numerator is fine, but a zero denominator never is.

What happens to

\frac{x - 3}{x - 3}

x = 3

It equals

1

everywhere else but is undefined at

x = 3

, since there it becomes

\frac{0}{0}

— an indeterminate, not

1

Can a rational expression have an empty set of restrictions?

Yes, if the denominator is never zero for real

x

, e.g.

\frac{x}{x ^{2} + 1}

— no real number makes

x^{2} + 1 = 0

, so the domain is all reals.

If both numerator and denominator are zero at

x = a

, is the expression automatically

0

1

Neither — it is

\frac{0}{0}

, undefined; you must factor to see whether the point is a removable hole and what value the function approaches.

What restriction survives when you simplify

\frac{x ( x + 3 )}{x}

It reduces to

x + 3

, but the cancelled

x

forces

x \neq = 0

, so the graph is the line

y = x + 3

with a hole at

x = 0

\frac{5}{7}

(constant over constant) a rational expression?

Yes — constants are degree-zero polynomials, so any numeric fraction is the simplest rational expression, with no restrictions.

For

\frac{1}{x - a}

, what does the value do as

x

creeps toward

a

from each side?

It grows without bound to

+ \infty

from one side and

- \infty

from the other, revealing a vertical asymptote — the denominator's approach to zero drives the blow-up (Limits and continuity).

Recall Fastest self-check for any "cancel" you attempt

Ask: is the thing I'm cancelling multiplied by everything else, or added to it? Multiplied → cancel is legal. Added → forbidden. Then ask: did that cancellation delete a denominator zero? If so, re-attach it as a restriction.