Multiplying & Dividing Rational Expressions

A rational expression is a fraction whose numerator and denominator are polynomials — things like $\dfrac{x^2 - 9}{x + 3}$ . Simplifying one works exactly like reducing $\dfrac{6}{8}$ to $\dfrac{3}{4}$ : find what the top and bottom have in common and divide it out. The difference is that with polynomials, the common pieces are factors, and you have to factor before you can see them.

The entire skill comes down to one rule: factor first, then cancel whole factors. Canceling individual terms — crossing out the $x^2$ 's in $\dfrac{x^2 - 9}{x^2 + 4x}$ — is the single most common algebra mistake there is, and it's always wrong.

Factor first, then cancel factors — never terms

Factor the numerator completely, factor the denominator completely, and cancel any factor that appears in both. A factor is a whole multiplied piece, like $(x + 3)$ or $x$ ; a term is a piece connected by addition or subtraction, like the $x^2$ inside $x^2 - 9$ .

Why terms can't cancel: canceling means dividing top and bottom by the same quantity, and division only distributes over multiplication, not addition. Try it with numbers: $\dfrac{2 + 6}{2} = 4$ , but if you 'cancel the 2s' you'd get $6$ . Wrong. Once everything is factored, though, the whole expression is a product, and matching factors divide out cleanly.

Bring your full factoring toolbox: greatest common factor first, then difference of squares ( $x^2 - 9 = (x - 3)(x + 3)$ ), then trinomials. If nothing on top matches anything on the bottom after factoring, the expression is already in simplest form — leave it alone.

Excluded values: the fine print

A fraction with zero in the denominator is undefined, so any $x$ -value that makes the original denominator zero must be excluded from the domain. Find them by setting each factor of the original denominator equal to zero — before canceling anything.

Here's the subtle part: the exclusions survive even when the factor cancels. $\dfrac{x^2 - 9}{x + 3}$ simplifies to $x - 3$ , but the original expression is still undefined at $x = -3$ . The simplified form is only equal to the original for $x \neq -3$ , so a complete answer says: $x - 3$ , where $x \neq -3$ .

The opposite-factors trick

Sometimes the top and bottom hold factors that are opposites rather than twins: $(3 - x)$ and $(x - 3)$ . They're not equal, but each is $-1$ times the other: $3 - x = -(x - 3)$ . Rewrite one of them by pulling out $-1$ , cancel the now-matching factors, and carry the leftover $-1$ into the answer. Any time a factor looks like a reversed version of another, this is the move.

Worked examples

Example 1: difference of squares

Simplify $\dfrac{x^2 - 9}{x + 3}$ .

Factor the numerator as a difference of squares

\dfrac{(x - 3)(x + 3)}{x + 3}

Cancel the common factor

(x + 3)

x - 3

State the excluded value from the original denominator

x \neq -3

Answer: $x - 3$ , where $x \neq -3$

Example 2: factor a GCF out of both

Simplify $\dfrac{4x^2 - 8x}{2x}$ .

Factor the numerator — the GCF is

4x

\dfrac{4x(x - 2)}{2x}

Cancel the common factor

x

and reduce

\dfrac{4}{2}

2

2(x - 2)

State the excluded value from the original denominator

x \neq 0

Answer: $2(x - 2)$ , where $x \neq 0$

Example 3: two trinomial-style factorizations

Simplify $\dfrac{x^2 + 5x + 6}{x^2 - 4}$ .

Factor the numerator

x^2 + 5x + 6 = (x + 2)(x + 3)

Factor the denominator as a difference of squares

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

Cancel the common factor

(x + 2)

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

Exclude every zero of the original denominator

x \neq -2, \; x \neq 2

Answer: $\dfrac{x + 3}{x - 2}$ , where $x \neq -2$ and $x \neq 2$

Example 4: opposite factors

Simplify $\dfrac{9 - x^2}{x^2 - x - 6}$ .

Factor the numerator

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

Factor the denominator

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

Rewrite

(3 - x)

-(x - 3)

to expose the match

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

Cancel the common factor

(x - 3)

and keep the

-1

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

Exclude the zeros of the original denominator

x \neq 3, \; x \neq -2

Answer: $-\dfrac{x + 3}{x + 2}$ , where $x \neq 3$ and $x \neq -2$

Try one yourself

Common questions

Why can't I just cancel the

x^2

terms on top and bottom?

Because canceling is division, and you can only divide out something that multiplies the entire numerator and the entire denominator. In $\dfrac{x^2 + 1}{x^2}$ , the $x^2$ on top is glued to the $+1$ by addition — it isn't a factor of the whole top. Factor first; if a piece doesn't appear as a full factor of both, it can't cancel.

Do the excluded values change after I simplify?

No — they come from the original denominator and they stay. If a factor cancels, its excluded value becomes a 'hole' in the graph rather than a vertical asymptote, but the expression is still undefined there. Always list exclusions from the denominator as it was before any canceling.

How do I know when a rational expression is fully simplified?

Factor both the numerator and denominator completely, cancel everything that matches, and then look again: if the top and bottom share no common factor other than $1$ , you're done. It's fine — and common — for an answer to still be a fraction.

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Multiplying & Dividing Rational Expressions

Factor first, then cancel factors — never terms

Excluded values: the fine print

The opposite-factors trick

Worked examples

Example 1: difference of squares

Example 2: factor a GCF out of both

Example 3: two trinomial-style factorizations

Example 4: opposite factors

Try one yourself

Common questions

Keep learning

Factoring Trinomials (a = 1)

Difference of Squares

Intro to Domain and Range