Multiplying and Dividing Rational Expressions

Intuition

A rational expression is a fraction whose numerator and denominator are polynomials, and it behaves like a fraction because it is one. So the rule for multiplying two of them is the rule you already use on $\frac{2}{3} \cdot \frac{5}{7}$ : multiply the tops, multiply the bottoms. That takes about ten seconds to learn, and it is not what this lesson is about. Two other things are, and both are where good students lose the marks. Multiplying before you factor turns a problem you could have finished on sight into a quartic you now have to factor from scratch. And dividing destroys evidence: flipping the divisor moves a polynomial out of a denominator and into a numerator. The “cannot be zero” condition riding on that polynomial stops being written anywhere on the page. The condition is still true. You just cannot see it any more.

Learning goals

Multiply straight across, with no common denominator needed
Flip the divisor, adding the condition that it is nonzero
Factor everything first, since expanding hides what cancels
Collect every restriction from the original expression
Write the restrictions even when the answer is a polynomial

The product rule, and why it is allowed

A rational expression is a quotient $\frac{A}{B}$ of polynomials with $B$ not the zero polynomial, and it is undefined at every number where its denominator evaluates to zero. To multiply two of them, do what you do with fractions:

\frac{A}{B} \cdot \frac{C}{D} = \frac{AC}{BD}.

The result is a rational expression again. Multiplying two polynomials always gives another polynomial, so $AC$ and $BD$ are polynomials, and $BD$ is not the zero polynomial as long as neither $B$ nor $D$ is. So multiplying two rational expressions always gives another rational expression, the same way multiplying two fractions always gives another fraction.

Multiplying straight across is not a convention someone picked. It follows from what a quotient means.

Why $\frac{A}{B}\cdot\frac{C}{D} = \frac{AC}{BD}$ #

Run the idea on numbers first. Take $\frac{2}{3} \cdot \frac{5}{7}$ , and write $p = \frac{2}{3}$ and $q = \frac{5}{7}$ . Those names mean exactly $p \cdot 3 = 2$ and $q \cdot 7 = 5$ , because that is what a quotient is: the number that gives back the top when you multiply it by the bottom. Multiply those two equations together and regroup, which is allowed because multiplication is commutative and associative:

(pq)(3 \cdot 7) = (p \cdot 3)(q \cdot 7) = 2 \cdot 5 = 10.

So $pq$ is a number that gives back $10$ when multiplied by $21$ . Only one number can do that, and it is $\frac{10}{21}$ , exactly $\frac{2}{3} \cdot \frac{5}{7}$ found the fast way. Nothing above depended on $2$ , $3$ , $5$ , and $7$ being those particular numbers. Rerun the same two moves with letters.

Fix a number $x$ at which both expressions are defined, so $B(x) \ne 0$ and $D(x) \ne 0$ . At that $x$ the two expressions are ordinary numbers, so it is enough to prove the ordinary fraction rule for numbers $a$ , $b$ , $c$ , $d$ with $b \ne 0$ and $d \ne 0$ . Write $p = \frac{a}{b}$ and $q = \frac{c}{d}$ , so that $pb = a$ and $qd = c$ , and multiply those equations together the same way:

(pq)(bd) = (pb)(qd) = ac.

So $pq$ is a number that gives back $ac$ when multiplied by $bd$ . Since $b \ne 0$ and $d \ne 0$ , the product $bd$ is nonzero, so exactly one number does that: if $u(bd) = v(bd) = ac$ then $(u - v)(bd) = 0$ , and a product of nonzero numbers is never zero, so $u = v$ . That one number is $\frac{ac}{bd}$ by definition, hence $pq = \frac{ac}{bd}$ .

The argument used nothing about $a$ , $b$ , $c$ , $d$ beyond $b \ne 0$ and $d \ne 0$ , so it runs at every $x$ with $B(x) \ne 0$ and $D(x) \ne 0$ . That is precisely the set where both factors are defined, and it is where the identity $\frac{A}{B} \cdot \frac{C}{D} = \frac{AC}{BD}$ holds.

Look at what the proof needed and what it never mentioned. It needed $B(x) \ne 0$ and $D(x) \ne 0$ . It said nothing about $A$ or $C$ , which are free to vanish. A product of rational expressions never picks up a restriction beyond the ones $B$ and $D$ already carry. Canceling a factor can still hide one of those restrictions from the page later, which the next section shows; multiplying itself adds nothing new. Division is a different story: it is about to add a restriction that was never in $B$ or $D$ at all.

Factor first, never last

Consider

\frac{x^2-4}{x^2+6x+9} \cdot \frac{x^2+2x-3}{x^2+x-6}.

The rule says multiply the numerators and multiply the denominators, and nothing stops you from doing that right away:

\frac{(x^2-4)(x^2+2x-3)}{(x^2+6x+9)(x^2+x-6)} = \frac{x^4+2x^3-7x^2-8x+12}{x^4+7x^3+9x^2-27x-54}.

Now look at what you are holding. To simplify it you must factor a quartic on top and a quartic on the bottom, hunting for rational roots and running synthetic division on each candidate. And the factorizations you need are the ones you were holding thirty seconds ago, before you multiplied them out. Expanding is the one operation that reliably destroys the structure you are about to need.

Factor first instead, and the cancellations are visible on sight:

\begin{aligned} \frac{(x-2)(x+2)}{(x+3)^2} \cdot \frac{(x+3)(x-1)}{(x+3)(x-2)} &= \frac{(x-2)(x+2)(x+3)(x-1)}{(x+3)^2(x+3)(x-2)} \\ &= \frac{(x+2)(x-1)}{(x+3)^2}. \end{aligned}

Same problem, no quartics, no root hunting, but look at what the last line quietly drops. The original carries $x \ne -3$ and $x \ne 2$ : the answer’s denominator $(x+3)^2$ still shows $x \ne -3$ , but $x \ne 2$ canceled away along with the $(x-2)$ that carried it, and it is no less true for having become invisible. You met the first idea when you graphed polynomials: the factored form is the one that shows you the roots, and expanding hides them. It is the same story here. The next section makes the second idea, that canceling erases restrictions from the page, precise. The factored form of a rational expression is the form in which its cancellations and its restrictions are legible. Every method in this chapter begins by putting everything into it.

Worked example 1 Multiply $\frac{3x^2-12}{x^2+7x+12} \cdot \frac{x^2+x-6}{x^2-4x+4}$

Factor all four polynomials before doing anything else. Pull the common factor of $3$ out first, then use the difference of squares:

3x^2 - 12 = 3(x^2 - 4) = 3(x-2)(x+2).

The other three are $x^2+7x+12 = (x+3)(x+4)$ , then $x^2+x-6 = (x+3)(x-2)$ , and finally $x^2-4x+4 = (x-2)^2$ . Read the restrictions off the two denominators right now, while they are factored: the first vanishes at $x = -3$ and $x = -4$ , the second at $x = 2$ .

Now multiply across and cancel the factors common to the top and the bottom:

\frac{3(x-2)(x+2)}{(x+3)(x+4)} \cdot \frac{(x+3)(x-2)}{(x-2)^2} = \frac{3(x-2)^2(x+2)(x+3)}{(x+3)(x+4)(x-2)^2} = \frac{3(x+2)}{x+4}.

The answer is $\dfrac{3(x+2)}{x+4}$ with $x \ne -4$ , $x \ne -3$ , and $x \ne 2$ . Only one of those three restrictions is still visible in the answer. The other two rode on factors that canceled, and they have to be carried along by hand.

Check your understanding

Simplify $\frac{x^2-1}{x+3} \cdot \frac{x+3}{x-1}$ , and give every restriction on $x$ .

x+1

, with

x \ne -3

and

x \ne 1

x+1

, with no restrictions C

x-1

, with

x \ne -3

and

x \ne 1

\frac{x+1}{x+3}

, with

x \ne -3

The restrictions live in the original

Canceling changes the domain, and that is a real change rather than a bookkeeping nuisance. Take

\frac{x+1}{x(x-4)} \cdot \frac{x-4}{x+6} = \frac{(x+1)(x-4)}{x(x-4)(x+6)} = \frac{x+1}{x(x+6)}.

At $x = 4$ the original expression asks you to divide by $x - 4 = 0$ , so it does not exist there. The simplified form is perfectly comfortable at $x = 4$ : it returns $\frac{5}{40} = \frac{1}{8}$ . The two are therefore not the same function. They agree at every number where the original is defined, and at $x = 4$ one of them exists and the other does not.

So the answer is not $\frac{x+1}{x(x+6)}$ . The answer is

\frac{x+1}{x(x+6)}, \qquad x \ne -6, \quad x \ne 0, \quad x \ne 4,

and the restriction $x \ne 4$ is as much a part of it as the algebra is. Restrictions are read off the original expression, before any canceling, because canceling is precisely the step that erases them.

Dividing means multiplying by the reciprocal

To divide by a fraction, flip it and multiply. Rational expressions are no different:

\frac{A}{B} \div \frac{C}{D} = \frac{A}{B} \cdot \frac{D}{C} = \frac{AD}{BC}.

Why flipping the divisor works, and what it costs#

Dividing by a quantity means undoing multiplication by it: $P \div Q$ is the quantity that gives back $P$ when you multiply it by $Q$ . That description only makes sense when $Q \ne 0$ . If $Q = 0$ then everything multiplied by $Q$ gives $0$ , so when $P \ne 0$ no such quantity exists at all, and when $P = 0$ every quantity qualifies. Either way there is no single value to call the quotient, which is what “division by zero is undefined” means.

Run the check on numbers first. Divide $\frac{2}{3}$ by $\frac{4}{5}$ : flip and multiply to get the candidate $\frac{2}{3} \cdot \frac{5}{4} = \frac{10}{12}$ . Multiply that candidate back by the divisor and see whether the dividend comes back:

\frac{10}{12} \cdot \frac{4}{5} = \frac{40}{60} = \frac{2}{3}.

It does, so $\frac{10}{12}$ really is $\frac{2}{3} \div \frac{4}{5}$ . Nothing above used $2$ , $3$ , $4$ , and $5$ specifically. Rerun the same check with letters.

Now fix a number $x$ at which $B(x)$ , $C(x)$ , and $D(x)$ are all nonzero. The divisor $\frac{C}{D}$ is then defined, and because its numerator is nonzero the divisor itself is nonzero, so the quotient we want exists. Test the candidate $\frac{AD}{BC}$ by multiplying it by the divisor, using the product rule proved above:

\frac{AD}{BC} \cdot \frac{C}{D} = \frac{ADC}{BCD} = \frac{A}{B},

where the last step cancels the factors $C$ and $D$ , both nonzero at $x$ . The candidate gives back the dividend when multiplied by the divisor, so the candidate is the quotient:

\frac{A}{B} \div \frac{C}{D} = \frac{AD}{BC} = \frac{A}{B} \cdot \frac{D}{C}.

Read off the hypotheses that argument used: $B(x) \ne 0$ , $D(x) \ne 0$ , and $C(x) \ne 0$ . The first two say the two rational expressions are defined. The third is new, and it is not a technicality. It is the statement that you are not dividing by zero.

So a quotient of rational expressions carries three conditions, one on each of three polynomials:

\frac{A}{B} \div \frac{C}{D} \quad \text{requires} \quad B \ne 0, \quad D \ne 0, \quad C \ne 0.

Of the four polynomials on the page, only $A$ , the dividend’s numerator, is free to be zero. The third condition is worth restating in the form you will actually use. Wherever the divisor $\frac{C}{D}$ is defined, it equals zero exactly when its numerator $C$ equals zero. If $C = 0$ and $D \ne 0$ then the fraction is $0$ , and conversely if the fraction is $0$ then multiplying both sides by $D$ gives $C = 0$ . So the rule is simply that the divisor’s numerator can never be zero, and that is the restriction almost nobody writes down.

The vanishing restriction

Here is the trap in its purest form. Divide:

\frac{x}{x-1} \div \frac{x-2}{x-5}.

Before touching anything, collect the three conditions. The dividend’s denominator gives $x \ne 1$ . The divisor’s denominator gives $x \ne 5$ . The divisor’s numerator gives $x \ne 2$ , because at $x = 2$ the divisor is $\frac{0}{-3} = 0$ and you would be dividing by zero. Now flip and multiply:

\frac{x}{x-1} \cdot \frac{x-5}{x-2} = \frac{x(x-5)}{(x-1)(x-2)}.

Nothing canceled. Nothing came close to canceling. And the answer has still lost a restriction: its denominators are $x-1$ and $x-2$ , so the page now announces $x \ne 1$ and $x \ne 2$ and says nothing whatever about $x \ne 5$ . Put $x = 5$ into the answer and you get $\frac{5 \cdot 0}{4 \cdot 3} = 0$ , an utterly ordinary value. Put $x = 5$ into the original and the divisor is $\frac{3}{0}$ , which does not exist.

The flip did not reveal a hidden restriction. It traded one hidden restriction for another. Before the flip, $C$ sits in a numerator, so $C \ne 0$ is invisible. After the flip, $D$ sits in a numerator, so $D \ne 0$ is invisible. No single form of the expression displays all three conditions at once.

Flipping the divisor does not reveal a restriction, it trades one for another. C and D swap places, and whichever of them lands in a numerator has its nonzero condition erased from the page. The dashed boxes mark a condition that is still true but is no longer written anywhere. Only the original expression, read with the division rule in mind, gives you all three.

The same story as a ledger, for the example above:

Restriction	Where it comes from	Shown by $\frac{x}{x-1} \div \frac{x-2}{x-5}$ ?	Shown by $\frac{x}{x-1} \cdot \frac{x-5}{x-2}$ ?
$x \ne 1$	the dividend’s denominator	yes	yes
$x \ne 5$	the divisor’s denominator	yes	no
$x \ne 2$	the divisor cannot be zero	no	yes

There is no stage after the original at which the page carries every restriction, so there is exactly one safe procedure. Collect the restrictions from the original expression, using the division rule, before you flip and before you cancel.

Canceling then makes things worse. Take the standard exercise

\frac{x+1}{x-2} \div \frac{x-3}{x-2}.

From the original: $x \ne 2$ , since $x-2$ is the denominator of both fractions, and $x \ne 3$ , since that is where the divisor is zero. Flip and multiply:

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

The answer is $\frac{x+1}{x-3}$ with $x \ne 2$ and $x \ne 3$ . The $x-2$ that was staring at you twice in the original has canceled itself out of existence. And the only restriction the answer still shows is $x \ne 3$ , which is the one that was invisible when you started. A polynomial answer can lose every trace the same way; the common mistakes below show exactly that case.

Check your understanding

Which values of $x$ must be excluded from $\frac{x-5}{x+2} \div \frac{x-4}{x+2}$ ?

x \ne -2

only B

x \ne 4

only C

x \ne -2

and

x \ne 4

x \ne -2

and

x \ne 5

Worked example 2 Divide $\frac{x^2-25}{x^2+2x-8} \div \frac{x^2+10x+25}{x^2-16}$

Factor all four polynomials first:

\frac{(x-5)(x+5)}{(x+4)(x-2)} \div \frac{(x+5)^2}{(x-4)(x+4)}.

Collect the restrictions from this original form, before flipping anything. The dividend’s denominator $(x+4)(x-2)$ gives $x \ne -4$ and $x \ne 2$ . The divisor’s denominator $(x-4)(x+4)$ gives $x \ne 4$ and $x \ne -4$ . The divisor’s numerator $(x+5)^2$ gives $x \ne -5$ , because the divisor is zero exactly where $(x+5)^2 = 0$ and you cannot divide by zero.

Now flip the divisor and multiply:

\begin{aligned} \frac{(x-5)(x+5)}{(x+4)(x-2)} \cdot \frac{(x-4)(x+4)}{(x+5)^2} &= \frac{(x-5)(x+5)(x-4)(x+4)}{(x+4)(x-2)(x+5)^2} \\ &= \frac{(x-5)(x-4)}{(x-2)(x+5)}. \end{aligned}

The answer is

\frac{(x-5)(x-4)}{(x-2)(x+5)}, \qquad x \ne -5, \quad x \ne -4, \quad x \ne 2, \quad x \ne 4.

Two of those four restrictions survive in the answer’s denominator, $x \ne 2$ and $x \ne -5$ . The other two do not. At $x = 4$ the answer returns $0$ , and at $x = -4$ it returns $-12$ , both perfectly finite, while the original expression exists at neither. Those two restrictions are true only because you wrote them down.

Worked example 3 Divide $\frac{x^2-9}{x^2+2x}$ by the polynomial $x+3$

A polynomial is a rational expression with denominator $1$ , so write the divisor as $\frac{x+3}{1}$ and the rule applies unchanged. Factor, then collect restrictions:

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

The dividend’s denominator gives $x \ne 0$ and $x \ne -2$ . The divisor’s denominator is the constant $1$ , which is never zero, so it costs nothing. The divisor’s numerator gives $x \ne -3$ : dividing by $x+3$ is illegal exactly where $x + 3 = 0$ .

That last restriction deserves a hard look. Nowhere in the expression as written is there a denominator containing $x+3$ , so nothing on the page hints at it. It comes from the division symbol alone.

Flip and multiply:

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

The answer is $\dfrac{x-3}{x(x+2)}$ with $x \ne -3$ , $x \ne -2$ , and $x \ne 0$ . Check the odd one out: at $x = -3$ the answer evaluates to $\frac{-6}{(-3)(-1)} = -2$ , a genuine number, while the original asks you to divide $\frac{0}{3}$ by $0$ . That is undefined, and only your notes remember it.

Check your understanding

The quotient $\frac{x^2-1}{x-2} \div \frac{x+1}{x-2}$ simplifies to $x-1$ . At which values of $x$ is the original quotient undefined?

A Nowhere, since

x-1

is defined for every real number B

x = 2

only C

x = -1

only D

x = 2

and

x = -1

Common mistakes

Multiplying out before factoring. Expanding two factored quadratics into a quartic throws away the only structure you had, and you will have to rebuild it with the rational root theorem. Factor, cancel, and expand at the very end only if the question demands it.
Canceling terms instead of factors. Only factors cancel. In $\frac{x+3}{x}$ the $x$ on top is a term added to $3$ , not a factor of the numerator, so nothing cancels: at $x = 1$ the expression is $4$ , not $3$ .
Reading the restrictions off the answer. The answer is what survives canceling, so it has already forgotten everything that canceled. Restrictions come from the original expression.
Forgetting that the divisor’s numerator cannot be zero. In $\frac{A}{B} \div \frac{C}{D}$ the condition $C \ne 0$ is written nowhere, and it is the one students lose. Dividing by an expression requires that expression to be nonzero.
Flipping the wrong fraction. The dividend stays exactly where it is. Only the divisor is inverted.
Believing a polynomial answer has no restrictions. $\frac{x^2-9}{x+1} \div \frac{x-3}{x+1}$ simplifies to $x+3$ , and it is still undefined at $x = -1$ and at $x = 3$ .
Hunting for a common denominator. Multiplication and division never need one. That is a tool for adding and subtracting, which is the next lesson.

Key takeaways

$\frac{A}{B} \cdot \frac{C}{D} = \frac{AC}{BD}$ , valid wherever $B \ne 0$ and $D \ne 0$ . No common denominator is needed, and multiplying alone never adds a restriction: numerators may vanish freely. Canceling can still hide one of $B$ ‘s or $D$ ‘s zeros from view, so the original denominators are still what you check.
$\frac{A}{B} \div \frac{C}{D} = \frac{A}{B} \cdot \frac{D}{C}$ , valid wherever $B \ne 0$ , $D \ne 0$ , and $C \ne 0$ . Of the four polynomials, only $A$ is allowed to be zero.
Factor everything first, then cancel. Expanding destroys the factorizations you need; factored form is where cancellations and restrictions are legible.
Every restriction comes from the original expression. Canceling erases restrictions from the page, and flipping trades a visible one for a hidden one, so the final answer is never a reliable record of the domain.
A simplified answer can be a polynomial, defined everywhere, and still describe an expression that is undefined at several points. The restrictions are part of the answer. Write them.

Multiplying and Dividing Rational Expressions

Learning goals

The product rule, and why it is allowed

Factor first, never last

The restrictions live in the original

Dividing means multiplying by the reciprocal

The vanishing restriction

Common mistakes

Practice

Multiple Choice Questions (MCQ)

Core practice