Multiplying and Dividing Rational Expressions: Free Response

Factor all four polynomials before reading anything off. Each of them factors on sight:

$x^2-49 = (x-7)(x+7), \quad x^2+4x-12 = (x+6)(x-2),$

$x^2-7x = x(x-7), \quad x^2-36 = (x-6)(x+6).$

Now take the three constrained polynomials in turn.

The dividend's denominator $(x+6)(x-2)$ must not vanish, or the first expression names no number at all. That gives $x \ne -6$ and $x \ne 2$.

The divisor's denominator $(x-6)(x+6)$ must not vanish, for the same reason applied to the second expression. That gives $x \ne 6$ and $x \ne -6$.

The divisor's numerator $x(x-7)$ must not vanish either, and this is the condition that is written nowhere. Wherever the divisor is defined it equals zero exactly when its numerator does, and dividing by zero is not an operation. That gives $x \ne 0$ and $x \ne 7$.

Collecting, the quotient excludes

$x = -6, \quad x = 0, \quad x = 2, \quad x = 6, \quad x = 7.$

The dividend's numerator $(x-7)(x+7)$ is the one polynomial under no constraint whatever: a numerator is free to vanish, and at $x = 7$ it does, harmlessly.

Flip the divisor and multiply, keeping everything factored:

$\frac{(x-7)(x+7)}{(x+6)(x-2)} \cdot \frac{(x-6)(x+6)}{x(x-7)} = \frac{(x-7)(x+7)(x-6)(x+6)}{(x+6)(x-2)\,x\,(x-7)}.$

The factors $x-7$ and $x+6$ appear top and bottom, so both divide out:

$\frac{(x+7)(x-6)}{x(x-2)}.$

The restrictions are the five collected in part A, unchanged: they belong to the expression that was written down, and no later step can repeal one. So the answer is

$\frac{(x+7)(x-6)}{x(x-2)}, \qquad x \ne -6, \ 0, \ 2, \ 6, \ 7.$

Now read the answer back. Its denominator is $x(x-2)$, which objects at $x = 0$ and at $x = 2$ and says nothing about the other three. Test one of them: at $x = 7$ the answer returns

$\frac{(14)(1)}{7 \cdot 5} = \frac{2}{5},$

an entirely ordinary number, while the printed quotient at $x = 7$ asks you to divide by $\frac{0}{13} = 0$. The two expressions are equal wherever the original is defined, and the answer is defined in three places where the original is not.

Compare the two rules by what each one needs.

$\frac{A}{B} \cdot \frac{C}{D} \ \text{ requires } \ B \ne 0, \ D \ne 0; \qquad \frac{A}{B} \div \frac{C}{D} \ \text{ requires } \ B \ne 0, \ D \ne 0, \ C \ne 0.$

The product's requirements are exactly the ones that make the two expressions exist. The quotient adds a third, and it is not about existence: it is the statement that you are not dividing by zero.

So replacing the sign keeps every exclusion that came from a denominator, which here is $x \ne -6$ and $x \ne 2$ from $(x+6)(x-2)$ together with $x \ne 6$ and $x \ne -6$ from $(x-6)(x+6)$. It drops the two that came from $x(x-7)$, namely $x \ne 0$ and $x \ne 7$.

Confirm the drop rather than asserting it. At $x = 7$ the product is

$\frac{0}{65} \cdot \frac{0}{13} = 0,$

and at $x = 0$ it is $\frac{-49}{-12} \cdot \frac{0}{-36} = 0$. Both are perfectly good values, so neither number is excluded.

The polynomials did not change, and neither did where any of them is written. What changed is which of them the operation demands be nonzero. That is why a restriction can belong to an operation rather than to an expression, and it is exactly the restriction students lose.

Check the algebra first, since a claim about someone else's work should not rest on trust. The dividend factors as written, because $x^3-16x = x(x^2-16) = x(x-4)(x+4)$ and $x^2+9x+20 = (x+4)(x+5)$. Flipping a polynomial divisor is legitimate: a polynomial is a rational expression over $1$, so its reciprocal is $\frac{1}{x(x-4)}$. Multiplying and cancelling $x$, $x-4$ and $x+4$ leaves

$\frac{x(x-4)(x+4)}{(x+4)(x+5)} \cdot \frac{1}{x(x-4)} = \frac{1}{x+5}.$

So the reduced form is right. What is missing is everything about the domain except one entry.

The dividend's denominator $(x+4)(x+5)$ forces $x \ne -4$ as well as $x \ne -5$. The student kept the second because it is still visible in $\frac{1}{x+5}$ and lost the first because the factor $x+4$ cancelled.

The divisor forces two more. Its denominator is the constant $1$, which is never zero and costs nothing, so nothing on the page looks like a denominator here at all. But the divisor is $x(x-4)$, and a divisor may not be zero, so $x \ne 0$ and $x \ne 4$.

The corrected line is

$\frac{1}{x+5}, \qquad x \ne -5, \ -4, \ 0, \ 4.$

Three of the four survive nowhere in the symbols the student ended on.

Evaluating the reported form is quick:

$\frac{1}{0+5} = \frac{1}{5}, \qquad \frac{1}{4+5} = \frac{1}{9}.$

Now evaluate the two halves of the printed expression separately, which is the only way to see what is being asked.

At $x = 0$ the dividend is $\frac{0-0}{0+0+20} = \frac{0}{20} = 0$, and the divisor is $0^2 - 4(0) = 0$.

At $x = 4$ the dividend is $\frac{64-64}{16+36+20} = \frac{0}{72} = 0$, and the divisor is $16 - 16 = 0$.

So at both values the instruction on the page is

$0 \div 0,$

which is not an instruction that names anything. The reported form is not reporting the value of that division; it is the value of a different expression, one built by cancelling the very factors that vanish there. The two agree everywhere the original is defined, and $x = 0$ and $x = 4$ are precisely two of the places it is not.

Go back to what a quotient is. To divide $P$ by $Q$ is to name the quantity that gives back $P$ when multiplied by $Q$. That description does the work in every case, so test the classmate's case against it.

With $Q = 0$ and $P \ne 0$, nothing multiplied by $Q$ is anything but $0$, so no quantity qualifies and the quotient does not exist. With $Q = 0$ and $P = 0$, the situation is the opposite and no better:

$q \cdot 0 = 0 \ \text{ for every } q,$

so every quantity qualifies. A description satisfied by everything picks out nothing. In neither case is there a value to report, which is why the condition is placed on the divisor alone and never mentions the dividend.

Make that concrete at $x = 0$, where the dividend and the divisor are both $0$. The quotient would have to be the number $q$ with $q \cdot 0 = 0$, and

$5 \cdot 0 = 0, \qquad \tfrac{1}{5} \cdot 0 = 0, \qquad (-3) \cdot 0 = 0,$

so $5$, $\tfrac{1}{5}$ and $-3$ all pass the test that was supposed to single one number out. Nothing in it prefers $0$ to any of them, and a requirement everything satisfies has singled out nothing.

The two values therefore stay excluded, and for the same reason as any other divisor zero.

How many times as fast means the first rate divided by the second, so the second model is the divisor. Factor both models first:

$R_1(x) = \frac{(x-4)(x+1)}{x-6}, \qquad R_2(x) = \frac{(x-8)(x+2)}{x(x-6)}.$

Collect the conditions from that form, before flipping. The first model needs $x \ne 6$. The second model needs $x(x-6) \ne 0$, giving $x \ne 0$ and $x \ne 6$. And the second model is the divisor, so it may not itself be zero, which rules out the zeros of its numerator: $x \ne 8$ and $x \ne -2$.

Now flip and multiply:

$\frac{(x-4)(x+1)}{x-6} \cdot \frac{x(x-6)}{(x-8)(x+2)} = \frac{x(x-4)(x+1)}{(x-8)(x+2)}.$

The comparison is therefore

$\frac{x(x-4)(x+1)}{(x-8)(x+2)}, \qquad x \ne -2, \ 0, \ 6, \ 8.$

The units cancel with the units: micrograms per minute divided by micrograms per minute leaves a bare number, which is what how many times as fast should be. And the reduced expression advertises only $x \ne 8$ and $x \ne -2$, the pair that happened to land in a denominator after the flip.

Evaluate the reduced expression at both temperatures:

$\frac{4 \cdot 0 \cdot 5}{(-4)(6)} = 0, \qquad \frac{6 \cdot 2 \cdot 7}{(-2)(8)} = -\frac{21}{4}.$

Now look at the models themselves, which is where the meaning is.

At $x = 4$ the first model gives $R_1(4) = \frac{(0)(5)}{-2} = 0$ and the second gives $R_2(4) = \frac{(-4)(6)}{4(-2)} = 3$. Dividing $0$ by $3$ is ordinary arithmetic, and the answer $0$ says the first process has stalled while the second is still running. A dividend is allowed to vanish, and this is what that permission looks like in the vessel.

At $x = 6$ both models have $x - 6$ underneath, so neither names a rate:

$R_1(6) = \frac{14}{0}, \qquad R_2(6) = \frac{(-2)(8)}{0}.$

Nothing is being measured, so no comparison exists. The reduced expression nevertheless reports $-\frac{21}{4}$, because the factor $x-6$ that objected cancelled during the flip. A confident number from a reduced form is not evidence that the situation it came from means anything at that input.

Build the reverse comparison the same way. The first model is now the divisor:

$\frac{(x-8)(x+2)}{x(x-6)} \cdot \frac{x-6}{(x-4)(x+1)} = \frac{(x-8)(x+2)}{x(x-4)(x+1)}.$

Collect its conditions from the printed form. The two denominators give $x \ne 0$ and $x \ne 6$, exactly as before, since those are the temperatures at which a model fails to exist and that has nothing to do with which way round the comparison runs. But the divisor is now $R_1$, so its numerator $(x-4)(x+1)$ may not vanish, giving $x \ne 4$ and $x \ne -1$.

Set the two lists side by side:

$\{-2, \ 0, \ 6, \ 8\} \qquad \text{against} \qquad \{-1, \ 0, \ 4, \ 6\}.$

They share only the two temperatures where a model is undefined. Take $x = 8$, where $R_2(8) = \frac{(0)(10)}{8 \cdot 2} = 0$ while $R_1(8) = \frac{(4)(9)}{2} = 18$. Asking how many times $18$ is $0$ has no answer; asking how many times $0$ is $18$ has the answer $0$. At $x = 4$ the same thing happens with the roles exchanged, since $R_1(4) = 0$.

The general statement is worth keeping, and worth stating carefully. Two comparisons that are reciprocals of each other need not be defined on the same set, because each one forbids the zeros of its own divisor and reversing the comparison exchanges the divisor for the dividend. They agree only when the two models vanish at the same values, apart from any value a denominator already forbids: two models with the same zeros, such as $\frac{x}{x-1}$ and $\frac{2x}{x-1}$, give a pair of comparisons excluding exactly the same set. A dividend may vanish; a divisor may not.

Write the quotient as $\frac{A}{B} \div \frac{C}{D}$, so that the three conditions are $B \ne 0$, $D \ne 0$ and $C \ne 0$, while $A$ is free to vanish.

Now read each line for what it shows underneath a bar.

$\frac{A}{B} \div \frac{C}{D} \ \text{ shows } \ B, D; \qquad \frac{A}{B} \cdot \frac{D}{C} \ \text{ shows } \ B, C.$

Completeness. The union of the two lists is $B$, $C$, $D$, which is exactly the set of constrained polynomials, so every excluded value is a zero of a denominator visible in one line or the other and the rule finds all of them.

No over-collection. Every polynomial in that union is genuinely constrained: $B$ and $D$ because the two expressions have to exist, $C$ because the divisor may not be zero. And $A$ never appears in a denominator in either line, so the rule never excludes a value on account of the one polynomial that is allowed to vanish.

Two conditions on the rule are doing real work, and both are stated in it. The lines must be read before cancelling, since cancelling removes factors from denominators and the removed factor's zeros are still excluded. And the flip must be a flip of the divisor only; flipping the dividend by mistake would put $A$ underneath a bar and manufacture restrictions that do not exist.

One special case is worth checking rather than assuming. If the divisor is a polynomial, $D$ is the constant $1$, which is never zero and contributes nothing; the rule still collects $B$ from both lines and $C$ from the product, which is the whole requirement.

Factor everything, which the rule demands first:

$\frac{(x+7)(x-3)}{(x-3)(x+1)} \div \frac{x+7}{x-8}, \qquad \frac{(x+7)(x-3)}{(x-3)(x+1)} \cdot \frac{x-8}{x+7}.$

The quotient line shows the denominators $(x-3)(x+1)$ and $x-8$, giving $x \ne 3$, $x \ne -1$ and $x \ne 8$. The product line shows $(x-3)(x+1)$ and $x+7$, giving $x \ne 3$, $x \ne -1$ and $x \ne -7$. The union is

$x \ne -7, \ -1, \ 3, \ 8.$

Cancelling $x-3$ and $x+7$ in the product leaves

$\frac{x-8}{x+1},$

whose own denominator mentions only $x \ne -1$.

Now the tidy first move. Reducing the dividend on sight turns it into $\frac{x+7}{x+1}$, and the factor $x - 3$ is gone from the page. The two lines then read $\frac{x+7}{x+1} \div \frac{x+7}{x-8}$ and $\frac{x+7}{x+1} \cdot \frac{x-8}{x+7}$, whose visible denominators are $x+1$, $x-8$ and $x+7$. The harvest is $x \ne -1$, $x \ne 8$ and $x \ne -7$, and $x \ne 3$ appears in neither line.

It is still a genuine restriction: at $x = 3$ the printed dividend is $\frac{0}{0}$. The rule was not wrong; it was fed a different expression from the one that was asked about.

Build the smallest thing that can fail. The product form's denominators are $B$ and $C$, so the polynomial it can never display is $D$, the divisor's denominator, which the flip carries upstairs. Any quotient whose $D$ has a zero not shared with $B$ or $C$ will do.

$\frac{1}{x} \div \frac{x+1}{x-9} = \frac{1}{x} \cdot \frac{x-9}{x+1} = \frac{x-9}{x(x+1)}.$

Read the product form on its own and it reports $x \ne 0$ and $x \ne -1$. But the printed quotient has $x - 9$ underneath a bar, so the divisor does not exist at $x = 9$ and the quotient excludes it. The product form is not merely silent about $x = 9$: it is cheerful there, returning $\frac{0}{90} = 0$.

So the flip does not reveal the hidden condition. It exchanges one hidden condition for another, taking $C$ out of a numerator and putting $D$ into one, which is why the rule needs both lines and no single line will do. The student's instinct is right that the flip exposes something; what it exposes is $C \ne 0$, and it conceals $D \ne 0$ in the same movement.

Decide where each forbidden value is to come from before writing a polynomial down. Three places can force one: the dividend's denominator $B$, the divisor's denominator $D$, and the divisor's numerator $C$.

There is a constraint on $D$ worth noticing first. Flipping puts $D$ in a numerator, where each of its factors either cancels against $B$ or $C$ or else survives as a factor of the reduced form. Surviving is not open here: the reduced form is $x+4$, so a survivor would have to be $x+4$ itself and would forbid $x = -4$, which is not on the list. Every factor of $D$ must therefore cancel, and $D$ forbids nothing that $B$ or $C$ does not forbid already. Put $D = x-12$ and let $B = x-12$ as well, so the shared denominator supplies $x = 12$.

That leaves $x = 11$ and $x = -8$ to come from the divisor's numerator, so take $C = (x-11)(x+8)$. Finally choose $A$ to make the reduction come out: with $A = (x+4)(x-11)(x+8)$,

$\frac{(x+4)(x-11)(x+8)}{x-12} \cdot \frac{x-12}{(x-11)(x+8)} = x+4.$

Check the second requirement as carefully as the first, since it is the one that can fail invisibly. The dividend's denominator is zero only at $x = 12$; the divisor's denominator is zero only at $x = 12$; the divisor's numerator is zero only at $x = 11$ and $x = -8$. The union is

$\{-8, \ 11, \ 12\},$

with nothing over, which is what the word exactly demanded. And the reduced form $x+4$ is a polynomial, so it forbids nothing at all: all three restrictions live outside the symbols anybody would end on.

A product constrains only $B$ and $D$, so all three values have to be zeros of those two denominators, and for the reduced form to be a polynomial both denominators have to cancel away entirely. Both demands can be met at once:

$\frac{(x+4)(x-12)}{(x-11)(x+8)} \cdot \frac{(x-11)(x+8)}{x-12} = \frac{(x+4)(x-12)(x-11)(x+8)}{(x-11)(x+8)(x-12)} = x+4.$

Its denominators are zero exactly at $x = 11$, $x = -8$ and $x = 12$, so the forbidden set is the one asked for.

So both operations can produce this domain, and it is worth being exact about what separates them. In the product, every excluded value is a zero of something printed underneath a bar, and a reader who scans the original for denominators finds all three. In the quotient of part A, $x = 11$ and $x = -8$ are zeros of a polynomial printed above a bar, and that reader finds only $x = 12$.

The two versions also fail differently under a careless first move. Cancelling in the product before recording anything destroys the very denominators the exclusions came from, whereas in the quotient the divisor's numerator is not something cancelling removes until the flip has happened. Neither is safer than the other in general. What is true of both is that the reduced form $x+4$ carries no trace of any of it.

Flipping and multiplying gives a single fraction, and reducing divides its numerator and its denominator by their greatest common factor $G$:

$\frac{A}{B} \div \frac{C}{D} = \frac{AD}{BC} = \frac{P}{Q}, \qquad P = \frac{AD}{G}, \quad Q = \frac{BC}{G}.$

So $BC = QG$ with $G$ a polynomial, which says exactly that $Q$ divides $BC$.

Now take any real number $r$ with $Q(r) = 0$. Then $B(r)C(r) = Q(r)G(r) = 0$, and a product of two numbers is zero only if one of them is, so $B(r) = 0$ or $C(r) = 0$. Consider the two cases.

If $B(r) = 0$, the dividend $\frac{A}{B}$ has a zero denominator at $r$ and names no number, so the original expression is undefined at $r$.

If $C(r) = 0$, look at the divisor $\frac{C}{D}$ at $r$. Either $D(r) = 0$ as well, in which case the divisor names no number, or $D(r) \ne 0$, in which case the divisor equals $\frac{0}{D(r)} = 0$ and the original expression asks you to divide by zero. Either way $r$ is forbidden.

So every zero of $Q$ is a forbidden value of the original, which is the claim. Reading it as a statement about domains: every value the reduced form rejects, the original rejects too, so

$\text{domain of } \tfrac{P}{Q} \ \supseteq \ \text{domain of the original}.$

The reverse fails, and part A is the counterexample: there the reduced form is $x+4$, whose domain is every real number, while the original is undefined at three of them. The containment can therefore be strict, and the reason is visible in the proof. Nothing in it mentions $D$, because flipping sends $D$ upstairs, where it need not appear in $Q$ at all, since $Q$ divides $BC$; and nothing in it says $Q$ carries every factor of $BC$, because $G$ may have taken some away. Those two gaps are precisely the two ways a restriction goes missing.