Simplifying Rational Expressions: Free Response

Nothing may be divided out while the two halves are sums, so factor both first. The numerator needs two numbers with product $-35$ and sum $2$, which are $7$ and $-5$, and the denominator is a difference of squares:

$\frac{x^2+2x-35}{x^2-49} = \frac{(x+7)(x-5)}{(x-7)(x+7)}$

Read the excluded values off that denominator now, before anything leaves the page. It is zero at $x=7$ and at $x=-7$, so the domain is every real number apart from those two.

At every input the domain allows, the factor $x+7$ is not zero, so it may be divided out of the whole numerator and the whole denominator:

$\frac{(x+7)(x-5)}{(x-7)(x+7)} = \frac{x-5}{x-7} \qquad (x \neq 7 \text{ and } x \neq -7)$

A check at an allowed input, $x=0$: the printed expression gives $\frac{-35}{-49} = \frac{5}{7}$, and the reduced one gives $\frac{-5}{-7} = \frac{5}{7}$.

Substitute into the reduced form first, where the arithmetic is short:

$\frac{(-7)-5}{(-7)-7} = \frac{-12}{-14} = \frac{6}{7}$

Now substitute into the printed expression, taking its numerator and its denominator separately:

$(-7)^2+2(-7)-35 = 49-14-35 = 0, \qquad (-7)^2-49 = 0$

So the printed expression asks for $\frac{0}{0}$ there, and that names nothing. A quotient $\frac{u}{v}$ is the one number $t$ with $tv = u$; when $u$ and $v$ are both zero, every real number satisfies $t \cdot 0 = 0$, so no single number is picked out.

The two therefore agree at every input except $x=-7$, where the reduced form alone has something to report. Its graph carries the point $\left(-7, \frac{6}{7}\right)$; the printed expression's graph is missing exactly that one point and nothing else.

Track each excluded input to the factor it came from. Both came from the printed denominator $(x-7)(x+7)$, so both are genuine restrictions on the printed expression, and neither is more real than the other.

What separates them is what happened next. The factor $x-7$ was not shared with the numerator, so it is still standing in the answer, and $\frac{x-5}{x-7}$ refuses $x=7$ without being told to. The factor $x+7$ was shared, so it was divided out, and the answer has no record of it: $\frac{x-5}{x-7}$ evaluates at $x=-7$ as cheerfully as anywhere else.

That asymmetry is a fact about the page, not about the mathematics. It is why the restriction is written beside the answer at the moment of cancelling, since afterwards there is nothing to reconstruct it from.

Now the cost of leaving it off. Writing

$\frac{x^2+2x-35}{x^2-49} = \frac{x-5}{x-7}$

with no condition attached asserts that the two sides agree at every real number. Test that assertion at $x=-7$: the right side is $\frac{6}{7}$ and the left side is not a number, so there is nothing there for it to equal. One case out of infinitely many is enough to make the unqualified claim false, and the qualified claim, restricted to $x \neq 7$ and $x \neq -7$, is true.

The numerator has four terms, which is the signal to pair them off and take a common factor out of each pair:

$x^3-5x^2+4x-20 = x^2(x-5) + 4(x-5) = (x-5)(x^2+4)$

The pairing was worth doing because both pairs left the same bracket behind. The denominator is a difference of squares:

$x^2-25 = (x-5)(x+5)$

Read the excluded values from that denominator before cancelling: it is zero at $x=5$ and at $x=-5$, so the domain is every real number apart from those two.

On that domain the shared factor $x-5$ is nonzero, so it divides out:

$\frac{(x-5)(x^2+4)}{(x-5)(x+5)} = \frac{x^2+4}{x+5} \qquad (x \neq 5 \text{ and } x \neq -5)$

Check at $x=0$: the printed expression gives $\frac{-20}{-25} = \frac{4}{5}$, and the reduced one gives $\frac{4}{5}$ as well.

A quadratic and a linear polynomial can share a non-constant factor in only one way: the linear one has to divide the quadratic. So the question is whether $x+5$ is a factor of $x^2+4$, and the Factor Theorem turns that into an evaluation.

The factor $x+5$ is $x-(-5)$, so it divides $x^2+4$ exactly when $x^2+4$ vanishes at $-5$:

$(-5)^2+4 = 29 \neq 0$

It does not, so $x+5$ is not a factor, and the expression is in lowest terms.

The stronger argument settles every linear factor at once instead of one at a time. The discriminant of $x^2+0x+4$ is

$0^2-4(1)(4) = -16 < 0$

so $x^2+4$ has no real zero whatever, and by the Factor Theorem it therefore has no factor $x-r$ with $r$ real at all. That covers $x+5$ and every other linear polynomial in one line, and it is the argument that survives a change of numbers: any numerator with a negative discriminant is safe from every linear denominator.

Take the two directions of the search's claim separately, because only one of them holds in general.

What it does settle. If $x-r$ with $r$ real divides both halves, then by the Factor Theorem both halves vanish at $r$, so the search finds $r$. Conversely, if both vanish at $r$, the Factor Theorem hands back $x-r$ as a factor of each. So the search is exactly a test for a shared LINEAR factor with a real root, and on that question it is complete.

What it does not settle. A common factor need not be linear, and a non-linear one need not have any real zero to be found. Take

$\frac{x^2+9}{x^4-81}$

The numerator has no real zero, so no number vanishes on both halves and the search reports lowest terms. But the denominator is a difference of squares:

$x^4-81 = (x^2-9)(x^2+9)$

so the numerator is a factor of the denominator outright, and the expression reduces to $\frac{1}{x^2-9}$, valid wherever the printed expression was defined, which is every real number except $3$ and $-3$.

The honest version of the classmate's rule is therefore narrower than they think: no shared real zero means no shared linear factor, and to rule out the rest you have to factor both halves completely and compare the factors themselves.

Take the two quantities separately. The areas are $x^2$ and $64$ square metres, so the extra ground is $x^2-64$. The fencing of a square is its perimeter, so the two fences are $4x$ and $32$ metres and the extra fencing is $4x-32$. The trade is one divided by the other:

$\frac{x^2-64}{4x-32}$

Factor before doing anything else. The numerator is a difference of squares and the denominator has a common factor of $4$:

$\frac{x^2-64}{4x-32} = \frac{(x-8)(x+8)}{4(x-8)}$

The denominator is zero exactly when $x=8$, so that value is excluded. Everywhere else $x-8$ is nonzero and divides out:

$\frac{(x-8)(x+8)}{4(x-8)} = \frac{x+8}{4} \qquad (x \neq 8)$

From the simplified expression:

$\frac{12+8}{4} = 5$

Directly from the paddocks: the areas are $144$ and $64$ square metres, so the extra ground is $80$ square metres; the perimeters are $48$ and $32$ metres, so the extra fencing is $16$ metres. Dividing,

$\frac{144-64}{48-32} = \frac{80}{16} = 5$

The two routes agree, as they must at any input the printed ratio accepts, and $12$ is such an input.

The unit is worth stating rather than assuming. The numerator is an area and the denominator a length, so the quotient is square metres per metre: each extra metre of fencing buys $5$ square metres of extra ground.

Put $x=8$ into the two differences separately, which is the only honest way to see what the ratio is being asked. The areas are $64$ and $64$, so the extra ground is $0$ square metres. The perimeters are $32$ and $32$, so the extra fencing is $0$ metres. The grower is asking how much ground is bought per metre of fencing when no fencing is bought at all, and that question has no answer:

$\frac{64-64}{32-32} = \frac{0}{0}$

A quotient is the unique number that multiplies the denominator back into the numerator. Here every real number multiplies $0$ into $0$, so uniqueness fails and nothing is picked out. This is not a gap in the arithmetic; it is the trade itself dissolving, because at $x=8$ there are not two paddocks to compare but one paddock described twice.

The simplified expression is untroubled, and it is worth being exact about why that is not a rescue. $\frac{x+8}{4}$ returns $\frac{16}{4} = 4$ at $x=8$. That number is the value of $\frac{x+8}{4}$, an expression which agrees with the ratio at every input the ratio accepts and is defined at one more besides. It is not the grower's trade at $x=8$, because there is no such trade. What it does describe is the trade at inputs near $8$, which the grower may well care about: a paddock of side $8.5$ metres buys about $4.1$ square metres per extra metre. But the value at $8$ itself has to be reported as no value, and the restriction $x \neq 8$ is what carries that fact once the factor $x-8$ has left the page.

Factor the first expression. Its numerator is a difference of squares, and its denominator needs two numbers with product $-24$ and sum $-2$, which are $-6$ and $4$:

$\frac{x^2-36}{x^2-2x-24} = \frac{(x-6)(x+6)}{(x-6)(x+4)}$

That denominator is zero at $x=6$ and at $x=-4$, so the first expression excludes both. Everywhere else $x-6$ is nonzero and divides out:

$\frac{(x-6)(x+6)}{(x-6)(x+4)} = \frac{x+6}{x+4} \qquad (x \neq 6 \text{ and } x \neq -4)$

The second printed expression is already a quotient of two linear polynomials. Its denominator is zero only at $x=-4$, so that is the only input it excludes. The two domains are therefore not the same, and they differ at exactly one number.

Two expressions in $x$ are the same when they accept the same inputs and return the same number at each of them. Both halves of that have to be checked, and here only one of them holds.

The values first. At every $x$ other than $6$ and $-4$, part A's cancelling is licensed, so the first expression returns exactly what the second does. There is no disagreement anywhere in that region.

The inputs second. At $x=6$ the first expression has denominator $36-12-24 = 0$ and numerator $36-36 = 0$, so it returns nothing, while the second returns

$\frac{6+6}{6+4} = \frac{12}{10} = \frac{6}{5}$

At $x=-4$ both refuse: the first because $(x-6)(x+4)$ vanishes there, the second because $x+4$ does.

So there is exactly one input, $x=6$, at which one expression answers and the other does not, and one input is enough. The second expression is defined on a strictly larger set, so the two are not the same expression, however closely their values agree on the smaller one. The honest statement is the equality with its condition attached: the first equals the second at every $x$ other than $6$ and $-4$.

Build the examples backwards. Start from a lowest-terms expression and multiply its numerator and its denominator by the same non-constant polynomial; the value is unchanged wherever everything is defined, but the new factor adds its own zeros to the denominator.

Multiplying top and bottom by $x-1$:

$\frac{(x+6)(x-1)}{(x+4)(x-1)}$

Its denominator vanishes at $-4$ and at $1$, so it excludes both, and dividing out $x-1$ returns $\frac{x+6}{x+4}$.

Multiplying top and bottom by $x$ instead:

$\frac{x(x+6)}{x(x+4)} = \frac{x^2+6x}{x^2+4x}$

Its denominator vanishes at $-4$ and at $0$, so it excludes those two, and it reduces to the same form.

The two originals have different domains and the same reduced form, so the reduced form cannot possibly determine the domain it came from. There are as many such originals as there are polynomials to multiply by.

What the reduced form does record is a floor, not the whole list. Its own denominator still refuses its own zeros, and those were excluded by every original too, since a surviving factor of the denominator was a factor of the original denominator. Everything above that floor is unrecoverable, which is exactly why the restrictions are written down at the moment of cancelling rather than reconstructed afterwards.

Let $x_0$ be any real number in the domain of the expression. By the definition of the domain, the denominator does not vanish there:

$B(x_0)\,d(x_0) \neq 0$

That is a statement about a product of two real numbers, $B(x_0)$ and $d(x_0)$. If either factor were zero the product would be zero, so neither is:

$d(x_0) \neq 0 \quad \text{and} \quad B(x_0) \neq 0$

Those are precisely the two hypotheses the cancellation law asks for, with $b = B(x_0)$ and $c = d(x_0)$, so the law applies at $x_0$ and gives

$\frac{A(x_0)\,d(x_0)}{B(x_0)\,d(x_0)} = \frac{A(x_0)}{B(x_0)}$

Since $x_0$ was an arbitrary member of the domain, the cancelling is licensed at every input the expression accepts, and there is no input in the domain where it has to be justified separately.

The same argument read backwards is the useful form: if $d$ did vanish at some $x_0$, then the denominator $B(x_0)\,d(x_0)$ would be zero as well, and $x_0$ would have been outside the domain from the start. The zeros of a cancelled factor were never allowed inputs.

Look at where the classmate's condition would have to come from. The printed denominator is $(3x+2)(x-4)$, and it is zero when either factor is:

$3x+2 = 0 \;\Rightarrow\; x = -\frac{2}{3}, \qquad x-4 = 0 \;\Rightarrow\; x = 4$

So the domain of the printed expression is every real number except $-\frac{2}{3}$ and $4$, and it was that before a single factor moved. The classmate's condition is one of those two, not a third: nothing has been added.

Part A says this can never go the other way. A cancelled factor is a factor of the denominator, so its zeros are zeros of the denominator, so they are already excluded. The cancelling hypothesis is therefore never new information about which inputs are allowed.

But the classmate is right about the writing, for a different reason than they give. After the cancelling the page reads $\frac{x+5}{x-4}$, which is perfectly happy at $x=-\frac{2}{3}$:

$\frac{-\frac{2}{3}+5}{-\frac{2}{3}-4} = \frac{13/3}{-14/3} = -\frac{13}{14}$

The printed expression has no value there, so the two are not the same object, and only the written restriction records the difference. The condition is old news about the domain and urgent news about the answer.

The restriction disappears from view when nothing left on the page objects to it, so to keep it visible the cancelled factor has to leave a copy of itself in the denominator. That happens when it appears more often below the bar than above it.

Take

$\frac{x^2+7x}{x^2} = \frac{x(x+7)}{x \cdot x}$

The printed denominator is zero only at $x=0$, so that is the only excluded input. The factor $x$ is common to both halves and divides out once:

$\frac{x(x+7)}{x \cdot x} = \frac{x+7}{x} \qquad (x \neq 0)$

The reduced form still has $x$ underneath, so it refuses $x=0$ on its own account. Every excluded input of the printed expression is still refused by the answer, and a reader who never saw the question would rule out exactly the right number.

The general statement behind the example is a comparison of counts. If a factor appears $m$ times above the bar and $n$ times below, cancelling leaves $n-m$ copies below when $n > m$, so any restriction that factor imposed is still enforced by the answer itself. When $m \ge n$ no copy survives, and a restriction it imposed is left with nothing on the page to enforce it. In this expression $m=1$ and $n=2$.

So the claim is false as stated, and the true version needs two conditions rather than one. A cancelled factor costs a restriction its visibility only if it had a real zero to forbid in the first place, and the cancelling exhausted it from the denominator. Fail either condition and nothing is lost: $\frac{x^2+7x}{x^2}$ fails the second, and $\frac{(x^2+9)(x+1)}{(x^2+9)(x-2)}$ fails the first, since $x^2+9$ has no real zero and so forbade nothing for the answer to forget.