The Rational Root Theorem: Free Response

The theorem reads two coefficients and nothing else. The constant term is $10$, and its positive divisors are the possible numerators:

$p \in \{1, 2, 5, 10\}$

The leading coefficient is $3$, and its positive divisors are the possible denominators:

$q \in \{1, 3\}$

Form every quotient $\frac{p}{q}$ and reduce it. Nothing reduces here, because no divisor of $10$ shares a factor with $3$, so all eight quotients are already in lowest terms and all eight are different:

$1, \ 2, \ 5, \ 10, \ \frac{1}{3}, \ \frac{2}{3}, \ \frac{5}{3}, \ \frac{10}{3}$

Each carries both signs, since neither divisibility condition cares about sign, so the finished list holds $16$ numbers. That is the whole service the theorem performs: infinitely many rational numbers replaced by sixteen, at the cost of no evaluation of $P$ at all.

Two candidates are cheaper than all the others, so spend those first. At $x = 1$ the value is simply the sum of the coefficients, and at $x = -1$ the signs alternate:

$P(1) = 3 + 8 - 33 + 10 = -12, \qquad P(-1) = -3 + 8 + 33 + 10 = 48$

Neither is zero. Take the next smallest integer candidate, $x = 2$:

$P(2) = 24 + 32 - 66 + 10 = 0$

That is a root, so divide it out. Synthetic division by $2$ on the coefficients $3, 8, -33, 10$ brings down the $3$, then $8 + 6 = 14$, then $-33 + 28 = -5$, then $10 - 10 = 0$:

$P(x) = (x - 2)\left(3x^2 + 14x - 5\right)$

The remainder $0$ confirms the root, and the quotient is a quadratic, which ends the hunt: no further candidate needs testing, because a quadratic can be finished outright. This one factors,

$3x^2 + 14x - 5 = (3x - 1)(x + 5),$

as expanding confirms. So $P(x) = (x - 2)(3x - 1)(x + 5)$, and a product is zero exactly when one of its factors is, so the roots are $2$, $\frac{1}{3}$ and $-5$.

Three evaluations settled a list of sixteen, and two of the three roots were never tested on their own: the factorization handed them over. Three is the cost of this particular order, not a target. Anyone who defended starting at $2$ pays one evaluation, and anyone who worked down the list from $-10$ pays a great many more. What the order has to be is defensible, not short.

The two divisibility conditions name exactly two coefficients: the numerator divides the constant term and the denominator divides the leading coefficient. No other coefficient appears anywhere in the statement. $P$ and $Q$ both have constant term $10$ and leading coefficient $3$, so the conditions read the same two numbers in both cases and produce the same sixteen candidates. Nothing in between was ever consulted.

Now evaluate:

$Q(10) = 3000 - 3200 + 190 + 10 = 0, \qquad P(10) = 3000 + 800 - 330 + 10 = 3480$

So $10$ is a root of $Q$ and is not a root of $P$, while sitting on the one list both polynomials share.

That fixes the status of a candidate list. It is built from two coefficients, and two coefficients cannot determine the roots, because the middle ones can be changed freely: they move the roots and leave the list untouched. What the list reports is where a rational root is permitted to sit. Which entries are roots is a separate question that only evaluation answers. The useful half is worth stating in the same breath: the list is complete, so any rational number absent from it is ruled out for both polynomials at once.

Take a rational root and write it as $\frac{p}{q}$ in lowest terms with $q \ge 1$, which is always possible: every rational number has such a form. The theorem then supplies two facts, and here it is the second that bites. Since the leading coefficient is $1$,

$q \mid 1$

A positive integer dividing $1$ has nowhere to go, because $1$ has exactly one positive divisor. So $q = 1$ and the root is

$\frac{p}{q} = \frac{p}{1} = p,$

an integer. The theorem's first condition then applies to that integer unchanged, giving $p \mid a_0$: the root divides the constant term. Both halves of the claim are out.

The step that leans on the hypothesis is the one that reads the divisors of the leading coefficient. Replace $1$ by $6$ and the condition becomes $q \mid 6$, which allows $q = 1$, $2$, $3$ or $6$. Three of those four leave a genuine fraction standing, so the conclusion collapses: nothing then forces a rational root to be an integer. The strength of the monic case is not that fractions are hard to find there. It is that the divisibility condition has already excluded every one of them.

The polynomial is monic with integer coefficients, so part A applies: every rational root is an integer dividing the constant term $-2$. The positive divisors of $2$ are $1$ and $2$, so the complete list is

$\pm 1, \quad \pm 2$

Four numbers, and not a fraction among them. That is the monic case paying for itself: without it, every denominator dividing the leading coefficient would have had to be checked as well, and the list would have been longer for no extra information.

Evaluate all four:

$P(1) = 4, \quad P(-1) = 2, \quad P(2) = 26, \quad P(-2) = 10$

Not one is zero. Every rational root of $P$ would have had to appear among those four, so $P$ has no rational root at all.

Now be exact about the size of that conclusion, because this is the point at which the theorem is most often overread. It says that no rational number is a root of $P$. It does not say that $P$ has no roots. It does not describe, locate or approximate any root that $P$ does have. The candidate list was never able to hold a number that is not a fraction, so its silence about such numbers is not evidence about them; it is the shape of the tool. An honest summary of where four evaluations leave you is this: $P$ has no rational roots, and from here that is all I know.

Turn the condition on $t$ into a polynomial equation by moving everything to one side:

$t^3 - 7 = 0$

So $t$ is a root of $P(x) = x^3 - 7$, which has integer coefficients and is monic. By part A, any rational root of $P$ is an integer dividing $-7$, so the entire candidate list is

$\pm 1, \quad \pm 7$

Evaluate all four:

$P(1) = -6, \quad P(-1) = -8, \quad P(7) = 336, \quad P(-7) = -350$

None is zero, so $P$ has no rational root.

The last step is the one worth writing out slowly, because it is where a fact about a polynomial becomes a fact about a number. $P$ has no rational root. $t$ is a root of $P$. If $t$ were rational it would be a rational root of $P$, and $P$ has none. So $t$ is not rational.

Nothing in that argument was special to $7$ or to cubes. Run the same four lines on $x^k - n$ and you get this: if no integer raised to the power $k$ gives $n$, then no fraction does either.

The requirement is that the concentration equals $20$, which is the equation

$2t^3 - 17t^2 + 38t = 20.$

The theorem reads the coefficients of a polynomial set equal to zero, so move the $20$ across:

$2t^3 - 17t^2 + 38t - 20 = 0$

That is standard form, and every coefficient is an integer, which is the hypothesis the theorem needs.

The interval comes from the situation, not from the algebra. The dose is given at time zero and the model is stated to hold for the first four hours only, so $0 < t < 4$. A solution of the equation outside that window is a number the model does not vouch for.

Now the constant terms, because the difference between them is the whole reason this equation is workable. Written as it stands, $C(t) = 2t^3 - 17t^2 + 38t$ has constant term $0$. Applied to that, the theorem would say that a rational root's numerator divides $0$, which every integer does, so it would filter nothing at all. Subtracting the $20$ is what puts a nonzero constant term in place for the theorem to grip.

The equation is $2t^3 - 17t^2 + 38t - 20 = 0$. Its constant term is $-20$, with positive divisors $1, 2, 4, 5, 10, 20$, and its leading coefficient is $2$, with positive divisors $1, 2$. Halving the even divisors of $20$ returns numbers already on the list, so after reducing and discarding duplicates the distinct positive quotients are

$1, \ 2, \ 4, \ 5, \ 10, \ 20, \ \frac{1}{2}, \ \frac{5}{2}$

Eight of them, so $16$ candidates once both signs are attached.

The window now does most of the work, and it does it before any arithmetic. A negative time is before the dose, and the model says nothing at or beyond four hours, so every candidate except four is gone:

$\frac{1}{2}, \ 1, \ 2, \ \frac{5}{2}$

Twelve of the sixteen died to the situation rather than to a calculation. Test the four survivors, integers first:

$C(1) - 20 = 2 - 17 + 38 - 20 = 3, \qquad C(2) - 20 = 16 - 68 + 76 - 20 = 4$

$C\left(\tfrac{1}{2}\right) - 20 = \tfrac{1}{4} - \tfrac{17}{4} + 19 - 20 = -5$

Three failures. Test the last survivor, $t = \frac{5}{2}$:

$2\left(\frac{125}{8}\right) - 17\left(\frac{25}{4}\right) + 38\left(\frac{5}{2}\right) - 20 = 31.25 - 106.25 + 95 - 20 = 0$

So the concentration is exactly $20$ milligrams per litre at $t = \frac{5}{2}$ hours, two and a half hours after the dose. Reading that back into the model rather than into the algebra: $C\left(\frac{5}{2}\right) = 31.25 - 106.25 + 95 = 20$, which is the required concentration in the required units.

Divide $2t^3 - 17t^2 + 38t - 20$ by $t - \frac{5}{2}$. Synthetic division on the coefficients $2, -17, 38, -20$ with corner $\frac{5}{2}$ gives $2$, then $-17 + 5 = -12$, then $38 - 30 = 8$, then $-20 + 20 = 0$:

$2t^3 - 17t^2 + 38t - 20 = \left(t - \frac{5}{2}\right)\left(2t^2 - 12t + 8\right)$

The hunt stops here, because a quadratic can be solved outright. Take the factor of $2$ out first and run the quadratic formula on $t^2 - 6t + 4$:

$t = \frac{6 \pm \sqrt{36 - 16}}{2} = \frac{6 \pm 2\sqrt{5}}{2} = 3 \pm \sqrt{5}$

Since $\sqrt{5}$ is a little over $2.23$, the two values are about $0.76$ and about $5.24$. Test them against the window $0 < t < 4$. The larger is past the four hours the model covers, so it is arithmetic the situation does not vouch for. The smaller is a genuine second time, and it survives a direct check: at $t \approx 0.7639$,

$2(0.7639)^3 - 17(0.7639)^2 + 38(0.7639) \approx 20$

So the concentration passes through $20$ milligrams per litre twice inside the window, once on the way up at about $0.76$ hours and once on the way down at $\frac{5}{2}$ hours, and only the second was ever within reach of the Rational Root Theorem. That time is rational and sat on the candidate list from the start. The time $3 - \sqrt{5}$ is irrational, so no list of fractions could have contained it, however long that list ran. What found it was the quadratic formula, working on what was left once the rational time had been divided out. That is the honest division of labour: the theorem supplies the rational candidates, testing decides which of them are roots, and about every other kind of root it says nothing at all.

Hold each number against the two divisibility conditions before evaluating anything. The constant term is $12$ and the leading coefficient is $2$.

For $3$, which is $\frac{3}{1}$ in lowest terms, the numerator $3$ divides $12$ and the denominator $1$ divides $2$. Both conditions hold, so the theorem does not exclude $3$. Nor does it endorse it. The theorem has said all it can and the question is still open, so there is nothing for it but arithmetic:

$R(3) = 54 - 45 - 12 + 12 = 9$

That is not zero, so $3$ is not a root, and only the evaluation could have told you.

For $\frac{4}{3}$, already in lowest terms, the denominator is $3$, and $3$ does not divide $2$. The theorem says every rational root has a denominator dividing the leading coefficient, so $\frac{4}{3}$ cannot be a root. No evaluation is needed. Substituting would give the same verdict and would spend exactly the effort the theorem exists to save.

The asymmetry between the two cases is the point. Failing a divisibility condition settles the matter outright. Passing both settles nothing.

A claim of the form "the roots are the numbers on the list" packs two implications into one sentence, and they have to be pulled apart before either can be judged.

The first is: if a rational number is a root of $R$, then it is on the candidate list. That is the theorem restated. It is proved, and it is the useful direction, because it is what lets a list of sixteen numbers stand in for every fraction there is.

The second is the converse: if a rational number is on the candidate list of $R$, then it is a root of $R$. One counterexample sinks it, and part A supplies one. The number $3$ is on the list, since $3$ divides $12$ and $1$ divides $2$, and

$R(3) = 9 \ne 0$

so it is not a root. The converse is false.

This is not a near miss. The list for $R$ holds sixteen numbers, and a cubic cannot have sixteen roots, so most entries on it are not roots whatever the polynomial turns out to factor into. That is what makes candidate the right word: the list names suspects, and evaluation is what convicts.

Priya's summary asserts both directions, so it claims strictly more than the theorem proves. Repaired, it reads: every rational root is on the list, and no rational number off the list is a root. Both halves of that are the one implication, read forwards and by contraposition.

The constant term is $3$, with positive divisors $1$ and $3$; the leading coefficient is $2$, with positive divisors $1$ and $2$. The candidate list is therefore

$\pm 1, \quad \pm 3, \quad \pm\frac{1}{2}, \quad \pm\frac{3}{2}$

Eight numbers, and a claim about the whole list needs the whole list evaluated:

$S(1) = 2, \quad S(-1) = 4, \quad S(3) = 48, \quad S(-3) = -42$

$S\left(\tfrac{1}{2}\right) = \tfrac{7}{4}, \quad S\left(-\tfrac{1}{2}\right) = \tfrac{17}{4}, \quad S\left(\tfrac{3}{2}\right) = \tfrac{21}{4}, \quad S\left(-\tfrac{3}{2}\right) = \tfrac{3}{4}$

Not one is zero. Every rational root of $S$ would have had to sit on that list, so $S$ has no rational root. Notice which direction is being used: this is the theorem, not its converse, and it is exactly what licenses the leap from eight failed evaluations to a statement about every fraction.

What it does not license is anything beyond that. "No rational root" is not "no root". The candidate list was never capable of holding a number that is not a fraction, so its silence about such numbers reports nothing whatever.

Now look again at the numbers already on the page, because the sweep produced more than a verdict. It was run to test divisibility candidates, but each test also produced a value, and two of those values have opposite signs:

$S(-3) = -42, \qquad S(-1) = 4$

A polynomial runs unbroken from one value to the next, so $S$ crosses zero somewhere between $-3$ and $-1$. That crossing is a real root, and since no rational number is a root of $S$, it is irrational. Notice which half of the work did that. The divisibility conditions cannot locate anything; they only ever rule fractions out. The evaluations located it, and you had already run them.

One more evaluation sharpens the interval rather than establishing anything new:

$S(-2) = -16 + 6 + 3 = -7$

Since $S(-2) < 0 < S(-1)$, the crossing lies between $-2$ and $-1$, a narrower trap than before. What none of this supplies is the number itself. The conditions say which fractions to rule out, the values say roughly where to look, and pinning the root down exactly is a job for neither.

Substituting $x = \frac{p}{q}$ into $P(x) = 0$ gives an equation cluttered with denominators:

$a\frac{p^3}{q^3} + b\frac{p^2}{q^2} + c\frac{p}{q} + d = 0$

Divisibility is a statement about integers, so the first move is to remove the fractions. Multiply every term by $q^3$, which is not zero. Each term picks up exactly the powers of $q$ it was missing:

$ap^3 + bp^2q + cpq^2 + dq^3 = 0$

Every term here is an integer, because $a$, $b$, $c$, $d$, $p$ and $q$ all are. Now isolate the single term carrying no factor of $p$, which is the last one, and pull $p$ out of everything else:

$dq^3 = -\left(ap^3 + bp^2q + cpq^2\right) = -p\left(ap^2 + bpq + cq^2\right)$

The bracket on the right is an integer, so $p$ divides $dq^3$.

One case has to be set aside before the next step, because that step compares prime factorizations and $0$ has none. If $d = 0$ then $p \mid d$ holds for every integer $p$, since $0 = p \times 0$, and there is nothing left to prove. So suppose from here that $d \ne 0$.

The final step is where the lowest-terms hypothesis earns its place. $p$ and $q$ have no prime factor in common, and the primes of $q^3$ are exactly the primes of $q$, so $p$ and $q^3$ have no prime factor in common either. A prime power sitting inside $p$ therefore cannot be accounted for by $q^3$, and must sit inside $d$. That holds for every prime power in $p$, so the whole of $p$ divides $d$:

$p \mid d$

Without coprimality the step simply fails, because $p$ would be free to hide inside $q^3$ instead.

The cleared equation is symmetric in a useful way. Written out again,

$ap^3 + bp^2q + cpq^2 + dq^3 = 0,$

every term after the first carries at least one factor of $q$, exactly as every term before the last carried at least one factor of $p$. So run the same manoeuvre from the other end. Isolate the first term and pull $q$ out of the rest:

$ap^3 = -\left(bp^2q + cpq^2 + dq^3\right) = -q\left(bp^2 + cpq + dq^2\right)$

The bracket is an integer, so $q$ divides $ap^3$.

The finish is the part A argument with $p$ and $q$ swapped. The two share no prime factor, so $q$ shares no prime factor with $p^3$, no prime power of $q$ can be accounted for by $p^3$, and each must sit inside $a$. Hence

$q \mid a$

Nothing else changed: the same cleared equation, the same coprimality, the same reasoning about where prime factors are allowed to live. That is why the theorem's two conditions look like mirror images. They are two readings of one equation, taken from its two ends.

It also explains something the rule alone never explains, which is why the middle coefficients appear in neither conclusion. Their terms carry a factor of $p$ and a factor of $q$ at once, so in both calculations they are swept into the bracket and never end up alone on the left.

Withdrawing lowest terms. The polynomial $4x^2 - 1$ has integer coefficients, and $\frac{1}{2}$ is genuinely a root:

$4\left(\frac{1}{2}\right)^2 - 1 = 1 - 1 = 0$

Written as $\frac{1}{2}$ the conclusions hold: the numerator $1$ divides the constant term $-1$, and the denominator $2$ divides the leading coefficient $4$. Now write the same number as $\frac{2}{4}$, so the numerator is $2$ and the denominator is $4$. The first conclusion demands that $2$ divide $-1$, and it does not. The number is still a root; all that changed was how it was written. Part A shows exactly where the damage is done: with a common factor of $2$ shared by numerator and denominator, that factor can be accounted for by the power of the denominator instead of by the constant term, and the divisibility never reaches $d$ at all.

Withdrawing integer coefficients. The polynomial $x^2 - \frac{5}{2}x + 1$ has $2$ as a root:

$4 - 5 + 1 = 0$

Here $2$ is $\frac{2}{1}$ in lowest terms, so the fraction is written correctly and the numerator is $2$. The conclusion demands that $2$ divide the constant term $1$, and it does not. The failure is upstream this time: the proof multiplied through by $q^3$ and declared every term an integer, which needs integer coefficients, and this polynomial has none such in the middle.

The repair is the standard one. Multiply the whole polynomial by $2$:

$2x^2 - 5x + 2$

Multiplying an equation by a nonzero constant moves no root, so $2$ is still a root, and the coefficients are now integers. Test the conclusions again: the numerator $2$ divides the constant term $2$, and the denominator $1$ divides the leading coefficient $2$. Both hold. Clearing fractional coefficients before applying the theorem is not housekeeping. It is what makes the theorem apply at all.