Vieta's Formulas: Free Response

Multiply the given cubic by $(x - r_4)$ and distribute. The $x$ reproduces the cubic one degree higher, and the $-r_4$ multiplies every term of the cubic by $-r_4$. Then collect like powers, one power at a time.

The coefficient of $x^3$ collects $-(r_1 + r_2 + r_3)$ from the first product and $-r_4$ from the second:

$-(r_1 + r_2 + r_3) - r_4 = -(r_1 + r_2 + r_3 + r_4)$

So it is the sum of all four roots, negated.

The coefficient of $x^2$ collects the cubic's pair sum, and $r_4$ times the cubic's root sum:

$(r_1r_2 + r_1r_3 + r_2r_3) + r_4(r_1 + r_2 + r_3)$

Multiplying the second bracket out turns it into $r_1r_4 + r_2r_4 + r_3r_4$, so the coefficient has six terms in all. Those six are exactly the ways to choose two of the four roots: three pairs that avoid $r_4$, which the cubic already had, and three that use it. Nothing is repeated and nothing is left out.

The coefficient of $x$ collects $-r_1r_2r_3$ and $-r_4$ times the cubic's pair sum:

$-r_1r_2r_3 - (r_1r_2r_4 + r_1r_3r_4 + r_2r_3r_4)$

Worth pausing here: the two contributions did not cancel, they combined, and both arrived negative. A term that uses three roots drags three minus signs out of three factors, and three minus signs multiply to a minus.

The constant term is the product of the four chosen roots, $(-r_1)(-r_2)(-r_3)(-r_4) = r_1r_2r_3r_4$, positive because four minus signs cancel in pairs.

Writing the four coefficients with the names the lesson gives them:

$(x - r_1)(x - r_2)(x - r_3)(x - r_4) = x^4 - e_1x^3 + e_2x^2 - e_3x + e_4$

Nothing was used but the distributive law, and the alternating sign is not a convention anyone chose. It counts how many roots each coefficient collects.

The factor theorem turns each of the four roots into a factor, and the Fundamental Theorem of Algebra says there are exactly four of them to turn, so $p$ splits completely:

$p(x) = a_4(x - r_1)(x - r_2)(x - r_3)(x - r_4)$

The identity from part A describes a product of four bare factors, whose leading coefficient is $1$. This $p$ has leading coefficient $a_4$, so the two are not yet comparable. Divide by $a_4$, which is legal because $a_4 \ne 0$, and both sides become monic:

$x^4 + \frac{a_3}{a_4}x^3 + \frac{a_2}{a_4}x^2 + \frac{a_1}{a_4}x + \frac{a_0}{a_4} = x^4 - e_1x^3 + e_2x^2 - e_3x + e_4$

Equal polynomials have equal coefficients, so read the four positions off in turn:

$e_1 = -\frac{a_3}{a_4}, \qquad e_2 = \frac{a_2}{a_4}, \qquad e_3 = -\frac{a_1}{a_4}, \qquad e_4 = \frac{a_0}{a_4}$

The division is the step that introduces $a_4$, and it is the step most often dropped, because for a monic polynomial it does nothing visible. Skip it and you are comparing $p$ with a monic product, which reports $a_4e_k$ where $e_k$ was wanted: every one of the four values comes out multiplied by the leading coefficient. On a quartic with leading coefficient $5$ that is an answer five times the true one, with nothing in its appearance to give it away.

First see why the real numbers are no help. For any real $x$ we have $x^2 + 4 \ge 4$, and $x^2 - 4x + 13 = (x - 2)^2 + 9 \ge 9$, so the product of the two is at least $36$ and never zero. The quartic has no real root at all.

That matters because the formulas are not statements about whichever roots happen to be convenient. They are statements about a list of four numbers whose product of linear factors rebuilds the polynomial. Over the real numbers there is no such list here: it would be empty, and an empty list adds to $0$, while the coefficients insist that $e_1 = -\frac{-4}{1} = 4$. The formula is not wrong. It was being handed the wrong list.

The Fundamental Theorem of Algebra supplies the right one, over the complex numbers. Solve each factor:

$x^2 + 4 = 0 \implies x = \pm 2i$

$x^2 - 4x + 13 = 0 \implies x = \frac{4 \pm \sqrt{16 - 52}}{2} = \frac{4 \pm 6i}{2} = 2 \pm 3i$

So the list is $2i$, $-2i$, $2 + 3i$, $2 - 3i$: four entries, arriving as two conjugate pairs, which is what real coefficients force.

Check the first formula. The imaginary parts cancel in pairs:

$2i + (-2i) + (2 + 3i) + (2 - 3i) = 4 = -\frac{a_3}{a_4}$

Check the last. Each conjugate pair has a real product, $(2i)(-2i) = 4$ and $(2 + 3i)(2 - 3i) = 4 + 9 = 13$:

$(2i)(-2i)(2 + 3i)(2 - 3i) = 4 \times 13 = 52 = \frac{a_0}{a_4}$

Both agree. The coefficients are real, every root is not, and the formulas never noticed, because what they were promised was a complete list and a complete list is what they were given.

Write the coefficients out first, including the position that has no term. An absent power is not a gap in the polynomial; it is a coefficient equal to zero:

$a_4 = 4, \quad a_3 = 0, \quad a_2 = -6, \quad a_1 = 3, \quad a_0 = 2$

Now apply $e_k = (-1)^k\frac{a_{4-k}}{a_4}$ four times, dividing by $4$ every time and reading the parity of $k$ every time:

$e_1 = -\frac{0}{4} = 0, \qquad e_2 = \frac{-6}{4} = -\frac{3}{2}$

$e_3 = -\frac{3}{4}, \qquad e_4 = \frac{2}{4} = \frac{1}{2}$

The absent $x^3$ term did not excuse you from the first formula. It supplied the numerator, and the numerator was $0$.

That is worth reading back into the roots, because it is a constraint and not an absence. $e_1$ is the sum of all four roots, so this quartic's four roots add to exactly $0$. Whatever they are, and none of them is obvious from the coefficients, they balance. Compare that with $e_2 = -\frac{3}{2}$, which is a genuine number rather than a coincidence of missing ink: both were read the same way, from the coefficient sitting $k$ places down from the leading one and a division by that leading $4$.

The factorization $2(x + 1)^2(x - 3)$ names the root $-1$ with multiplicity two and the root $3$ with multiplicity one, so the complete list has three entries:

$-1, \quad -1, \quad 3$

From that list, treating the two copies of $-1$ as two separate roots that happen to be equal:

$e_1 = -1 + (-1) + 3 = 1$

$e_2 = (-1)(-1) + (-1)(3) + (-1)(3) = 1 - 3 - 3 = -5$

$e_3 = (-1)(-1)(3) = 3$

From the coefficients, with $a_3 = 2$, $a_2 = -2$, $a_1 = -10$ and $a_0 = -6$:

$e_1 = -\frac{-2}{2} = 1, \qquad e_2 = \frac{-10}{2} = -5, \qquad e_3 = -\frac{-6}{2} = 3$

The two routes agree in all three places, and note that they had to agree by different-looking arithmetic: one route multiplied roots, the other divided coefficients by $2$.

Now shorten the list to one entry per distinct value, $-1$ and $3$, and run the first route again. The sum becomes $-1 + 3 = 2$. There is now only one pair, so the pair sum becomes $(-1)(3) = -3$. And the product of everything on the list becomes $(-1)(3) = -3$.

So the shortened list reports $2$, $-3$, $-3$ where the truth is $1$, $-5$, $3$. Every one of the three is wrong, and the third has even changed sign. The place where the damage is easiest to see is $e_2$: the full list contributes the term $(-1)(-1)$, the product of the two copies of the repeated root, and that term simply has nowhere to come from once one copy has been thrown away. Multiplicity is not a bookkeeping nicety here. It is the difference between a true statement and a false one.

Run the test as it stands. The polynomial is $(x^2 - 9)^2$, so its root list is $3$, $3$, $-3$, $-3$, and the coefficients give

$e_1 = -\frac{0}{1} = 0$

since there is no $x^3$ term. The distinct values are $3$ and $-3$, and they add to $0$ as well. The two readings agree, so the test returns a verdict of no repeated root, and it is wrong four times over: every entry on that list is one of a repeated pair.

Now ask why it failed, because the failure is not bad luck. $e_1$ is a single number, and it is being asked to report on a list of four entries. The two extra copies changed the total by $3 + (-3) = 0$, so the very arithmetic meant to detect them destroyed the evidence. Any test built on a symmetric total can be defeated the same way: arrange the repeats so that what they contribute to that particular total is nothing. A total compresses a list into one number, and a repeat is a fact about the list that the compression is free to discard.

What a test for a repeated root has to compare is counts, not totals. The Fundamental Theorem of Algebra fixes the length of the complete list at $n$, the degree, and a repeated root is exactly the situation where fewer than $n$ distinct values are spread over those $n$ places. So the honest comparison is $n$ against the number of distinct values, and both sides of it are counts, neither of which any symmetric sum reports.

One half of the classmate's test does survive, and it is worth keeping. If the two readings disagree there must be a repeat, because with no repeats the root list holds each value exactly once, so the two readings are sums of the very same collection and cannot differ. A disagreement is therefore real evidence. An agreement is no evidence at all, and the better habit is not to depend on either: the shortened list is not an approximation of the true one, so build the complete list with multiplicity from the start and there is nothing left to detect.

Turn each figure into an equation. A rectangular box has twelve edges, four of each length, so the edging is

$4(x + y + z) = 52$

It has six faces in three matching pairs, so the glass is

$2(xy + xz + yz) = 108$

And the capacity is the product of the three lengths:

$xyz = 72$

Look at what the left sides are. They are the sum of the three lengths, the sum of their three pairwise products, and their product: precisely $e_1$, $e_2$ and $e_3$ for the list $x, y, z$. That is why the order sheet can be read at all. Every figure on it is symmetric, unchanged if two sides are swapped, which is exactly the kind of quantity these formulas handle.

Divide the first equation by $4$ and the second by $2$:

$e_1 = 13, \qquad e_2 = 54, \qquad e_3 = 72$

Now run the formulas backwards. A monic cubic with roots $x$, $y$, $z$ is $t^3 - e_1t^2 + e_2t - e_3$, with the signs alternating, so

$t^3 - 13t^2 + 54t - 72$

The three side lengths are the roots of that cubic. Nothing has been solved yet; what has happened is that three equations in three unknowns, two of them not linear, have become one polynomial in one unknown.

The cubic $t^3 - 13t^2 + 54t - 72$ is monic with whole-number coefficients, so any rational root is a whole number dividing $72$. Start at the small divisors, since a side length of a case this size is unlikely to be large:

$1 - 13 + 54 - 72 = -30, \qquad 8 - 52 + 108 - 72 = -8$

Neither $1$ nor $2$ is a root. Try $3$:

$27 - 117 + 162 - 72 = 0$

So $3$ is a root, and the hunt is nearly over: dividing it out leaves a quadratic, which can be finished outright. Synthetic division by $3$ on the coefficients $1, -13, 54, -72$ brings down the $1$, then $-13 + 3 = -10$, then $54 - 30 = 24$, then $-72 + 72 = 0$:

$t^3 - 13t^2 + 54t - 72 = (t - 3)(t^2 - 10t + 24)$

The remainder of $0$ confirms the root, and the quadratic factors as $(t - 4)(t - 6)$, as multiplying it back out confirms. So the three roots are $3$, $4$ and $6$, and those are the three side lengths in centimetres.

Check them against the whole order sheet rather than one line of it. Edging: $4(3 + 4 + 6) = 4 \times 13 = 52$ cm. Glass: $2(12 + 18 + 24) = 2 \times 54 = 108$ square centimetres. Capacity:

$3 \times 4 \times 6 = 72 \text{ cubic centimetres}$

All three figures are accounted for, which is the check worth running: the cubic was built from all three, so an error in any one of them would have shown up here.

Set the second sheet up exactly as the first. Dividing the edging by $4$ and the glass by $2$:

$e_1 = 7, \qquad e_2 = 11, \qquad e_3 = 4$

so the cubic whose roots are the three side lengths is

$t^3 - 7t^2 + 11t - 4$

It is monic with whole-number coefficients, so a rational root has to be a whole number dividing $4$, and the entire candidate list is $\pm 1$, $\pm 2$ and $\pm 4$. The three negative entries are already ruled out by the situation, since a side length is positive, but test all six so that the verdict rests on the complete list:

$1 - 7 + 11 - 4 = 1, \qquad 8 - 28 + 22 - 4 = -2$

$64 - 112 + 44 - 4 = -8$

and the three negative candidates give $-23$, $-62$ and $-224$. Not one of the six is zero, and every rational root of the cubic would have had to appear among them. So the cubic has no rational root at all, and therefore no side length meeting this sheet is a rational number of centimetres. Nothing the framer can set out on that scale will fit, and the failure is exact rather than a matter of tolerance: it is not that the numbers come out awkward, it is that no fraction whatever is a solution.

Compare that with the first sheet, where the same sweep found a root at the third candidate it tried. The two sheets were read the same way and the machinery behaved identically. All that differs is what the candidate list happened to contain.

Now be exact about the size of the conclusion, because this is where a sweep is most often overread. A candidate list is a statement about fractions and about nothing else. It says no rational number is a side length here. It does not say the three figures are impossible, and it does not locate, approximate or describe any length that might meet them. The Fundamental Theorem of Algebra still puts three roots on the list, and Vieta's formulas still describe them: what those roots are, and whether they are real, is a question this argument never asked and cannot answer.

Each factorization hands over its roots, so the true product needs no formula at all, which is what makes these two cases a fair test of one.

The cubic has roots $1$, $-2$ and $5$, so their product is

$1 \times (-2) \times 5 = -10$

The rule predicts $\frac{a_0}{a_n} = \frac{20}{2} = 10$. Right size, wrong sign. The rule fails here.

The quartic has roots $1$, $-1$, $2$ and $-3$, so their product is

$1 \times (-1) \times 2 \times (-3) = 6$

The rule predicts $\frac{a_0}{a_n} = \frac{12}{2} = 6$. It succeeds here.

So the rule is not simply wrong, which is what makes it durable: one of these two tests would have confirmed it. Now notice how little separates the two polynomials. Both have leading coefficient $2$, so that cannot be what decides the verdict, and dividing by it was done the same way in both. The one thing that differs is the number of roots. The product of all the roots is the symmetric sum that collects every one of them, so it drags $n$ minus signs out of the $n$ factors, and $n$ minus signs multiply to $(-1)^n$: minus for the cubic, plus for the quartic. The corrected rule carries that factor:

$r_1r_2\cdots r_n = (-1)^n\frac{a_0}{a_n}$

Check it on both. For the cubic, $(-1)^3\frac{20}{2} = -10$, matching. For the quartic, $(-1)^4\frac{12}{2} = 6$, matching too.

Choose the value of $x$ that kills everything but the constant term. In standard form every term except the last carries a factor of $x$, so

$p(0) = a_0$

Evaluate the factored form at the same place. Each factor $(x - r_i)$ becomes $-r_i$:

$p(0) = a_n(-r_1)(-r_2)\cdots(-r_n)$

There are $n$ factors and each contributes one minus sign, so pull all $n$ of them out at once:

$p(0) = (-1)^na_nr_1r_2\cdots r_n$

The two evaluations are of the same polynomial at the same point, so they are equal:

$a_0 = (-1)^na_nr_1r_2\cdots r_n$

Multiply both sides by $(-1)^n$, which is legal because $(-1)^n(-1)^n = 1$, and divide by $a_n$, which is legal because $a_n \ne 0$:

$r_1r_2\cdots r_n = (-1)^n\frac{a_0}{a_n}$

That is the corrected rule, for every degree at once, and the whole proof was one substitution.

Now the diagnosis. The sign is not decided by the degree as such; it is decided by how many roots the term collects, and the constant term happens to collect all of them, which is why the degree ends up controlling it here. A quadratic has two roots, two minus signs come out, and two minus signs cancel. So on quadratics the original rule and the corrected rule give the same answer every single time, and a student could test it on a hundred of them without ever seeing a discrepancy. The rule is not half remembered. It is a true statement about even degrees that was quietly promoted to a statement about all of them.

An "exactly when" is two claims, and here they have very different fates.

Forwards, if $n$ is even then $(-1)^n = 1$, so part B reads $r_1r_2\cdots r_n = \frac{a_0}{a_n}$ directly. That half is sound.

Backwards, the claim is that equality forces the degree to be even, and one polynomial breaks it. Take

$x^3 + 2x^2 - 8x = x(x + 4)(x - 2)$

Its roots are $0$, $-4$ and $2$, so their product is $0$. Its constant term is $0$ and its leading coefficient is $1$, so $\frac{a_0}{a_n} = 0$ as well. The two sides agree, and the degree is $3$, which is odd. The reverse direction is false.

What happened is that both sides were nothing at all, so the sign in front of one of them had nothing to act on. Chasing that down settles the general case too. At odd degree part B gives $r_1r_2\cdots r_n = -\frac{a_0}{a_n}$, and this can equal $\frac{a_0}{a_n}$ only when

$\frac{a_0}{a_n} = -\frac{a_0}{a_n}, \quad \text{that is, when} \quad a_0 = 0$

So the only odd-degree escapes are the polynomials with constant term $0$, which are exactly the polynomials having $0$ as a root, since $p(0) = a_0$.

Add the condition $a_0 \ne 0$ and the repaired claim holds both ways. Forwards is unchanged. Backwards, equality plus $a_0 \ne 0$ forces $(-1)^n = 1$, so $n$ is even. Both directions, tested separately, which is the only way an "exactly when" can honestly be checked.

There is a second repair, and it is worth having because it throws nothing away. Rather than excluding the awkward case, absorb it: equality holds exactly when $n$ is even OR $a_0 = 0$. Both halves of that are already proved above. Forwards, an even degree gives equality by part B, and $a_0 = 0$ gives it by making both sides $0$ whatever the degree. Backwards, equality means $\frac{a_0}{a_n}\bigl((-1)^n - 1\bigr) = 0$, so one of the two factors vanishes, which is exactly the two cases named. Either repair is correct; the first narrows the claim and the second widens it.