The Fundamental Theorem of Algebra: Free Response

Read the degree off the factored form. Degrees add when polynomials are multiplied, and an exponent multiplies the degree of whatever it sits on, so the constant $6$ contributes nothing, $(x + 4)^3$ contributes $3$, and $(x - 9)^2$ contributes $2$.

$\deg p = 3 + 2 = 5$

Written out with nothing collected, $p(x) = 6(x + 4)(x + 4)(x + 4)(x - 9)(x - 9)$. That is five linear factors, and each one carries exactly one root, so counted with multiplicity there are five roots:

$-4,\quad -4,\quad -4,\quad 9,\quad 9$

Only two different numbers appear on that list, so $p$ has two distinct roots.

Watch the sign while reading a root off a factor. A factor $(x - r)$ vanishes at $r$, and $(x + 4)$ is $\bigl(x - (-4)\bigr)$, so it carries the root $-4$ and not $4$.

The root $-4$ has multiplicity $3$ and the root $9$ has multiplicity $2$, and the check that always applies is that the multiplicities of the distinct roots add up to the degree: $3 + 2 = 5$.

A factorization is complete only when every factor is linear, so the quadratic has to be split. Solve $x^2 + 49 = 0$, which is $x^2 = -49$. Now $(7i)^2 = 49i^2 = -49$, and $(-7i)^2 = 49i^2 = -49$ as well, so the two values are $7i$ and $-7i$:

$x^2 + 49 = (x - 7i)(x + 7i)$

Substituting that in gives the complete factorization.

$q(x) = 6(x + 4)^3(x - 9)^2(x - 7i)(x + 7i)$

Count the linear factors: $3 + 2 + 1 + 1 = 7$. So the degree is $7$ and there are seven roots counted with multiplicity, namely $-4$ three times, $9$ twice, and then $7i$ and $-7i$ once each. That list names four different numbers, so $q$ has four distinct roots.

The multiplicities $3$, $2$, $1$, $1$ add to $7$, matching the degree, which is the arithmetic check worth running every time.

One feature of the answer is worth noticing before you leave it: of those seven roots, only the ones at $-4$ and $9$ are real numbers.

Take the two repairs one at a time.

The first is counted with multiplicity, and the polynomial $p$ from part A is its witness. The degree of $p$ is $5$, but if you ask how many different numbers are roots of it the answer is $2$. Nothing has gone wrong: the five linear factors are all present, but three of them name the same number and the other two name another. So a sentence promising $n$ roots is false the moment "roots" is read as "different numbers", and it becomes true again once the count is taken over the factors, repeating each root as often as its factor appears.

The second repair is over the complex numbers, and the polynomial $q$ from part B is its witness. Its degree is $7$, but if a root is required to be a real number then $q$ has only $-4$ and $9$, since $7i$ and $-7i$ live nowhere on the real line. Even counting those real ones with multiplicity leaves five against a degree of $7$. The two missing roots did not fail to exist; they failed to be real.

The repairs are independent, which is exactly why both are needed. Every root of $p$ is real, so $p$ needs only the first repair. The roots of $x^2 + 49$ are two different numbers, so a claim about it needs only the second. A single sentence has to survive both situations at once, and the repaired form is the statement that the multiplicities of the distinct roots, taken over the complex numbers, add up to the degree.

$m_1 + m_2 + \cdots + m_k = n$

The constant $c(x) = 6$ is a different matter, and it is worth being precise about what fails. The Fundamental Theorem of Algebra opens with "every polynomial of degree at least $1$", and $c$ has degree $0$, so the theorem's promise of a root is simply not addressed to it. That hypothesis is load bearing rather than decorative: $6 = 0$ is never true, so $c$ has no root at all, and a version of the theorem that dropped "degree at least $1$" would be false on its first line. Notice that the counting sentence itself survives here in a hollow way, since degree $0$ predicts $0$ roots and a nonzero constant indeed has none. It is the existence claim, not the counting claim, that the hypothesis is protecting.

Confirm the given root first, because every later step inherits the mistake if it is wrong. Mind the odd powers of a negative number.

$p(-4) = 256 - 384 + 160 - 192 + 160 = 0$

So $-4$ is a root, and the factor theorem promises that $(x + 4)$ divides $p$. That, on its own, says the multiplicity is at least $1$ and nothing more.

Divide. Synthetic division of the coefficients $1,\, 6,\, 10,\, 48,\, 160$ by $-4$ brings down $1$, then gives $6 - 4 = 2$, then $10 - 8 = 2$, then $48 - 8 = 40$, then $160 - 160 = 0$. The remainder is $0$ and the quotient is

$x^3 + 2x^2 + 2x + 40.$

The question now is whether $(x + 4)$ goes into that quotient, so test the quotient at $-4$, not the original polynomial. Synthetic division of $1,\, 2,\, 2,\, 40$ by $-4$ gives $2 - 4 = -2$, then $2 + 8 = 10$, then $40 - 40 = 0$. Remainder $0$ again, with quotient

$x^2 - 2x + 10.$

Test once more. This is the step that decides the answer, so it has to be done rather than assumed.

$(-4)^2 - 2(-4) + 10 = 16 + 8 + 10 = 34$

That is not zero, so $(x + 4)$ does not divide a third time and the count is settled: the multiplicity of $-4$ is exactly $2$. In the shape the definition asks for, $p(x) = (x + 4)^2\bigl(x^2 - 2x + 10\bigr)$ with the leftover factor taking the nonzero value $34$ at $-4$.

What is left after the divisions is a quadratic, and a quadratic is always finishable. Its discriminant is $(-2)^2 - 4(1)(10) = 4 - 40 = -36$, which is negative, so both of its roots are non-real.

$x = \frac{2 \pm \sqrt{-36}}{2} = \frac{2 \pm 6i}{2} = 1 \pm 3i$

Halving the whole numerator is the step most often fumbled: the $2$ and the $6i$ both get halved. Those two values give the two remaining linear factors, and assembling everything,

$p(x) = (x + 4)^2\bigl(x - (1 + 3i)\bigr)\bigl(x - (1 - 3i)\bigr).$

Count what that says. There are four linear factors, so the degree is $4$ and there are four roots counted with multiplicity: $-4$, $-4$, $1 + 3i$, $1 - 3i$. Three different numbers appear, so there are three distinct roots, and the multiplicities $2 + 1 + 1 = 4$ agree with the degree.

The factorization also closes the search. A product is zero only when one of its factors is zero, so any number that kills $p$ has to kill one of these four factors, which makes $-4$, $1 + 3i$ and $1 - 3i$ the only roots $p$ has. There is nothing further to look for, and knowing when to stop looking is most of what a complete factorization buys you.

Dividing one polynomial by another has exactly one outcome: there is a single quotient and a single remainder. So running the division a second time on the same two polynomials is not a new experiment, it is the first experiment repeated, and it is guaranteed to return the same remainder.

The remainder theorem makes the point sharper. Dividing $p(x)$ by $(x + 4)$ leaves the remainder $p(-4)$, and $p(-4) = 0$ was given at the start. So the classmate's second remainder was fixed at zero before they began, and it would have been zero for a polynomial whose multiplicity at $-4$ was $1$ just as surely as for one whose multiplicity was $5$. A test that returns the same answer in every case distinguishes nothing.

A one-line illustration settles it. For $g(x) = (x + 4)(x - 1)$, the root $-4$ has multiplicity $1$, and yet dividing $g$ by $(x + 4)$ twice in the classmate's sense gives remainder $0$ both times. Their procedure would report "at least twice" for a polynomial where the truth is once.

What a real second test looks at is the quotient. Write the first division as an identity,

$p(x) = (x + 4)\,Q(x), \qquad Q(x) = x^3 + 2x^2 + 2x + 40,$

and ask whether $(x + 4)$ divides $Q$. By the factor theorem that happens exactly when $Q(-4) = 0$, and here it does:

$Q(-4) = -64 + 32 - 8 + 40 = 0$

so a second copy of the factor comes out, and only now has anything new been learned. Had that value been nonzero, the count would have stopped at $1$. Every step of the measurement works this way: each division moves the question into a smaller polynomial, and the answer arrives when one of those smaller polynomials finally refuses.

A rational root written in lowest terms has a numerator dividing the constant term and a denominator dividing the leading coefficient. The constant term is $2$, whose divisors are $1$ and $2$, and the leading coefficient is $1$, whose only divisor is $1$. So the whole candidate list is

$\pm 1, \qquad \pm 2.$

Before evaluating, notice that $p$ has no $x^4$, $x^3$ or $x^2$ term. Those coefficients are $0$, not missing, so they contribute nothing and there is no place for them to hide a sign error.

Test all four, and test all four rather than stopping at the first miss, since the conclusion depends on the list being exhausted.

$p(1) = 1 - 4 + 2 = -1, \qquad p(-1) = -1 + 4 + 2 = 5$

$p(2) = 32 - 8 + 2 = 26, \qquad p(-2) = -32 + 8 + 2 = -22$

Not one of the four is zero. The Rational Root Theorem guarantees that any rational root would have appeared on that list, and none of the list is a root, so the conclusion is exact: $p$ has no rational root. That is a complete answer to the question the theorem was built to answer, and it is worth stating in those words rather than as "nothing worked".

The counting is fixed by the degree and nothing else. The Fundamental Theorem of Algebra supplies a root, the factor theorem converts it into a linear factor, and repeating that on each quotient in turn breaks $p$ into linear factors. Since the leading coefficient is $1$,

$p(x) = (x - r_1)(x - r_2)(x - r_3)(x - r_4)(x - r_5)$

for complex numbers $r_1, \ldots, r_5$ that need not be different from one another. So counted with multiplicity there are exactly $5$ roots, and since those five factors can name at most five different numbers, there are at most $5$ distinct roots. Both counts hold over the complex numbers, which is where the theorem operates.

Now compare that with what part A did. The sweep asked a membership question about a much smaller collection of numbers: it asked whether any root is rational. The answer was no, and that answer is exact and final.

What it did not ask is whether any root is real. Being non-rational and being non-real are different things, and the sweep cannot tell them apart, so it leaves the real count entirely open.

What it also did not touch is existence. The five roots were guaranteed before a single candidate was tested, and no amount of failed testing can revoke that guarantee, because the theorem's promise does not depend on any search succeeding. Reading "no candidate worked" as "there is nothing there" confuses the outcome of a method with a fact about the polynomial.

Start with what the theorem says, word for word: every polynomial of degree at least $1$ with complex coefficients has at least one complex root. In symbols, all it delivers is

$\deg p \ge 1 \implies p(z) = 0 \ \text{for some complex } z.$

Read the verb. It asserts that a certain number is out there. It does not name that number, it does not narrow it down, and there is no procedure anywhere in the sentence.

The proof explains why the statement has that shape. The argument watches what $p$ does to a circle of radius $R$ in the complex plane. For large $R$ the leading term dominates and the image loops around the origin. As $R$ shrinks to $0$ the image collapses to the single point $p(0)$. If that point is the origin then $0$ is already a root and there is nothing left to prove; otherwise the collapsed image sits away from the origin, and since a loop cannot stop enclosing a point without crossing it, somewhere in between the image passes through the origin. Either way a root is produced. What the argument never says is where, and an argument built that way could not, because at no step does it ever hold a specific number in its hand.

So the classmate's step is "exists, therefore expressible", and it is that step, not the theorem, that has to be defended. At this degree it cannot be. Ruffini and then Abel proved that no general formula in radicals solves polynomial equations of degree $5$ and above, so there is no expression built from the coefficients by arithmetic and root extraction that returns the roots of an arbitrary quintic. The guarantee and the formula come apart, permanently, and not for want of effort.

None of that makes the roots of $p$ any less real, in the sense of existing. They exist exactly as firmly as the roots of a quadratic do. What you lose is a general recipe, and what remains is what remained before: the Rational Root Theorem when a root happens to be rational, factoring when the polynomial happens to cooperate, and numerical approximation to whatever accuracy you need. The theorem's contribution is not a method. It is the assurance that a method, when one is available, is not hunting for something that was never there.

Let the distinct roots of $p$ be $s_1, s_2, \ldots, s_k$ with multiplicities $m_1, m_2, \ldots, m_k$. Two facts about that data drive everything, and both are worth stating before either direction begins. First, each $m_i$ is a positive integer: a number that is a root has at least one linear factor carrying it. Second, the complete factorization has exactly $n$ linear factors and every one of them carries exactly one of the $s_i$, so

$m_1 + m_2 + \cdots + m_k = n.$

Direction one: $n$ distinct roots forces every root simple. Suppose $k = n$. Then the display above is $n$ positive integers adding to $n$. Suppose one of them, say $m_j$, were at least $2$. Each of the other $n - 1$ multiplicities is at least $1$, so the total would be at least

$2 + (n - 1) = n + 1,$

which exceeds $n$ and contradicts the identity. So no multiplicity can be $2$ or more, and every $m_i = 1$: every root is simple. This direction uses the identity as a constraint that squeezes the multiplicities from above.

Direction two: every root simple forces $n$ distinct roots. Suppose every $m_i = 1$. Then the left side of the identity is $1$ added to itself $k$ times, which is $k$, so

$k = n,$

and $p$ has exactly $n$ distinct roots. This direction uses the identity in the opposite way, as an evaluation that computes $k$ outright.

Both directions hold, so the biconditional is genuine. It is worth noticing that neither direction is a restatement of the other: the first argues from a count of multiplicities to their sizes, the second from their sizes to their count, and reversing an implication is never free.

Take

$p(x) = (x - 5)^4(x + 3)^2(x - 1).$

Its degree is $4 + 2 + 1 = 7$. Its distinct roots are $5$, $-3$ and $1$, so the number of distinct roots is $3$. Its repeated roots, those of multiplicity at least $2$, are $5$ and $-3$, so the classmate's count of repeated roots is $2$ and their formula predicts

$7 - 2 = 5.$

The true number of distinct roots is $3$, so the formula is wrong on this polynomial, and one polynomial is all it takes to sink a general claim.

The reason it fails is that it charges a flat fee. A root of multiplicity $m$ uses up $m$ of the $n$ linear factors while contributing only $1$ to the count of distinct roots, so it costs $m - 1$, not $1$. Here the root $5$ costs $3$, the root $-3$ costs $1$, and the simple root $1$ costs nothing, a total of $4$, while the classmate charged $1$ for each of the two repeated roots and nothing for the simple one, a total of $2$. Summing the correct fee over the distinct roots gives the relation

$n - k = (m_1 - 1) + (m_2 - 1) + \cdots + (m_k - 1),$

which is just the identity $m_1 + \cdots + m_k = n$ rearranged, since subtracting $1$ from each of the $k$ terms subtracts $k$ in total. Checking it on the same polynomial: the left side is $7 - 3 = 4$ and the right side is $(4 - 1) + (2 - 1) + (1 - 1) = 4$.

It is worth seeing exactly when the classmate's version survives, because that explains how it got proposed. A simple root contributes $0$ to the right side either way, and a repeated root contributes $m - 1$ where the classmate contributes $1$, so the two agree precisely when every repeated root has multiplicity exactly $2$. Run that both ways to be sure. If every repeated root has multiplicity $2$, then each contributes $2 - 1 = 1$ and the two totals match. Conversely, if the totals match, then since each repeated root contributes at least $1$ and no simple root contributes anything, no repeated root can contribute more than $1$, which forces $m - 1 = 1$ and so $m = 2$ for every one of them. So the formula is correct exactly on the polynomials whose repetitions are all doubles, which is a common enough situation to make a false rule look reliable. The polynomial above breaks that condition at the root $5$, whose multiplicity is $4$.

Two things have to be shown, and they are shown in different ways: that $1$ is attainable, and that $0$ is not.

Attainable. Fix any complex number $r$ and take $p(x) = (x - r)^n$, which has degree $n$. A product is zero only when one of its factors is zero, and every factor here is the same one, so

$(x - r)^n = 0 \iff x - r = 0 \iff x = r.$

That is exactly one distinct root, with multiplicity $n$, and the construction works at every degree $n \ge 1$. So the floor is reached, not merely approached.

Not zero. A polynomial with no roots at all would contradict the Fundamental Theorem of Algebra, which says that every polynomial of degree at least $1$ with complex coefficients has at least one complex root. So the count of distinct roots is at least $1$ for every such polynomial, and the floor is $1$.

Notice which hypothesis that second argument leans on. It is the degree condition, and it is not decoration: a nonzero constant has degree $0$ and no roots whatsoever, and the floor of $1$ simply does not apply to it, because the theorem never covered it. Take the condition away and the claim is false immediately.

So the number of distinct roots of a polynomial of degree $n \ge 1$ lies between $1$ and $n$, and both ends are attained. The two ends come from different places, which is the point worth carrying away: the ceiling comes from the factorization, since $n$ linear factors can name at most $n$ different numbers, and the floor comes from existence, which is the one thing the factorization argument cannot supply on its own.

Put each polynomial into the shape the definition provides. That the multiplicity of $r$ in $f$ is $2$ means precisely that there is a polynomial $F$ with

$f(x) = (x - r)^2 F(x), \qquad F(r) \neq 0,$

and likewise for $g$ there is a polynomial $G$ with

$g(x) = (x - r)^3 G(x), \qquad G(r) \neq 0.$

Multiply the two and collect the copies of the repeated factor, using that powers of the same base add:

$f(x)g(x) = (x - r)^{2 + 3} F(x)G(x) = (x - r)^5 F(x)G(x).$

That much shows $(x - r)^5$ divides $fg$, so the multiplicity is at least $5$. Stopping here would be the whole error the word "exactly" is guarding against, because nothing so far rules out a sixth copy hiding inside $FG$.

The definition tells you what closes the gap: the leftover factor must not vanish at $r$. The leftover here is $FG$, and evaluating a product of polynomials at a number is the same as multiplying their values there:

$(FG)(r) = F(r)\,G(r).$

This is the single step where the two hypotheses are spent. Both $F(r)$ and $G(r)$ are nonzero numbers, and a product of two nonzero complex numbers cannot be zero: if $uv = 0$ and $u \neq 0$, then multiplying by $\frac{1}{u}$ forces $v = 0$. So $(FG)(r) \neq 0$.

With both requirements met, $fg$ has been written as $(x - r)^5$ times a polynomial that does not vanish at $r$, which is the definition of multiplicity $5$. Nothing about the numbers $2$ and $3$ was used beyond their being the two multiplicities, so the same argument shows in general that multiplicities add across a product.

Part A's multiplicities of $2$ and $3$ are set aside here: the claim to demonstrate is about arbitrary $f$ and $g$, so the two pairs below are free to use whichever multiplicities make the point.

To show a quantity is not determined by given data, hold the data fixed and make the quantity move. So build two pairs that agree on both input multiplicities and disagree on the output. Take $r = 0$ throughout, which costs nothing and keeps the algebra visible, since the repeated factor is then just $x$.

First pair. Let $f_1(x) = x^2$ and $g_1(x) = 2x^2$. Check each against the definition rather than reading an exponent off: $f_1(x) = x^2 \cdot 1$ with the leftover factor $1$, which is $1$ at $0$ and so nonzero, giving multiplicity exactly $2$; and $g_1(x) = x^2 \cdot 2$ with leftover $2$, again nonzero at $0$, so multiplicity exactly $2$.

Second pair. Let $f_2(x) = x^2$ and $g_2(x) = x^3 - x^2$. The first is as before. For the second, factor it as $g_2(x) = x^2(x - 1)$, whose leftover factor takes the value $-1$ at $0$, which is not zero, so the multiplicity of $0$ in $g_2$ is exactly $2$ as well.

So both pairs present the same data: multiplicity $2$ in the first polynomial and multiplicity $2$ in the second. Now add.

$x^2 + 2x^2 = 3x^2, \qquad x^2 + \bigl(x^3 - x^2\bigr) = x^3$

The first sum is $3x^2$, with leftover factor $3$ at $x = 0$, so the multiplicity of $0$ in it is $2$. The second sum is $x^3$, with leftover factor $1$, so the multiplicity of $0$ in it is $3$.

Same inputs, different outputs, so the multiplicity in a sum is not a function of the two given multiplicities. The mechanism is visible in the second pair: the $x^2$ terms cancelled, and cancellation is something addition permits and multiplication does not.

The definition pins a multiplicity with two requirements: how many copies of $(x - r)$ come out, and that the leftover factor is not zero at $r$. Trace both requirements through each operation.

For a product, the copies add and the leftover factors multiply, so the leftover value at $r$ is $F(r)G(r)$. A product of nonzero numbers is never zero, so the second requirement is inherited automatically. That is why part A came out exactly, with no cases to consider.

For a sum, only as many copies as the smaller supply can be guaranteed, and the leftover values are added rather than multiplied. Two nonzero numbers can perfectly well add to zero, and when they do the guaranteed count stops being the whole story. That is precisely what happened in the second pair of part B, and nothing about the given multiplicities predicts whether it will happen.

There is nonetheless a condition that settles the matter. Write

$f(x) = (x - r)^a F(x), \qquad g(x) = (x - r)^b G(x)$

with $F(r) \neq 0$ and $G(r) \neq 0$, and suppose the two multiplicities are different, say $a < b$. Factor out the smaller power, which is all that both terms are guaranteed to contain:

$f(x) + g(x) = (x - r)^a\Bigl(F(x) + (x - r)^{b-a}G(x)\Bigr)$

Now evaluate the bracket at $r$. Because $a < b$, the exponent $b - a$ is at least $1$, so the factor $(x - r)^{b-a}$ is zero at $r$ and the second term disappears entirely, whatever $G(r)$ happens to be. What survives is

$F(r) + 0 = F(r) \neq 0.$

So $f + g$ has been written as $(x - r)^a$ times something that does not vanish at $r$, which is the definition's shape with $m = a$. The multiplicity is exactly $a$, the smaller of the two, and it was forced by the two multiplicities alone: nothing about $F$ or $G$ beyond their nonvanishing entered the argument. Nothing needs to be assumed about $f + g$ either, since a product of two nonzero polynomials is nonzero and both factors above are nonzero, so $f + g$ automatically has a multiplicity to report.

The condition cannot be dropped, and part B is the reason. When $a = b$, the same factoring leaves the bracket equal to $F(r) + G(r)$ at $r$, and two nonzero numbers may or may not cancel. Three outcomes are possible, and they are worth keeping apart.

If $F(r) + G(r) \neq 0$, the multiplicity of $r$ in $f + g$ is exactly $a$, just as in the unequal case.

If $F(r) + G(r) = 0$ while $f + g$ is not the zero polynomial, then at least one further copy of $(x - r)$ comes out and the multiplicity is some larger finite number, but $a$ and $b$ do not say which.

If $g$ is exactly $-f$, then $f + g$ is the zero polynomial, and that case is not a larger multiplicity at all: it is the absence of one. No $m$ can satisfy $0 = (x - r)^m q(x)$ with $q(r) \neq 0$, because $q(r) \neq 0$ makes $q$ a nonzero polynomial and a product of two nonzero polynomials is never the zero polynomial.

Part B exhibited the first two outcomes with $a = b = 2$, so equal multiplicities genuinely determine nothing. The same construction works at any common value $m$: the pair $(x - r)^m$ and $2(x - r)^m$ sums to multiplicity $m$, while the pair $(x - r)^m$ and $(x - r)^{m+1} - (x - r)^m$ sums to $(x - r)^{m+1}$, of multiplicity $m + 1$, and both members of that second pair have multiplicity exactly $m$ since their leftover factors take the values $1$ and $-1$ at $r$. Taking $g = -f$ with $f = (x - r)^m$ gives the third outcome, both inputs again having multiplicity exactly $m$.

So the honest summary runs both ways: unequal multiplicities determine the answer and force it to be the smaller one, and equal multiplicities determine nothing at all, not even whether there is a multiplicity to determine.