Complex and Irrational Roots: Free Response

Real coefficients are exactly what the conjugate root theorem asks for, so the conjugate of $4 - 3i$ is a root of $p$ as well. Conjugating flips the sign of the imaginary part and leaves the real part alone:

$\overline{4 - 3i} = 4 + 3i$

The two are different numbers, because the imaginary part is not zero, so both linear factors divide $p$ and therefore so does their product. Multiply them by taking the sum and the product of the pair separately, since those are the two quantities in which the $i$ cancels:

$(4 - 3i) + (4 + 3i) = 8, \qquad (4 - 3i)(4 + 3i) = 16 + 9 = 25$

The product used $(a - bi)(a + bi) = a^2 + b^2$ with $a = 4$ and $b = 3$: the two cross terms cancel, and $-9i^2$ becomes $+9$. Now assemble the quadratic, whose middle coefficient is the negative of the sum and whose constant term is the product:

$\big(x - (4 - 3i)\big)\big(x - (4 + 3i)\big) = x^2 - 8x + 25$

Both coefficients came out real, as the general shape $x^2 - 2ax + (a^2 + b^2)$ promises. Notice that the sign of $b$ never mattered: the sum used $2a$ and the product used $b^2$, so the pair $4 + 3i$ and $4 - 3i$ would have produced the same factor in the same order.

The quadratic from part A is the tool. Since $4 - 3i$ and $4 + 3i$ are two distinct roots of $p$, the product of their linear factors divides $p$ exactly, so $x^2 - 8x + 25$ divides $x^3 - 14x^2 + 73x - 150$ with no remainder.

Carry the division out, staying inside the real numbers throughout, since both the divisor and the dividend are real. The first step is $x^3 \div x^2 = x$, and $x(x^2 - 8x + 25) = x^3 - 8x^2 + 25x$; subtracting leaves $-6x^2 + 48x - 150$. The next step is $-6x^2 \div x^2 = -6$, and $-6(x^2 - 8x + 25) = -6x^2 + 48x - 150$; subtracting leaves nothing at all:

$x^3 - 14x^2 + 73x - 150 = (x^2 - 8x + 25)(x - 6)$

The remainder of zero is a check on part A as much as on the arithmetic here. The quotient $x - 6$ contributes the root $6$, and a cubic has exactly three roots counted with multiplicity, so the list is finished:

$4 - 3i, \qquad 4 + 3i, \qquad 6$

The third root had no choice about being real. The pair consumed two of the three slots, and a nonreal root in the last slot would have needed a partner of its own, with nowhere left to put it.

Test the claim where it is being made, which means evaluating $g$ at both numbers rather than reasoning about them.

First confirm the premise. Squaring gives $(4 - 3i)^2 = 16 - 24i + 9i^2 = 7 - 24i$, and the middle term is $-(6 - i)(4 - 3i) = -(24 - 18i - 4i + 3i^2) = -(21 - 22i) = -21 + 22i$:

$g(4 - 3i) = (7 - 24i) + (-21 + 22i) + (14 + 2i) = 0$

So $4 - 3i$ really is a root of $g$, and the classmate's claim is being applied to a polynomial that satisfies its premise. Now do the same work at the partner. Here $(4 + 3i)^2 = 16 + 24i - 9 = 7 + 24i$, and $-(6 - i)(4 + 3i) = -(24 + 18i - 4i + 3) = -27 - 14i$:

$g(4 + 3i) = (7 + 24i) + (-27 - 14i) + (14 + 2i) = -6 + 12i$

That is not zero, so $4 + 3i$ is not a root of $g$, and the claim fails on the very first polynomial it is asked about.

No theorem was violated. The conjugate root theorem is a conditional, and its hypothesis is that every coefficient is real. The coefficients of $g$ are $1$, $-(6 - i)$ and $14 + 2i$, and the last two are not real numbers, so the theorem was silent about $g$ from the start. Factoring makes the situation visible:

$g(x) = \big(x - (2 + 2i)\big)\big(x - (4 - 3i)\big)$

Two nonreal roots, and they are not conjugates of each other: $\overline{4 - 3i} = 4 + 3i$ is nowhere on that list, and neither is $\overline{2 + 2i} = 2 - 2i$. Nothing about $g$ requires either of them to be there. So the pairing is not a property of the number $4 - 3i$. It is a property of the polynomial that has it as a root, and specifically of that polynomial's coefficients: the same number is paired inside $p$ and unpaired inside $g$.

Every coefficient is rational and $\sqrt{10}$ is irrational, so the surd conjugate theorem applies. Its swap changes the sign of the $\sqrt{10}$ term and nothing else:

$(3 - \sqrt{10})^{*} = 3 + \sqrt{10}$

The two partners are different numbers, so both linear factors divide $q$, and so does their product. Take the sum and the product of the pair:

$(3 - \sqrt{10}) + (3 + \sqrt{10}) = 6, \qquad (3 - \sqrt{10})(3 + \sqrt{10}) = 9 - 10 = -1$

The product is where the radical dies, by the difference-of-squares pattern. Assembling the quadratic,

$\big(x - (3 - \sqrt{10})\big)\big(x - (3 + \sqrt{10})\big) = x^2 - 6x - 1,$

and both of its coefficients are rational, as they had to be. This is the general shape $x^2 - 2ax + (a^2 - b^2 d)$ with $a = 3$, $b = -1$ and $d = 10$, since $3 - \sqrt{10} = 3 + (-1)\sqrt{10}$. Only $b^2$ ever enters the factor, so the sign of $b$ makes no difference to it. Mind the sign in that constant term. The complex pair produced $a^2 + b^2$, a sum, because $i^2 = -1$ flipped a sign on the way; here $(\sqrt{d})^2 = d$ flips nothing, so a difference is what appears.

Part A supplies the divisor. The two partners are distinct roots of $q$, so $x^2 - 6x - 1$ divides $q(x)$ exactly, and every number the division touches is rational.

The first step is $x^3 \div x^2 = x$, and $x(x^2 - 6x - 1) = x^3 - 6x^2 - x$; subtracting leaves $4x^2 - 24x - 4$. The next step is $4x^2 \div x^2 = 4$, and $4(x^2 - 6x - 1) = 4x^2 - 24x - 4$; subtracting leaves nothing:

$x^3 - 2x^2 - 25x - 4 = (x^2 - 6x - 1)(x + 4)$

The remaining factor gives the third root, and the cubic has exactly three roots counted with multiplicity, so the list is complete:

$3 - \sqrt{10}, \qquad 3 + \sqrt{10}, \qquad -4$

That third root had no choice about being rational. The surd pair used two of the three slots, and what was left came from dividing one rational polynomial by another, so the leftover factor has rational coefficients, and a rational linear factor has a rational root.

The question is whether the weaker hypothesis is strong enough on its own, so try to build something that satisfies it and defies the conclusion. Nothing says the polynomial has to be complicated:

$h(x) = x - 3 + \sqrt{10}$

Its coefficients are $1$ and $-3 + \sqrt{10}$. Both are real numbers, since $\sqrt{10}$ is real, so $h$ meets the classmate's hypothesis exactly. A polynomial of degree $1$ has one root, and that root is $3 - \sqrt{10}$. The proposed partner is not it:

$h(3 + \sqrt{10}) = (3 + \sqrt{10}) - 3 + \sqrt{10} = 2\sqrt{10} \neq 0$

So real coefficients do not force the surd partner, and the proposal fails.

Nothing has gone wrong with the conjugate root theorem, either. It applies perfectly well to $h$, and what it says is empty here: $3 - \sqrt{10}$ is a real number, a real number is its own complex conjugate, so the theorem hands back the root you already had. All of its content lives in the nonreal roots.

The two hypotheses buy two different things because the two swaps fix two different sets of numbers. Complex conjugation fixes exactly the real numbers, so real coefficients pass through it untouched and the complex pairing follows. The surd swap fixes exactly the rational numbers, and it does not fix every real number: it moves the coefficient $-3 + \sqrt{10}$ of $h$ to $-3 - \sqrt{10}$, which is precisely the step at which the proof refuses to run for $h$.

One caution in the other direction, because it is easy to overcorrect into a second false claim. Rational coefficients are sufficient for the partner, not necessary: $\pi\,(x^2 - 6x - 1)$ has irrational coefficients and carries both partners among its roots. What is true is that with merely real coefficients there is no guarantee either way, and the polynomial in part B and the polynomial $h$ here show the two outcomes occurring side by side.

Rational coefficients are real coefficients as well, so both pairing theorems fire at once, and the roots they force have to be counted before any polynomial is written down.

The surd theorem applies to $2 + \sqrt{13}$, since $2$, $1$ and $13$ are all rational and $\sqrt{13}$ is irrational. It forces the partner $2 - \sqrt{13}$: one further root.

The conjugate root theorem applies to $5i$ and forces $-5i$. It forces more than a bare appearance, though. A root and its conjugate occur with the same multiplicity, because every nonreal root of a real polynomial comes from a real quadratic factor $(x - z)(x - \overline{z})$, and each appearance of that factor supplies one copy of each. So $-5i$ is a root of multiplicity $2$ as well.

Now add up the forced roots, counting each as often as its multiplicity requires:

$1 + 1 + 2 + 2 = 6$

A polynomial cannot carry more roots counted with multiplicity than its degree, so the degree is at least $6$. A lower bound is only worth what meets it, and part B builds a polynomial of degree exactly $6$ satisfying every condition, so the least possible degree is $6$.

Build each pair into its own quadratic first. That is the move that clears the radical in one case and the $i$ in the other.

The surd pair has sum $4$ and product $2^2 - 13 = -9$:

$\big(x - (2 + \sqrt{13})\big)\big(x - (2 - \sqrt{13})\big) = x^2 - 4x - 9$

The conjugate pair has sum $0$ and product $0^2 + 5^2 = 25$:

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

That second quadratic supplies $5i$ once and $-5i$ once. The requirement is two of each, so it appears squared. Multiplying the pieces gives a polynomial of degree $2 + 4 = 6$:

$p(x) = (x^2 - 4x - 9)(x^2 + 25)^2$

Every factor has rational coefficients, so the product does too, and the leading coefficients multiply to $1$, so it is monic. The factored form is also the informative one: it displays all six roots and the multiplicity of each, which any expansion would bury.

Weakening the hypothesis switches off exactly one of the two theorems, so the honest way to answer is to ask which.

The conjugate root theorem asks only for real coefficients, so it still applies in full. The root $5i$ of multiplicity $2$ still forces $-5i$ of multiplicity $2$, and $(x^2 + 25)^2$ is still a factor, consuming degree $4$.

The surd theorem is the casualty. It asks for rational coefficients, and there is now no rational hypothesis to appeal to. So $2 + \sqrt{13}$ has to be present once and forces no partner at all, consuming degree $1$:

$1 + 2 + 2 = 5$

A lower bound needs a polynomial that meets it, so exhibit one:

$p(x) = \big(x - 2 - \sqrt{13}\big)(x^2 + 25)^2$

Its coefficients are real, since $2 + \sqrt{13}$ is a real number, and its roots are $2 + \sqrt{13}$ once and $5i$, $-5i$ twice each: every condition met, at degree $5$.

The gap between $6$ and $5$ is the distinction the lesson keeps drawing, priced in degrees. Real coefficients pair up nonreal roots and say nothing whatever about irrational ones. The surd pairing was what the rational hypothesis was buying, and the cost of giving it up is exactly one root.

Sort the roots into two piles and let the degree pay for each. The Fundamental Theorem of Algebra says $p$ has exactly $7$ roots counted with multiplicity, and every one of them is real or nonreal.

The real pile is settled by the hypothesis. The only real root is $3$ and it is simple, so it contributes the single linear factor $x - 3$, and the real pile consumes $1$ of the $7$.

The nonreal pile is where the pairing works. Each nonreal root $z$ brings $\overline{z}$ with it, and the two multiply into a quadratic with real coefficients:

$(x - z)(x - \overline{z}) = x^2 - 2ux + (u^2 + v^2), \qquad z = u + vi, \ v \neq 0$

So the nonreal roots arrive packaged as real quadratics, each consuming $2$ of the degree. Writing $m$ for how many such quadratics there are,

$7 = 1 + 2m \quad\Longrightarrow\quad m = 3.$

The factorization therefore has the shape

$p(x) = a\,(x - 3)\,d_1(x)\,d_2(x)\,d_3(x),$

with $a \neq 0$ the leading coefficient: one linear factor and three quadratic ones. Each $d_j$ is monic, since the display above builds it as a product of two monic linear factors, which is what leaves $a$ as the leading coefficient of the whole product. Each $d_j$ must also have a negative discriminant, and that is forced rather than assumed. A real quadratic whose discriminant is zero or more splits into real linear factors, which would give $p$ real linear factors beyond the single one it is allowed, whether they repeat $3$ or introduce a new real root. The hypothesis leaves room for neither.

The count of nonreal roots follows immediately: three quadratics with two roots each gives $2m = 6$ nonreal roots out of the $7$. That number is even, as the pairing demands, and the one root left over is the real one.

Let $p$ have real coefficients and odd degree $n$. Everything turns on showing that the nonreal roots can only be counted in twos.

Suppose $z$ is a nonreal root, so $z = u + vi$ with $v \neq 0$. Two facts about $\overline{z}$ matter. First, $\overline{z}$ is a root as well, by the conjugate root theorem, which applies because the coefficients are real. Second, $\overline{z} \neq z$: conjugation fixes a number exactly when that number is real, and $z$ is not. So a nonreal root never partners itself, and the nonreal roots fall into genuine two-element pairs.

The multiplicities match too, so the pairing survives being counted with multiplicity. Each pair contributes the real quadratic $(x - z)(x - \overline{z})$, and every appearance of that quadratic in the factorization supplies one copy of $z$ and one copy of $\overline{z}$. Writing $m$ for the number of such quadratic factors and $k$ for the number of real linear factors, the Fundamental Theorem of Algebra accounts for the entire degree:

$n = k + 2m$

Now read the parity rather than the size. The term $2m$ is even whatever $m$ is, so

$k = n - 2m$

has the same parity as $n$. Since $n$ is odd, $k$ is odd. Now $k$ is a count, so it is a whole number that is at least $0$, and an odd whole number is never $0$:

$k \geq 1$

So at least one factor $x - r$ with $r$ real appears in the factorization, and that $r$ is a real root of $p$. Nothing about the size of $n$ was used beyond its parity, so the argument holds at degree $1$, at degree $101$, and everywhere odd in between.

Edit one, rational coefficients. The statement still holds, and it needs no new proof at all. Every rational number is a real number, so a polynomial with rational coefficients is in particular a polynomial with real coefficients, and part B already speaks about it. Narrowing a hypothesis can never break a theorem: it only makes the theorem address fewer polynomials, and every one of those was already being addressed. The conclusion does not get stronger either. It still promises a real root and nothing more.

Edit two, no condition on the coefficients. Now the statement fails, and one polynomial is enough to say so. Take

$f(x) = (x - 3i)(x^2 + 4) = x^3 - 3ix^2 + 4x - 12i.$

Its degree is $3$, which is odd. Its roots are $3i$ from the first factor, and from $x^2 = -4$ the two numbers $2i$ and $-2i$. Not one of the three is real, so the conclusion of part B is false for $f$, and the coefficients $-3i$ and $-12i$ are exactly what allow it to be false.

It is worth seeing where the proof of part B stopped working, since a counterexample is far more useful when you can point at the line it breaks. The nonreal roots of $f$ are $3i$, $2i$ and $-2i$, and they do not pair off: $3i$ has no $-3i$ beside it, because the conjugate root theorem never applied. Once the pairing is gone, the count of nonreal roots is free to be odd, and here it is odd: there are three of them. The parity argument had nothing left to stand on.

The converse asks a root list to control the coefficients, and the quickest way to break a claim of that shape is a nonzero constant multiple: multiplying by a nonzero constant $c$ changes every coefficient while leaving every root where it was, since $c\,p(r) = 0$ happens precisely when $p(r) = 0$.

Start from a real cubic whose nonreal roots already pair and whose roots are easy to read:

$(x - 4)(x^2 + 9) = x^3 - 4x^2 + 9x - 36$

Its roots are $4$ from the first factor and, from $x^2 = -9$, the pair $3i$ and $-3i$. Now scale it by the nonreal constant $-i$:

$g(x) = -i(x - 4)(x^2 + 9) = -ix^3 + 4ix^2 - 9ix + 36i$

The roots have not moved, so $g$ has degree $3$, the real root $4$, and the conjugate pair $3i$ and $-3i$, each occurring once. Check one of them straight from the factored form:

$g(3i) = -i\,(3i - 4)\big((3i)^2 + 9\big) = -i\,(3i - 4)(0) = 0$

And read the coefficients: $-i$, $4i$, $-9i$ and $36i$. Not one of them is real, so $g$ satisfies the converse's hypothesis and fails its conclusion as completely as a polynomial can. The converse is false.

The choice of $-i$ was not special. Any nonreal constant would have done, and so would any real cubic with one real root and one conjugate pair.

Read the root list with its multiplicities, which is exactly what the factored form is telling you: $4i$ occurs once, $-4i$ occurs twice, and the degree is $3$ in total.

Suppose, for contradiction, that $q = c\,r$ for some nonzero constant $c$ and some polynomial $r$ with real coefficients. Dividing by $c$ changes neither where a polynomial vanishes nor how many times a linear factor divides it, so $r$ has precisely the same roots as $q$ with precisely the same multiplicities: $4i$ once and $-4i$ twice.

But a polynomial with real coefficients cannot list a conjugate pair unevenly. Its nonreal roots all come from real quadratic factors $(x - z)(x - \overline{z})$, and every appearance of such a factor supplies one copy of $z$ and one copy of $\overline{z}$, so the two finish with the same multiplicity:

$\text{multiplicity of } 4i \ = \ \text{multiplicity of } -4i$

Here those multiplicities would have to be $1$ and $2$. No such $r$ exists, so $q$ is not a nonzero constant times a polynomial with real coefficients.

Notice how narrowly this misses. The two nonreal roots of $q$ genuinely are conjugates of each other, so the root list passes any inspection that reads only the values. What it fails is the count, and the count is not a technicality: it is the difference between a matched pair and an unmatched leftover.

Let $p$ be nonzero, of degree $n$, with leading coefficient $a$. If $n = 0$ then $p$ is the constant $a$, which is $a$ times the real polynomial $1$, so assume $n \geq 1$.

The Fundamental Theorem of Algebra, applied repeatedly, factors $p$ completely over the complex numbers:

$p(x) = a\,(x - w_1)(x - w_2)\cdots(x - w_n)$

Here $w_1, \ldots, w_n$ is the complete root list, each root written as often as its multiplicity. Now sort those $n$ factors into two heaps.

A factor $x - w$ with $w$ real is already a polynomial with real coefficients, so it can be set aside exactly as it stands.

A factor $x - z$ with $z$ nonreal is where the hypothesis is spent. It says $\overline{z}$ appears in the list with the same multiplicity as $z$, so the copies of $x - z$ and the copies of $x - \overline{z}$ can be matched one to one, with none of either left over. Multiply each matched couple together:

$(x - z)(x - \overline{z}) = x^2 - 2ux + (u^2 + v^2), \qquad z = u + vi$

Both of those coefficients are real, because $z + \overline{z} = 2u$ and $z\,\overline{z} = u^2 + v^2$ are the two quantities a conjugate pair makes real.

Every factor has now been absorbed into a real linear factor or a real quadratic one, and a product of polynomials with real coefficients has real coefficients. Calling that product $r$,

$p(x) = a\,r(x), \qquad a \neq 0,$

which is the claim, with $r$ monic into the bargain.

The equal multiplicities were spent at exactly one step: the matching. Without them the heap of nonreal factors cannot be exhausted two at a time, and a factor such as $x + 4i$ is left standing alone, unabsorbed and carrying an $i$ that nothing cancels. Part B is that failure in its smallest form.