Complex and Irrational Roots: Practice

Conjugation respects addition and multiplication for all complex numbers, so those two steps run no matter what the coefficients are. They are facts about conjugation, not about $p$.

The hypothesis enters at exactly one point, when the bars are stripped off the coefficients.

$\overline{a_k} = a_k \quad\text{because } a_k \text{ is real}$

That is the hinge. Take the hypothesis away and this line fails, which is precisely why $x - i$, whose coefficient $-i$ satisfies $\overline{-i} = i \neq -i$, has no conjugate pairing.

The coefficients are real, so $2 - i$ is a root as well. The pair has sum $4$ and product $2^2 + 1^2 = 5$, giving the factor $x^2 - 4x + 5$.

Divide. The first step gives $x^2$, and $x^2(x^2 - 4x + 5) = x^4 - 4x^3 + 5x^2$; subtracting leaves $9x^2 - 36x + 45$, which is $9$ times the divisor, so the remainder is zero.

$x^4 - 4x^3 + 14x^2 - 36x + 45 = (x^2 - 4x + 5)(x^2 + 9)$

The quotient gives $x^2 = -9$, so $x = 3i$ or $x = -3i$: a second conjugate pair, and the polynomial has no real roots.

Real coefficients force $1 - i$ to be a root too, so the four roots are $1+i$, $1-i$, $-3$, $-3$. The conjugate pair has sum $2$ and product $1^2 + 1^2 = 2$, giving the factor $x^2 - 2x + 2$.

$p(x) = 2(x^2 - 2x + 2)(x + 3)^2$

Evaluate at $x = 0$ without expanding anything.

$p(0) = 2 \cdot (0 - 0 + 2) \cdot (0 + 3)^2 = 2 \cdot 2 \cdot 9 = 36$

Rational coefficients force the partner $3 + \sqrt{7}$ to be a root, so the cubic has all three roots $3 - \sqrt{7}$, $3 + \sqrt{7}$, and $2$. The surd pair has sum $6$ and product $3^2 - 7 = 2$.

$\big(x - (3-\sqrt{7})\big)\big(x - (3+\sqrt{7})\big) = x^2 - 6x + 2$

Multiply by the last factor.

$(x^2 - 6x + 2)(x - 2) = x^3 - 8x^2 + 14x - 4$

The theorem needs every coefficient to be rational. Scan the four.

Three of them carry a coefficient with a $\sqrt{3}$ in it, so the theorem does not apply to them, and in fact $1 - \sqrt{3}$ really is not a root of any of those three. Only one polynomial has rational coefficients throughout.

$x^2 - 2x - 2 \quad\text{has coefficients}\quad 1,\ -2,\ -2$

Check the conclusion: $(1 - \sqrt{3})^2 - 2(1 - \sqrt{3}) - 2 = (4 - 2\sqrt{3}) - 2 + 2\sqrt{3} - 2 = 0$, so $1 - \sqrt{3}$ is indeed a root of it.

Real coefficients make $-i$ a root as well, so the three roots are $i$, $-i$, and $2$. The conjugate pair gives the factor $(x - i)(x + i) = x^2 + 1$.

$p(x) = (x^2 + 1)(x - 2)$

Expanding,

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

A quick check: $p(i) = i^3 - 2i^2 + i - 2 = -i + 2 + i - 2 = 0$.

The two remaining roots are nonreal and the coefficients are real, so each one's conjugate must also be a root. With only two slots left, the two roots must be conjugates of each other.

$\overline{1 + i} = 1 - i \quad\checkmark$

The other three pairs are not conjugate pairs: $\overline{i} = -i$, not $2i$; $\overline{1+i} = 1 - i$, not $-1 - i$ (that flips the real part too); and $\overline{2i} = -2i$, not $3i$.

There are no real roots, since $x^4 + 4 \geq 4$ for every real $x$. Manufacture a difference of squares by adding and subtracting $4x^2$.

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

Each factor has discriminant $4 - 8 = -4$, so the quadratic formula gives

$x = \frac{2 \pm 2i}{2} = 1 \pm i \qquad\text{and}\qquad x = \frac{-2 \pm 2i}{2} = -1 \pm i.$

Four roots in two conjugate pairs. Note that $2i$ is not a root: $(2i)^4 + 4 = 16 + 4 = 20$.

The coefficients are rational, so the surd theorem gives $3 - \sqrt{2}$ as a root. The pair has sum $6$ and product $3^2 - 2 = 7$.

$\big(x - (3+\sqrt{2})\big)\big(x - (3-\sqrt{2})\big) = x^2 - 6x + 7$

Divide the cubic by that factor. The first step gives $x$, leaving $-2x^2 + 12x - 14$, which is $-2$ times the divisor, so the remainder is zero.

$x^3 - 8x^2 + 19x - 14 = (x^2 - 6x + 7)(x - 2)$

The third root is $2$.

The proof works by pushing a swap through the polynomial and demanding that every coefficient survive it unchanged. Complex conjugation fixes exactly the reals, so real coefficients are enough for it. The surd swap fixes exactly the rationals, so it needs rational coefficients.

A merely real coefficient can be knocked around by the surd swap, and then the argument collapses.

$x^2 - (2 + \sqrt{2})\,x + 2\sqrt{2} = (x - 2)(x - \sqrt{2})$

This has real coefficients and the root $\sqrt{2}$, yet $-\sqrt{2}$ is not a root. The surd swap does respect multiplication, and irrational numbers certainly can be roots of real polynomials, so the other explanations are simply false.

Degree $6$ means exactly $6$ roots counted with multiplicity. The real ones account for $2$ of them, since $x = 1$ is the only real root and it is double.

$6 - 2 = 4$

The answer $4$ passes the parity check as well: nonreal roots come in conjugate pairs, so their count must be even, and $4$ is.

A root and its conjugate always occur with the same multiplicity, because each real quadratic factor $(x - z)(x - \overline{z})$ supplies one of each, and every nonreal root comes from such a factor.

$\text{multiplicity of } 2 + i \ = \ \text{multiplicity of } 2 - i$

Saying $2 + i$ is double while $2 - i$ is simple breaks that rule, so it cannot happen. The other three are fine: $(x^2+1)^2$ has the roots $i, -i, i, -i$; $(x^2 - 6x + 10)^3$ gives $3 \pm i$ each three times; and $(x-5)^3$ is a legitimate real cubic.