Complex and Irrational Roots: Practice

The Fundamental Theorem of Algebra says a degree-$6$ polynomial has exactly $6$ roots counted with multiplicity, and every root is either real or nonreal.

$6 - 4 = 2$

So $2$ of them are real. The count of nonreal roots is even, as the pairing requires, and that leaves an even number of real roots too.

The conjugate $-3 - 2i$ is a root as well. Multiply the two linear factors, using $z + \overline{z} = 2a$ and $z\,\overline{z} = a^2 + b^2$ with $a = -3$ and $b = 2$.

The sum is $-6$ and the product is $(-3)^2 + 2^2 = 13$.

$\big(x - (-3+2i)\big)\big(x - (-3-2i)\big) = x^2 + 6x + 13$

Watch the sign: subtracting a sum of $-6$ produces $+6x$.

Rational coefficients force the partner $1 - \sqrt{6}$ to be a root, so multiply the two linear factors.

The sum is $2$, and the product is $(1 + \sqrt{6})(1 - \sqrt{6}) = 1 - 6 = -5$.

$\big(x - (1+\sqrt{6})\big)\big(x - (1-\sqrt{6})\big) = x^2 - 2x - 5$

The general shape is $x^2 - 2ax + (a^2 - b^2 d)$, which here is $x^2 - 2x + (1 - 6)$.

Nonreal roots of a real polynomial come in conjugate pairs, so their number is twice the number of pairs, say $2m$.

$2m = 5 \;\Longrightarrow\; m = \tfrac{5}{2}$

There is no whole number of pairs that produces $5$, so an odd count is impossible. The other three counts all occur: $0$ for $(x-1)^7$, $2$ for $(x^2+1)(x-1)^5$, and $6$ for $(x^2+1)(x^2+4)(x^2+9)(x-1)$.

The coefficients are real, so $-i$ is a root too, and the pair forces the factor $(x - i)(x + i) = x^2 + 1$.

Divide. The steps give $x^2$, then $x$, then $-6$, and the remainder is zero.

$x^4 + x^3 - 5x^2 + x - 6 = (x^2 + 1)(x^2 + x - 6)$

Factor the quotient: $x^2 + x - 6 = (x + 3)(x - 2)$, so the real roots are $2$ and $-3$.

Each nonreal root drags its conjugate along, and the real root $4$ needs no partner.

$\overline{2i} = -2i, \qquad \overline{1 - i} = 1 + i$

That makes five roots in all ($2i$, $-2i$, $1-i$, $1+i$, $4$), which exactly fills a degree of $5$.

The discriminant is $(-2)^2 - 4(1)(10) = -36$, which is negative, so the two roots form a conjugate pair.

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

Dividing both parts by $2$ is what turns $6i$ into $3i$, and the pair is $1 + 3i$ and $1 - 3i$ as the theorem predicts.

The first two roots already form a conjugate pair, so they need nothing extra. The root $3i$ is nonreal and unpaired, so its conjugate must join the list.

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

That gives four roots ($1+i$, $1-i$, $3i$, $-3i$), so the degree is at least $4$. Degree $4$ works, since $(x^2 - 2x + 2)(x^2 + 9)$ has real coefficients and exactly those roots.

Real coefficients make $1 + 2i$ a root as well. The pair has sum $2$ and product $1^2 + 2^2 = 5$, so it forces the factor $x^2 - 2x + 5$.

Divide the cubic by it. The first step gives $x$, leaving $-2x^2 + 4x - 10$, which is $-2$ times the divisor, so nothing remains.

$x^3 - 4x^2 + 9x - 10 = (x^2 - 2x + 5)(x - 2)$

The third root is therefore $2$.

If $z$ is a nonreal root of a real polynomial, then $\overline{z}$ is a root as well, and $\overline{z} \neq z$ precisely because $z$ is not real. So nonreal roots arrive two at a time.

$\text{number of nonreal roots} = 0,\ 2,\ \text{or } 4$

One is odd, so it is impossible, no matter what the other roots do.

Factor as a difference of squares, then factor again.

$x^4 - 16 = (x^2 - 4)(x^2 + 4) = (x - 2)(x + 2)(x^2 + 4)$

The factor $x^2 + 4$ gives $x^2 = -4$, so $x = 2i$ or $x = -2i$. The four roots are $2$, $-2$, $2i$, $-2i$: two real roots and one conjugate pair, which is exactly the pattern real coefficients allow.

Factor it to see the roots: $x^2 - (3+i)x + 3i = (x - 3)(x - i)$, so the roots are $3$ and $i$, and indeed $-i$ is not among them.

$\text{coefficients: } 1, \ -(3 + i), \ 3i$

The coefficient $-(3+i)$ is not real, so the hypothesis of the theorem fails and it says nothing here. The theorem is a conditional, and a conditional makes no claim when its hypothesis is false.