Complex and Irrational Roots: Additional Practice (Set 2)

A root and its conjugate occur with the same multiplicity, since every real quadratic factor $(x - z)(x - \overline{z})$ contributes one of each.

So $-2i$ is also a double root, and $1 - i$ is also simple.

$2 + 2 + 1 + 1 = 6$

That fills the degree exactly, leaving no room for a real root. The list with $1 + i$ twice is wrong because it leaves $1 - i$ unpaired, and the list giving $2i$ multiplicity $2$ but $-2i$ multiplicity $1$ breaks the equal-multiplicity rule.

Degree $8$ means $8$ roots counted with multiplicity, and none of them is real, so all $8$ are nonreal.

Nonreal roots of a real polynomial arrive two at a time, one conjugate pair at a time.

$\frac{8}{2} = 4$

So there are $4$ pairs, and the polynomial factors into $4$ real quadratics with negative discriminant.

Conjugation sends each nonreal root to another nonreal root, and it never fixes one, because $\overline{z} = z$ happens only for real $z$.

So the distinct nonreal roots split into pairs $\{z, \overline{z}\}$ with no leftovers, and their number is twice the number of pairs, say $2m$.

$2m = 3 \;\Longrightarrow\; m = \tfrac{3}{2}$

No whole number of pairs produces $3$, so no such polynomial exists. Multiplicity cannot rescue it, because this is a count of distinct values, not of roots with multiplicity.

The coefficients are rational, so the partner $-1 - \sqrt{2}$ is a root. Note the swap flips the sign of the $\sqrt{2}$ term only, so the partner is not $1 - \sqrt{2}$.

The pair has sum $-2$ and product $(-1)^2 - 2 = -1$, giving the factor $x^2 + 2x - 1$.

$x^3 + x^2 - 3x + 1 = (x^2 + 2x - 1)(x - 1)$

The third root is $1$, which checks out directly: $1 + 1 - 3 + 1 = 0$.

The conjugate root theorem is a one-way street, and its converse is false. Multiplying a polynomial by a constant cannot move any root, so

$i\,(x^2 + 1) = ix^2 + i$

has the perfectly paired roots $i$ and $-i$ while its coefficients are nonreal. So paired roots do not force real coefficients.

What they do force is this: group the nonreal roots into real quadratic factors and the real roots into real linear factors, and the product is a polynomial with real coefficients. Only the leading constant is left free, so $p$ is a constant times a real polynomial. Odd degree and a guaranteed real root both fail on $x^2 + 1$.

Real coefficients make $-i$ a root as well, so $x^2 + 1$ divides the quartic. The long division gives $x^2$, then $-2x$, then $5$, with no remainder.

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

The quotient has discriminant $4 - 20 = -16$, so

$x = \frac{2 \pm 4i}{2} = 1 \pm 2i.$

All four roots are nonreal, in two conjugate pairs.

Rational coefficients are also real coefficients, so both theorems apply. The root $\sqrt{2}$ drags in $-\sqrt{2}$, and the root $i$ drags in $-i$, for four roots in all.

$(x - \sqrt{2})(x + \sqrt{2}) = x^2 - 2, \qquad (x - i)(x + i) = x^2 + 1$

Multiply the two quadratics, each of which has rational coefficients.

$(x^2 - 2)(x^2 + 1) = x^4 - x^2 - 2$

Degree $4$ is the least possible, since four roots are forced.

A counterexample needs a nonreal root whose conjugate is not a root. The last three all have real coefficients, so their roots do pair up ($\pm i$, then $1 \pm i$, then $\pm 2i$), and none of them can serve.

$p(x) = x - i \ \Rightarrow \ p(i) = 0, \qquad p(-i) = -2i \neq 0$

Its coefficient $-i$ is not real, which is exactly why the pairing is free to fail.

Nonreal roots come in pairs, so their count is even. An odd degree therefore always leaves an odd number of real roots, which cannot be zero.

$3 = (\text{real count}) + (\text{even count}) \ \Rightarrow\ \text{real count is odd}$

A real cubic with no real root does not exist. The others all occur: $x^4 + 1$ and $x^2 + 1$ have no real roots (both even degree), and $x(x^2+1)(x^2+4)$ has degree $5$ with the single real root $0$.

The coefficients are real, so $-2i$ is a root too, and the pair gives the real factor $(x - 2i)(x + 2i) = x^2 + 4$.

Divide. The steps of the long division give $x^2$, then $2x$, then $2$, with no remainder.

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

Both factors have negative discriminants ($-16$ and $4 - 8 = -4$), so neither splits any further over the reals, exactly as the factorization theorem promises.

The Rational Root Theorem offers $\pm 1, \pm 2, \pm 5, \pm 10$, and $x = 2$ works, since $8 + 2 - 10 = 0$. Dividing by $x - 2$ leaves a quadratic.

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

Apply the quadratic formula to the quotient, whose discriminant is $4 - 20 = -16$.

$x = \frac{-2 \pm 4i}{2} = -1 \pm 2i$

The two nonreal roots are $-1 + 2i$ and $-1 - 2i$, a conjugate pair as required. The pair $-1+2i$ and $1-2i$ is not a conjugate pair, since conjugation never flips the real part.

Rational coefficients are real coefficients too, so both pairings fire. The root $2 + \sqrt{3}$ brings $2 - \sqrt{3}$, and $1 + i$ brings $1 - i$. That is four roots, filling the degree.

$\big(x - (2+\sqrt{3})\big)\big(x - (2-\sqrt{3})\big) = x^2 - 4x + (4 - 3) = x^2 - 4x + 1$

$\big(x - (1+i)\big)\big(x - (1-i)\big) = x^2 - 2x + 2$

Multiply the two quadratics.

$(x^2 - 4x + 1)(x^2 - 2x + 2) = x^4 - 6x^3 + 11x^2 - 10x + 2$