Complex and Irrational Roots: Additional Practice (Set 1)

The coefficients are real, so $1 + 3i$ is a root as well. The pair has sum $2$ and product $1^2 + 3^2 = 10$, so it forces the factor $x^2 - 2x + 10$.

Divide. The first step gives $x$, leaving $-2x^2 + 4x - 20$, which is $-2$ times the divisor, so the remainder is zero.

$x^3 - 4x^2 + 14x - 20 = (x^2 - 2x + 10)(x - 2)$

The real root is $2$.

Rational coefficients are also real, so both theorems apply: $\sqrt{5}$ brings $-\sqrt{5}$, and $2i$ brings $-2i$. Each pair supplies a quadratic factor.

$(x - \sqrt{5})(x + \sqrt{5}) = x^2 - 5, \qquad (x - 2i)(x + 2i) = x^2 + 4$

Multiply them.

$(x^2 - 5)(x^2 + 4) = x^4 + 4x^2 - 5x^2 - 20 = x^4 - x^2 - 20$

The partner $4 + \sqrt{11}$ is a root as well. Their sum is $8$, and their product cancels the surd.

$(4 - \sqrt{11})(4 + \sqrt{11}) = 16 - 11 = 5$

So the forced factor is

$x^2 - 8x + 5.$

The general shape is $x^2 - 2ax + (a^2 - b^2 d)$ with $a = 4$, $b = 1$, $d = 11$.

The conjugate $-2 - 5i$ must be the other root. Their sum is $-4$ and their product is $(-2)^2 + 5^2 = 29$.

$\big(x - (-2+5i)\big)\big(x - (-2-5i)\big) = x^2 - (-4)x + 29 = x^2 + 4x + 29$

Subtracting a negative sum is what produces the $+4x$, and the product of a conjugate pair is $a^2 + b^2$, never $a^2 - b^2$.

The conjugate root theorem guarantees the second item, the factorization argument guarantees the third, and the pair $z, \overline{z}$ builds the real quadratic factor $x^2 - 2ax + (a^2+b^2)$, which guarantees the fourth.

A real root, however, is guaranteed only for odd degree.

$p(x) = x^2 + 1 \quad\text{has roots } i \text{ and } -i \text{ and no real root at all}$

Even-degree real polynomials can avoid the real axis entirely.

The $4$ distinct nonreal roots form two conjugate pairs (which is consistent, since the distinct nonreal roots always come in an even number), and each occurs twice.

$4 \times 2 = 8 \ \text{ nonreal roots counted with multiplicity}$

A degree-$9$ polynomial has $9$ roots in all, so

$9 - 8 = 1$

leaves exactly one real root. That agrees with the odd-degree rule, which promised at least one.

A root and its conjugate share a multiplicity, so $2 - i$ is a double root as well.

$2 + 2 = 4 \ \text{ nonreal roots counted with multiplicity}$

The degree is $5$, so

$5 - 4 = 1$

leaves one real root. It could not have been zero: the degree is odd, so a real root is guaranteed.

Apply the quadratic formula to each factor. The first has discriminant $36 - 52 = -16$.

$x = \frac{6 \pm 4i}{2} = 3 \pm 2i$

The second has discriminant $4 - 20 = -16$.

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

Dividing by the $2$ in the denominator is what turns $4i$ into $2i$. Both factors deliver a conjugate pair, which is what real coefficients demand.

Both proofs push a swap through the polynomial. The swap slides past every $+$ and every $\times$ for free, so the only thing that can go wrong is a coefficient being altered on the way.

$\overline{a_k} = a_k \ \text{ needs } a_k \text{ real}, \qquad a_k^{*} = a_k \ \text{ needs } a_k \text{ rational}$

Complex conjugation fixes exactly the real numbers, and the surd swap fixes exactly the rational ones. In each case the coefficients must live inside the swap's fixed set, and that one requirement produces both hypotheses.

The fourth root is forced: $1 - i$, the conjugate of $1 + i$. The nonreal pair has sum $2$ and product $1^2 + 1^2 = 2$, so it gives the factor $x^2 - 2x + 2$.

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

Evaluate at $0$ without expanding.

$p(0) = (-2)(-3)(2) = 12$

Real coefficients force $1 - i$ to be the third root. That pair has sum $2$ and product $1^2 + 1^2 = 2$, giving the factor $x^2 - 2x + 2$.

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

Expand.

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

The $x^2$ terms cancel, which is why the cubic has no quadratic term. Check: $p(-2) = -8 + 4 + 4 = 0$.

Rational coefficients are also real, so both pairings must hold at once: every surd root needs its partner, and every nonreal root needs its conjugate.

In the first list the surds are paired, but $1 + i$ has no $1 - i$ beside it, and a complete root list has no room to hide one.

$1 + i \text{ present}, \qquad 1 - i \text{ missing}$

That is exactly what the conjugate root theorem forbids, so no rational polynomial has exactly those three roots. The other three lists are legitimate, and each is realized by a rational polynomial:

$(x^2 - 2x - 1)(x - 3), \qquad (x^2 - 3)(x^2 + 1), \qquad (x - 4)^3$