The Fundamental Theorem of Algebra: Additional Practice (Set 2)

Group the terms in pairs and factor each pair.

$x^3 - 3x^2 + 4x - 12 = x^2(x - 3) + 4(x - 3) = (x - 3)(x^2 + 4)$

The first factor gives $x = 3$. For the second, $x^2 = -4$, so $x = \pm 2i$, since $(2i)^2 = 4i^2 = -4$.

The complete factorization is $(x - 3)(x - 2i)(x + 2i)$: three roots for a cubic, exactly as promised.

Divide by $(x - 1)$ and keep going until a remainder is nonzero. Synthetic division of $1, -2, 2, -2, 1$ by $1$ gives the row $1, -1, 1, -1$ with remainder $0$, so the quotient is $x^3 - x^2 + x - 1$.

Divide again. Synthetic division of $1, -1, 1, -1$ by $1$ gives $1, 0, 1$ with remainder $0$, so the new quotient is $x^2 + 1$.

Test a third time.

$1^2 + 1 = 2 \ne 0$

The third division fails, so the multiplicity is exactly $2$ and $p(x) = (x - 1)^2(x^2 + 1)$.

Factor out the $2$, then use the difference of squares twice.

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

The four roots $1$, $-1$, $i$, and $-i$ are all different, so every root is simple and there are $4$ distinct roots, matching the degree.

Factor the quadratic piece over the complex numbers, since $x^2 + 1 = (x - i)(x + i)$.

$p(x) = (x - 1)(x - 1)(x - i)(x + i)$

Four linear factors give four roots counted with multiplicity, but the question asks for distinct roots. The two copies of $(x-1)$ name the same number, so the distinct roots are $1$, $i$, and $-i$: three of them, with $1$ carrying multiplicity $2$.

Two successful divisions mean $p(x) = (x - 2)^2 q(x)$, where $q$ is the second quotient.

The third division has remainder $7$, and by the Remainder Theorem that remainder is the value $q(2)$.

$p(x) = (x - 2)^2 q(x), \qquad q(2) = 7 \ne 0$

That is exactly the definition of multiplicity $2$: the factor comes out twice and no more.

Counted with multiplicity, the number of roots is the degree, and degrees add when factors multiply.

$\deg p = 3 \cdot 2 + 1 = 7$

In full, $p(x) = (x - i)^3(x + i)^3(x - 5)$, which is seven linear factors: the roots $i$ and $-i$ each with multiplicity $3$, and $5$ with multiplicity $1$.

An argument that uses only the ordinary rules of arithmetic never asks what kind of numbers the coefficients are, so it would run word for word over the rationals. But the statement fails there.

$x^2 - 2 \quad \text{has rational coefficients and no rational root}$

A proof built from algebraic rules alone would therefore prove something false. Any correct proof must use a property the reals have and the rationals lack, namely completeness, which is what makes continuity work.

The complete factorization of a degree-$4$ polynomial has exactly four linear factors, and every root of $p$ must be the root of one of them.

$p(x) = a(x - r_1)(x - r_2)(x - r_3)(x - r_4)$

Four factors can name at most four different numbers, so five distinct roots is impossible. All the others happen: $(x-1)^4$ has one, $(x-1)^2(x-2)^2$ has two, and $(x-1)(x-2)(x-3)(x-4)$ has four.

First factor the inside, since $x^2 + 4 = (x - 2i)(x + 2i)$, then square.

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

That is four linear factors, so the degree is $4$ and there are $4$ roots counted with multiplicity. Only two different numbers appear, $2i$ and $-2i$, each with multiplicity $2$, so there are $2$ distinct roots.

The complete factorization has exactly $n$ linear factors, and each factor carries exactly one root, so the multiplicities count the factors.

$m_1 + m_2 + \cdots + m_k = n$

The count $k$ of distinct roots is at most $n$ but need not equal it, and the multiplicities are free to differ from one another, so the other three equations can all fail.

Substitute $x = 1 + i$ and use $(1 + i)^2 = 1 + 2i + i^2 = 2i$.

$(1+i)^2 - 2(1+i) + 2 = 2i - 2 - 2i + 2 = 0$

So $1 + i$ is a root of $x^2 - 2x + 2$. The others miss: $x^2 + 2x + 2$ has roots $-1 \pm i$, $x^2 - 2x - 2$ has roots $1 \pm \sqrt{3}$, and $x^2 + 2$ has roots $\pm i\sqrt{2}$.

The five linear factors must all name the same root, since there is only one distinct root, and its multiplicity therefore has to be $5$.

$p(x) = 1 \cdot \bigl(x - (-2)\bigr)^5 = (x + 2)^5$

The second choice has the root $2$, the third has degree $1$, and the fourth does not vanish at $-2$ at all, since $(-2 + 2)^5 + 1 = 1$.