The Fundamental Theorem of Algebra: Additional Practice (Set 1)

The five multiplicities are positive integers that add to the degree.

$m_1 + m_2 + m_3 + m_4 + m_5 = 8$

To make one of them as large as possible, make the other four as small as possible, so set each of those to $1$. That uses up $4$, leaving $8 - 4 = 4$ for the last one. A multiplicity of $5$ or more would force the total above $8$.

Group the terms in pairs and pull out the common factor from each pair.

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

The first factor gives $x = -2$. The second gives $x^2 = -4$, so $x = \pm 2i$. A cubic has three roots, and $-2$, $2i$, $-2i$ are all of them.

Factor the inside first, then cube.

$p(x) = \bigl((x-1)(x+1)\bigr)^3 = (x - 1)^3(x + 1)^3$

The degree is $6$, so there are six roots counted with multiplicity. But the factors name only two different numbers, $1$ and $-1$, each with multiplicity $3$. So there are $2$ distinct roots.

Add the exponents for the degree, then count the different numbers the factors name.

$(x-1)^4(x-2)(x-3): \quad 4 + 1 + 1 = 6, \quad \text{roots } 1,\ 2,\ 3$

The second choice has degree $4$, the third has six distinct roots, and the fourth has only one. Only the first has degree $6$ with exactly three distinct roots.

Find the roots first. The discriminant is $(-6)^2 - 4(3)(15) = 36 - 180 = -144$.

$x = \frac{6 \pm \sqrt{-144}}{6} = \frac{6 \pm 12i}{6} = 1 \pm 2i$

The complete factorization keeps the leading coefficient out front: $p(x) = 3\bigl(x - (1 + 2i)\bigr)\bigl(x - (1 - 2i)\bigr)$. The first choice drops the $3$ and so expands to $x^2 - 2x + 5$, not $p$.

The Rational Root Theorem gives the candidates $\pm 1, \pm 2, \pm 3, \pm 6$. Testing them, $p(3) = 27 - 45 + 24 - 6 = 0$, while $p(2) = 8 - 20 + 16 - 6 = -2$.

Synthetic division of $1, -5, 8, -6$ by $3$ gives $1, -2, 2$ with remainder $0$, so the quotient is $x^2 - 2x + 2$. Its discriminant is $4 - 8 = -4$.

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

The three roots are $3$, $1 + i$, and $1 - i$.

Multiply out the leading term of every factor. Each contributes one $x$, so the highest-degree term of the product is $a x^n$.

$a \cdot x \cdot x \cdots x = a x^n$

So $a$ is the leading coefficient. The constant term is instead $a(-r_1)(-r_2)\cdots(-r_n)$, which is a different number, and $a$ is $1$ only when $p$ happens to be monic.

Suppose $p(c) = 0$ for some complex number $c$. Substituting into the factorization gives a product equal to zero.

$a(c - r_1)(c - r_2)\cdots(c - r_n) = 0$

The zero-product property then forces one factor to vanish. The leading coefficient $a$ is never zero, so $c - r_i = 0$ for some $i$, meaning $c = r_i$. The numbers $r_i$ need not be distinct, so the second choice is not even true in general.

Since $3$ is a root, $(x - 3)$ is a factor, and the other factor is $(x - r)$ for the remaining root $r$. Match the constant terms: $(-3)(-r) = 3i$, so $3r = 3i$ and $r = i$.

Check by expanding.

$(x - 3)(x - i) = x^2 - ix - 3x + 3i = x^2 - (3 + i)x + 3i$

The other root is $i$. This polynomial has non-real coefficients, so there is no reason for its roots to come in conjugate pairs, and indeed they do not.

The Fundamental Theorem of Algebra applies to polynomials of degree at least $1$, and $p(x) = -7$ has degree $0$. The equation $-7 = 0$ has no solutions, so it has no roots at all.

Every other choice is non-constant and therefore does have a root: $x$ has the root $0$, and $x^2 + 3$ has the roots $\pm i\sqrt{3}$. Even the one with a non-real coefficient works out.

$ix - 2 = 0 \;\Rightarrow\; x = \frac{2}{i} = -2i, \qquad \text{since} \qquad i(-2i) - 2 = 2 - 2 = 0$

With two distinct roots there are two multiplicities, and they must add up to the degree.

$4 + 1 = 5$

That works, giving a factorization like $a(x - s_1)^4(x - s_2)$. The other pairs sum to $4$, $6$, and $6$, so none of them can belong to a degree-$5$ polynomial.

Degrees add when polynomials are multiplied; they do not multiply.

$\deg p = 10 + 7 = 17$

So there are $17$ roots counted with multiplicity. The factors name only two different numbers, $2$ (multiplicity $10$) and $-3$ (multiplicity $7$), so there are just $2$ distinct roots. As a check, $10 + 7 = 17$ is the degree, exactly as the counting theorem requires.