The Fundamental Theorem of Algebra: Practice

A product is zero exactly when one of its factors is zero, so set each factor to zero in turn.

$x - 3 = 0 \;\Rightarrow\; x = 3, \qquad x + 2 = 0 \;\Rightarrow\; x = -2, \qquad x - 5 = 0 \;\Rightarrow\; x = 5$

The root of the factor $(x - r)$ is $+r$, so watch the signs: $(x + 2)$ gives the root $-2$, not $2$.

The multiplicity of a root is the number of times its factor appears in the complete factorization.

$p(x) = (x - 4)(x - 4)(x + 1)$

The factor $(x - 4)$ appears twice, so the root $4$ has multiplicity $2$. The multiplicity is not the value of the root, and it is not the degree.

Set the polynomial to zero. A product is zero only when a factor is zero, and every factor here is the same one.

$(x-1)^3 = 0 \;\Rightarrow\; x - 1 = 0 \;\Rightarrow\; x = 1$

There are three linear factors, so there are three roots counted with multiplicity, but they all name the same number. The count of distinct roots is $1$.

For every real number $x$, the square $x^2$ is at least $0$, so $x^2 + 1$ is at least $1$ and never reaches zero.

$x^2 + 1 = 0 \;\Rightarrow\; x^2 = -1$

No real number squares to $-1$, so there are no real roots. This does not mean there are no roots: over the complex numbers $p$ has the two roots $i$ and $-i$.

Look for two numbers whose product is $6$ and whose sum is $-5$. Those numbers are $-2$ and $-3$.

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

The other choices expand to $x^2 + 5x + 6$, $x^2 - 7x + 6$, and $x^2 + x - 6$, none of which is $p$.

The theorem is an existence statement and nothing more: a non-constant polynomial with complex coefficients has at least one complex root.

It supplies no formula (for degree $5$ and above no general formula in radicals exists), and it does not force roots to be real or rational.

$x^2 + 1 = 0 \;\Rightarrow\; x = \pm i$

That example alone rules out the other three claims.

The leading term comes from taking the $x$ out of every factor and multiplying by the number in front.

$5 \cdot x \cdot x \cdot x \cdot x = 5x^4$

So the leading coefficient is $5$. The number $120$ is the constant term, $5(-1)(-2)(-3)(-4)$, which is a different thing entirely.

A degree-$5$ polynomial factors into exactly five linear factors, and every root of $p$ must be the root of one of them.

$p(x) = a(x - r_1)(x - r_2)(x - r_3)(x - r_4)(x - r_5)$

That gives at most five different numbers, so at most $5$ distinct roots. It can have fewer, if some of the $r_i$ repeat.

The second choice is the Fundamental Theorem of Algebra, stated correctly.

The first fails on $x^2 + 1$, which has real coefficients and no real root. The third fails on $(x-1)^2$, which has degree $2$ and only one distinct root. The fourth fails because four linear factors cannot supply five different roots.

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

Set each factor to zero. The factor $(x - r)$ vanishes at $x = r$.

$x - i = 0 \;\Rightarrow\; x = i, \qquad x + i = 0 \;\Rightarrow\; x = -i, \qquad x - 2 = 0 \;\Rightarrow\; x = 2$

The degree is $3$, so there are three roots and the list is complete. Dropping $-i$ would leave the list short.

A root $r$ of multiplicity $m$ contributes the factor $(x - r)$ exactly $m$ times, and the leading coefficient sits out front.

$p(x) = 1 \cdot (x - 7)(x - 7)(x - 7) = (x - 7)^3$

The second choice has the root $-7$, the third has degree $1$, and the fourth has two different roots.

The Fundamental Theorem of Algebra requires degree at least $1$, and $p(x) = 4$ is a nonzero constant of degree $0$.

$4 = 0 \text{ is never true}$

So it has no roots. The others all do: $x^2 + 1$ has $i$ and $-i$, $x^3$ has the root $0$ with multiplicity $3$, and $x - 5$ has the root $5$.