The Fundamental Theorem of Algebra: Practice

Since $1$ is a root, the Factor Theorem says $(x - 1)$ divides $p$. Synthetic division of $1, -6, 11, -6$ by $1$ gives the row $1, -5, 6$ with remainder $0$, so the quotient is $x^2 - 5x + 6$.

Factor the quotient by finding two numbers with product $6$ and sum $-5$, namely $-2$ and $-3$.

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

The last choice has the wrong quotient. Its quadratic is $x^2 - 5x - 6$, and $(x - 1)(x^2 - 5x - 6)$ expands to $x^3 - 6x^2 - x + 6$, which is not $p$. Even with the correct quotient $x^2 - 5x + 6$ it would still not be a complete factorization, because that quadratic factors further as $(x - 2)(x - 3)$.

Divide out the known root. Synthetic division of $1, -4, 9, -10$ by $2$ gives $1, -2, 5$ with remainder $0$, so the quotient is $x^2 - 2x + 5$.

Apply the quadratic formula to the quotient.

$x = \frac{2 \pm \sqrt{4 - 20}}{2} = \frac{2 \pm \sqrt{-16}}{2} = \frac{2 \pm 4i}{2} = 1 \pm 2i$

The three roots are $2$, $1 + 2i$, and $1 - 2i$, which is three roots for a degree-$3$ polynomial, exactly as required.

Each of the five linear factors carries exactly one root, so the multiplicities of the distinct roots count the factors and must account for all of them.

$m_1 + m_2 + m_3 = 5$

For example the multiplicities could be $3, 1, 1$ or $2, 2, 1$. They cannot all be $1$, since three ones sum to $3$, not $5$.

The candidates are the fractions (factor of $6$) over (factor of $2$), so $\pm 1, \pm 2, \pm 3, \pm 6, \pm \frac{1}{2}, \pm \frac{3}{2}$. Test the four offered.

$p(3) = 2(27) - 3(9) - 11(3) + 6 = 54 - 27 - 33 + 6 = 0$

The others miss: $p(1) = -6$, $p(-3) = -42$, and $p(2) = -12$. So $3$ is the root, and dividing it out leaves $2x^2 + 3x - 2$.

Multiplicity counts how many times $(x - 2)$ divides $p$, so one successful division proves only that the multiplicity is at least $1$. Test the quotient at $2$.

$2^3 - 3(2)^2 + 4 = 8 - 12 + 4 = 0$

The remainder is zero, so $(x - 2)$ goes in again and the multiplicity is at least $2$. Keep dividing until a remainder comes out nonzero. (The quotient is a cubic, so the quadratic formula does not apply to it.)

Degrees add, so the degree is $3 + 1 = 4$, and counted with multiplicity a degree-$4$ polynomial has exactly $4$ roots.

$\text{roots with multiplicity: } 2,\ 2,\ 2,\ -1$

Only two different numbers appear in that list, so there are $2$ distinct roots. The two counts are $4$ and $2$.

A factorization is complete when every factor is linear, since the theorem guarantees a degree-$n$ polynomial splits into $n$ linear factors.

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

is three linear factors. The others all still contain a quadratic that factors further: $x^2 + 1 = (x-i)(x+i)$, $x^2 - 4 = (x-2)(x+2)$, and $x^2 + 4 = (x - 2i)(x + 2i)$.

Each root $r$ contributes the factor $(x - r)$, and the root $0$ contributes the factor $x$ itself. The leading coefficient goes in front.

$p(x) = 2\bigl(x - (-1)\bigr)(x - 0)(x - 4) = 2x(x + 1)(x - 4)$

The first choice has degree $2$ and no root at $0$, the third has roots $1$ and $-4$ (wrong signs), and the last has leading coefficient $1$.

The discriminant is $(-6)^2 - 4(1)(13) = 36 - 52 = -16$, which is negative, so both roots are non-real.

$x = \frac{6 \pm \sqrt{-16}}{2} = \frac{6 \pm 4i}{2} = 3 \pm 2i$

Dividing the whole numerator by $2$ is the step most often fumbled: both the $6$ and the $4i$ get halved.

The first factor gives $x = -2$ (twice). For the second, solve $x^2 + 9 = 0$, so $x^2 = -9$.

$x^2 = -9 \;\Rightarrow\; x = \pm 3i \qquad \text{since} \qquad (3i)^2 = 9i^2 = -9$

The complete factorization is $(x+2)(x+2)(x - 3i)(x + 3i)$: four roots with multiplicity, and three distinct ones, namely $-2$, $3i$, and $-3i$.

The definition of multiplicity is exactly this shape: $r$ has multiplicity $m$ when $p(x) = (x - r)^m q(x)$ with $q(r) \ne 0$.

$p(x) = (x - 1)^2 q(x), \qquad q(1) = 5 \ne 0$

Since the leftover factor does not vanish at $1$, no further copy of $(x-1)$ can be pulled out, so the multiplicity is exactly $2$. The value $5$ is a value of $q$, not a multiplicity.

Degrees add, so the degree is $2 + 2 = 4$, and there are four linear factors.

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

The factors name only two different numbers, $3$ and $-3$, each with multiplicity $2$. So the degree is $4$ and there are $2$ distinct roots.