The Fundamental Theorem of Algebra: Practice

Divide by $(x + 3)$ repeatedly and stop when a remainder is nonzero. Synthetic division of $1, 3, -7, -15, 18$ by $-3$ gives the row $1, 0, -7, 6$ with remainder $0$, so the quotient is $x^3 - 7x + 6$.

Divide again. Synthetic division of $1, 0, -7, 6$ by $-3$ gives $1, -3, 2$ with remainder $0$, so the new quotient is $x^2 - 3x + 2$.

Now test a third time.

$(-3)^2 - 3(-3) + 2 = 9 + 9 + 2 = 20 \ne 0$

The third division fails, so $(x+3)$ divides exactly twice and the multiplicity is $2$. In full, $p(x) = (x + 3)^2(x - 1)(x - 2)$.

A root $i$ of multiplicity $3$ means the factor $(x - i)$ appears three times, so $p(x) = 2(x - i)^3$.

Expand using $i^2 = -1$ and $i^3 = -i$, so that $(-i)^2 = -1$ and $(-i)^3 = i$.

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

Multiplying by $2$ gives $2x^3 - 6ix^2 - 6x + 2i$. Checking at $x = i$: $2i^3 - 6i(i^2) - 6i + 2i = -2i + 6i - 6i + 2i = 0$, as it must be.

The complete factorization $p(x) = a(x - r_1)\cdots(x - r_n)$ has exactly $n$ linear factors, which makes the first, second, and fourth statements true.

The third fails whenever a factor repeats.

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

This has degree $2$ but only one distinct root. A polynomial has exactly $n$ distinct roots precisely when every root is simple, which is a special case rather than the rule.

The multiplicities of the distinct roots add up to the degree, since each of the six linear factors carries exactly one root.

$3 + 2 + m = 6$

So $m = 1$. The multiplicity cannot be $0$, because $i$ was given as a root, and a multiplicity of $0$ would mean the factor $(x - i)$ never appears.

Add the exponents to get the degree, then count the different numbers named by the factors.

$(x - 1)^4(x + 6): \quad \text{degree } 4 + 1 = 5, \quad \text{roots } 1 \text{ and } -6$

The second choice has five distinct roots, the third has only one, and the fourth has three. Only the first has degree $5$ with exactly $2$ distinct roots.

Write one factor per root, repeated as many times as its multiplicity, and put the leading coefficient in front. The root $0$ contributes the factor $x$.

$p(x) = 3 \cdot x \cdot (x - 2)^2 \cdot (x + 1)$

The degree is $1 + 2 + 1 = 4$, as required. The first choice has degree $3$ and no root at $0$, the third gives $0$ multiplicity $2$ and $2$ multiplicity $1$ (the multiplicities are swapped), and the last has leading coefficient $1$.

Since $i$ is a root, $(x - i)$ is a factor, and the other factor must be $(x - r)$ for the second root $r$. Compare the constant terms: $(-i)(-r) = -i$ forces $r = -1$.

Check by expanding.

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

That is exactly $p$, so the other root is $-1$. Notice this polynomial has non-real coefficients, and its two roots are not conjugates of each other.

The theorem is an existence statement. It says a root is there; it never says how to find it.

The first choice is the theorem itself, the second follows by dividing out one root at a time, and the third is the closure statement those two together deliver.

$p(x) = a(x - r_1)(x - r_2)\cdots(x - r_n)$

No formula is included, and for degree $5$ and above Ruffini and Abel proved that no general formula in radicals exists at all.

The first factor gives the root $1$, twice. Solve the quadratic factor with the quadratic formula, whose discriminant is $2^2 - 4(1)(5) = -16$.

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

So the distinct roots are $1$, $-1 + 2i$, and $-1 - 2i$. The degree is $4$, and the multiplicities $2 + 1 + 1$ add to $4$, as they must.

Factor out the largest common power of $x$, then use the difference of squares.

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

The degree is $5$, so there are five roots counted with multiplicity: $0$ three times, plus $1$ and $-1$. Only three different numbers appear, so there are $3$ distinct roots.

Substitute $x = 0$ into the factored form and solve for $r$.

$p(0) = (0 - 2)^2(0 - r) = 4(-r) = -4r$

Setting $-4r = 12$ gives $r = -3$. As a check, $p(x) = (x-2)^2(x+3)$ has $p(0) = 4 \cdot 3 = 12$.

Since $p(1) = 0$, the Factor Theorem gives the factor $(x - 1)$. Synthetic division of $1, 0, 0, -1$ by $1$ produces the row $1, 1, 1$ with remainder $0$, so the quotient is $x^2 + x + 1$.

Apply the quadratic formula to the quotient, whose discriminant is $1 - 4 = -3$.

$x = \frac{-1 \pm \sqrt{-3}}{2} = \frac{-1 \pm i\sqrt{3}}{2}$

A cubic has three roots, and here they are: $1$ and the two non-real ones. Only one of them is real, which is why "the cube root of $1$" is a misleading phrase.