# Linear Algebra/Factoring and Complex Numbers: A Review

This subsection is a review only and we take the main results as known. For proofs, see (Birkhoff & MacLane 1965) or (Ebbinghaus 1990).
Just as integers have a division operation— e.g., "$4$ goes $5$ times into $21$ with remainder $1$"— so do polynomials.

## Theorem 1.1 (Division Theorem for Polynomials)

Let $c(x)$ be a polynomial. If $m(x)$ is a non-zero polynomial then there are quotient and remainder polynomials $q(x)$ and $r(x)$ such that
    $c(x)=m(x)\cdot q(x)+r(x)$
where the degree of $r(x)$ is strictly less than the degree of $m(x)$.
In this book constant polynomials, including the zero polynomial, are said to have degree $0$. (This is not the standard definition, but it is convienent here.)
The point of the integer division statement "$4$ goes $5$ times into $21$ with remainder $1$" is that the remainder is less than $4$— while $4$ goes $5$ times, it does not go $6$ times. In the same way, the point of the polynomial division statement is its final clause.

## Example 1.2

If $c(x)=2x^{3}-3x^{2}+4x$ and $m(x)=x^{2}+1$ then $q(x)=2x-3$ and $r(x)=2x+3$. Note that $r(x)$ has a lower degree than $m(x)$.

## Corollary 1.3

The remainder when $c(x)$ is divided by $x-\lambda$ is the constant polynomial $r(x)=c(\lambda )$.

## Proof

The remainder must be a constant polynomial because it is of degree less than the divisor $x-\lambda$, To determine the constant, take $m(x)$ from the theorem to be $x-\lambda$ and substitute $\lambda$ for $x$ to get $c(\lambda )=(\lambda -\lambda )\cdot q(\lambda )+r(x)$.
If a divisor $m(x)$ goes into a dividend $c(x)$ evenly, meaning that $r(x)$ is the zero polynomial, then $m(x)$ is a factor of $c(x)$. Any root of the factor (any $\lambda \in \mathbb {R}$ such that $m(\lambda )=0$) is a root of $c(x)$ since $c(\lambda )=m(\lambda )\cdot q(\lambda )=0$. The prior corollary immediately yields the following converse.

## Corollary 1.4

If $\lambda$ is a root of the polynomial $c(x)$ then $x-\lambda$ divides $c(x)$ evenly, that is, $x-\lambda$ is a factor of $c(x)$.
Finding the roots and factors of a high-degree polynomial can be hard. But for second-degree polynomials we have the quadratic formula: the roots of $ax^{2}+bx+c$ are
    $\lambda _{1}={\frac {-b+{\sqrt {b^{2}-4ac}}}{2a}}\qquad \lambda _{2}={\frac {-b-{\sqrt {b^{2}-4ac}}}{2a}}$
(if the discriminant $b^{2}-4ac$ is negative then the polynomial has no real number roots). A polynomial that cannot be factored into two lower-degree polynomials with real number coefficients is irreducible over the reals.

## Theorem 1.5

Any constant or linear polynomial is irreducible over the reals. A quadratic polynomial is irreducible over the reals if and only if its discriminant is negative. No cubic or higher-degree polynomial is irreducible over the reals.

## Corollary 1.6

Any polynomial with real coefficients can be factored into linear and irreducible quadratic polynomials. That factorization is unique; any two factorizations have the same powers of the same factors.
Note the analogy with the prime factorization of integers. In both cases, the uniqueness clause is very useful.

## Example 1.7

Because of uniqueness we know, without multiplying them out, that $(x+3)^{2}(x^{2}+1)^{3}$ does not equal $(x+3)^{4}(x^{2}+x+1)^{2}$.

## Example 1.8

By uniqueness, if $c(x)=m(x)\cdot q(x)$ then where $c(x)=(x-3)^{2}(x+2)^{3}$ and $m(x)=(x-3)(x+2)^{2}$, we know that $q(x)=(x-3)(x+2)$.
While $x^{2}+1$ has no real roots and so doesn't factor over the real numbers, if we imagine a root— traditionally denoted $i$ so that $i^{2}+1=0$— then $x^{2}+1$ factors into a product of linears $(x-i)(x+i)$.
So we adjoin this root $i$ to the reals and close the new system with respect to addition, multiplication, etc. (i.e., we also add $3+i$, and $2i$, and $3+2i$, etc., putting in all linear combinations of $1$ and $i$). We then get a new structure, the complex numbers, denoted $\mathbb {C}$.
In $\mathbb {C}$ we can factor (obviously, at least some) quadratics that would be irreducible if we were to stick to the real numbers. Surprisingly, in $\mathbb {C}$ we can not only factor $x^{2}+1$ and its close relatives, we can factor any quadratic.
    $ax^{2}+bx+c=a\cdot {\bigx-{\frac {-b+{\sqrt {b^{2}-4ac}}}{2a}}{\big )}\cdot {\bigx-{\frac {-b-{\sqrt {b^{2}-4ac}}}{2a}}{\big )}$

## Example 1.9

The second degree polynomial $x^{2}+x+1$ factors over the complex numbers into the product of two first degree polynomials.
    ${\bigx-{\frac {-1+{\sqrt {-3}}}{2}}{\big )}{\bigx-{\frac {-1-{\sqrt {-3}}}{2}}{\big )}={\bigx-(-{\frac {1}{2}}+{\frac {\sqrt {3}}{2}}i){\big )}{\bigx-(-{\frac {1}{2}}-{\frac {\sqrt {3}}{2}}i){\big )}$

## Corollary 1.10 (Fundamental Theorem of Algebra)

Polynomials with complex coefficients factor into linear polynomials with complex coefficients. The factorization is unique.

## References

* Ebbinghaus, H. D. (1990), Numbers, Springer-Verlag.
* Birkhoff, Garrett; MacLane, Saunders (1965), Survey of Modern Algebra (Third ed.), Macmillian.