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.

$ax^{2}+bx+c=a\cdot {\big (}x-{\frac {-b+{\sqrt {b^{2}-4ac}}}{2a}}{\big )}\cdot {\big (}x-{\frac {-b-{\sqrt {b^{2}-4ac}}}{2a}}{\big )}$

Example 1.9

The second degree polynomial $x^{2}+x+1$ factors over the complex numbers into the product of two first degree polynomials.

${\big (}x-{\frac {-1+{\sqrt {-3}}}{2}}{\big )}{\big (}x-{\frac {-1-{\sqrt {-3}}}{2}}{\big )}={\big (}x-(-{\frac {1}{2}}+{\frac {\sqrt {3}}{2}}i){\big )}{\big (}x-(-{\frac {1}{2}}-{\frac {\sqrt {3}}{2}}i){\big )}$

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.