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 or .''

Just as integers have a division operation&mdash; e.g., "$$ 4 $$ goes $$ 5 $$ times into $$ 21 $$ with remainder $$ 1 $$"&mdash; so do polynomials.

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 $$ 4 $$&mdash; 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.

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.

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.

Note the analogy with the prime factorization of integers. In both cases, the uniqueness clause is very useful.

While $$ x^2+1 $$ has no real roots and so doesn't factor over the real numbers, if we imagine a root&mdash; traditionally denoted $$ i $$ so that $$ i^2+1=0 $$&mdash; 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) $$