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 Template:Harv or Template:Harv.

Just as integers have a division operation— e.g., "$4$ goes $5$ times into $21$ with remainder $1$"— 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 $4$— 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 $\lambda \in ℝ$ 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 $a{x}^2+bx+c$ are

${\lambda }_1=\frac{-b+\sqrt{b^2-4ac}}{2a}{\lambda }_2=\frac{-b-\sqrt{b^2-4ac}}{2a}$

(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— 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 $ℂ$.

In $ℂ$ we can factor (obviously, at least some) quadratics that would be irreducible if we were to stick to the real numbers. Surprisingly, in $ℂ$ we can not only factor $x^2+1$ and its close relatives, we can factor any quadratic.

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

References

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