Factoring a Polynomial

Remember ``factoring polynomials''? Consider the second-order polynomial

$$p(x) = x^2-5x+6.$$

It is second-order because the highest power of $x$ is $2$ (only non-negative integer powers of $x$ are allowed in this context). The polynomial is also monic because its leading coefficient, the coefficient of $x^2$ , is $1$ . By the fundamental theorem of algebra (discussed further in §2.4), there are exactly two roots (or zeros) of any second order polynomial. These roots may be real or complex (to be defined). For now, let's assume they are both real and denote them by $r_1$ and $r_2$ . Then we have $p(r_1)=0$ and $p(r_2)=0$ , and we can write

$$p(x) = (x-r_1)(x-r_2).$$

This is the factored form of the monic polynomial $p(x)$ . (For a non-monic polynomial, we may simply divide all coefficients by the first to make it monic, and this doesn't affect the zeros.) Multiplying out the symbolic factored form gives

$$p(x) = (x-r_1)(x-r_2) = x^2 - (r_1 + r_2)x + r_1 r_2.$$

Comparing with the original polynomial, we find we must have

$$\begin{eqnarray*} r_1+r_2 &=& 5 \\ r_1 r_2 &=& 6. \end{eqnarray*}$$

This is a system of two equations in two unknowns. Unfortunately, it is a nonlinear system of two equations in two unknowns.^2.1 Nevertheless, because it is so small, the equations are easily solved. In beginning algebra, we did them by hand. However, nowadays we can use a software tool such as Matlab or Octave to solve very large systems of linear equations.

The factored form of this simple example is

$$p(x) = x^2-5x+6 = (x-r_1)(x-r_2) = (x-2)(x-3).$$

Note that polynomial factorization rewrites a monic $n$ th-order polynomial as the product of $n$ first-order monic polynomials, each of which contributes one zero (root) to the product. This factoring business is often used when working with digital filters [71].