Suppose you have a polynomial that is factored: that is the polynomial is written in the form $p(x) = c(a_1 x + b_1)(a_2 x + b_2)\cdots (a_n x + b_n)$ where $c$ is the leading coefficient and the degree of the polynomial is $n$. The coefficients $a_k$ are assumed to be non-zero. What are the roots of this polynomial?

This is a simpler algorithm, for a root of a polynomial is when that polynomial equals zero, so if $p(x) = 0$, then you know that:

• If a product of $n$ numbers is zero, then at least one of those numbers must be zero.
• If a product of $n$ numbers is non-zero, then none of the numbers can be zero.

Right from the start, we know that if $c = 0$, then we have the zero polynomial, so every point is a "root" (although, its just easier to say it is the zero polynomial). If $c \ne 0$, then the only way for the polynomial to be zero is if any one of the terms in product $(a_k x + b_k)$ equals zero.

You know from algebra that $a_k x + b_k = 0$ if and only if $a_k x = -b_k$, which equals zero if and only if $x = -\frac{b_k}{a_k}$, and you will recall that we assumed that all of the $a_k$s were non-zero, so this division is valid.

Thus, to find all the roots, we simply solve each term in the product as an equation. For example, suppose we have factor the polynomial and we find that the factored polynomial is $-3(2x - 9)(3x + 1)(x + 5)(3x - 7)(2x - 9)$, so the roots are the solution to $2x - 9 = 0$, $3x + 1 = 0$, $x + 5 = 0$, $3x - 7 = 0$, and $2x - 9 = 0$; and thus the roots are $x = \frac{9}{2}$, $x = -\frac{1}{3}$, $x = -5$, $x = \frac{7}{3}$ and the last root is a repeat of one we've already found, and so we say that $x = \frac{9}{2}$ is a double root.

When factoring a polynomial, if we find that the discriminant of a quadratic is negative, this means that the roots are complex conjugate pairs. In this case, the term in the factorization is a quadratic and not a linear polynomial. Thus, the polynomial may be of the form $p(x) = c(a_1 x + b_1)\cdots (a_{n_1} x + b_{n_1})(d_1x^2 + e_1x + f_1)\cdots (d_{n_2}x^2 + e_{n_2}x + f_{n_2})$ where the degree of the polynomial is $n_1 + 2n_2$.

In this case, the roots are: $-\frac{b_k}{a_k}$ for $k = 1, \ldots, n_1$ and $-\frac{e_k}{2d_k} + \frac{\sqrt{4d_kf_k - e_k^2}}{2d_k}j$ and and $-\frac{e_k}{2d_k} - \frac{\sqrt{4d_kf_k - e_k^2}}{2d_k}j$ for $k = 1, \ldots, n_2$.