Given a polynomial $p(x)$, one question that may be asked is "What is the multiplicity of a root $x_0$?" We will describe two algorithms that will yeild that multiplicty.

Polynomial division

Here is an algorithm that determines the multiplicity of a root using polynomial division:

Count the number of times that you can repeatedly divide $p(x)$ by $x - x_0$ and still get a remainder of zero.

• If after the first division, the remainder is not zero, then $x_0$ is not a root and we could say that the multiplicity is zero.
• Otherwise, if we were able to divide out $x - x_0$ a total of $k$ times, and each of the $k$ times the remainder was zero, but with the $(k + 1)$^th polynomial division, the remainder was non-zero, then the multiplicity of the root $x_0$ is $k$.

Given the polynomial

$p(x) = x^6 + 2x^5 - 8x^4 - 14x^3 + 11x^2 + 28x + 12$,

demonstrate that $0$, $-3$, $2$ and $-1$ have multiplicty 0, 1, 2 and 3, respectively.

Differentiation

The previous algorithm requires that polynomial devision be performed $k + 1$ times to determine that the polynomial has a root of multiplicity $k$ at a given point. We will now look at an algorithm that requires the user to simply calculate derivatives and to evaluate those derivatives at the point in question.

Given a polynomial $p(x)$, if $p(x_0) \ne 0$ then $x_0$ is not a root and therefore the multiplicty is said to be zero.
Otherwise, if $p(x_0) = 0$ and we are able to differentiate the polynomial and evaluate the derivative at $x_0$, with each derivative up to the $k$^th evaluated at $x_0$ being zero, but the $k$^th derivative evaluated at $x_0$ is non-zero, then the polynomial has a root of multiplicty $k$ at the pointer $x_0$.

Given the polynomial

$p(x) = x^6 + 2x^5 - 8x^4 - 14x^3 + 11x^2 + 28x + 12$,

demonstrate that $0$, $-3$, $2$ and $-1$ have multiplicty 0, 1, 2 and 3, respectively.

1. The point $x = 0$ has multiplicty zero because $p(0) = 12$, which is not zero.
2. The point $x = -3$ has multiplicty one because $p(-3) = 0$ but $\frac{d}{dx}p(-3) = -200 \ne 0$.
3. The point $x = 2$ has multiplicty two because $p(2) = 0$, $\frac{d}{dx}p(2) = 0, but $\frac{d^2}{dx^2}p(2) = 270 \ne 0$.
4. The point $x = -1$ has multiplicty three because $p(-1) = 0$, $\frac{d}{dx}p(-1) = \frac{d^2}{dx^2}p(-1) = 0, but $\frac{d^3}{dx^3}p(-1) = 108 \ne 0$.