Ruffini’s technique is a practical tool that can be used to factorize polynomials. It’s particularly useful when solving equations that are of a degree higher than the second and cannot be directly reduced to quadratic equations, monomials, or general trinomials. This technique allows us to factorize a polynomial \(P(x)\) of degree \(n \gt 1\) into a polynomial of the form:

\[P(x) = D(x) \cdot Q(x)\]

Here, \(D (x)\) and \(Q(x) \) are two polynomials respectively of degree \(1\) and degree \(n-1\). The process of reducing the degree of a polynomial \ ( P(x)) using Ruffini’s technique is a step-by-step one, which can be carried out mechanically. However, it’s important to note that this method is not the only one available for factorization. Other forms, such as special products or simpler methods, can also be applied in some cases.


The process

Consider a polynomial of the form:

\[a_{n}x^{n}+a_{n-1}x^{n-1}+\dotsb +a_{2}x^{2}+a_{1}x+a_{0}\] \[a_n, a_{n-1},…,a_{0} \in \mathbb{Z} \]

If exist any rational roots, they are in the form:

\[r= \frac{p}{q}\]

  • \(p\) is a divisor of the constant term \(a_0\).
  • \(q\) is a divisor of the maximum grade coefficient \(a_n\).

  • With that being said, let us proceed directly to an illustrative example. Let’s consider the polynomial \(P(x)= x^3-6x^2 + 11x-6\) and look for all roots that satisfy \(r=\frac{p}{q}\). In this case \(p=-6\) and \(q=1\). \[r=\frac{-6}{1} = -6\]

  • Now let’s look for divisors of the term \(-6\): \[r \in \lbrace \pm 1, \pm 2, \pm 3, \pm 6 \rbrace \] To determine which of these values constitute a root of the polynomial, let’s substitute them one by one for the variable \(x\) in the original polynomial and check if the result is \(0\). \(P(x)\). For \(x = 1\) we obtain: \[P(1) = 1^3-6 \cdot 1^2+11 \cdot 1 -6 = 0\] In this case \(P(1)\) it’s a root.

  • Now that we have found a root, we need to construct the following table. In the first row, we insert the coefficients of the initial polynomial \(P(x)\) sorted by degree (if a coefficient of a certain degree \(k\) is missing, remember to insert \(0\) in its place). In the second row, instead, we insert the value of the root we just found, which is \(1\). We have: \begin{array}{c|ccc|c} & 1 & -6 & 11 & -6 \\ 1 &\\ \hline & & & & \ \end{array}

  • We put at the first position of the third row the value of the leading coefficient (third degree in this case): \begin{array}{c|ccc|c} & 1 & -6 & 11 & -6 \\ 1 &\\ \hline & 1 & & & \ \end{array}

  • Now let’s multiply the element marked with \( \overset{a}{1} \) by the element marked with \( \overset{b}{1} \) to obtain the element marked with \( \overset{c}{1} \): \begin{array}{c|ccc|c} & 1 & -6 & 11 & -6 \\ \overset{a}{1} & & \overset{c}{1} \\ \hline & \overset{b}{1} & & & \ \end{array}

  • Sum the element marked with \( \overset{a}{-6} \) with the element marked with \( \overset{b}{1} \) to obtain the element marked with \( \overset{c}{-5} \): \begin{array}{c|ccc|c} & 1 & \overset{a}{-6} & 11 & -6 \\ 1 & & \overset{b}{1} \\ \hline & 1 & \overset{c}{-5} & & \ \end{array}

  • Iterate the process and multiply the element marked with \( \overset{a}{1} \) by the element marked with \( \overset{b}{-5} \) to obtain the element marked with \( \overset{c}{-5} \) :\begin{array}{c|ccc|c} & 1 & -6 & 11 & -6 \\ \overset{a}{1} & & 1 & \overset{c}{-5} \\ \hline & 1 & \overset{b}{-5} & & \\ \end{array}

  • Sum the element marked with \( \overset{a}{11} \) with the element marked with \( \overset{b}{-5} \) to obtain the element marked with \( \overset{c}{-6} \): \begin{array}{c|ccc|c} & 1 & -6 & \overset{a}{11} & -6 \\ 1 & & 1 & \overset{b}{-5} \\ \hline & 1 & -5 & \overset{c}{-6} & \ \end{array}

  • Multiply the element marked with \( \overset{a}{1} \) by the element marked with \( \overset{b}{6} \) to obtain the element marked with \( \overset{c}{6} \): \begin{array}{c|ccc|c} & 1 & -6 & 11 & -6 \\ \overset{a}{1} & & 1 & -5 & \overset{c}{6} \\ \hline & 1 & -5 & \overset{b}{6} & \\ \end{array}

  • Sum the element marked with \( \overset{a}{-6} \) with the element marked with \( \overset{b}{6} \) to obtain the element marked with \( \overset{c}{0} \): \begin{array}{c|ccc|c} & 1 & -6 & 11 & \overset{a}{-6} \\ 1 & & 1 & -5 & \overset{b}{6} \\ \hline & 1 & -5 & 6 & \overset{c}{0} \\ \end{array}


When we reach the end of the table and find that the remainder is zero, then we have found a root \(r\) of the polynomial such that we can factorize it as follows:

\[P(x) = (x-r) \cdot Q(x) \]

  • \(r= 1\) is the root we found at the beginning. \(Q(x)\) is a polynomial of degree \(n-1\), with coefficients arranged by degree in the order they appear in the third row. In our example we have: \[P(x) = (x-1)(x^2-5x+6)\]

  • This polynomial is easily factorizable into: \[(x-1)(x-2)(x-3)\] In this way, we can solve the associated equation by immediately identifying its solutions.


Notes

The application of Ruffini’s rule provides a systematic and efficient method for polynomial division, which leads to a rapid factorization of polynomials and the solution of equations of a higher degree. This mechanical process makes it easier to solve equations beyond the second degree. Overall, Ruffini’s rule facilitates the solution of higher-degree equations in a structured and effective manner.