Ruffini's Rule

The mechanics of Ruffini's rule

Write the coefficients of $P(x)$ in a row (with a $0$ for any missing degrees) and the value $a$ separately. The first coefficient is brought down unchanged; each following result is obtained by multiplying the last number in the bottom row by $a$ and adding the coefficient in its column.

Ruffini also evaluates and confirms roots

The last number in the Ruffini row is the remainder, and the remainder of dividing by $(x-a)$ is, by the remainder theorem, $P(a)$. So Ruffini can be used to calculate the numerical value of a polynomial without substituting term by term, and to check whether a number is a root: if the remainder is $0$, it is.

Factorising with a known root

When a root $a$ is already known, Ruffini gives the quotient directly: a factor of one degree lower. By repeating the process with the quotient (if another obvious root appears) you arrive at the complete factorisation.

Unknown parameters from a Ruffini condition

When the polynomial has an undetermined coefficient, a divisibility condition ("it is divisible by $x-a$", "the remainder when dividing by $x-a$ is $c$") turns into an equation with Ruffini: the last number in the row must equal whatever the condition says.

Ruffini in modelling problems

Volumes, profits and trajectories are expressed as polynomials in which it is useful to find factors: a volume that is a product of dimensions, or the moments at which a profit passes through zero.

When the divisor is not exactly $x-a$

Ruffini is designed for dividing by $x-a$, with coefficient $1$ on the $x$. Dividing by $x-\frac12$ is no problem: $a=\frac12$. But to divide by a binomial such as $2x-3$, you must first take out the $2$ as a factor: $2x-3=2\left(x-\frac32\right)$, apply Ruffini with $a=\frac32$, and then divide the quotient obtained by $2$ to compensate.

Designing, testing and comparing an algorithm

Ruffini also allows you to evaluate a polynomial. For $P(x)=a_nx^n+\cdots+a_1x+a_0$, the Horner scheme chains together multiplications and additions. First design how you would do it with pencil and paper; then compare your procedure with this pseudocode:

entrada: coeficientes [a_n, ..., a_0] y valor t

v ← a_n

para cada coeficiente c restante, de mayor a menor grado:

    v ← v * t + c

devolver v

Each arrow represents the next multiplication and addition:

For a polynomial of degree n, Horner performs n multiplications and n additions. If your direct algorithm calculates each power from scratch by repeated products and then multiplies by its coefficient, it uses $n(n+1)/2$ multiplications and n additions. If it reuses powers, the count changes: always state which algorithm you are comparing. For the cubic above, the general counts are 3 versus 6 multiplications, without simplifying zero terms.

Modify the algorithm to evaluate the same polynomial at a list of values, resetting v for each value. Write down the modification, run it by hand or in a program, and keep a table of input, expected output and actual output. Also check $t=1$: the result must be the sum of the coefficients. Explain why, after processing every coefficient, v represents exactly the nested polynomial $((2t-3)t+0)t+5$.

Summary