Ruffini’s Rule
What Ruffini’s Rule Says
If you’re given a polynomial P(x) and one of its roots, say x1, Ruffini’s rule lets you divide P(x) by the binomial (x − x1) to find the quotient polynomial Q(x): $$ Q(x) = \frac{P(x)}{(x - x_1)} $$ This means the original polynomial can be expressed as: $$ P(x) = (x - x_1) \cdot Q(x) $$ Here, Q(x) is a polynomial of one degree lower than P(x).
Why Use It?
Ruffini’s method is a handy shortcut for factoring polynomials or solving polynomial equations by breaking them down into simpler, lower-degree components.
It’s particularly useful for reducing the degree of a higher-degree equation like P(x) = 0, making it more manageable.
Keep in mind: to apply Ruffini’s rule, you need to know at least one root of the polynomial.
How Do You Find the Roots? If the polynomial P(x) has integer coefficients and the leading coefficient is 1, any rational roots (if they exist) must be integer divisors of the constant term.
A Step-by-Step Example
Suppose we have this third-degree polynomial:
$$ x^3 + 4x^2 + 5x + 2 $$
Heads up: Make sure the polynomial is written in standard form, with terms arranged in descending powers of x.
We’re looking for a factor of the form (x − a),
where a can be any integer, positive or negative:
$$ (x - a) $$
To find such a factor, we look for a root - a number that makes P(x) equal to zero.
$$ (x + 1),\ (x - 1),\ (x + 2),\ (x - 2),\ (x + 3),\ (x - 3),\ \ldots $$
We start by testing a = 1 by plugging it into P(x). It’s not a root:
$$ P(1) = 1^3 + 4(1)^2 + 5(1) + 2 = 12 $$
Trying a = -1 instead, we find that it is a root:
$$ P(-1) = (-1)^3 + 4(-1)^2 + 5(-1) + 2 = 0 $$
Note: Since the leading coefficient is 1, we can test only the integer divisors of the constant term (2): ±1, ±2. In this case, -1 works.
So now that we know a = -1 is a root, our divisor is:
$$ (x - a) = (x - (-1)) = (x + 1) $$
Next, we divide the polynomial by (x + 1):
$$ \frac{x^3 + 4x^2 + 5x + 2}{x + 1} = Q(x) $$
We do this using Ruffini’s synthetic division:
$$ \begin{array}{c|lcc|r} & 1 & 4 & 5 & 2 \\ -1 & & -1 & -3 & -2 \\ \hline & 1 & 3 & 2 & 0 \end{array} $$
How Ruffini’s Division Works: Start by listing the coefficients of the polynomial: 1, 4, 5, 2. $$ \begin{array}{c|lcc|r} & 1 & 4 & 5 & 2 \\ & & & & \\ \hline & & & & \end{array} $$ Now place a = -1 in the margin. That’s the opposite of the constant in the divisor (x + 1). $$ \begin{array}{c|lcc|r} & 1 & 4 & 5 & 2 \\ -1 & & & & \\ \hline & & & & \end{array} $$ Bring down the first coefficient (1) into the result row: $$ \begin{array}{c|lcc|r} & 1 & 4 & 5 & 2 \\ -1 & & & & \\ \hline & 1 & & & \end{array} $$ Multiply 1 by -1 and write it above the second column: $$ \begin{array}{c|lcc|r} & 1 & 4 & 5 & 2 \\ -1 & & -1 & & \\ \hline & 1 & & & \end{array} $$ Add: 4 + (-1) = 3 $$ \begin{array}{c|lcc|r} & 1 & 4 & 5 & 2 \\ -1 & & -1 & & \\ \hline & 1 & 3 & & \end{array} $$ Continue: 3·(-1) = -3 $$ \begin{array}{c|lcc|r} & 1 & 4 & 5 & 2 \\ -1 & & -1 & -3 & \\ \hline & 1 & 3 & & \end{array} $$ Add: 5 + (-3) = 2 $$ \begin{array}{c|lcc|r} & 1 & 4 & 5 & 2 \\ -1 & & -1 & -3 & -2 \\ \hline & 1 & 3 & 2 & \end{array} $$ Then: 2·(-1) = -2 And 2 + (-2) = 0 $$ \begin{array}{c|lcc|r} & 1 & 4 & 5 & 2 \\ -1 & & -1 & -3 & -2 \\ \hline & 1 & 3 & 2 & 0 \end{array} $$ If the remainder is 0, the division worked. The bottom row gives the coefficients of the quotient: x2 + 3x + 2
So we’ve found the quotient polynomial Q(x):
$$ x^2 + 3x + 2 $$
Therefore, we can write:
$$ \frac{x^3 + 4x^2 + 5x + 2}{x + 1} = x^2 + 3x + 2 $$
And the fully factored form becomes:
$$ x^3 + 4x^2 + 5x + 2 = (x + 1)(x^2 + 3x + 2) $$
Note: A polynomial can be expressed as the product of its divisor and quotient: $$ P(x) = Q(x) \cdot (x - a) $$
This gives us the complete factorization of the polynomial:
$$ (x + 1)(x^2 + 3x + 2) $$
Now we can quickly find all the roots.
For ( $ x + 1 $ ), the root is: $$ x = -1 $$
For ( $ x^2 + 3x + 2 $ ), the roots are: $$ \frac{-3 \pm \sqrt{9 - 8}}{2} = \frac{-3 \pm 1}{2} = \begin{cases} x = -1 \\ x = -2 \end{cases} $$
So, the roots of the polynomial x3 + 4x2 + 5x + 2 are -1 (twice) and -2.
How to Find the Roots of a Polynomial
Given a polynomial of degree n:
$$ P(x) = a_nx^n + a_{n-1}x^{n-1} + \ldots + a_0 $$
You can determine its roots (also known as zeros) using different strategies depending on whether the leading coefficient is equal to 1 or not.
- When the leading coefficient is 1 (an = 1)
If the leading coefficient of P(x) is 1, then any possible integer roots must be among the integer divisors of the constant term a0.Example: Here, the leading coefficient (of x3) is 1: $$ P(x) = x^3 + 4x^2 + 5x + 2 $$ Therefore, any integer roots must be among the divisors of the constant term a0 = 2: $$ S = \{ 1, -1, 2, -2 \} $$ We evaluate P(x) at each candidate: $$ P(1) = 1^3 + 4(1)^2 + 5(1) + 2 = 12 $$ $$ P(-1) = (-1)^3 + 4(-1)^2 + 5(-1) + 2 = 0 $$ So x = -1 is a root of the polynomial.
Note: This method only detects integer roots. If a polynomial has non-integer roots, they won’t be identified this way. For instance, $$ P(x) = x^2 - 4x + 2 $$ has two irrational roots: x1 = 2 + √2 and x2 = 2 − √2. None of the integer divisors of the constant term (±1, ±2) is a root of the polynomial, so this method fails in such cases.
- When the leading coefficient is not 1 (an ≠ 1)
If a rational number N/D (in lowest terms) is a root of a polynomial P(x) with integer coefficients, then:- The numerator N must divide the constant term a0.
- The denominator D must divide the leading coefficient an.
Example: Consider the polynomial $$ P(x) = 3x^4 + 4x^2 + 5x - 2 $$ The leading coefficient is 3, and its divisors are {±1, ±3}. The constant term is -2, whose divisors are {±1, ±2}. The set of possible rational roots is: $$ \frac{\{1, -1, 2, -2\}}{\{1, -1, 3, -3\}} $$ which simplifies to: $$ S = \left\{ 1, -1, \frac{1}{3}, -\frac{1}{3}, 2, -2, \frac{2}{3}, -\frac{2}{3} \right\} $$ We test these candidates: $$ P(1) = 3(1)^4 + 4(1)^2 + 5(1) - 2 = 10 $$ $$ P(-1) = 3(-1)^4 + 4(-1)^2 + 5(-1) - 2 = 0 $$ So x = -1 is a rational root, and we can proceed with Ruffini’s rule to factor the polynomial further.
Note: This method identifies only rational roots. If the polynomial has irrational roots, they won’t appear in this list. For example: $$ P(x) = x^2 - 4x + 2 $$ has irrational roots x1 = 2 + √2 and x2 = 2 − √2. The constant term 2 has divisors {±1, ±2}, and the leading coefficient 1 has divisors {±1}, but none of the resulting fractions is a root of P(x).
The Proof
Ruffini’s rule tells us that a polynomial P(x) is divisible by (x − k) if and only if k is a root of the polynomial. In that case, the division produces a quotient Q(x) and no remainder:
$$ \frac{P(x)}{(x - k)} = Q(x), \quad \text{with remainder } R = 0 $$
Therefore, the polynomial can be factored as:
$$ P(x) = (x - k) \cdot Q(x) $$
To justify Ruffini’s rule, we need to show that any polynomial P(x) is divisible by (x − k) whenever k is a root of P(x).
Why is P(x) divisible by (x − k) if k is a root?
The division of polynomials always yields a quotient Q(x) and a remainder R:
$$ \frac{P(x)}{(x - k)} = Q(x) + R $$
By definition, P(x) is divisible by (x − k) if and only if R = 0.
Now, suppose k is a root of P(x), i.e., P(k) = 0:
$$ P(k) = 0 $$
According to the Remainder Theorem, when a polynomial P(x) is divided by (x − k), the remainder is equal to P(k):
$$ R = P(k) $$
Since P(k) = 0, it follows that:
$$ R = 0 $$
This confirms that P(x) is divisible by (x − k) if k is a root.
Thus, Ruffini’s rule is valid:
$$ P(x) = (x - k) \cdot Q(x) $$
And so on.
