Ruffini's rule is a procedure for dividing a polynomial by a first-degree binomial of the form \(x-a\) quickly and efficiently. At first glance it may appear to be nothing more than a shorthand algorithm, but its significance runs much deeper: it is a compact form of polynomial long division and, at the same time, a tool that connects the value of a polynomial at a point, the remainder of the division, and the presence of linear factors.
For this reason, Ruffini's rule should not be studied as a mere mechanical technique. It provides a concrete understanding of three fundamental ideas in algebra: polynomial division, the Remainder Theorem, and the Factor Theorem.
- Dividing a polynomial by \(x-a\)
- The key idea behind Ruffini's rule
- Statement of Ruffini's rule
- A worked example
- Proof of Ruffini's rule
- The Remainder Theorem
- The Factor Theorem
- Using Ruffini's rule to factor a polynomial
- Missing terms
- Ruffini's rule and rational roots
- Ruffini's rule does not find all roots
- Division by \(ax+b\)
- Example with a non-monic divisor
Dividing a polynomial by \(x-a\)
Let \(P(x)\) be a polynomial with real coefficients, or more generally with coefficients in a field, and let \(a\) be a fixed number. To divide \(P(x)\) by \(x-a\) means to find a polynomial \(Q(x)\) and a constant \(r\) such that
\[ P(x)=(x-a)Q(x)+r. \]
Since \(x-a\) has degree \(1\), the remainder must have degree less than \(1\), so it is necessarily a constant. Ruffini's rule provides an efficient way to determine both the quotient \(Q(x)\) and the remainder \(r\) without carrying out the full long division each time, provided the divisor is a monic first-degree binomial.
The key idea behind Ruffini's rule
Consider a polynomial of degree \(n\):
\[ P(x)=c_nx^n+c_{n-1}x^{n-1}+\cdots+c_1x+c_0, \]
with \(c_n\neq 0\). Dividing \(P(x)\) by \(x-a\) yields a quotient of degree \(n-1\), which we write as
\[ Q(x)=b_{n-1}x^{n-1}+b_{n-2}x^{n-2}+\cdots+b_1x+b_0. \]
Substituting this expression for \(Q(x)\) into the relation \(P(x)=(x-a)Q(x)+r\) and expanding the product gives:
\[ \begin{aligned} P(x)= {} & b_{n-1}x^n+(b_{n-2}-ab_{n-1})x^{n-1} \\ & +(b_{n-3}-ab_{n-2})x^{n-2}+\cdots \\ & +(b_0-ab_1)x+(r-ab_0). \end{aligned} \]
Matching coefficients with those of \(P(x)\) yields the system of recurrence relations
\[ \begin{cases} b_{n-1}=c_n,\\ b_{n-2}=c_{n-1}+ab_{n-1},\\ b_{n-3}=c_{n-2}+ab_{n-2},\\ \quad \vdots\\ b_0=c_1+ab_1,\\ r=c_0+ab_0. \end{cases} \]
These relations are the heart of Ruffini's rule. The procedure amounts to bringing down the leading coefficient and then, at each step, multiplying the last result by \(a\) and adding it to the next coefficient.
Statement of Ruffini's rule
Let \(P(x)=c_nx^n+c_{n-1}x^{n-1}+\cdots+c_1x+c_0\) be a polynomial of degree \(n\). To divide \(P(x)\) by \(x-a\), list the coefficients \(c_n,\ c_{n-1},\ \ldots,\ c_1,\ c_0\) in order and build the sequence
\[ \begin{cases} b_{n-1}=c_n,\\ b_{k-1}=c_k+ab_k \quad \text{for } k=n-1,n-2,\ldots,1,\\ r=c_0+ab_0. \end{cases} \]
Then \(P(x)=(x-a)Q(x)+r\), where
\[ Q(x)=b_{n-1}x^{n-1}+b_{n-2}x^{n-2}+\cdots+b_1x+b_0. \]
The number \(r\) is the remainder of the division.
A worked example
Divide \(P(x)=2x^3-3x^2+4x-5\) by \(x-2\). Here \(a=2\) and the coefficients of the polynomial are \(2,\ -3,\ 4,\ -5\). Applying the rule:
\[ \begin{array}{c|rrrr} 2 & 2 & -3 & 4 & -5\\ & & 4 & 2 & 12\\ \hline & 2 & 1 & 6 & 7 \end{array} \]
The first three entries in the bottom row are the coefficients of the quotient; the last entry is the remainder. Therefore \(Q(x)=2x^2+x+6\) and \(r=7\), that is,
\[ 2x^3-3x^2+4x-5=(x-2)(2x^2+x+6)+7. \]
Proof of Ruffini's rule
Ruffini's rule is not a trick: it is the shorthand form of the coefficient-matching argument in the division \(P(x)=(x-a)Q(x)+r\).
By the polynomial division theorem, there exist a unique quotient \(Q(x)\) and a unique remainder \(r\) such that \(P(x)=(x-a)Q(x)+r\). Since the divisor has degree \(1\), the quotient has degree \(n-1\). Write
\[ Q(x)=b_{n-1}x^{n-1}+b_{n-2}x^{n-2}+\cdots+b_1x+b_0. \]
Expanding \((x-a)Q(x)\):
\[ \begin{aligned} (x-a)Q(x) ={}& xQ(x)-aQ(x)\\ ={}& b_{n-1}x^n+b_{n-2}x^{n-1}+\cdots+b_1x^2+b_0x\\ &-ab_{n-1}x^{n-1}-ab_{n-2}x^{n-2}-\cdots-ab_1x-ab_0. \end{aligned} \]
Adding the remainder \(r\) gives
\[ \begin{aligned} P(x)= {} & b_{n-1}x^n+(b_{n-2}-ab_{n-1})x^{n-1}\\ & +(b_{n-3}-ab_{n-2})x^{n-2}+\cdots\\ & +(b_0-ab_1)x+(r-ab_0). \end{aligned} \]
Since two polynomials are equal if and only if all their corresponding coefficients are equal, comparing with \(P(x)=c_nx^n+c_{n-1}x^{n-1}+\cdots+c_0\) yields:
\[ \begin{cases} c_n=b_{n-1},\\ c_{n-1}=b_{n-2}-ab_{n-1},\\ c_{n-2}=b_{n-3}-ab_{n-2},\\ \quad \vdots\\ c_1=b_0-ab_1,\\ c_0=r-ab_0. \end{cases} \]
Solving for the \(b_k\) recovers precisely the recurrence relations of Ruffini's rule.
The Remainder Theorem
The Remainder Theorem states that the remainder of the division of \(P(x)\) by \(x-a\) equals \(P(a)\).
Proof. From \(P(x)=(x-a)Q(x)+r\), setting \(x=a\) gives \(P(a)=(a-a)Q(a)+r=r\). Hence the remainder equals the value of the polynomial at \(a\).
The Factor Theorem
The Factor Theorem follows immediately from the Remainder Theorem: \(x-a\) divides \(P(x)\) if and only if \(P(a)=0\).
\[ x-a \text{ divides } P(x) \quad \Longleftrightarrow \quad P(a)=0. \]
Proof. From \(P(x)=(x-a)Q(x)+r\) and the Remainder Theorem we know that \(r=P(a)\). If \(x-a\) divides \(P(x)\), then \(r=0\), so \(P(a)=0\). Conversely, if \(P(a)=0\), then \(r=0\), and therefore \(P(x)=(x-a)Q(x)\), meaning \(x-a\) is a factor of \(P(x)\). Together, the two implications establish the equivalence.
Using Ruffini's rule to factor a polynomial
One of the principal applications of Ruffini's rule is polynomial factorization. If one finds a number \(a\) such that \(P(a)=0\), then by the Factor Theorem \(x-a\) divides \(P(x)\). Ruffini's rule then yields the quotient, that is, the complementary factor in the factorization.
Consider \(P(x)=x^3-4x^2+x+6\). Testing \(x=2\):
\[ P(2)=8-16+2+6=0. \]
Therefore \(x-2\) is a factor. Applying Ruffini's rule:
\[ \begin{array}{c|rrrr} 2 & 1 & -4 & 1 & 6\\ & & 2 & -4 & -6\\ \hline & 1 & -2 & -3 & 0 \end{array} \]
This gives \(P(x)=(x-2)(x^2-2x-3)\). Since \(x^2-2x-3=(x-3)(x+1)\), the complete factorization is
\[ x^3-4x^2+x+6=(x-2)(x-3)(x+1). \]
Missing terms
When applying Ruffini's rule, it is essential to include the coefficients of every power of \(x\), even those that do not appear explicitly in the polynomial. A missing term corresponds to a zero coefficient.
Consider the polynomial \(P(x)=x^4-3x^2+2x-1\), in which the cubic term is absent. To apply Ruffini's rule correctly, one must write
\[ P(x)=x^4+0x^3-3x^2+2x-1, \]
with coefficient list \(1,\ 0,\ -3,\ 2,\ -1\). Omitting the zero would shift all subsequent coefficients out of position, producing an incorrect result.
Ruffini's rule and rational roots
In practice, Ruffini's rule is often used in conjunction with a systematic search for rational roots of a polynomial. When a polynomial has integer coefficients, any rational root it may possess is far from arbitrary: it is constrained by the coefficients themselves.
Specifically, if \(P(x)=c_nx^n+c_{n-1}x^{n-1}+\cdots+c_0\) has integer coefficients and \(\dfrac{p}{q}\), written in lowest terms, is a rational root of \(P(x)\), then \(p\) divides the constant term \(c_0\) and \(q\) divides the leading coefficient \(c_n\). In the monic case (\(c_n=1\)), every rational root must be an integer divisor of the constant term.
Proof (Rational Root Theorem). Suppose \(P\!\left(\dfrac{p}{q}\right)=0\), where \(p\) and \(q\) are coprime integers with \(q\neq 0\). Then
\[ c_n\left(\frac{p}{q}\right)^n+c_{n-1}\left(\frac{p}{q}\right)^{n-1}+\cdots+c_1\frac{p}{q}+c_0=0. \]
Multiplying through by \(q^n\):
\[ c_np^n+c_{n-1}p^{n-1}q+\cdots+c_1pq^{n-1}+c_0q^n=0. \]
Moving the last term to the right-hand side, the left-hand side is divisible by \(p\), so \(c_0q^n\) must be as well. Since \(p\) and \(q^n\) are coprime, it follows that \(p\mid c_0\). An analogous argument, isolating \(c_np^n\) on the right, shows that \(q\mid c_n\).
Complete Example (Factoring with Ruffini's rule). Factor \(P(x)=x^3-6x^2+11x-6\). Since the polynomial is monic, any integer root must divide the constant term \(-6\):
\[ \pm1,\ \pm2,\ \pm3,\ \pm6. \]
Computing \(P(1)=1-6+11-6=0\), we see that \(x-1\) is a factor. Applying Ruffini's rule:
\[ \begin{array}{c|rrrr} 1 & 1 & -6 & 11 & -6\\ & & 1 & -5 & 6\\ \hline & 1 & -5 & 6 & 0 \end{array} \]
This gives \(P(x)=(x-1)(x^2-5x+6)\). Since \(x^2-5x+6=(x-2)(x-3)\), the complete factorization is
\[ x^3-6x^2+11x-6=(x-1)(x-2)(x-3). \]
Ruffini's rule does not find all roots
It is important to dispel a common misconception: Ruffini's rule is not a universal method for finding all roots of a polynomial. It allows one to divide a polynomial by a binomial of the form \(x-a\) and to factor it once a root \(a\) is known, but it provides no automatic means of locating irrational or complex roots.
If a polynomial has no rational roots, a systematic search through the divisors of the constant term yields nothing. For instance, \(x^2+1\) has no real roots and cannot be factored into real linear factors.
Division by \(ax+b\)
Ruffini's rule applies to division by a monic binomial \(x-a\). A general binomial \(ax+b\) with \(a\neq 0\) can always be rewritten as
\[ ax+b=a\!\left(x-\left(-\frac{b}{a}\right)\right), \]
whose zero is \(x=-\dfrac{b}{a}\). To test whether \(ax+b\) divides \(P(x)\), it therefore suffices to check whether \(P\!\left(-\dfrac{b}{a}\right)=0\).
One must be careful, however: dividing by \(ax+b\) is not the same as dividing by \(x+\dfrac{b}{a}\), since the two divisors differ by the constant factor \(a\). The root is the same, but the quotient changes accordingly.
Example with a non-monic divisor
Divide \(P(x)=2x^2-3x-2\) by \(2x+1\). The divisor vanishes at \(x=-\dfrac{1}{2}\). Checking:
\[ P\!\left(-\frac{1}{2}\right)=2\cdot\frac{1}{4}-3\cdot\left(-\frac{1}{2}\right)-2=\frac{1}{2}+\frac{3}{2}-2=0. \]
Hence \(2x+1\) divides \(P(x)\), and indeed \(2x^2-3x-2=(2x+1)(x-2)\). To apply Ruffini's rule one works with the monic binomial \(x+\dfrac{1}{2}\): the resulting quotient differs from the one obtained by dividing by \(2x+1\), but the divisibility check takes place at the same point \(x=-\dfrac{1}{2}\).
Ruffini's rule is far more than a computational shortcut. It grows directly out of polynomial division and encodes in operational form the coefficient-matching argument for division by a first-degree binomial.
Its theoretical significance emerges most clearly through the Remainder Theorem and the Factor Theorem. Dividing a polynomial by \(x-a\), evaluating \(P(a)\), deciding whether \(a\) is a root, and checking whether \(x-a\) is a factor are all different facets of the same algebraic structure.
Ruffini's rule, therefore, should be remembered not merely as a table to fill in, but as a bridge between computation and theory: on one hand it streamlines polynomial division; on the other it allows the factorization of a polynomial to be read directly from its roots.