Factoring a Polynomial with Ruffini's Method
Ruffini’s method is a shortcut for factoring polynomials: once you find a root, you divide by the corresponding factor $(x-r)$.
If a polynomial $P(x)$ vanishes at $x=r$, then $(x-r)$ must be a factor of $P(x)$. In other words, $(x-r)$ divides $P(x)$.
$$ \frac{P(x)}{(x-r)} = Q(x) $$
This means that by carrying out the division, we can rewrite the original polynomial as the product of two lower-degree polynomials.
$$ P(x) = Q(x) (x-r) $$
Note. This follows directly from the Remainder Theorem: $$ P(x) = (x-r)\,Q(x) + P(r) $$. If $P(r)=0$, the remainder drops out, leaving $$ P(x) = (x-r)\,Q(x) $$. So the problem boils down to finding at least one root $r$.
How do we actually find the roots?
When the coefficients of the polynomial $ a_n x^n + .... + a_1 x + a_0 $ are integers, the Rational Root Theorem narrows down the possibilities:
Consider fractions of the form $ \frac{p}{q} $
- The numerator $p$ must divide the constant term $a_0$
- The denominator $q$ must divide the leading coefficient $a_n$
This gives us a finite set of candidates $ \frac{p}{q} $ to test.
For each candidate, evaluate $P(\tfrac{p}{q})$. If the result is zero, i.e. $ P(\tfrac{p}{q})=0$, then $ r=\tfrac{p}{q} $ is a root.
$$ P(x=\tfrac{p}{q}) = 0 \;\;\Rightarrow\;\; r = \tfrac{p}{q} $$
Once we’ve pinned down a root $r$, we divide $P(x)$ by $(x-r)$ using Ruffini’s rule to get the quotient $Q(x)$. This gives us the polynomial in factored form:
$$ P(x) = Q(x) (x-r) $$
Note. A polynomial doesn’t always have rational or integer roots. That doesn’t mean it’s irreducible: it can still be factored using other tools, such as the quadratic formula, factoring by grouping, or algebraic identities.
A Worked Example
Take the cubic polynomial:
$$ P(x)=2x^3+5x^2-x-6 $$
The divisors of the constant term $-6$ are:
$$ p = \{ \pm1, \pm2, \pm3, \pm6 \} $$
The divisors of the leading coefficient $2$ are:
$$ q = \{ \pm1, \pm2 \} $$
So the possible rational roots are all combinations $ \frac{p}{q} $:
$$ \frac{p}{q} = \frac{\{ \pm1, \pm2, \pm3, \pm6 \}}{\{ \pm1, \pm2 \}} = \{ \pm1, \pm2, \pm3, \pm6, \pm \tfrac{1}{2}, \pm \tfrac{2}{3}, \pm \tfrac{3}{2} \} $$
Now let’s test these values by plugging them into $P(x)$:
$$ P(-1) = 2 (-1)^3+ 5 (-1)^2 -(-1) - 6 = -2+5+1-6 = -2 $$
$$ P(1) = 2 (1)^3+ 5 (1)^2 -(1) - 6 = 2+5-1-6 = 0 $$
Perfect - we’ve found a root at $ r=1 $.
This means the polynomial is divisible by $(x-r)$, i.e. $(x-1)$.
$$ \frac{ 2x^3+5x^2-x-6 }{x-1} = Q(x) $$
Let’s carry out the division using Ruffini’s method:
$$ \begin{array}{c|lcc|r} & 2 & 5 & -1 & -6 \\ r=1 & & 2 & 7 & 6 \\ \hline & 2 & 7 & 6 & 0 \end{array} $$
The quotient $Q(x)$ comes out to $2x^2+7x+6$.
$$ \frac{ 2x^3+5x^2-x-6 }{x-1} = 2x^2+7x+6 $$
Multiplying both sides by $(x-1)$ gives the factorization:
$$ 2x^3+5x^2-x-6 = (2x^2+7x+6)(x-1) $$
So we’ve expressed the cubic as the product of two lower-degree polynomials.
$$ (2x^2+7x+6)(x-1) $$
If needed, we can repeat the process on the quadratic factor, applying Ruffini (or other methods) to reduce it further.
Note. The quadratic factor $ 2x^2+7x+6 $ is reducible. We need two numbers $p$ and $q$ such that $pq=ac=2\cdot6=12$ and $p+q=b=7$. Here $p=4$ and $q=3$. Rewrite: $$ 2x^2+4x+3x+6 $$ Group terms: $$ 2x(x+2)+3(x+2) $$ Factor: $$ (x+2)(2x+3) $$ So the full irreducible factorization of the original cubic is: $$ (x+2)(2x+3)(x-1) $$
And so on.
