Ruffini’s Theorem
A polynomial P(x) is divisible by a binomial (x − k) if and only if the polynomial evaluates to zero at x = k. $$ P(k)=0 $$
This allows us to determine whether a polynomial is divisible by a binomial of the form (x − k) without performing long division.
It's especially helpful when factoring a polynomial using Ruffini’s method.
Note. To factor a polynomial - that is, to express it as the product of two lower-degree polynomials - we need to confirm that dividing P(x) by (x − k) yields a remainder of zero, R = 0. Only in this case is P(x) divisible by (x − k) and thus factorable.
A practical example
Let’s factor the following cubic polynomial:
$$ 2x^3 - 9x^2 + 11x - 6 $$
We begin by evaluating the polynomial at several integer values near zero:
$$ \begin{array}{c|cr} x & \text{P(x)} \\ \hline -3 & -174 \\ -2 & -80 \\ -1 & -28 \\ 0 & -6 \\ 1 & -2 \\ 2 & -4 \\ 3 & \color{red}0 \end{array} $$
Since P(3) = 0, the polynomial vanishes at x = 3.
By Ruffini’s theorem, this tells us that P(x) is divisible by (x − 3).
$$ P(x) \div (x - 3) = Q(x) \quad \text{with remainder } R = 0 $$
We can therefore write the factorization as:
$$ P(x) = (x - 3) \cdot Q(x) $$
Verification. Let’s apply Ruffini’s method to factor the polynomial. $$ \begin{array}{c|lcc|r} & 2 & -9 & 11 & -6 \\ 3 & & 6 & -9 & 6 \\ \hline & 2 & -3 & 2 & 0 \end{array} $$ This confirms the factorization: $$ P(x) = (x - 3) \cdot (2x^2 - 3x + 2) $$ The result is correct.
Understanding the theorem
If a polynomial P(x) is divisible by (x − k), then the remainder from the division must be zero:
$$ R = 0 $$
According to the Remainder Theorem, when we divide a polynomial P(x) by a binomial (x − k), the remainder equals the value of the polynomial at x = k:
$$ P(k) = R $$
In this case, we’ve assumed that R = 0, which gives us:
$$ P(k) = 0 $$
So, if a polynomial P(x) is divisible by (x − k), then it must satisfy P(k) = 0.
The converse is also true.
If P(k) = 0 for some value x = k, then dividing P(x) by (x − k) yields no remainder (R = 0).
And if the remainder is zero, then P(x) is divisible by (x − k).
Therefore, if P(k) = 0, then P(x) is divisible by the binomial (x − k).
And so the logic flows both ways.
