多項式と因数定理

Dividing a polynomial $f(x)$ by $(x-a)$ gives a quotient $q(x)$ and remainder $r$, so $f(x)=(x-a)q(x)+r$. Dividing by a linear factor like $(x-a)$ always leaves a constant remainder. You can carry this out by long division or by synthetic division on the coefficients — but if you only need the remainder, the remainder theorem below is far faster.

The remainder theorem states that the remainder when $f(x)$ is divided by $(x-a)$ equals $f(a)$. Substituting $x=a$ into the identity $f(x)=(x-a)q(x)+r$ kills the $(x-a)$ term and leaves $f(a)=r$. So you get the remainder simply by evaluating $f(a)$ — no division needed. For division by $(ax-b)$, substitute $x=\frac{b}{a}$.

The factor theorem is the special case of the remainder theorem where the remainder is $0$, corresponding to exact division. That is, $f(a)=0$ if and only if $(x-a)$ is a factor of $f(x)$. To factorise a cubic, start by testing divisors of the constant term — $\pm 1,\pm 2,\dots$ — substituting them into $f$ until you find one value $a$ with $f(a)=0$.

Once the factor theorem gives one factor $(x-a)$, divide $f(x)$ by $(x-a)$ to obtain a quadratic quotient. Factorise that quadratic (or use the formula) to get all three factors, or all three roots. To solve $f(x)=0$, read the roots from the factorised form $(x-a)(x-b)(x-c)=0$ by setting each factor to zero.