What Is the Factor Theorem and the Remainder Theorem?

The remainder theorem states that the remainder when a polynomial $f(x)$ is divided by (x-a) is $f(a)$. The factor theorem follows: (x-a) is a factor of $f(x)$ if and only if $f(a)=0$. These results test divisibility without full polynomial division.

This is IB Mathematics: Analysis and Approaches HL content from AHL 2.12: Polynomial functions. It can appear on Paper 1 or Paper 2.

Why the Theorems Work

When $f(x)$ is divided by (x-a), polynomial division gives

$$f(x)=(x-a)q(x)+r,$$

where $q(x)$ is the quotient and $r$ is the constant remainder. Substituting $x=a$ gives $f(a)=r$ because . Therefore, if , the division is exact and (x-a) is a factor.

Worked Examples

Let $f(x)=x^3-4x^2+x+6$. To test $$, calculate

$$f(2)=8-16+2+6=0.$$

Therefore, $x-2$ is a factor. For division by $x-1$, $f(1)=4$; the remainder is $4$, so is not a factor.

The method also finds unknown coefficients. Suppose $g(x)=x^3+kx-6$ and $x-2$ is a factor. Then , so , giving .

A common misconception is ignoring the divisor's sign. For $x+3=x-(-3)$, substitute $x=-3$, not $x=3$. Also, proves only the corresponding linear factor; it does not give all remaining factors.

On Paper 1, evaluate the polynomial exactly and show each substitution clearly. On Paper 2, a GDC can check a result, but the algebraic condition should still be written. If the divisor is (bx-c), first identify its zero $x=\frac{c}{b}$; the remainder is found by evaluating the polynomial at that zero.

IB Exam Technique

For show that, calculate $f(a)$ and explicitly conclude that it equals zero. For find the remainder, state $f(a)$ as the remainder. If finding a parameter, form $f(a)=0$, solve it, and check the result. Do not merely write “by the factor theorem”; the substitution provides the required evidence. Full polynomial division is needed only when the quotient or complete factorization is requested.