# What is the difference between the remainder theorem and the factor theorem?

##### 1 Answer
Sep 12, 2015

The two theorems are similar, but refer to different things.
See explanation.

#### Explanation:

The remainder theorem tells us that for any polynomial $f \left(x\right)$, if you divide it by the binomial $x - a$, the remainder is equal to the value of $f \left(a\right)$.

The factor theorem tells us that if $a$ is a zero of a polynomial $f \left(x\right)$, then $\left(x - a\right)$ is a factor of $f \left(x\right)$, and vice-versa.

For example, let's consider the polynomial

$f \left(x\right) = {x}^{2} - 2 x + 1$

Using the remainder theorem

We can plug in $3$ into $f \left(x\right)$.

$f \left(3\right) = {3}^{2} - 2 \left(3\right) + 1$
$f \left(3\right) = 9 - 6 + 1$
$f \left(3\right) = 4$

Therefore, by the remainder theorem, the remainder when you divide ${x}^{2} - 2 x + 1$ by $x - 3$ is $4$.

You can also apply this in reverse. Divide ${x}^{2} - 2 x + 1$ by $x - 3$, and the remainder you get is the value of $f \left(3\right)$.

Using the factor theorem

The quadratic polynomial $f \left(x\right) = {x}^{2} - 2 x + 1$ equals $0$ when $x = 1$.
This tells us that $\left(x - 1\right)$ is a factor of ${x}^{2} - 2 x + 1$.

We can also apply the factor theorem in reverse:

We can factor ${x}^{2} - 2 x + 1$ into ${\left(x - 1\right)}^{2}$, therefore $1$ is a zero of $f \left(x\right)$.

Basically, the remainder theorem links the remainder of division by a binomial with the value of a function at a point, while the factor theorem links the factors of a polynomial to its zeros.