The Remainder Theorem

Dividing a polynomial $LaTeX: f\left(x\right)\:$ $f\left(x\right)\:$ by a binomial of the form $LaTeX: x-k\:$ $x-k\:$ (perhaps using synthetic division?), gives a special case of the division theorem which ends up being useful.

Special Case of the Division Theorem

For any polynomial f(x) and any complex number k, there exists a unique polynomial q(x) and number r such that the following holds.

$LaTeX: f\left(x\right)=\left(x-k\right)\cdot q\left(x\right)+r$ $f\left(x\right)=\left(x-k\right)\cdot q\left(x\right)+r$

If we use this to find $LaTeX: f\left(k\right)\:$ $f\left(k\right)\:$ we get:

$LaTeX: f\left(k\right)=\left(k-k\right)\cdot q\left(k\right)+r=\left(0\right)\cdot q\left(k\right)+r=0+r=r$ $f\left(k\right)=\left(k-k\right)\cdot q\left(k\right)+r=\left(0\right)\cdot q\left(k\right)+r=0+r=r$

So, when we divide a polynomial $LaTeX: f\left(x\right)\:$ $f\left(x\right)\:$ by a binomial of the form $LaTeX: x-k,\:$ $x-k,\:$ the remainder is $LaTeX: f\left(k\right).\:$ $f\left(k\right).\:$

The Remainder Theorem

If the polynomial f(x) is divided by x − k, the remainder is equal to f(k).

Examples

Here is a video showing several examples of using the remainder theorem to calculate f(k) by dividing by x-k and knowing that f(k) will be the remainder.