But let’s look at the picture of $x^3-3x+3$ carefully, taking note of the loops.

We see one loop at the $x^2$ level, two at the $x$ level, and one at the ones level. Plus we see a remainder of $5$. As each loop represents the quantity $x-2$, this means that

$p\left(x\right)=x^3-3x+3=\left(x-2\right) \times x^2 + 2\left(x-2\right) \times x + \left(x-2\right) \times 1 + 5$.

(This is one $x-2$ at the $x^2$ level, two at the $x$ level, and one at the ones level, and $5$.)

This shows that $p\left(x\right)$ is a combination of $\left(x-2\right)$s plus an extra $5$.

$p\left(x\right)=$ multiples of $\left(x-2\right) + 5$.

That “$+5$” is standing out like a sore thumb. If you put in $x=2$ we get

$p\left(2\right)=$ multiples of $0 + 5 = 0+5 = 5$.

In general, dividing a polynomial $p\left(x\right)$ by a term of the form $x-h$ will give

$p\left(x\right)=$ multiples of $\left(x-h\right) + r$

where $r$ is a remainder. Putting $x=h$ shows that $p\left(h\right)=r$.

This is the Remainder Theorem for polynomials.

*Dividing a polynomial $p\left(x\right)$ by a term $x-h$ gives a remainder that is a single number equal to $p\left(h\right)$, the value of the polynomial at $x=h$.*

People like this theorem because it shows that if $p\left(h\right)=0$ for some number $h$, then $p\left(x\right)$ is an multiple of $x-h$. (The remainder is zero.) This gives the Factor Theorem for polynomials.

*A polynomial $p$ has a factor $x-h$ precisely when $h$ is a zero of the polynomial, that is, precisely when $p\left(h\right)=0$.*

This is a big deal for people interested in factoring.