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# Station  RThe Remainder Theorem

High school teachers have asked me if the dots and boxes approach can be used to explain the “Remainder Theorem.” This optional station is for anyone interested in learning about the mathematics of this piece of extra-advanced polynomial algebra.

WARNING: This passage is not for the faint hearted!

### Question 1

Let’s examine $\dfrac{x^3-3x+3}{x-2}$. This is the polynomial $p\left(x\right)=x^3-3x+3$ divided by the simple (linear) polynomial $x-2$.

Here’s what I get on the $1 \leftarrow x$ machine. (I had to add in some of dot/antidot pairs.)

Check this. Try it on the$1 \leftarrow x$ machine too.

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.

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