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Station  R
The 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 x33x+3x2\dfrac{x^3-3x+3}{x-2}. This is the polynomial p(x)=x33x+3p\left(x\right)=x^3-3x+3 divided by the simple (linear) polynomial x2x-2.

Here’s what I get on the 1x1 \leftarrow x machine. (I had to add in some of dot/antidot pairs.)
I6SR-image170

Check this. Try it on the1x1 \leftarrow x machine too.

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But let’s look at the picture of x33x+3x^3-3x+3 carefully, taking note of the loops.

I6SR-image170

We see one loop at the x2x^2 level, two at the xx level, and one at the ones level. Plus we see a remainder of 55. As each loop represents the quantity x2x-2, this means that

p(x)=x33x+3=(x2)×x2+2(x2)×x+(x2)×1+5p\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 x2x-2 at the x2x^2 level, two at the xx level, and one at the ones level, and 55.)

This shows that p(x)p\left(x\right) is a combination of (x2)\left(x-2\right)s plus an extra 55.

p(x)=p\left(x\right)= multiples of (x2)+5\left(x-2\right) + 5.

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

p(2)=p\left(2\right)= multiples of 0+5=0+5=50 + 5 = 0+5 = 5.

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

p(x)=p\left(x\right)= multiples of (xh)+r\left(x-h\right) + r
where rr is a remainder. Putting x=hx=h shows that p(h)=rp\left(h\right)=r.

This is the Remainder Theorem for polynomials.

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

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

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

This is a big deal for people interested in factoring.

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