So there I was, merrily teaching the factor and remainder theorems, and my student asked me one of my favourite questions: “I accept that the method works, but why does it?”

(I like that kind of question because it makes me think on my feet in class, and that makes me feel alive!)

Let’s get everyone up to speed, shall we?

The factor theorem says that $(x-a)$ is a factor of a polynomial $p(x)$ if and only if $p(a)=0$. That means, if a question says “show that $x-7$ is a factor of $8x^3 -58x^2 - x + 105$”, all you need to do is work out what happens to that expression when you stick a 7 in. You get $8 \times 343 - 58 \times 49 - 7 + 105$, which is $2744 - 2842 - 7 + 105 = 0$. Boom, it’s a factor.

The remainder theorem is a close cousin of the factor theorem, and says that when you divide $p(x)$ by $(x-a)$, the remainder you get is $p(a)$. Notice that this fits perfectly well with the factor theorem: if the remainder when you divide by something is zero, what you divided by is a factor!

What is a remainder?

Let’s think about dividing numbers, which are much less weird things to divide by. Suppose I need to split 39 items between seven people. I can go one-for-you, one-for-you, and so on - at least until I hand out the 35th item. Everyone now has five items, and there are four left over.

That means, I can write $39 = 7 \times 5 + 4$. The five items everyone got is the quotient, and the four left over is the remainder.

Any positive integer divided by any other can be expressed this way: the quotient is how many times the second number goes completely into the first, and the remainder is what’s left over. More formally, $a \div b$ gives an answer of $q$ with remainder $r$ if and only if $a = b \times q + r$, with $0 \le r \lt b$.

The same goes for polynomials… with a few wrinkles

If I want to divide $p(x)$ by $(x-a)$ – and frankly, who doesn’t? – I’m working with polynomials rather than just numbers. ((‘Just numbers’ are a subset of the polynomials, by the way, but polynomial division and integer division are different things. Roll with it.))

My quotient $q(x)$ and my remainder $r(x)$ will both be polynomials; the degree of $q$ will be one smaller than the degree of $p$ in this context (for example, if $p$ is a quartic, $q$ will be a cubic); $r$, meanwhile, will be of degree 0 - which is to say, a constant.

Putting it together, $\frac{p(x)}{x-a}$ gives a quotient $q(x)$ and remainder $r$ if and only if $p(x) = (x-a) q(x) + r$, where $r$ is a constant.

Do you see where this is going?

Now, if you put $x=a$ into this equation, you get $p(a) = (a-a) q(a) + r$. The first term on the right-hand side is clearly 0, so $p(a) = r$.

If you stick $a$ into your polynomial, you get the remainder out. And, as a bonus, if there’s no remainder, you’ve found a factor.

I think that’s pleasing, don’t you?