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*!)

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!

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 $0 \le r \lt b$.

If I want to divide $p(x)$ by $(x-a)$ -- and frankly, who doesn't? -- I'm working with polynomials rather than just numbers.1

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.

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?

- 'Just numbers' are a subset of the polynomials, by the way, but polynomial division and integer division are different things. Roll with it. [↩]

## J. Hall

Thank you. This answers my question exactly.

## Jack

Dude, thank you … I’ve been searching for this answer for a long time.now I can die in peace

## Ben

But why does it work? You’ve shown how we arrive at the remainder theorem using the division algorithm. But why does dividing a polynomial by (x-a) give us the same value as a remainder, as when we substitute ‘a’ for ‘x’ in that polynomial? What is the going on when we substitute ‘a’ for ‘x’ that relates to dividing by (x-a)?

## Colin

I shall have to have a think about this and see if I can clear it up!

## Ash cooke

The x-a is just an example of a factor of x – a constant. If you were to divide through a polynomial by just x , It would amount to a zero value remainder. So if you have a dividing value in terms of x, ie, x-2 or x-5, you can use his dividing algorithm as he described to trace it back to x-a. It kind of works that way as that’s how it works. Sorry if I butchered this explaination.

## Joshua

since p(x) = (x-a) x q (x) + r

if you substitute x with a

p(a) = (a-a) x q(a) + r

p(a) = 0 x q(a) + r

p(a) = 0 + r

P(a) = r

the only reason you substitute x with a is so you can get a zero value on the right side of the equation which will leave you with just the remainder.