The Remainder Theorem

In algebra, the polynomial remainder theorem or little Bézout's theorem, is an application of polynomial long division. It states that the remainder of a polynomial $f(x)$ divided by a linear divisor $(x-a)$ is equal to $f(a)$.

For example, take the polynomial:

$f(x)=x^3-12x^2-42$ 

Then divide it by $x-3$. 

This gives the quotient $x^2-9x-27$ and the remainder $-123$. Therefore, $f(3)=-123$. We can check this by plugging $3$ into the equation, which yields:

$\begin{aligned} 3^3 - 12\cdot 3^2 - 42 &= 27-108-42 \\ &= -123 \end{aligned}$

In particular, $f(a)=0$ if and only if $(x-a)$ divides $f(x).$

Synthetic division

To use the remainder theorem, one must first perform division, which is a bit of work. A shorthand way to perform long division is synthetic division. It uses less writing and fewer calculations. It also takes significantly less space than long division. Most importantly, the subtractions in long division are converted to additions by switching the signs at the very beginning, preventing sign errors. Synthetic division only works for polynomials divided by linear expressions with a leading coefficient equal to $1.$ 

Let's use synthetic division to solve the example above $f(x)=x^3−12x^2−42 $ divided by $x-3$: 

We start by writing down the coefficients from the dividend and the negative second coefficient of the divisor. Note that we explicitly write out all zero terms!

$\ \ |1 \ -12 \ 0 \ -42 \\ 3 |\underline{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ } \\ \ \ | $

Bring down the first coefficient and multiply it by the divisor. Place the resulting $3$ under the $-12$.

$\ \ |1 \ -12 \ 0 \ -42 \\ 3 |\underline{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ } \\ \ \ |1$

Then add the next column of coefficients, get the result and multiply that by the divisor to find the third coefficient $-27$:

$$ $\ \ |1 \ -12 \ 0 \ -42 \\ 3 |\underline{\ \ \ \ \ \ \ \ \ 3 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ } \\ \ \ |1 \ \ \ - 9$

$\ \ |1 \ -12 \ \ \ \ \ 0 \ -42 \\ 3 |\underline{\ \ \ \ \ \ \ \ \ 3 \ - 27 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ } \\ \ \ |1 \ \ \ -9 \ - 27$

So the quotient must be the second degree polynomial $x^2 - 9x - 27$. Now we can also see what the remainder is, just by repeating the procedure:

$\ \ |1 \ -12 \ \ \ \ \ 0 \ -42 \\ 3 |\underline{\ \ \ \ \ \ \ \ \ 3 \ - 27 \ -81 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ } \\ \ \ |1 \ \ \ - 9 \ - 27 \ \textcolor{red}{-123}$

In particular, the number we write on the left is a root of the upper polynomial if and only if the last number we obtain is $0$. 

A special case of this is when the left number is $1$: then the last number equals the sum of all coefficients! Thus $1$ is a zero of a polynomial if and only if its coefficients add to $0.$

Other Leading Coefficients

When we divide by $ax-b$ and $a \not = 1$, we can divide by $(x-b/a)$ and then divide the result by $a$. This way we can still use synthetic division.