Polynomial Long Division
Consider the function
f (x) = d(x)q(x) + r(x)
where d(x) is the divisor, q(x) is the quotient, and r(x) is the remainder after division. We
want to solve
f (x)
q(x) =
d(x)
with a remainder r(x) = 0.
Let's say we have
f (x) = x
4
+ 6x
3
+ x
2
− 24x − 20
One root of f (x) is x = −1
Standard Approach
Since we have the root x = −1, we can divide f (x) by x + 1
x
x + 1 x
3
4
+ 5x
+ 6x
4
2
3
− 4x − 20
+ x
2
− 24x − 20
3
x + x
–
0 + 5x
3
+ x
2
3
2
5x + 5x
–
0 − 4x
2
− 24x
2
−4x − 4x
–
0 − 20x − 20
−20x − 20
–
0
∴ f (x) = (x + 1) (x
3
+ 5x
2
− 4x − 20)
The standard approach takes up a lot of real-estate on your page, and can be a problem
when limited paper is available.
Synthetic Division
When dividing by a first-order polynomial, the same division can be written in a short-hand
way called Synthetic Division.
Reminder:
f (x) = 1x
4
+ 6x
3
+ 1x
2
− 24x − 20
x 1 = −1
Using synthetic division:
−1
1
1
6
1
−24
−20
−1
−5
4
20
5
−4
−20
0
So how does Synthetic Division work?
We normally have d(x) = x + 1. We now only care about x's root, i.e. x = −1.
Place this x value to the left of the vertical line.
Place all coefficients of f (x) in the top row to the vertical lines' right.
The highest power of x typically has a coefficient of 1 for ease of calculations. Therefore
it can brought down to the first column of the bottom row. This is the first coefficient of
q(x).
For each subsequent coefficient of q(x), multiply q(x)'s previous coefficient with the
divisor root and into the next column, below the top row. Sum the column values, place in
bottom row, and repeat.
∴ q(x) = x
3
+ 5x
2
− 4x − 20