Dividing Polynomials

Dividing polynomials is simply the polynomial version of long division from grade school: how many times does one thing go into another, and what's left over? The formula $P(x) = (x - r)Q(x) + R$ is the same as saying $17 = 5\cdot3 + 2$, with $P(x)$ (the thing you're dividing), $x - r$ (what you divide by), $Q(x)$ (the quotient, how many times it goes in) and $R$ (the remainder). That's all of it. It's useful because it's how you break a polynomial into factors once you know one of them.

Long division works exactly like it does with numbers: divide the leading terms, multiply back, subtract, bring down the next term, and repeat.

Synthetic division is a shortcut for the most common case, dividing by $x - r$. It throws away the $x$'s and keeps only the coefficients. For $(2x^3 - 3x^2 + 4x - 5) \div (x - 1)$:

1. Write $r = 1$ and the coefficients $2, -3, 4, -5$.
2. Bring down the 2. Multiply by $r$ and add to the next coefficient: $-3 + 2 = -1$. Repeat: $4 + (-1) = 3$, then $-5 + 3 = -2$.
3. Read off quotient $2x^2 - x + 3$ and remainder $-2$.

If a power is missing, put a zero in its place, so $x^3 + 1$ has coefficients $1, 0, 0, 1$. Otherwise the columns slide out of line.

Why the remainder is $P(r)$. Here's the neat part. Plug $x = r$ into $P(x) = (x - r)Q(x) + R$. The first term becomes $0\cdot Q(r)$, which is $0$ no matter what $Q$ is, so you're left with $P(r) = R$. Notice what that means: you can find a remainder without dividing at all, just by evaluating. And if $P(r) = 0$, the remainder is zero, so $x - r$ divides evenly and is a factor. That's the factor theorem, and it's what the rest of this unit runs on.

Polynomial division powers error-correcting codes (CRC checksums divide a message polynomial and keep the remainder) and the partial fractions used for integration and Laplace transforms.