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College Algebra · Chapter 10

Synthetic Division: One Column at a Time

Long division of polynomials is mostly bookkeeping — and most of it cancels. Synthetic division throws away everything that always cancels and leaves three moves: bring down, multiply, add. Step through a column at a time and watch why the very last number is the value of the polynomial.

Set up the division

Top row — the coefficients of the dividend, highest power first, with a 0 for every missing power.   Middle row — each product, k × the number diagonally below-left.   Bottom row — the running result: quotient coefficients, then the remainder.

Where the remainder lives on the graph

Synthetic divisionchoose a problem, then press Next

Why the three moves are legal

It is long division with the guaranteed clutter deleted

Divide 2x3 − 5x2 + 3x − 7 by x − 3 the long way, and watch what each move turns into.

1. 2x3 ÷ x = 2x2. The x's carried no information — the answer's coefficient is just the 2 you started with. So the first coefficient simply comes down. Which column a number sits in already records its power.

2. Multiply back: 2x2(x − 3) = 2x3 − 6x2. That leading 2x3 is guaranteed to cancel — we chose 2x2 precisely to make it cancel. So never write it. Only the second piece survives, and its number is 2 × (−3). That is the multiply row.

3. Subtract: −5x2 − (−6x2) = +x2. Subtracting 2 × (−3) is exactly the same as adding 2 × (+3). Flip the divisor's sign once, at the very start, and every subtraction for the rest of the problem becomes an addition.

That is the entire method: drop the x's (column position replaces them), drop the term that always cancels, and flip the sign once so you add instead of subtract. Nothing was approximated and nothing was lost — the answer is identical to long division, every time.

Why the last number equals P(k)

Every division can be written as a multiplication:

P(x) = (x − k) · Q(x) + R

Now substitute x = k. The factor (k − k) is 0, so the whole quotient term vanishes no matter how ugly Q is, and you are left with P(k) = R. That is the Remainder Theorem, and it is what the graph above is showing you: the height of the curve at x = k is the number in the bottom-right corner.

It also gives you the Factor Theorem for free: R = 0 exactly when (x − k) divides evenly, which is exactly when k is a zero of P. Landing on a 0 in that last box means you have found a root and factored the polynomial in the same three seconds.

There is one more thing hiding in the bottom row. Written as a nest, P(3) for that first example is ((2·3 − 5)·3 + 3)·3 − 7 = 11 — and 2, 1, 6, 11 are precisely the numbers in the bottom row. Each “multiply, then add” is one layer of the nest. Synthetic division and evaluating the polynomial are literally the same arithmetic.

Two rules that catch almost every mistake

Write the zeros. Synthetic division reads meaning into column position, so a missing power must still occupy its column. x4 − 16 is 1, 0, 0, 0, −16 — five entries, not two. Skipping the placeholders shifts every power and silently ruins the answer.
Only for x − k. The divisor must be linear with leading coefficient 1. For x + 4, rewrite as x − (−4) and use k = −4. For 2x − 1, first factor: 2x − 1 = 2(x − ½). Synthetically divide by x − ½, then halve every quotient coefficient; the remainder is unchanged. For a divisor of degree 2 or higher, synthetic division does not apply — use long division.