Synthetic Division and the Remainder Theorem: Practice

The corner holds the value of $x$ that makes the divisor zero, the $c$ in $x - c$.

$x - 5 = 0 \quad \text{at} \quad x = 5$

So the corner number is $5$. The divisor's sign is already flipped in the corner, which is why the tableau adds instead of subtracts.

The top row lists the coefficients of the dividend in descending order of degree, keeping each coefficient's own sign.

$x^3 + 2x^2 - 5x + 1 \;\longrightarrow\; 1,\; 2,\; -5,\; 1$

Every power from $x^3$ down to the constant is present, so no placeholder zeros are needed.

The dividend has no $x^2$ term, and every missing power needs a placeholder zero in its column.

$x^3 + 0x^2 - 4x + 7 \;\longrightarrow\; 1,\; 0,\; -4,\; 7$

The zero sits in the $x^2$ column, between the $x^3$ coefficient and the $x$ coefficient. Skipping it shifts every later column one place left and wrecks the whole division.

The corner is $1$ and the top row is $1,\; 7,\; 10$. Bring down the $1$; then $1 \times 1 = 1$ and $7 + 1 = 8$; then $8 \times 1 = 8$ and $10 + 8 = 18$.

$\begin{array}{c|rrr} 1 & 1 & 7 & 10 \\ & & 1 & 8 \\ \hline & 1 & 8 & 18 \end{array}$

The bottom row gives quotient $x + 8$ and remainder $18$. As a check, the Remainder Theorem gives $P(1) = 1 + 7 + 10 = 18$.

By the Remainder Theorem, the remainder equals $P(2)$, the polynomial evaluated where the divisor is zero.

$P(2) = 2^2 + 3(2) + 5 = 4 + 6 + 5 = 15$

Evaluating at $-2$ instead would give $4 - 6 + 5 = 3$, which is the remainder for the different divisor $x + 2$.

The divisor is $x + 1 = x - (-1)$, so the remainder is $P(-1)$.

$P(-1) = (-1)^3 - 2(-1) + 4 = -1 + 2 + 4 = 5$

The odd power keeps its minus sign, and $-2$ times $-1$ turns positive. Evaluating at $+1$ by mistake gives $1 - 2 + 4 = 3$.

By the Remainder Theorem the remainder is $P(4)$.

$P(4) = 16 - 24 + 8 = 0$

A remainder of $0$ means the division is exact: $x - 4$ divides $x^2 - 6x + 8$ with nothing left over, matching the factorization $(x - 2)(x - 4)$.

Synthetic division only ever adds down each column.

$\text{new bottom entry} = \text{coefficient above} + c \times \text{previous bottom entry}$

The subtraction that long division performs is already baked in, because the corner stores $c$ with the divisor's sign flipped: subtracting a $-c$ multiple is the same as adding a $+c$ multiple. Subtracting again would undo that and compute a different division.

In the identity $P(x) = (x - c)Q(x) + r$, the degrees on both sides must match, and the remainder is a constant that cannot affect the leading term.

$\deg P = 1 + \deg Q \quad \Longrightarrow \quad \deg Q = 4 - 1 = 3$

The divisor spends one power of $x$, so the quotient sits exactly one degree below the dividend. That is why the tableau's bottom row starts the quotient one degree down.

The divisor $x + 2 = x - (-2)$ puts $-2$ in the corner, with top row $1,\; 2,\; -1,\; -2$. Bring down the $1$; then $1 \times (-2) = -2$ and $2 + (-2) = 0$; then $0 \times (-2) = 0$ and $-1 + 0 = -1$; then $-1 \times (-2) = 2$ and $-2 + 2 = 0$.

$\begin{array}{c|rrrr} -2 & 1 & 2 & -1 & -2 \\ & & -2 & 0 & 2 \\ \hline & 1 & 0 & -1 & 0 \end{array}$

The bottom row $1,\; 0,\; -1$ with remainder $0$ gives quotient $x^2 - 1$. Using corner $+2$ by mistake produces $x^2 + 4x + 7$ with remainder $12$, the division by $x - 2$ instead.

The tableau's recurrence carries one running number per column, which encodes division by a monic linear divisor.

$\text{divisor} = x - c$

Divisors like $x + 4$ still qualify, since $x + 4 = x - (-4)$. A non-monic linear divisor such as $2x - 1$ needs the extra step of factoring out the $2$, and a quadratic divisor needs long division, because its remainder can have degree $1$.

Run synthetic division with corner $3$ and top row $1,\; -2,\; 1,\; -5$: bring down $1$; then $1 \times 3 = 3$ and $-2 + 3 = 1$; then $1 \times 3 = 3$ and $1 + 3 = 4$; then $4 \times 3 = 12$ and $-5 + 12 = 7$.

$\begin{array}{c|rrrr} 3 & 1 & -2 & 1 & -5 \\ & & 3 & 3 & 12 \\ \hline & 1 & 1 & 4 & 7 \end{array}$

The final number is the remainder, and by the Remainder Theorem it equals $P(3)$. Direct evaluation confirms it: $27 - 18 + 3 - 5 = 7$.