Synthetic Division and the Remainder Theorem: Practice

The corner is $1$ with top row $1,\; -6,\; 11,\; -6$. Bring down the $1$; then $1 \times 1 = 1$ and $-6 + 1 = -5$; then $-5 \times 1 = -5$ and $11 - 5 = 6$; then $6 \times 1 = 6$ and $-6 + 6 = 0$.

$\begin{array}{c|rrrr} 1 & 1 & -6 & 11 & -6 \\ & & 1 & -5 & 6 \\ \hline & 1 & -5 & 6 & 0 \end{array}$

The quotient is $x^2 - 5x + 6$ and the remainder is $0$, so the division is exact. The check $P(1) = 1 - 6 + 11 - 6 = 0$ agrees.

Three powers are missing, so the top row is $1,\; 0,\; 0,\; 0,\; -16$ with corner $2$. Bring down the $1$; then each step doubles and adds: $0 + 2 = 2$, $0 + 4 = 4$, $0 + 8 = 8$, and finally $-16 + 16 = 0$.

$\begin{array}{c|rrrrr} 2 & 1 & 0 & 0 & 0 & -16 \\ & & 2 & 4 & 8 & 16 \\ \hline & 1 & 2 & 4 & 8 & 0 \end{array}$

The quotient is $x^3 + 2x^2 + 4x + 8$ with remainder $0$. Dividing by $x + 2$ instead would give the sign-alternating quotient $x^3 - 2x^2 + 4x - 8$.

With placeholders for the missing $x^3$ and $x$ terms, the top row is $1,\; 0,\; -5,\; 0,\; 4$ and the corner is $-1$. Bring down the $1$; then $0 - 1 = -1$; then $-5 + 1 = -4$; then $0 + 4 = 4$; then $4 - 4 = 0$.

$\begin{array}{c|rrrrr} -1 & 1 & 0 & -5 & 0 & 4 \\ & & -1 & 1 & 4 & -4 \\ \hline & 1 & -1 & -4 & 4 & 0 \end{array}$

The quotient is $x^3 - x^2 - 4x + 4$ with remainder $0$. The check $P(-1) = 1 - 5 + 4 = 0$ confirms the exactness.

By the Remainder Theorem the remainder is $P(3)$.

$P(3) = 2(27) - 7(9) + 5 = 54 - 63 + 5 = -4$

Forgetting the constant term gives $-9$, and flipping the sign of the middle term gives $54 + 63 + 5 = 122$; keeping every sign straight gives $-4$.

The divisor vanishes at $x = -2$, so the remainder is $P(-2)$. The even power stays positive and the odd power keeps its minus sign.

$P(-2) = 16 - 3(-8) + 2(-2) - 1 = 16 + 24 - 4 - 1 = 35$

The middle term is the trap: $-3 \times (-8) = +24$. Treating $(-2)^3$ as $+8$ gives $-13$, and evaluating at $+2$ gives $-5$.

The bottom row says the quotient is $x^2 - 3x + 2$ and the remainder is $0$, so the division identity reconstructs the dividend.

$P(x) = (x - 4)(x^2 - 3x + 2) = x^3 - 3x^2 + 2x - 4x^2 + 12x - 8 = x^3 - 7x^2 + 14x - 8$

Collecting like terms gives $P(x) = x^3 - 7x^2 + 14x - 8$.

The divisor vanishes at $x = -1$, so the remainder is $P(-1)$. Set it equal to $-8$.

$P(-1) = -1 - k - 4 = -5 - k = -8 \quad \Longrightarrow \quad k = 3$

Checking: $P(x) = x^3 + 3x - 4$ gives $P(-1) = -1 - 3 - 4 = -8$ as required. The sign trap is $k(-1) = -k$, not $+k$.

The $x$ term is missing, so the top row is $1,\; -2,\; -7,\; 0,\; 11$ with corner $5$. Bring down the $1$; then $-2 + 5 = 3$; then $-7 + 15 = 8$; then $0 + 40 = 40$; then $11 + 200 = 211$.

$\begin{array}{c|rrrrr} 5 & 1 & -2 & -7 & 0 & 11 \\ & & 5 & 15 & 40 & 200 \\ \hline & 1 & 3 & 8 & 40 & 211 \end{array}$

By the Remainder Theorem $P(5) = 211$, and the direct check $625 - 250 - 175 + 11 = 211$ agrees.

Two independent checks are needed. The corner holds the value that makes $x + 3$ zero, which is $-3$, and the top row must include a placeholder zero for each missing power.

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

So the correct setup pairs corner $-3$ with the five-entry row. A three-entry row would divide a quadratic, not a quartic.

The top row needs placeholders for the missing $x^2$ and $x$ terms: $1,\; 0,\; 0,\; 27$ with corner $-3$. Bring down the $1$; then $0 - 3 = -3$; then $0 + 9 = 9$; then $27 - 27 = 0$.

$\begin{array}{c|rrrr} -3 & 1 & 0 & 0 & 27 \\ & & -3 & 9 & -27 \\ \hline & 1 & -3 & 9 & 0 \end{array}$

The quotient is $x^2 - 3x + 9$ with remainder $0$, which is the sum-of-cubes factorization $x^3 + 27 = (x + 3)(x^2 - 3x + 9)$.

The corner is $2$ with top row $2,\; 1,\; -13,\; 6$. Bring down the $2$; then $2 \times 2 = 4$ and $1 + 4 = 5$; then $5 \times 2 = 10$ and $-13 + 10 = -3$; then $-3 \times 2 = -6$ and $6 - 6 = 0$.

$\begin{array}{c|rrrr} 2 & 2 & 1 & -13 & 6 \\ & & 4 & 10 & -6 \\ \hline & 2 & 5 & -3 & 0 \end{array}$

The quotient is $2x^2 + 5x - 3$ with remainder $0$, so $x - 2$ divides this cubic exactly.

The final entry of the bottom row is the remainder of the division, and the Remainder Theorem says that remainder equals the polynomial's value at the corner number.

$P(2) = \text{remainder} = 13$

No further computation is needed: the tableau has already evaluated the polynomial at $2$ while dividing. That double duty is exactly why synthetic division doubles as fast evaluation.