Synthetic Division and the Remainder Theorem: Practice

Factor the divisor: $2x - 1 = 2\left(x - \tfrac{1}{2}\right)$, and run the tableau with corner $\tfrac{1}{2}$.

$\begin{array}{c|rrrr} \tfrac{1}{2} & 2 & 5 & -7 & 6 \\ & & 1 & 3 & -2 \\ \hline & 2 & 6 & -4 & 4 \end{array}$

So $P(x) = \left(x - \tfrac{1}{2}\right)(2x^2 + 6x - 4) + 4 = (2x - 1)(x^2 + 3x - 2) + 4$. Dividing by $2x - 1$ gives quotient $x^2 + 3x - 2$ and remainder $4$: the quotient gets divided by $2$, but the remainder stays put.

Synthetic division on $76$ columns would be absurd, but the Remainder Theorem needs only one evaluation, at $x = -1$. Odd powers of $-1$ give $-1$ and even powers give $+1$.

$P(-1) = (-1)^{75} - 2(-1)^{40} + 5 = -1 - 2 + 5 = 2$

The remainder is $2$. This is the theorem's real power: the remainder costs one evaluation no matter how large the degree.

The identity holds for every $x$, not just at $x = 3$, so substitute $x = 1$ directly into the divided form.

$P(1) = (1 - 3)(1 + 2 + 5) + 4 = (-2)(8) + 4 = -16 + 4 = -12$

No expansion is needed. The value $4$ answers a different question: it is $P(3)$, the remainder on dividing by $x - 3$.

By the Remainder Theorem the remainder is $P(c) = c^2 - 5c + 6$, so the division is exact exactly when this vanishes.

$c^2 - 5c + 6 = 0 \quad \Longrightarrow \quad (c - 2)(c - 3) = 0 \quad \Longrightarrow \quad c = 2 \text{ or } c = 3$

Both values work and no others do: the divisions by $x - 2$ and $x - 3$ are exact, matching $x^2 - 5x + 6 = (x - 2)(x - 3)$.

The $x^2$ term is missing, so the top row is $2,\; 5,\; 0,\; -1,\; 21$ with corner $-3$. Bring down the $2$; then $5 - 6 = -1$; then $0 + 3 = 3$; then $-1 - 9 = -10$; then $21 + 30 = 51$.

$\begin{array}{c|rrrrr} -3 & 2 & 5 & 0 & -1 & 21 \\ & & -6 & 3 & -9 & 30 \\ \hline & 2 & -1 & 3 & -10 & 51 \end{array}$

By the Remainder Theorem $P(-3) = 51$. The direct check agrees: $2(81) + 5(-27) - (-3) + 21 = 162 - 135 + 3 + 21 = 51$.

Divide by $x - 1$ first: corner $1$, top row $1,\; -2,\; -5,\; 6$ gives bottom row $1,\; -1,\; -6$ with remainder $0$, so $P(x) = (x - 1)(x^2 - x - 6)$.

$x^2 - x - 6 = (x - 3)(x + 2)$

Dividing the quotient by $x - 3$ (corner $3$, row $1,\; -1,\; -6$, bottom $1,\; 2,\; 0$) confirms the second exact division and leaves $x + 2$. Altogether $P(x) = (x - 1)(x - 3)(x + 2)$, and expanding this product restores $x^3 - 2x^2 - 5x + 6$.

Factor the divisor: $2x - 4 = 2(x - 2)$. Dividing by $x - 2$ gives $P(x) = (x - 2)Q(x) + r$ with $r = P(2)$, and absorbing the $2$ into the quotient, $P(x) = (2x - 4)\cdot\tfrac{1}{2}Q(x) + r$, leaves the remainder unchanged.

$r = P(2) = 8 + 8 - 4 = 12$

The remainder is $12$, not $6$: only the quotient gets divided by the leading factor, never the remainder.

Since $3x - 2 = 3\left(x - \tfrac{2}{3}\right)$, the remainder is $P\left(\tfrac{2}{3}\right)$.

$P\left(\tfrac{2}{3}\right) = 3 \cdot \tfrac{8}{27} - 2 \cdot \tfrac{4}{9} + 2 - 2 = \tfrac{8}{9} - \tfrac{8}{9} + 0 = 0$

The remainder is $0$, so $3x - 2$ divides this cubic exactly; indeed $P(x) = (3x - 2)\left(x^2 + 1\right)$, as expanding confirms.

Each long-division stage subtracts a multiple of the whole divisor $x - c$. The $x$ part of that multiple cancels the leading term by design, and the remaining part subtracts $-c$ times the newest quotient coefficient.

$a_k - (-c)\,b_k = a_k + c\,b_k$

Subtracting a negative multiple is adding a positive one, and the corner holds $c$ (the divisor's constant already negated) so that the tableau can add. That is also why dividing by $x + 5$ needs corner $-5$: the sign flip happens once, in the corner, and never again.

The tableau's recurrence $b_{k-1} = a_k + c\,b_k$ is the coefficient bookkeeping of dividing by $x - c$ specifically: one number multiplied by one corner value, column by column.

$\deg(\text{remainder}) < \deg(x^2 - 4) = 2$

Dividing by a quadratic leaves a remainder that can be linear, not just constant, and each stage of that long division updates two coefficients at once. The single-corner, single-carry tableau has no place to store that. (Dividing by $x^2 - 4 = (x-2)(x+2)$ can still be done by two successive synthetic divisions, tracking both remainders carefully.)

Write the division identity and evaluate it at $x = 6$; the identity holds at every input, not only at $5$.

$P(6) = (6 - 5)\left(6^2 + 3(6) + 2\right) + 9 = (1)(36 + 18 + 2) + 9 = 56 + 9 = 65$

So $P(6) = 65$. The remainder $9$ alone answers only the question about $P(5)$.

The remainder is $P(-2)$, so set it equal to $25$ and solve for $k$.

$P(-2) = (-2)^4 + k(-2)^2 - 3 = 16 + 4k - 3 = 25 \quad \Longrightarrow \quad 4k = 12 \quad \Longrightarrow \quad k = 3$

Both powers of $-2$ are even, so no sign flips occur. Checking: $16 + 12 - 3 = 25$ as required.