Synthetic Division and the Remainder Theorem

Intuition

Divide $2x^3 - 5x^2 + 3x - 7$ by $x - 3$ with long division and watch what your pen actually writes. It writes the letter $x$ dozens of times, along with exponents, plus signs, and a staircase of subtractions, while the only numbers doing any work are the coefficients. Synthetic division is what survives when every symbol that carries no information is deleted: three short rows of arithmetic that produce the same quotient and the same remainder. Hiding in its bottom-right corner is one of the workhorse facts of this chapter, the Remainder Theorem. That theorem says the remainder of dividing $P(x)$ by $x - c$ is exactly the number $P(c)$ . One small tableau hands you the quotient, the remainder, and a fast evaluation of the polynomial, all at once.

Learning goals

Run the tableau: bring down, multiply, add, repeat
Match it to long division through the coefficient recurrence
Apply the Remainder Theorem, since the remainder is $P(c)$
Evaluate in nested form, which is Horner's method
Adjust for a divisor $ax - b$ by dividing the quotient

What long division keeps rewriting

Run the long division of $2x^3 - 5x^2 + 3x - 7$ by $x - 3$ from the last lesson, but keep an eye on which parts of the writing are decisions and which parts are ritual. Each stage picks the next quotient term, multiplies it by $x - 3$ , and subtracts:

\begin{aligned} 2x^3 - 5x^2 + 3x - 7 &= (x - 3)\cdot 2x^2 + \bigl(x^2 + 3x - 7\bigr), \\[2pt] x^2 + 3x - 7 &= (x - 3)\cdot x + \bigl(6x - 7\bigr), \\[2pt] 6x - 7 &= (x - 3)\cdot 6 + 11. \end{aligned}

Stacking the three stages gives $2x^3 - 5x^2 + 3x - 7 = (x - 3)(2x^2 + x + 6) + 11$ , so the quotient is $2x^2 + x + 6$ and the remainder is $11$ .

Now look at what each subtraction really did. Every quotient term was chosen so that the leading terms cancel exactly, so that half of the step carries no information at all. The other half subtracts $-3$ times the newest quotient coefficient, and subtracting a $-3$ multiple is the same as adding a $+3$ multiple. Strip away the guaranteed cancellation and here is the entire computation, written in bare numbers. The leading coefficient $2$ came straight down, and after that each new number was the next coefficient of the dividend plus $3$ times the previous new number:

-5 + 3 \cdot 2 = 1, \qquad 3 + 3 \cdot 1 = 6, \qquad -7 + 3 \cdot 6 = 11.

The numbers $2, 1, 6$ are precisely the quotient coefficients, and the final number $11$ is the remainder. Long division computed nothing else. Synthetic division is a layout that computes exactly this chain and refuses to write anything more.

The tableau: bring down, multiply, add

To divide a polynomial $P(x)$ by $x - c$ , build a three-row tableau:

In the corner, write $c$ , the number that makes the divisor zero. For $x - 3$ that is $3$ ; for $x + 2$ , which is $x - (-2)$ , it is $-2$ . The corner always holds the divisor’s sign already flipped, and that is why the tableau will add where long division subtracted.
Across the top row, write the coefficients of $P(x)$ in descending order of degree, inserting a $0$ for every missing power. Skipping a placeholder shifts every later column into the wrong place, the same way dropping a zero digit corrupts a base-ten numeral.
Then follow the rhythm: bring down the leading coefficient; multiply it by $c$ and write the product under the next coefficient; add down that column. Next, multiply the sum by $c$ again, and repeat that pattern until the last column is added.

For the division above, the finished tableau is

\begin{array}{c|rrrr} 3 & 2 & -5 & 3 & -7 \\[2pt] & & 6 & 3 & 18 \\[2pt] \hline & 2 & 1 & 6 & 11 \end{array}

Read the bottom row from the left. The last number is the remainder. Everything before it lists the quotient coefficients, starting one degree below the dividend: here $2, 1, 6$ means $Q(x) = 2x^2 + x + 6$ , with remainder $11$ . The degree drops by one because the divisor spent one power of $x$ : a cubic divided by a linear polynomial has a quadratic quotient.

The rhythm of the tableau for dividing 2x cubed minus 5x squared plus 3x minus 7 by x minus 3. Bring the leading 2 straight down, multiply by the corner number 3, write the product under the next coefficient, add down the column, and repeat. The bottom row holds the quotient coefficients 2, 1, 6 and, after the tick mark, the remainder 11.

Worked example 1 Divide $3x^3 + 4x^2 - 11x + 2$ by $x - 2$

The divisor is $x - 2$ , so the corner number is $2$ , and the top row is $3,\; 4,\; -11,\; 2$ . Bring down the $3$ ; then $3 \times 2 = 6$ and $4 + 6 = 10$ ; then $10 \times 2 = 20$ and $-11 + 20 = 9$ ; then $9 \times 2 = 18$ and $2 + 18 = 20$ :

\begin{array}{c|rrrr} 2 & 3 & 4 & -11 & 2 \\[2pt] & & 6 & 20 & 18 \\[2pt] \hline & 3 & 10 & 9 & 20 \end{array}

The quotient is $Q(x) = 3x^2 + 10x + 9$ and the remainder is $20$ . To confirm it, expand the division identity:

\begin{aligned} (x - 2)(3x^2 + 10x + 9) + 20 &= 3x^3 + 4x^2 - 11x - 18 + 20 \\ &= 3x^3 + 4x^2 - 11x + 2. \end{aligned}

The check costs one multiplication and catches nearly every slip, so make it a habit.

Worked example 2 Divide $x^4 - 3x^2 + 5x - 6$ by $x + 2$

Two traps live in this one. First, the divisor is $x + 2 = x - (-2)$ , so the corner number is $-2$ , not $2$ . Second, the dividend has no $x^3$ term, so the top row needs a placeholder zero: $1,\; 0,\; -3,\; 5,\; -6$ . Now run the rhythm, keeping careful track of signs. Bring down the $1$ ; then $1 \times (-2) = -2$ and $0 + (-2) = -2$ ; then $(-2) \times (-2) = 4$ and $-3 + 4 = 1$ ; then $1 \times (-2) = -2$ and $5 + (-2) = 3$ ; finally $3 \times (-2) = -6$ and $-6 + (-6) = -12$ :

\begin{array}{c|rrrrr} -2 & 1 & 0 & -3 & 5 & -6 \\[2pt] & & -2 & 4 & -2 & -6 \\[2pt] \hline & 1 & -2 & 1 & 3 & -12 \end{array}

The quotient is $Q(x) = x^3 - 2x^2 + x + 3$ and the remainder is $-12$ . As a spot check, the dividend at $x = -2$ is $16 - 12 - 10 - 6 = -12$ , which matches the remainder. Why that check is guaranteed to work is exactly where this lesson is headed.

Check your understanding

Use synthetic division to divide $x^3 - 7x + 6$ by $x - 2$ .

Q(x) = x - 5

with remainder

-4

Q(x) = x^2 + 2x - 3

with remainder

0

Q(x) = x^2 - 2x - 3

with remainder

0

Q(x) = x^2 + 2x + 3

with remainder

12

Why the shortcut is exactly long division

A pattern spotted in one example is not a proof. Here is the argument that the tableau’s rhythm produces the true quotient and remainder for every polynomial and every $c$ .

The tableau computes the true quotient and remainder#

Let $P(x) = a_n x^n + a_{n-1}x^{n-1} + \cdots + a_1 x + a_0$ with $n \ge 1$ , and fix any number $c$ . Define the bottom-row numbers exactly as the rhythm does: the bring-down step sets $b_{n-1} = a_n$ , each multiply-and-add step sets

b_{k-1} = a_k + c\,b_k \qquad \text{for } k = n-1,\, n-2,\, \ldots,\, 1,

and the final addition sets $r = a_0 + c\,b_0$ . Collect the bottom row into the polynomial $Q(x) = b_{n-1}x^{n-1} + \cdots + b_1 x + b_0$ . The claim is that $P(x) = (x - c)\,Q(x) + r$ .

Expand the product. Multiplying $Q(x)$ by $x$ shifts every coefficient up one degree, and multiplying by $-c$ scales it, so

(x - c)\,Q(x) = b_{n-1}x^{n} + (b_{n-2} - c\,b_{n-1})x^{n-1} + \cdots + (b_0 - c\,b_1)x - c\,b_0.

Now add $r$ and compare against $P(x)$ one coefficient at a time. The $x^n$ coefficient is $b_{n-1}$ , which the bring-down step made equal to $a_n$ . For each $k$ from $1$ to $n - 1$ , the coefficient of $x^k$ is $b_{k-1} - c\,b_k$ , and the recurrence says $b_{k-1} = a_k + c\,b_k$ , so this coefficient is $(a_k + c\,b_k) - c\,b_k = a_k$ . The constant term is $r - c\,b_0$ , and the final addition says $r = a_0 + c\,b_0$ , so the constant term is $a_0$ . Every coefficient matches, which proves the identity $P(x) = (x - c)\,Q(x) + r$ .

It remains to say why $Q$ and $r$ are the quotient and remainder rather than merely a way of rewriting $P$ . The number $r$ is a constant, so it is either zero or of degree less than $1$ , the degree of the divisor $x - c$ . The division algorithm from the last lesson has a uniqueness half. By that half, only one pair consisting of a quotient and a remainder meeting that condition can satisfy such an identity. The tableau produced that pair.

Read the recurrence $b_{k-1} = a_k + c\,b_k$ once more and the whole design is visible. Long division’s leading-term cancellations are the terms this formula never bothers to write, and long division’s subtraction of a $-c$ multiple is the $+\,c\,b_k$ sitting inside the addition. The tableau is not a different method that happens to agree with long division. It is long division, holding only the numbers that matter.

The Remainder Theorem

The spot check in Worked Example 2 was no coincidence, and the reason is short enough to fit in three lines.

The Remainder Theorem: dividing $P(x)$ by $x - c$ leaves remainder $P(c)$ #

Divide any polynomial $P(x)$ by $x - c$ . The division algorithm from the last lesson produces a quotient $Q(x)$ and a remainder that is zero or of degree less than the divisor’s degree $1$ . In either case the remainder is a constant polynomial. Call that constant $r$ , so that

P(x) = (x - c)\,Q(x) + r

holds as an identity: the two sides are the same polynomial, so they agree at every single value of $x$ . In particular they agree at the one value chosen to make the first term vanish. Substitute $x = c$ :

P(c) = (c - c)\,Q(c) + r = 0 \cdot Q(c) + r = r.

Whatever number $Q(c)$ happens to be, multiplying it by zero erases it. The remainder therefore equals $P(c)$ .

Two remarks keep this proof honest. First, nothing was divided by zero: no fraction ever appears in the argument. We evaluated a polynomial identity, valid for all $x$ , at one convenient input. That is also why the theorem needed no tableau: it follows from the bare existence and uniqueness of polynomial division. Synthetic division is simply the fastest way to find that quotient $Q$ and that remainder $r$ . Second, the theorem is an equality of numbers, not an approximation, and it runs in both directions. Want the remainder without dividing? Evaluate $P(c)$ . Want $P(c)$ without plugging in powers? Divide and read the remainder. The next section takes that second direction seriously.

Check your understanding

Without dividing, find the remainder when $x^3 - 4x^2 + 1$ is divided by $x + 2$ .

-23

-7

-24

9

Division as evaluation

Unwind the tableau’s recurrence for a cubic $P(x) = a_3x^3 + a_2x^2 + a_1x + a_0$ and something familiar appears. The bottom row is $b_2 = a_3$ , then $b_1 = a_2 + c\,b_2$ , then $b_0 = a_1 + c\,b_1$ , then $r = a_0 + c\,b_0$ . Substitute each line into the next:

r = a_0 + c\Bigl(a_1 + c\bigl(a_2 + c\,a_3\bigr)\Bigr) = \Bigl(\bigl(a_3\,c + a_2\bigr)\,c + a_1\Bigr)\,c + a_0.

Multiply the nest out and you get $a_3c^3 + a_2c^2 + a_1c + a_0$ , which is $P(c)$ on the nose. So the tableau’s final number computes $P(c)$ by direct arithmetic, with no division algorithm in sight. That direct computation is a second, independent proof of the Remainder Theorem, and the same unwinding works in every degree. This nested way of evaluating a polynomial is called Horner’s method, and the tableau performs it one multiply-and-add per coefficient. So a degree- $n$ polynomial costs just $n$ multiplications, instead of building each power of $c$ from scratch. As a bonus, the intermediate numbers $b_k$ are not waste products. They are the quotient coefficients, so evaluating $P(c)$ this way hands you the whole division for free.

Worked example 3 Evaluate $P(4)$ for $P(x) = x^5 - 4x^4 + 2x^3 + x - 10$

Computing $4^5$ , $4 \cdot 4^4$ , and $2 \cdot 4^3$ separately invites arithmetic errors. Run the tableau instead. The top row needs a placeholder for the missing $x^2$ term: $1,\; -4,\; 2,\; 0,\; 1,\; -10$ , with $4$ in the corner:

\begin{array}{c|rrrrrr} 4 & 1 & -4 & 2 & 0 & 1 & -10 \\[2pt] & & 4 & 0 & 8 & 32 & 132 \\[2pt] \hline & 1 & 0 & 2 & 8 & 33 & 122 \end{array}

Each step is a single multiply and add: $1 \times 4 = 4$ and $-4 + 4 = 0$ ; then $0 \times 4 = 0$ and $2 + 0 = 2$ ; then $2 \times 4 = 8$ and $0 + 8 = 8$ ; then $8 \times 4 = 32$ and $1 + 32 = 33$ ; finally $33 \times 4 = 132$ and $-10 + 132 = 122$ . By the Remainder Theorem,

P(4) = 122,

and the brute-force check agrees: $1024 - 1024 + 128 + 4 - 10 = 122$ . Five multiplications replaced the whole tower of powers, and the bottom row even reports the quotient $x^4 + 2x^2 + 8x + 33$ from dividing $P(x)$ by $x - 4$ .

Check your understanding

To find the remainder when $x^3 + 4x^2 - 11$ is divided by $x + 3$ , which corner number do you use, and what is the remainder?

A corner

3

, remainder

52

B corner

-3

, remainder

52

C corner

-3

, remainder

-2

D corner

3

, remainder

-2

When the divisor is not monic

The tableau, as built, divides only by $x - c$ : a linear divisor whose leading coefficient is $1$ . A divisor like $2x - 1$ does not fit the corner directly, but factoring out the leading coefficient rescues it, because $2x - 1 = 2\left(x - \tfrac{1}{2}\right)$ . Divide by the monic part $x - \tfrac{1}{2}$ first, then hand the leftover factor of $2$ to the quotient.

Worked example 4 Divide $2x^3 + 5x^2 - 7x + 6$ by $2x - 1$

Since $2x - 1 = 2\left(x - \tfrac{1}{2}\right)$ , run the tableau with corner number $\tfrac{1}{2}$ . Bring down the $2$ ; then $2 \times \tfrac{1}{2} = 1$ and $5 + 1 = 6$ ; then $6 \times \tfrac{1}{2} = 3$ and $-7 + 3 = -4$ ; then $-4 \times \tfrac{1}{2} = -2$ and $6 + (-2) = 4$ :

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

The tableau says $2x^3 + 5x^2 - 7x + 6 = \left(x - \tfrac{1}{2}\right)(2x^2 + 6x - 4) + 4$ . Now rewrite the monic factor as $\tfrac{1}{2}(2x - 1)$ and let the $\tfrac{1}{2}$ act on the quotient:

2x^3 + 5x^2 - 7x + 6 = (2x - 1)\left(x^2 + 3x - 2\right) + 4.

Dividing by $2x - 1$ gives quotient $x^2 + 3x - 2$ and remainder $4$ . Notice exactly what changed in that rewrite and what did not. The quotient got divided by $2$ , while the remainder stayed put, because the factor of $2$ was absorbed inside the product term only. Halving the remainder as well is the standard mistake here. Consistently with the Remainder Theorem, the remainder is $P\left(\tfrac{1}{2}\right) = \tfrac{1}{4} + \tfrac{5}{4} - \tfrac{14}{4} + \tfrac{24}{4} = 4$ .

The same reasoning in general: to divide $P(x)$ by $ax - b$ , run the tableau with corner $c = \tfrac{b}{a}$ , divide the resulting quotient by $a$ , and leave the remainder alone. In particular the Remainder Theorem extends verbatim: the remainder on dividing by $ax - b$ is $P\left(\tfrac{b}{a}\right)$ , still a single evaluation.

When the remainder is zero

In the first checkpoint the remainder came out $0$ , and the division identity collapsed to a pure product: $x^3 - 7x + 6 = (x - 2)(x^2 + 2x - 3)$ , with nothing left over. The division is exact, and $x - 2$ is a factor of the cubic. The Remainder Theorem tells you precisely when this exact division happens. Since the remainder is $P(c)$ , the remainder is zero exactly when $P(c) = 0$ , that is, exactly when $c$ is a root of the polynomial. Dividing out $x - c$ then trades a degree- $n$ polynomial for a degree- $(n-1)$ quotient, one full degree of difficulty cheaper. That two-way link between roots and factors is the Factor Theorem, and it is the entire subject of the next lesson. Here you have already done all the hard work of proving that theorem.

Common mistakes

Wrong sign in the corner. The corner holds the value of $x$ that makes the divisor zero. For $x - 5$ that is $5$ , but for $x + 5$ it is $-5$ . Writing $+5$ for $x + 5$ silently computes a different division.
Skipping placeholder zeros. The top row for $x^3 - 7x + 6$ is $1,\; 0,\; -7,\; 6$ . Omitting the $0$ shifts every later column one place left and wrecks both quotient and remainder.
Subtracting inside the tableau. Synthetic division only ever adds. The subtraction from long division is already baked into the corner’s flipped sign, so subtracting again undoes it.
Misreading the bottom row. The last number is the remainder, not a coefficient, and the quotient starts one degree below the dividend. A bottom row of $2, 1, 6, 11$ from dividing a cubic means $2x^2 + x + 6$ remainder $11$ , never $2x^3 + x^2 + 6x + 11$ .
Sign slips when $c$ is negative. Odd powers keep the minus, even powers lose it: $(-2)^3 = -8$ but $(-2)^4 = 16$ . Write each product out rather than trusting momentum.
Mishandling non-monic divisors. For $2x - 1$ , run the corner $\tfrac{1}{2}$ , then divide the quotient by $2$ and leave the remainder unchanged. Halving the remainder, or feeding $2x - 1$ into the corner directly, both give wrong answers.

Key takeaways

The rhythm: corner holds $c$ from $x - c$ ; top row lists coefficients with $0$ placeholders; bring down, multiply by $c$ , add, repeat. Bottom row gives the quotient coefficients (one degree lower) and, last, the remainder.
It is long division: matching coefficients in $P(x) = (x - c)Q(x) + r$ forces exactly the tableau’s recurrence $b_{k-1} = a_k + c\,b_k$ , so the shortcut and long division are the same computation.
Remainder Theorem: dividing $P(x)$ by $x - c$ leaves remainder $P(c)$ , because evaluating the identity $P(x) = (x - c)Q(x) + r$ at $x = c$ kills the product term.
Division is evaluation: the tableau computes $P(c)$ in the nested form $\bigl((a_3c + a_2)c + a_1\bigr)c + a_0$ , one multiply-and-add per coefficient, and yields the quotient as a byproduct. This is Horner’s method.
Divisor $ax - b$ : run corner $c = \tfrac{b}{a}$ , divide the quotient by $a$ , keep the remainder as is; the remainder equals $P\left(\tfrac{b}{a}\right)$ .
Remainder $0$ exactly when $P(c) = 0$ : exact division certifies a factor, the doorway to the Factor Theorem in the next lesson.

Synthetic Division and the Remainder Theorem

Learning goals

What long division keeps rewriting

The tableau: bring down, multiply, add

Why the shortcut is exactly long division

The Remainder Theorem

Division as evaluation

When the divisor is not monic

When the remainder is zero

Common mistakes

Practice

Multiple Choice Questions (MCQ)

Free Response Questions (FRQ)