Polynomial long division and synthetic division: Worked examples

This page is the drill room for Polynomial long division and synthetic division. The parent note taught the rules; here we hit every kind of division problem you can meet, one worked example per case, so nothing on an exam is new.

Before we start, one anchor from the parent so this page stands alone:

Reminders of the two tools (defined fully in the parent):

Long division — works for any divisor.
Synthetic division — a shortcut for divisors of the form $(x - c)$ only; you feed it the zero $c$ (the value making $x - c = 0$ ), never the whole binomial.

The scenario matrix

Every polynomial-division problem falls into one of these case classes. Below the table, one worked example covers each cell. The "Ex" column tells you which example hits it.

#	Case class	What's tricky about it	Ex
A	Linear divisor $x - c$ , $c > 0$ , remainder $\neq = 0$	plain synthetic division, positive $c$	1
B	Linear divisor $x + c$ , so $c < 0$	sign trap: use $- c$	2
C	Missing terms (gaps in powers)	must insert $0$ coefficients	3
D	Remainder $= 0$ (exact division / root found)	connects to factor theorem	4
E	Divisor not monic ( $a x - b$ , $a \neq = 1$ )	synthetic needs a rescale	5
F	Quadratic (or higher) divisor	synthetic forbidden, long division only; remainder has degree $\geq 1$	6
G	Degree of $P <$ degree of $D$ (degenerate)	quotient is $0$ ; nothing to divide	7
H	Real-world word problem (area ÷ length)	translate words → polynomials	8
I	Exam twist: find unknown coefficient from a given remainder	run the algorithm backwards	9

Where the "opposite over adjacent"-style why matters, I make you forecast the answer first — guessing sharpens what to watch for.

Case A — positive $c$ , nonzero remainder

Worked example Example 1 ·

$(3 x^{3} - 2 x^{2} + 4 x - 5) \div (x - 2)$ Forecast: the quotient drops one degree, so it's a quadratic. Will the remainder be big or small? (Hint: it equals $P (2)$ — you'll see why.)

1. Write the coefficients in descending order. $3 x^{3} - 2 x^{2} + 4 x - 5 \Rightarrow [3, - 2, 4, - 5]$ . Why this step? Synthetic division works only on the coefficient list — the powers of $x$ are implied by position, so no term may be skipped.

2. Find $c$ . Divisor $x - 2 = 0 \Rightarrow c = 2$ . Why this step? Synthetic division feeds on the zero of the divisor, not the coefficient. Here the sign is easy: $x - 2$ already looks like $x - c$ .

3. Run bring-down → (multiply by 2) → add. See the figure — each red arrow is a "multiply by $c$ ", each column sum is an "add".

Column by column: $3 \to 3 \cdot 2 = 6 \to - 2 + 6 = 4 \to 4 \cdot 2 = 8 \to 4 + 8 = 12 \to 12 \cdot 2 = 24 \to - 5 + 24 = 19.$ Bottom row: $[3, 4, 12 ∣ 19]$ .

4. Read off the answer. Last number is the remainder; the rest are quotient coefficients, one degree lower: $Q (x) = 3 x^{2} + 4 x + 12, R = 19.$

Verify: Plug $x = 2$ into $P$ : $3 (8) - 2 (4) + 4 (2) - 5 = 24 - 8 + 8 - 5 = 19$ . That matches $R$ exactly — the Remainder Theorem in action: dividing by $x - c$ leaves remainder $P (c)$ .

Case B — the sign trap ( $x + c$ )

Worked example Example 2 ·

$(x^{3} + 4 x^{2} + x - 6) \div (x + 3)$ Forecast: the divisor is $x + 3$ . Do we feed $+ 3$ or $- 3$ into synthetic division? Commit to an answer, then check.

1. Rewrite the divisor as $x - c$ . $x + 3 = x - (- 3)$ , so $c = - 3$ . Why this step? This is where most marks are lost. The zero of $x + 3$ is $x = - 3$ ; that is the number the algorithm needs. Using $+ 3$ would divide by the wrong thing.

2. Coefficients: $[1, 4, 1, - 6]$ . No gaps.

3. Bring-down / multiply by $- 3$ / add: $1 \to 1 \cdot (- 3) = - 3 \to 4 - 3 = 1 \to 1 \cdot (- 3) = - 3 \to 1 - 3 = - 2 \to - 2 \cdot (- 3) = 6 \to - 6 + 6 = 0.$ Bottom row: $[1, 1, - 2 ∣ 0]$ .

4. Read off: $Q (x) = x^{2} + x - 2$ , $R = 0$ .

Verify: $R = 0$ means $x = - 3$ is a root. Check $P (- 3) = - 27 + 36 - 3 - 6 = 0$ ✓. And $x^{2} + x - 2 = (x + 2) (x - 1)$ , so $P (x) = (x + 3) (x + 2) (x - 1)$ — fully factored, a bonus.

Case C — missing terms (gaps)

Worked example Example 3 ·

$(x^{4} - 5 x^{2} + 4) \div (x + 1)$ Forecast: there is no $x^{3}$ term and no $x$ term. If you write only the visible numbers $[1, - 5, 4]$ , your quotient will come out the wrong degree. Guess: how many coefficients should the list have for a degree-4 polynomial?

1. Fill every power with a coefficient, including zeros. $x^{4} + 0 x^{3} - 5 x^{2} + 0 x + 4 \Rightarrow [1, 0, - 5, 0, 4]$ — five entries for degrees $4, 3, 2, 1, 0$ . Why this step? Position = power. A missing slot silently shifts every later coefficient into the wrong power, like writing $105$ as $15$ .

2. $c$ : $x + 1 = x - (- 1)$ , so $c = - 1$ .

3. Run it: $1 \to (- 1) \to 0 - 1 = - 1 \to (- 1) (- 1) = 1 \to - 5 + 1 = - 4 \to (- 4) (- 1) = 4 \to 0 + 4 = 4 \to 4 (- 1) = - 4 \to 4 - 4 = 0.$ Bottom row: $[1, - 1, - 4, 4 ∣ 0]$ .

4. Read off: $Q (x) = x^{3} - x^{2} - 4 x + 4$ , $R = 0$ .

Verify: $P (- 1) = 1 - 5 + 4 = 0$ ✓ (so $x = - 1$ is a root, $R = 0$ consistent). Sanity-check the degree: $P$ was degree 4, divisor degree 1, so $Q$ must be degree 3 — and it is. Had we dropped the zeros, $Q$ would have wrongly come out degree 1.

Case D — exact division (remainder zero, root found)

Worked example Example 4 · Is

$x = \frac{1}{2}$ a root of $2 x^{3} - 3 x^{2} + 1$ ? Forecast: the Rational Root Theorem allows $\pm \frac{1}{2}$ as a candidate. If $x = \frac{1}{2}$ is a root, the remainder is $0$ . Guess before dividing.

1. Coefficients: $[2, - 3, 0, 1]$ (note the $0$ for the missing $x$ term). $c = \frac{1}{2}$ . Why this step? We test a root by dividing by $x - \frac{1}{2}$ and watching the remainder; zero remainder $\Leftrightarrow$ root, by the factor theorem.

2. Run it: $2 \to 2 \cdot \frac{1}{2} = 1 \to - 3 + 1 = - 2 \to - 2 \cdot \frac{1}{2} = - 1 \to 0 - 1 = - 1 \to - 1 \cdot \frac{1}{2} = - \frac{1}{2} \to 1 - \frac{1}{2} = \frac{1}{2} .$ Bottom row: $[2, - 2, - 1 ∣ \frac{1}{2}]$ .

3. Interpret. Remainder is $\frac{1}{2} \neq = 0$ , so $x = \frac{1}{2}$ is not a root.

Verify: $P (\frac{1}{2}) = 2 \cdot \frac{1}{8} - 3 \cdot \frac{1}{4} + 1 = \frac{1}{4} - \frac{3}{4} + 1 = \frac{1}{2}$ ✓. (The Remainder Theorem again — the remainder is $P (c)$ .) Lesson: a case where the answer is "no", so you don't get a factorisation. That's a valid, common outcome.

Case E — non-monic divisor $a x - b$

Worked example Example 5 ·

$(6 x^{2} + 7 x - 3) \div (2 x - 1)$ Forecast: synthetic division as taught wants a monic divisor ( $x - c$ , leading coefficient 1). $2 x - 1$ has leading coefficient 2. Do we abandon synthetic division, or can we rescale?

1. Factor the leading coefficient out of the divisor. $2 x - 1 = 2 (x - \frac{1}{2})$ . Why this step? We turn the divisor into (number) × (monic). Dividing by $2 (x - \frac{1}{2})$ is the same as dividing by $x - \frac{1}{2}$ and then dividing the quotient by $2$ .

2. Synthetic-divide by $x - \frac{1}{2}$ ( $c = \frac{1}{2}$ ). Coefficients $[6, 7, - 3]$ : $6 \to 6 \cdot \frac{1}{2} = 3 \to 7 + 3 = 10 \to 10 \cdot \frac{1}{2} = 5 \to - 3 + 5 = 2.$ Bottom row: $[6, 10 ∣ 2]$ . So dividing by $x - \frac{1}{2}$ : quotient $6 x + 10$ , remainder $2$ .

3. Rescale the quotient (not the remainder). Because $2 x - 1$ is twice $x - \frac{1}{2}$ , the true quotient is $\frac{1}{2} (6 x + 10) = 3 x + 5$ . The remainder $2$ is unchanged — it is what's genuinely left over. Why this step? $P = (x - \frac{1}{2}) (6 x + 10) + 2 = (2 x - 1) \cdot \frac{1}{2} (6 x + 10) + 2 = (2 x - 1) (3 x + 5) + 2$ .

4. Final: $Q (x) = 3 x + 5$ , $R = 2$ .

Verify: $(2 x - 1) (3 x + 5) + 2 = 6 x^{2} + 10 x - 3 x - 5 + 2 = 6 x^{2} + 7 x - 3$ ✓.

Case F — quadratic divisor (long division mandatory)

Worked example Example 6 ·

$(x^{4} + 0 x^{3} + 0 x^{2} + 3 x - 2) \div (x^{2} + x + 1)$ Forecast: the divisor has degree 2, so synthetic division is off the table. The remainder must have degree less than 2, i.e. at most $a x + b$ — a linear leftover, not a single number. Watch for that.

Long division, following Divide–Multiply–Subtract–Bring down (from the parent). The figure tracks the descending remainders.

1. Divide leading terms: $\frac{x ^{4}}{x ^{2}} = x^{2}$ . Multiply: $x^{2} (x^{2} + x + 1) = x^{4} + x^{3} + x^{2}$ . Subtract: $(x^{4} + 0 x^{3} + 0 x^{2}) - (x^{4} + x^{3} + x^{2}) = - x^{3} - x^{2}$ . Bring down $+ 3 x$ : current dividend $- x^{3} - x^{2} + 3 x$ . Why this step? We kill the highest term ( $x^{4}$ ) by choosing the quotient term that reproduces it.

2. Divide: $\frac{- x ^{3}}{x ^{2}} = - x$ . Multiply: $- x (x^{2} + x + 1) = - x^{3} - x^{2} - x$ . Subtract: $(- x^{3} - x^{2} + 3 x) - (- x^{3} - x^{2} - x) = 4 x$ . Bring down $- 2$ : current dividend $4 x - 2$ .

3. Divide: $\frac{4 x}{x ^{2}}$ — degree of $4 x$ (which is 1) is less than degree of divisor (2). Stop. Why this step? Once the leftover's degree drops below the divisor's, no divisor can fit inside it — that's the definition of "done".

4. Read off: $Q (x) = x^{2} - x$ , $R (x) = 4 x - 2$ .

Verify: $(x^{2} + x + 1) (x^{2} - x) + (4 x - 2)$ . Expand the product: $x^{4} - x^{3} + x^{3} - x^{2} + x^{2} - x = x^{4} - x$ . Add remainder: $x^{4} - x + 4 x - 2 = x^{4} + 3 x - 2$ ✓ (matches the original with its zero terms).

Case G — degenerate: dividend smaller than divisor

Case H — real-world word problem

Worked example Example 8 · A rectangle's area is

$A (x) = 2 x^{2} + 11 x + 12$ square metres and one side has length $(x + 4)$ metres. Find the other side. Forecast: area = length × width. If the width divides the area exactly, the "shape closes" — expect remainder $0$ . Guess the degree of the answer.

1. Translate. Other side $= \frac{A ( x )}{x + 4}$ . Divide $2 x^{2} + 11 x + 12$ by $x + 4$ . Why this step? Dividing area by one side gives the other side — the polynomial version of $width = area \div length$ .

2. Synthetic division, $c = - 4$ , coefficients $[2, 11, 12]$ : $2 \to 2 \cdot (- 4) = - 8 \to 11 - 8 = 3 \to 3 \cdot (- 4) = - 12 \to 12 - 12 = 0.$ Bottom row: $[2, 3 ∣ 0]$ .

3. Interpret with units. Quotient $= 2 x + 3$ , remainder $= 0$ . Other side $= (2 x + 3)$ metres. Why this step? Remainder $0$ confirms $(x + 4)$ is a genuine factor — the rectangle's dimensions are exact, as physical lengths should be.

Verify: $(x + 4) (2 x + 3) = 2 x^{2} + 3 x + 8 x + 12 = 2 x^{2} + 11 x + 12$ ✓. Units: metres × metres = square metres ✓. Sanity: at $x = 1$ , sides $5$ m and $5$ m give $25$ m², and $A (1) = 2 + 11 + 12 = 25$ ✓.

Case I — exam twist: find an unknown coefficient

Worked example Example 9 · When

$x^{3} + k x^{2} - 4 x + 6$ is divided by $(x - 2)$ , the remainder is $10$ . Find $k$ . Forecast: we don't know $k$ yet, so we can't just divide numbers — we'll carry $k$ as a symbol and let the remainder equation pin it down. Guess whether $k$ is positive.

1. Use the Remainder Theorem instead of grinding division. Remainder of division by $x - 2$ is $P (2)$ . Why this step? The theorem turns a division problem into a single substitution — far faster than symbolic long division, and it's the whole point of this connection.

2. Write the equation. $P (2) = 2^{3} + k \cdot 2^{2} - 4 \cdot 2 + 6 = 8 + 4 k - 8 + 6 = 4 k + 6.$

3. Set equal to the given remainder and solve. $4 k + 6 = 10 \Rightarrow 4 k = 4 \Rightarrow k = 1.$ Why this step? The problem tells us the remainder is $10$ ; matching our expression to it isolates $k$ .

Verify: with $k = 1$ , $P (x) = x^{3} + x^{2} - 4 x + 6$ and $P (2) = 8 + 4 - 8 + 6 = 10$ ✓. If you insist on dividing, synthetic division of $[1, 1, - 4, 6]$ by $c = 2$ gives bottom row $[1, 3, 2 ∣ 10]$ — remainder $10$ ✓.

[!recall]- Quick self-test (cloze)

The number fed into synthetic division for divisor $x + 7$ is == $- 7$ ==.

If $de g (P) < de g (D)$ then the quotient is == $0$ and the remainder is $P$ itself==.

Dividing by a quadratic must use long division, and its remainder can be as large as degree == $1$ == (linear).

Synthetic-division remainder equals

P (c)

, the value of the dividend at the divisor's zero (Remainder Theorem).

For a non-monic divisor $a x - b$ , after synthetic division by $x - \frac{b}{a}$ you divide the quotient by $a$ but leave the remainder unchanged.

[!mnemonic] One line to carry out the door

"Zero, not the sign" — always feed synthetic division the value that makes the divisor zero. Everything else (gaps → zeros, stop when the leftover shrinks, rescale non-monic quotients) follows from the single identity $P = D Q + R$ .

See also: 2.1.8-Factoring-polynomials, 3.2.4-Partial-fraction-decomposition, 2.1.1-Polynomials-definition-and-degree.