Polynomial long division and synthetic division: Question bank

These are thinking questions, not grinding-arithmetic questions (that's the worked-example child pages). Each line below is a trap that catches a real misconception about polynomial division. Read the prompt, answer out loud before peeking, then click to reveal the reasoning.

Vocabulary reminders, so no symbol is used unexplained:

Dividend $P (x)$ = the thing being split up. Divisor $D (x)$ = the thing we split it into. Quotient $Q (x)$ = how many times it fits. Remainder $R (x)$ = what's left over that no longer fits.
$de g$ means degree = the highest power of $x$ present (see 2.1.1-Polynomials-definition-and-degree). A constant like $7$ has degree $0$ ; the zero polynomial has no degree at all.
The master equation the whole topic hangs on: $P (x) = D (x) Q (x) + R (x), de g (R) < de g (D) or R = 0.$

True or false — justify

The rule "== $de g (R) < de g (D)$ ==" is the referee for almost every question here. Keep it in view.

Dividing a degree-5 polynomial by a degree-2 polynomial always gives a degree-3 quotient.

True. The leading term of

P

is (leading of

D

) times (leading of

Q

), so

5 = 2 + de g (Q)

forces

de g (Q) = 3

. Degrees subtract exactly like exponents.

If the remainder is

0

, then the divisor is a factor of the dividend.

True — that is the definition of "factor".

P = D \cdot Q + 0

means

P = D \cdot Q

, so

D

divides

P

cleanly (link: 2.1.13-Factor-theorem-and-remainder-theorem).

Synthetic division can divide

x^{3} + 1

x^{2} + 1

False. Synthetic division only handles linear divisors

x - c

. A degree-2 divisor has no single "zero

c

" to feed the shortcut; you must use long division.

Long division of polynomials can fail if the divisor has higher degree than the dividend.

False, it doesn't fail — it just stops immediately. If

de g (D) > de g (P)

, then

P

itself already satisfies

de g (R) < de g (D)

, so

Q = 0

and

R = P

The remainder when dividing any polynomial by

x - c

is always a single number.

True. The divisor has degree

1

, and the referee demands

de g (R) < 1

, i.e.

de g (R) = 0

— a constant. This constant equals

P (c)

(Remainder Theorem).

Dividing by

2 x - 6

is the same as dividing by

x - 3

, so I can just use

c = 3

in synthetic division.

False in general.

2 x - 6 = 2 (x - 3)

; the quotient changes by a factor of

2

and the remainder changes too. You may divide out the

2

first, but you must then account for it, so raw synthetic division with

c = 3

gives the wrong quotient.

Every polynomial can be divided by

x - 1

with some quotient and remainder.

True. Division with remainder always exists and is unique for any divisor of degree

\geq 1

. Whether the remainder is zero is a separate question.

If the remainder is a nonzero constant, the dividend has no rational roots.

False — a nonzero remainder only tells you that particular divisor isn't a factor. Other divisors could still divide it (see 2.15-Rational-root-theorem).

Spot the error

Each item shows a step a student wrote. Find the mistake and name why it's wrong.

Dividing

x^{3} + 5

x - 2

, a student writes coefficients as

[1, 5]

Wrong: the missing

x^{2}

and

x

terms need placeholders. Correct coefficient list is

[1, 0, 0, 5]

— omitting zeros scrambles place value exactly like writing

15

for

1005

For

P (x) \div (x + 4)

, a student uses

c = + 4

in synthetic division.

Wrong sign. We use the zero of the divisor:

x + 4 = 0

gives

x = - 4

, so

c = - 4

. The synthetic scheme adds instead of subtracts, which is why the sign flips.

Long-dividing, a student gets remainder

- 2 x + 7

while dividing by

x^{2} + 1

and keeps going.

Wrong: they should have stopped.

de g (- 2 x + 7) = 1 < 2 = de g (x^{2} + 1)

, so the referee says done. A degree-1 leftover can't contain a degree-2 divisor.

After subtracting, a student writes the new term

- x^{2}

in the

x^{3}

column.

Wrong: alignment error. Like terms must sit in the same column (

x^{2}

under

x^{2}

). Misaligning means you'd later add an

x^{2}

to an

x^{3}

, which is not allowed — they aren't like terms.

A student computes

2 x^{3} \div x = 2 x^{2}

then writes

2 x^{2}

times

D = x + 2

2 x^{3} + 2 x^{2}

Wrong multiply:

2 x^{2} \cdot (x + 2) = 2 x^{3} + 4 x^{2}

, not

2 x^{3} + 2 x^{2}

. Every term of the divisor must be multiplied, including the constant.

Using synthetic division by

x - 3

, the last number came out

0

, and the student reports quotient of the same degree as the dividend.

Wrong: the quotient is always one degree lower than the dividend, because we divided out a degree-1 factor. The remaining coefficients (all but the last) are the quotient's.

Why questions

These probe the machinery, not the arithmetic.

Why must we fill missing terms with

0

coefficients?

Because column position encodes the power of

x

. A blank would shift every following coefficient into the wrong power, just like dropping a zero in a decimal number ruins its value.

Why does dividing by

(x - c)

leave a remainder equal to

P (c)

From

P (x) = (x - c) Q (x) + R

. Substitute

x = c

: the

(x - c)

factor becomes

0

, killing the

Q

term, leaving

P (c) = R

. This is the Remainder Theorem (see 2.1.13-Factor-theorem-and-remainder-theorem).

Why does synthetic division use

+ c

even though the divisor is

x - c

Long division subtracts each product; synthetic division rewrites "subtract the product of

- c

" as "add the product of

+ c

". Folding the sign into the constant is exactly what compresses the algorithm.

Why is the quotient one degree lower than the dividend when dividing by a linear factor?

Degrees add in multiplication:

de g (P) = de g (D) + de g (Q)

. With

de g (D) = 1

, we get

de g (Q) = de g (P) - 1

Why do we stop the algorithm once

de g (R) < de g (D)

Because the leading term of

D

can no longer divide the leading term of what's left — there's no quotient term to extract. Continuing would produce fractions in

x

, leaving the world of polynomials.

Why does polynomial division underpin 3.2.4-Partial-fraction-decomposition?

Partial fractions require a proper fraction (numerator degree < denominator degree). If it's improper, you first divide to split off a polynomial part, leaving a proper remainder to decompose.

Why can factoring via division (2.1.8-Factoring-polynomials) shrink a hard polynomial?

Each factor found (remainder

0

) peels off one degree, turning a big polynomial into (linear factor) × (smaller quotient) you can attack again — a divide-and-conquer ladder.

Why is the quotient–remainder pair unique?

If two pairs both worked, subtracting the equations gives

D \cdot (Q_{1} - Q_{2}) = R_{2} - R_{1}

. The right side has degree

< de g (D)

but the left (if

Q_{1} \neq = Q_{2}

) has degree

\geq de g (D)

— a contradiction, so they must be equal.

Edge cases

The scenarios that "the formula" quietly assumes away.

Dividing

5

(a constant) by

x - 1

: what are

Q

and

R

Q = 0

R = 5

. Since

de g (5) = 0 < 1 = de g (x - 1)

, the constant already satisfies the referee; nothing fits, so the whole thing is remainder.

Dividing the zero polynomial

0

by any

D (x)

: what happens?

Q = 0

and

R = 0

. Nothing to distribute, everything trivially zero — a valid, if boring, division.

Dividing

x^{2} - 4

x - 2

gives remainder

0

. What does that tell you beyond "it divides evenly"?

That

x = 2

is a root:

P (2) = 0

. Zero remainder and "

c

is a root" are the same statement wearing different clothes.

Can the remainder ever have the same degree as the divisor?

No. If it did, the divisor's leading term would still divide it, meaning we could extract one more quotient term. The referee

de g (R) < de g (D)

forbids equality.

Divide by

x - c

where

P (c) = 0

but

P^{'} (c)

also

= 0

(a repeated root). What do you expect after one division?

Remainder

0

, and the resulting quotient still has

x - c

as a factor. You can divide by

x - c

again — the multiplicity of the root equals how many times

(x - c)

divides cleanly.

What is the quotient and remainder of

P (x) \div P (x)

(dividing by itself,

P \neq = 0

Q = 1

R = 0

. Anything nonzero fits into itself exactly once with nothing left over — same as

7 \div 7 = 1

If a divisor of degree

3

leaves a remainder, what is the largest degree that remainder can have?

Degree

2

. The referee caps it at

de g (D) - 1 = 3 - 1 = 2

; a general remainder here looks like

a x^{2} + b x + d

Recall One-line summary of every trap above

There are really only three referees running the whole show: (1) missing terms need zero placeholders, (2) synthetic division uses the divisor's zero $c$ , and (3) you stop the moment $de g (R) < de g (D)$ . Almost every mistake violates one of these three.