Remainder theorem and factor theorem — proof and applications: Question bank

This is a rapid-fire trap page for the parent topic. Each line is a claim, a question, or a broken bit of reasoning — read the left side, decide your answer, then reveal. The reasoning on the right is what actually matters.

True or false — justify

Every claim below is either fully true or has a hidden flaw. The reveal explains the reasoning, not just the verdict.

The remainder when

p (x)

is divided by

(x - a)

is always a single constant, never a polynomial in

x

True — the remainder's degree must be less than the divisor's degree; since

(x - a)

has degree 1, the only allowed remainder degree is 0, i.e. a constant (or zero).

p (a) = 0

then

(x - a)

is guaranteed to be a factor of

p (x)

True — this is the reverse direction of the Factor Theorem:

p (a) = 0

forces the remainder to be zero, so

p (x) = (x - a) q (x)

exactly.

The Remainder Theorem only works if

p (x)

has integer coefficients.

False — the proof uses only "substitute

x = a

to kill the

(x - a)

factor", which is valid for any real (or complex) coefficients.

Dividing by

(x - a)

and evaluating

p (a)

can give different remainders.

False — the Remainder Theorem proves they are the same value; if you got different answers, one of the two computations has an arithmetic error.

(x - a)

is a factor of

p (x)

, then

a

is a root of the equation

p (x) = 0

True — "factor" means remainder zero, which by the Remainder Theorem means

p (a) = 0

, and

p (a) = 0

is exactly the statement "

a

is a root".

The remainder from dividing by

(x - a)

tells you the quotient too.

False — the theorem gives you only the remainder (

p (a)

); to get the quotient

q (x)

you still need long or synthetic division.

p (x)

has degree 3, then

(x - a) q (x)

has

q (x)

of degree 2.

True — dividing a degree-3 polynomial by a degree-1 factor drops the degree by exactly one, so

q (x)

has degree 2.

(x - a)

being a factor of

p (x)

means every value of

p

near

x = a

is zero.

False — it only means

p (a) = 0

at the single point

x = a

; the graph crosses (or touches) the axis there, but is generally nonzero elsewhere.

Spot the error

Each line contains a piece of plausible-sounding reasoning. Find where it breaks.

"To find the remainder of

p (x) \div (x + 3)

, I evaluate

p (3)

Error — the value you plug in is the root of the divisor.

x + 3 = 0

gives

x = - 3

, so you must evaluate

p (- 3)

, not

p (3)

(2 x - 1)

has the number

1

in it, so I evaluate

p (1)

to get the remainder."

Error — solve

2 x - 1 = 0

first, giving

x = \frac{1}{2}

; the remainder is

p (\frac{1}{2})

. The visible constant is not the root.

p (2) = 5 \neq = 0

, but I can still divide

p (x)

(x - 2)

, so

(x - 2)

is a factor."

Error — you can divide anything, but "factor" demands remainder

0

. Since

p (2) = 5 \neq = 0

(x - 2)

is not a factor.

(x - a)

is a factor because the long division 'came out nicely'."

Error — "nice-looking" is not a criterion; the only criterion is a remainder of exactly

0

(equivalently

p (a) = 0

). Verify the last line, not the vibe.

"If the remainder is a fraction like

- \frac{1}{2}

, I made a mistake — remainders should be whole numbers."

Error — polynomial remainders are just constants and can be any real number, including fractions; only integer division forces whole-number remainders.

p (x) = x^{2} + 1

divided by

(x - i)

makes no sense because we only allow real

a

Error — the theorem holds for complex

a

too; here

p (i) = i^{2} + 1 = 0

, so

(x - i)

genuinely is a factor over the complex numbers.

"Since

(x - 1)

and

(x - 2)

are both factors, the polynomial must have degree exactly 2."

Error — it must have degree at least 2; a degree-3 or higher polynomial can carry both these factors plus others.

Why questions

Each answer is the reason, not a restatement of the fact.

Why do we substitute exactly

x = a

in the proof, not some other value?

Because

x = a

makes

(x - a) = 0

, which annihilates the entire

(x - a) q (x)

term and leaves the equation reading

p (a) = r

— isolating the remainder cleanly.

Why must the remainder have degree less than the divisor?

If the remainder's degree equalled or exceeded the divisor's, you could divide once more and pull another term into the quotient — so a "true" remainder is what's left when no further division is possible.

Why is the Factor Theorem just a special case of the Remainder Theorem?

The Factor Theorem is the Remainder Theorem with the extra condition "remainder

= 0

"; setting

r = p (a) = 0

turns "remainder equals

p (a)

" into "

(x - a)

divides evenly".

Why does testing small integer roots (like

x = 1, 2, 3

) work so often when factoring?

The Rational Root Theorem says any rational root must divide the constant term, so small integer divisors of the constant are the natural first candidates to test with the Factor Theorem.

Why does the Remainder Theorem give a shortcut over Polynomial Long Division?

Long division computes both quotient and remainder step by step; if you only want the remainder, evaluating

p (a)

replaces the whole process with one substitution.

Why does the generalisation use

p (\frac{b}{a})

for a divisor

(a x - b)

Because

a x - b = 0

x = \frac{b}{a}

; the remainder is always

p

evaluated at the value that zeroes the divisor, whatever its coefficient.

Why can't

(x - a)

appear as a factor of a nonzero constant polynomial?

A nonzero constant has degree

0

; a factor

(x - a)

has degree

1

, and you can't fit a degree-1 factor inside a degree-0 object — no root exists.

Edge cases

Boundary and degenerate inputs the theorem quietly still covers.

What is the remainder when

p (x)

is divided by

(x - 0) = x

It is

p (0)

, which is simply the constant term of

p (x)

— dividing by

x

reads off the value at zero.

p (a) = 0

and

p^{'}

ideas aside, can

(x - a)

be a factor twice (a repeated root)?

Yes — if

(x - a)^{2}

divides

p (x)

, then

a

is a repeated root;

p (a) = 0

alone confirms

(x - a)

is a factor at least once, not how many times. ::: See Solving Polynomial Equations for multiplicity.

(x - a)

a factor of the zero polynomial

p (x) = 0

Yes, vacuously —

0 = (x - a) \cdot 0

for every

a

, so every linear term "divides" the zero polynomial; the zero polynomial is a degenerate case.

p (x)

is a constant

c \neq = 0

, what is the remainder on division by

(x - a)

The remainder is

c

itself, since

p (a) = c

for all

a

; the quotient is

0

and nothing gets divided out.

For a degree-1 polynomial like

p (x) = x - a

, what does the Factor Theorem give?

p (a) = 0

, so

(x - a)

is a factor of itself with quotient

1

— the smallest nontrivial case, confirming consistency.

By the Fundamental Theorem of Algebra, how many linear factors of the form

(x - a)

can a degree-

n

polynomial have (counting multiplicity)?

Exactly

n

over the complex numbers — so the Factor Theorem, applied repeatedly, fully splits any polynomial into

n

linear pieces. ::: Tools: Synthetic Division, Factoring Polynomials.

Can two different values

a \neq = b

both give the same remainder

p (a) = p (b)

Yes — a polynomial can take the same output at different inputs (e.g. a parabola), so equal remainders do not mean equal divisors.

Recall One-line summary of every trap here

Always plug in the root of the divisor (not the visible number), a factor requires remainder zero (not just "you can divide"), and the theorems hold even at the weird inputs — zero, constants, complex roots, and repeated factors.