What is the difference between the Remainder Theorem and the Factor Theorem?

The Remainder Theorem says the remainder when P(x) is divided by (x - c) equals P(c). The Factor Theorem is a special case: if P(c) = 0, then (x - c) is a factor of P(x) and divides it evenly with no remainder.

Can I use the Remainder Theorem if the divisor is not (x - c)?

No, the divisor must be a linear binomial of the form (x - c). If the divisor is (2x - 4), (x^2 + 1), or anything else, you need Polynomial Long Division or Synthetic Division instead.

Do I have to use the Remainder Theorem to find remainders?

No, you can always use Polynomial Long Division or Synthetic Division. The Remainder Theorem is just much faster when all you need is the remainder, not the quotient.

What if P(c) turns out to be a large negative number?

That is perfectly fine—remainders can be any real number, positive or negative. The remainder is simply whatever value P(c) evaluates to.

Remainder Theorem

The Remainder Theorem lets you find the remainder when a polynomial is divided by (x - c) by simply evaluating the polynomial at c.

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

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What each symbol means

What Remainder Theorem takes

Symbol	Meaning
$P$	The polynomial being divided; a function of x that you evaluate at specific values to find the remainder.
$Q$	The quotient polynomial resulting from the division; it has degree one less than P, and you do not need to find it to use the Remainder Theorem.
$x$	The variable in the polynomial; the unknown quantity you substitute numbers in for to evaluate the polynomial.
$c$	The constant in the divisor (x - c); you substitute this value for x to instantly calculate the remainder as P(c).

When to use it

Use this theorem when you need to find a remainder after dividing a polynomial by a linear factor, without doing the full division.

Level

Usually taught in: Algebra II

Worked examples

1. Find the remainder: divide by $(x - 2)$

Problem

Find the remainder when

P(x) = x^2 + 3x - 5

is divided by

(x - 2)

$c = 2$
The divisor is $(x - 2)$ , so $c = 2$ .
$P(2) = (2)^2 + 3(2) - 5$
Substitute $x = 2$ into $P(x)$ .
$P(2) = 4 + 6 - 5 = 5$
Evaluate: the remainder is $5$ .

Answer: $\text{remainder} = 5$

The Remainder Theorem says the remainder is $P(c)$ , so we just evaluate $P$ at $x = 2$ . There is no long division needed.

2. Find the remainder: negative c and cubic polynomial

Problem

Find the remainder when

P(x) = 2x^3 - 5x + 1

is divided by

(x + 3)

$x + 3 = x - (-3), \text{ so } c = -3$
Rewrite the divisor to identify $c = -3$ .
$P(-3) = 2(-3)^3 - 5(-3) + 1$
Substitute $x = -3$ into $P(x)$ .
$P(-3) = 2(-27) + 15 + 1 = -54 + 15 + 1$
Calculate each term: $(-3)^3 = -27$ , and $-5 \times (-3) = 15$ .
$P(-3) = -38$
Add: $-54 + 15 + 1 = -38$ .

Answer: $\text{remainder} = -38$

When the divisor has a plus sign, $c$ is negative. We substitute carefully and evaluate the cubic. The remainder is $-38$ , which is negative because the polynomial's cubic term dominates.

3. Word problem: concert ticket revenue

Problem

A concert venue models its revenue (in thousands of dollars) after

d

days as

R(d) = d^3 + 2d^2 - 4d + 15

. The venue sells both standard ($25) and premium ($40) tickets. To find the revenue after 4 days using the Remainder Theorem, find the remainder when

R(d)

is divided by

(d - 4)

$c = 4$
The divisor is $(d - 4)$ , so $c = 4$ .
$R(4) = (4)^3 + 2(4)^2 - 4(4) + 15$
Substitute $d = 4$ into $R(d)$ .
$R(4) = 64 + 32 - 16 + 15$
Calculate each term: $(4)^3 = 64$ , $2(4)^2 = 32$ , and $-4(4) = -16$ .
$R(4) = 95$
Add all terms: $64 + 32 - 16 + 15 = 95$ .

Answer: $\text{revenue after 4 days} = 95 \text{ thousand dollars}$

Instead of polynomial long division, the Remainder Theorem gives us the answer instantly by evaluating $R(4)$ . The revenue is $95{,}000$ dollars, combining income from both ticket tiers.

Common mistakes

Where Remainder Theorem usually goes wrong

The mistake	Why it is wrong	The fix
Writing the remainder as $(x - c)$ instead of evaluating $P(c)$ .	The remainder is a single number, namely $P(c)$ , not the divisor itself or any polynomial.	Always substitute the value $c$ into $P$ and calculate a single number: $P(c)$ is your final answer.
When dividing by $(x + 5)$ , using $c = 5$ instead of $c = -5$ .	The divisor must be in the form $(x - c)$ ; writing $(x + 5)$ as $(x - (-5))$ shows that $c = -5$ , not $+5$ .	Rewrite $(x + k)$ as $(x - (-k))$ first to see the true value of $c$ , then substitute $x = -k$ .
Using the Remainder Theorem to find the quotient $Q(x)$ .	The Remainder Theorem only gives you the remainder $P(c)$ ; it tells you nothing about what $Q(x)$ is.	Use Polynomial Long Division or Synthetic Division if you need to find both the quotient and remainder.

Tips and when to use something else

The Remainder Theorem is a shortcut for finding remainders only; if you need both the quotient and remainder, use Synthetic Division instead.
If $P(c) = 0$ (the remainder is zero), then $(x - c)$ divides $P(x)$ evenly—this connection is the Factor Theorem.
The Remainder Theorem works for any polynomial $P(x)$ and any constant $c$ , even if $c$ is negative, a fraction, or irrational.
Always rewrite the divisor as $(x - c)$ first; if you see $(x + 7)$ , recognize that it is $(x - (-7))$ , so substitute $x = -7$ .

Related formulas

Synthetic DivisionA fast method for dividing a polynomial by a binomial of the form (x - c), giving quotient and remainder efficiently without long division.Factor TheoremThe Factor Theorem tells you when a linear binomial divides a polynomial: (x - c) is a factor of P(x) exactly when P(c) = 0.Polynomial Long DivisionPolynomial long division breaks a fraction into quotient and remainder, letting you simplify rational expressions or test polynomial divisibility.

Frequently asked questions

What is the difference between the Remainder Theorem and the Factor Theorem?: The Remainder Theorem says the remainder when $P(x)$ is divided by $(x - c)$ equals $P(c)$ . The Factor Theorem is a special case: if $P(c) = 0$ , then $(x - c)$ is a factor of $P(x)$ and divides it evenly with no remainder.
Can I use the Remainder Theorem if the divisor is not $(x - c)$ ?: No, the divisor must be a linear binomial of the form $(x - c)$ . If the divisor is $(2x - 4)$ , $(x^2 + 1)$ , or anything else, you need Polynomial Long Division or Synthetic Division instead.
Do I have to use the Remainder Theorem to find remainders?: No, you can always use Polynomial Long Division or Synthetic Division. The Remainder Theorem is just much faster when all you need is the remainder, not the quotient.
What if $P(c)$ turns out to be a large negative number?: That is perfectly fine—remainders can be any real number, positive or negative. The remainder is simply whatever value $P(c)$ evaluates to.

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Reviewed 2026-09-18