How is the Factor Theorem different from the Remainder Theorem?

The Remainder Theorem says the remainder when you divide P(x) by (x - c) equals P(c). Factor Theorem is a special case: if P(c) = 0, then the remainder is zero, meaning (x - c) is an exact factor. So Remainder Theorem is about remainders in general, while Factor Theorem tells you when there is no remainder.

Does Factor Theorem only work for real numbers?

No. You can use Factor Theorem with complex numbers too. For example, if P(x) = x^2 + 1 and c = i (the imaginary unit), then P(i) = -1 + 1 = 0, so (x - i) is a complex factor. This is useful for finding all roots of a polynomial, real or complex.

Can Factor Theorem help me solve polynomial equations?

Absolutely. If you find one root c where P(c) = 0, Factor Theorem tells you (x - c) is a factor. Then you can divide P(x) by (x - c) to get a quotient of lower degree, reducing a hard problem to simpler pieces. Repeat this process to solve completely.

What should I do after I confirm (x - c) is a factor?

Once Factor Theorem confirms that (x - c) is a factor, use polynomial long division or synthetic division to divide P(x) by (x - c) and find the quotient. This quotient is a polynomial of one degree lower, which you can then analyze, graph, or factor further.

Factor Theorem

The Factor Theorem tells you when a linear binomial divides a polynomial: (x - c) is a factor of P(x) exactly when P(c) = 0.

(x - c) \mid P(x) \iff P(c) = 0

Solve a problem with Factor Theorem

Type the problem. The solver will use Factor Theorem where Factor Theorem is the right tool, and tell you when it is not.

What each symbol means

What Factor Theorem takes

Symbol	Meaning
$P$	$P$ denotes a polynomial function (e.g., $P(x) = x^2 + 3x - 1$ ); treating it as a single number instead of a function prevents you from computing $P(c)$ .
$x$	$x$ is the variable in polynomial $P(x)$ , and $c$ is the number you substitute for it in the Factor Theorem—confusing them will give you the wrong test.
$c$	$c$ is the specific number whose polynomial value you check (if $P(c) = 0$ , then $(x - c)$ is a factor); confusing $c$ with the binomial $(x - c)$ causes errors.

When to use it

When you need to test whether a given binomial is a factor of a polynomial, or to find and confirm roots quickly.

Level

Usually taught in: Algebra II

Worked examples

1. Check if a binomial is a factor

Problem

Check whether

(x + 2)

is a factor of

P(x) = x^3 + 4x^2 + 5x + 2

$(x + 2) = (x - (-2))$
We rewrite the binomial in the form $(x - c)$ to identify $c = -2$ .
$P(-2) = (-2)^3 + 4(-2)^2 + 5(-2) + 2$
We substitute $c = -2$ into the polynomial $P(x) = x^3 + 4x^2 + 5x + 2$ .
$P(-2) = -8 + 16 - 10 + 2 = 0$
We evaluate by computing each term: $(-2)^3 = -8$ , $4(-2)^2 = 16$ , $5(-2) = -10$ , so the sum is zero.

Answer: $(x + 2) \text{ is a factor of } P(x)$

The Factor Theorem says that $(x - c)$ is a factor of $P(x)$ exactly when $P(c) = 0$ . Since $P(-2) = 0$ , we know $(x + 2)$ is a factor.

2. Test a rational value as a root

Problem

(x - \tfrac{1}{2})

a factor of

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

$c = \frac{1}{2}$
From the binomial $(x - \tfrac{1}{2})$ , we identify $c = \tfrac{1}{2}$ .
$P\left(\frac{1}{2}\right) = 2\left(\frac{1}{2}\right)^3 - \left(\frac{1}{2}\right)^2 - 1$
We substitute $c = \tfrac{1}{2}$ into $P(x) = 2x^3 - x^2 - 1$ .
$P\left(\frac{1}{2}\right) = 2 \cdot \frac{1}{8} - \frac{1}{4} - 1 = \frac{1}{4} - \frac{1}{4} - 1 = -1$
We evaluate: $2 \cdot (\tfrac{1}{2})^3 = \tfrac{1}{4}$ , and $\tfrac{1}{4} - \tfrac{1}{4} - 1 = -1 \neq 0$ .

Answer: $(x - \tfrac{1}{2}) \text{ is not a factor of } P(x)$

Since $P(\tfrac{1}{2}) = -1 \neq 0$ , the binomial $(x - \tfrac{1}{2})$ is not a factor. The Factor Theorem requires $P(c) = 0$ for $(x - c)$ to divide the polynomial.

3. Verify a factor in a real-world context

Problem

A school bake sale revenue model is

R(t) = t^3 - 5t^2 + 6t

dollars, where

t

is hours after opening. Show that

(t - 2)

is a factor of

R(t)

by using the Factor Theorem.

$c = 2$
From the binomial $(t - 2)$ , we identify $c = 2$ .
$R(2) = (2)^3 - 5(2)^2 + 6(2)$
We substitute $t = 2$ into the revenue function.
$R(2) = 8 - 20 + 12 = 0$
We evaluate: $2^3 = 8$ , $5(2)^2 = 20$ , and $6(2) = 12$ , giving us $8 - 20 + 12 = 0$ .

Answer: $(t - 2) \text{ is a factor of } R(t)$

By the Factor Theorem, since $R(2) = 0$ , the binomial $(t - 2)$ divides $R(t)$ evenly. In context, this means the revenue returns to zero at the 2-hour mark, which corresponds to a natural boundary in the revenue cycle.

Common mistakes

Where Factor Theorem usually goes wrong

The mistake	Why it is wrong	The fix
If $(2x - 4)$ is a factor of $P(x)$ , then $P(2) = 0$ .	Factor Theorem applies only to monic binomials (those with leading coefficient 1); the binomial $(2x - 4) = 2(x - 2)$ is not monic, so you cannot apply Factor Theorem directly.	Factor out the coefficient: $(2x - 4) = 2(x - 2)$ . For $(x - 2)$ to be a factor, check $P(2) = 0$ . Then $(2x - 4)$ is a factor if and only if $(x - 2)$ is.
Since $P(5) = 0$ , we know $(5 - x)$ is a factor of $P(x)$ .	The binomial must be in the form $(x - c)$ , not $(c - x)$ . Although $(5 - x) = -(x - 5)$ , Factor Theorem directly tells you that $(x - 5)$ is a factor, not $(5 - x)$ .	When $P(5) = 0$ , the factor is $(x - 5)$ , not $(5 - x)$ . The order in $(x - c)$ matters.
I found that $P(-3) = 7 \neq 0$ , so $(x - 3)$ is not a factor.	You tested the wrong value: to check whether $(x - 3)$ is a factor, you must evaluate $P(3)$ , not $P(-3)$ . The test value must always match the constant in the binomial.	Always match the test value to the binomial: for $(x - 3)$ check $P(3)$ ; for $(x + 4) = (x - (-4))$ check $P(-4)$ .

Tips and when to use something else

Always write the binomial in the form $(x - c)$ first. If you have $(x + 5)$ , rewrite it as $(x - (-5))$ so you correctly identify $c = -5$ .
The Remainder Theorem states that when you divide $P(x)$ by $(x - c)$ , the remainder is $P(c)$ —Factor Theorem is the special case where the remainder is zero.
Factor Theorem is a shortcut: instead of performing polynomial division to test divisibility, you just evaluate one number. This is much faster for checking individual factors.
Once you confirm $(x - c)$ is a factor using Factor Theorem, use synthetic division or polynomial long division to find the quotient polynomial and continue factoring.

Related formulas

Remainder TheoremThe Remainder Theorem lets you find the remainder when a polynomial is divided by (x - c) by simply evaluating the polynomial at c.Polynomial Long DivisionPolynomial long division breaks a fraction into quotient and remainder, letting you simplify rational expressions or test polynomial divisibility.Synthetic DivisionA fast method for dividing a polynomial by a binomial of the form (x - c), giving quotient and remainder efficiently without long division.

Frequently asked questions

How is the Factor Theorem different from the Remainder Theorem?: The Remainder Theorem says the remainder when you divide $P(x)$ by $(x - c)$ equals $P(c)$ . Factor Theorem is a special case: if $P(c) = 0$ , then the remainder is zero, meaning $(x - c)$ is an exact factor. So Remainder Theorem is about remainders in general, while Factor Theorem tells you when there is no remainder.
Does Factor Theorem only work for real numbers?: No. You can use Factor Theorem with complex numbers too. For example, if $P(x) = x^2 + 1$ and $c = i$ (the imaginary unit), then $P(i) = -1 + 1 = 0$ , so $(x - i)$ is a complex factor. This is useful for finding all roots of a polynomial, real or complex.
Can Factor Theorem help me solve polynomial equations?: Absolutely. If you find one root $c$ where $P(c) = 0$ , Factor Theorem tells you $(x - c)$ is a factor. Then you can divide $P(x)$ by $(x - c)$ to get a quotient of lower degree, reducing a hard problem to simpler pieces. Repeat this process to solve completely.
What should I do after I confirm $(x - c)$ is a factor?: Once Factor Theorem confirms that $(x - c)$ is a factor, use polynomial long division or synthetic division to divide $P(x)$ by $(x - c)$ and find the quotient. This quotient is a polynomial of one degree lower, which you can then analyze, graph, or factor further.

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Reviewed 2026-09-18