Remainder and Factor Theorems: Examples

The remainder and factor theorems connect dividing a polynomial with evaluating it. The remainder theorem says that when $f(x)$ is divided by $x - c$ , the remainder is $f(c)$ . The factor theorem says that $x - c$ is a factor of $f(x)$ exactly when $f(c) = 0$ . Together they let you test factors without long division.

Before this lesson: You should know Synthetic Division, Polynomial Long Division, Function Notation and Factoring Trinomials.

Why the Remainder Equals f(c)

Any division can be written as dividend = divisor · quotient + remainder. For numbers, $17 = 5 \cdot 3 + 2$ . For a polynomial divided by $x - c$ , it looks like this:

f(x) = (x - c) \cdot q(x) + r

Now put $x = c$ into both sides. The factor $x - c$ becomes $c - c = 0$ , so the whole product becomes 0. All that is left is $f(c) = r$ . The remainder is the value of the function at $c$ .

The Two Theorems

Theorem	What it says	Use it to
Remainder theorem	When $f(x)$ is divided by $x - c$ , the remainder is $f(c)$ .	Find a remainder by plugging in, or find $f(c)$ by synthetic division
Factor theorem	$x - c$ is a factor of $f(x)$ if and only if $f(c) = 0$ .	Test whether $x - c$ is a factor, and find factors from known zeros

The factor theorem is a special case of the remainder theorem. A factor divides evenly, so its remainder is 0.

On a graph, $f(c) = 0$ means the graph crosses or touches the x-axis at $x = c$ . So every x-intercept $c$ gives a factor $x - c$ (see key features of functions).

Graph of f(x) = x^3 − 2x^2 − 5x + 6. It crosses the x-axis at x = −2, 1 and 3, marked with dots and labeled as factors x + 2, x − 1 and x − 3. A separate point at (2, −4) is labeled f(2) = −4, the remainder when f(x) is divided by x − 2. — Each zero gives a factor. At x = 2 the graph is at −4, so dividing by x − 2 leaves a remainder of −4.

Worked Examples

Example 1: Find a remainder by plugging in

Find the remainder when $x^3 - 4x^2 + 2x + 5$ is divided by $x - 3$ .

The divisor is $x - 3$ , so $c = 3$ . By the remainder theorem, the remainder is $f(3)$ .
Substitute 3 (see function notation): $f(3) = 3^3 - 4(3)^2 + 2(3) + 5$
Evaluate the powers: $27 - 4(9) + 2(3) + 5$
Multiply: $27 - 36 + 6 + 5$
Add from left to right: $2$

Answer: The remainder is 2.

Example 2: Find f(c) with synthetic division

Find $f(3)$ for $f(x) = 2x^4 - 5x^3 + x - 8$ .

Plugging in works, but big powers get messy. Synthetic division by $x - 3$ gives the same number as its remainder (see synthetic division). This is called synthetic substitution.

$c = 3$ . The coefficients are 2, −5, 0, 1, −8 (0 for the missing $x^2$ term).
Bring down 2.
Multiply: $2(3) = 6$
Add: $-5 + 6 = 1$
Multiply: $1(3) = 3$
Add: $0 + 3 = 3$
Multiply: $3(3) = 9$
Add: $1 + 9 = 10$
Multiply: $10(3) = 30$
Add: $-8 + 30 = 22$ . This is the remainder.

Answer: $f(3) = 22$ . Check: $2(81) - 5(27) + 3 - 8 = 162 - 135 + 3 - 8 = 22$ .

Example 3: Test a factor, then factor completely

Is $x + 2$ a factor of $f(x) = x^3 - 2x^2 - 5x + 6$ ? If so, factor $f(x)$ completely.

Write $x + 2$ as $x - (-2)$ , so $c = -2$ . Find $f(-2)$ : $(-2)^3 - 2(-2)^2 - 5(-2) + 6$
Evaluate the powers (see multiplying integers): $-8 - 2(4) - 5(-2) + 6$
Multiply: $-8 - 8 + 10 + 6$
Add: $0$ . By the factor theorem, $x + 2$ is a factor.
Divide by $x + 2$ with synthetic division. Coefficients 1, −2, −5, 6 and $c = -2$ . Bring down 1.
Multiply: $1(-2) = -2$
Add: $-2 + (-2) = -4$
Multiply: $-4(-2) = 8$
Add: $-5 + 8 = 3$
Multiply: $3(-2) = -6$
Add: $6 + (-6) = 0$ . The quotient is $x^2 - 4x + 3$ .
Factor the trinomial (see factoring trinomials): $x^2 - 4x + 3 = (x - 1)(x - 3)$

Answer: Yes. $f(x) = (x + 2)(x - 1)(x - 3)$ . These factors match the x-intercepts −2, 1 and 3 on the graph above.

Example 4: Find a missing coefficient

Find $k$ so that $x - 2$ is a factor of $x^3 + kx^2 - 4x + 12$ .

By the factor theorem, $f(2)$ must equal 0: $2^3 + k(2)^2 - 4(2) + 12 = 0$
Evaluate the powers: $8 + 4k - 4(2) + 12 = 0$
Multiply: $8 + 4k - 8 + 12 = 0$
Combine the numbers: $4k + 12 = 0$
Subtract 12 from both sides: $4k = -12$
Divide both sides by 4: $k = -3$

Answer: $k = -3$ . Check: $8 - 12 - 8 + 12 = 0$ .

Example 5: Word problem with break-even points

A company’s profit, in thousands of dollars, is $P(x) = x^3 - 12x^2 + 44x - 48$ , where $x$ is the number of items made, in hundreds. The company breaks even (profit 0) at $x = 2$ . Find the other break-even points.

Check $x = 2$ : $P(2) = 8 - 12(4) + 44(2) - 48$
Multiply: $8 - 48 + 88 - 48$
Add: $0$ . So $x - 2$ is a factor.
Divide by $x - 2$ . Coefficients 1, −12, 44, −48 and $c = 2$ . Bring down 1.
Multiply: $1(2) = 2$
Add: $-12 + 2 = -10$
Multiply: $-10(2) = -20$
Add: $44 + (-20) = 24$
Multiply: $24(2) = 48$
Add: $-48 + 48 = 0$ . The quotient is $x^2 - 10x + 24$ .
Factor the trinomial: $(x - 4)(x - 6)$
So $P(x) = (x - 2)(x - 4)(x - 6)$ , which is 0 at $x = 2$ , 4 and 6.

Answer: The company also breaks even at 400 items and 600 items.

Common Mistakes

Mistake	Example of the error	Fix
Using the wrong sign for $c$	Testing $x + 2$ with $f(2)$	$x + 2 = x - (-2)$ , so test $f(-2)$ .
Sign errors with negative powers	$(-2)^3 = 8$ or $-2^2 = 4$ when substituting	$(-2)^3 = -8$ and $(-2)^2 = 4$ . Keep the parentheses.
Mixing up the zero and the factor	$f(3) = 0$ , so $x + 3$ is a factor	A zero at 3 gives the factor $x - 3$ .
Calling a nonzero remainder a factor	$f(-3) = 9$ , so $x + 3$ is a factor	A factor needs remainder 0. Since 9 is not 0, $x + 3$ is not a factor.
Using it with a divisor like $2x - 1$	Remainder is $f(1)$	$2x - 1 = 0$ at $x = \frac{1}{2}$ , so the remainder is $f\left(\frac{1}{2}\right)$ .

Practice Problems

Find the remainder when $x^3 + 2x^2 - x + 4$ is divided by $x - 2$ .
Find the remainder when $2x^3 - x^2 + 3$ is divided by $x + 1$ .
Find the remainder when $x^4 - 3x^3 + 5$ is divided by $x - 3$ .
Is $x - 4$ a factor of $x^3 - 5x^2 + 2x + 8$ ?
Is $x + 3$ a factor of $x^3 + 4x^2 - 2x - 6$ ?
$x + 1$ is a factor of $x^3 - 7x - 6$ . Factor it completely.
$x = 1$ is a zero of $x^3 + 2x^2 - 13x + 10$ . Factor it completely.
Find $k$ so that $x - 1$ is a factor of $2x^3 - x^2 + kx + 4$ .
Use synthetic substitution to find $f(-2)$ for $f(x) = x^4 + 3x^3 - x + 2$ .
A box has a volume of $x^3 + 6x^2 + 11x + 6$ cubic inches. Show that $x + 1$ is a factor, then find the other two edges.

Show answers

Substitute 2: $2^3 + 2(2)^2 - 2 + 4$
Powers: $8 + 2(4) - 2 + 4$
Multiply: $8 + 8 - 2 + 4$
Add: $18$
Answer: the remainder is 18
$c = -1$ . Substitute: $2(-1)^3 - (-1)^2 + 3$
Powers: $2(-1) - 1 + 3$
Multiply: $-2 - 1 + 3$
Add: $0$
Answer: the remainder is 0 (so $x + 1$ is a factor)
Substitute 3: $3^4 - 3(3)^3 + 5$
Powers: $81 - 3(27) + 5$
Multiply: $81 - 81 + 5$
Add: $5$
Answer: the remainder is 5
Substitute 4: $4^3 - 5(4)^2 + 2(4) + 8$
Powers: $64 - 5(16) + 2(4) + 8$
Multiply: $64 - 80 + 8 + 8$
Add: $0$
Answer: yes, $x - 4$ is a factor
$c = -3$ . Substitute: $(-3)^3 + 4(-3)^2 - 2(-3) - 6$
Powers: $-27 + 4(9) - 2(-3) - 6$
Multiply: $-27 + 36 + 6 - 6$
Add: $9$
Answer: no. The remainder is 9, not 0
Synthetic division with $c = -1$ and coefficients 1, 0, −7, −6. Bring down 1
Multiply: $1(-1) = -1$
Add: $0 + (-1) = -1$
Multiply: $-1(-1) = 1$
Add: $-7 + 1 = -6$
Multiply: $-6(-1) = 6$
Add: $-6 + 6 = 0$
Quotient: $x^2 - x - 6$
Factor: $(x - 3)(x + 2)$
Answer: $(x + 1)(x - 3)(x + 2)$
Synthetic division with $c = 1$ and coefficients 1, 2, −13, 10. Bring down 1
Multiply: $1(1) = 1$
Add: $2 + 1 = 3$
Multiply: $3(1) = 3$
Add: $-13 + 3 = -10$
Multiply: $-10(1) = -10$
Add: $10 + (-10) = 0$
Quotient: $x^2 + 3x - 10$
Factor: $(x + 5)(x - 2)$
Answer: $(x - 1)(x + 5)(x - 2)$
Set $f(1) = 0$ : $2(1)^3 - (1)^2 + k(1) + 4 = 0$
Simplify: $2 - 1 + k + 4 = 0$
Combine: $k + 5 = 0$
Subtract 5: $k = -5$
$c = -2$ ; coefficients 1, 3, 0, −1, 2. Bring down 1
Multiply: $1(-2) = -2$
Add: $3 + (-2) = 1$
Multiply: $1(-2) = -2$
Add: $0 + (-2) = -2$
Multiply: $-2(-2) = 4$
Add: $-1 + 4 = 3$
Multiply: $3(-2) = -6$
Add: $2 + (-6) = -4$
Answer: $f(-2) = -4$
Test $c = -1$ : $(-1)^3 + 6(-1)^2 + 11(-1) + 6$
Powers: $-1 + 6(1) + 11(-1) + 6$
Multiply: $-1 + 6 - 11 + 6$
Add: $0$ , so $x + 1$ is a factor
Synthetic division, coefficients 1, 6, 11, 6. Bring down 1
Multiply: $1(-1) = -1$
Add: $6 + (-1) = 5$
Multiply: $5(-1) = -5$
Add: $11 + (-5) = 6$
Multiply: $6(-1) = -6$
Add: $6 + (-6) = 0$
Quotient: $x^2 + 5x + 6$
Factor: $(x + 2)(x + 3)$
Answer: the other edges are $x + 2$ and $x + 3$ inches

FAQ

What is the remainder theorem?

When a polynomial $f(x)$ is divided by $x - c$ , the remainder is $f(c)$ . For example, dividing $x^2 + 1$ by $x - 2$ leaves a remainder of $f(2) = 5$ .

What is the factor theorem?

$x - c$ is a factor of $f(x)$ if and only if $f(c) = 0$ . In other words, $c$ is a zero of the polynomial exactly when $x - c$ divides it evenly.

How do you use the factor theorem to factor a polynomial?

Find a number $c$ with $f(c) = 0$ , then divide by $x - c$ with synthetic division. Factor the quotient, often a quadratic, to finish.

What is the difference between a zero and a factor?

A zero is a number, like 3, where $f(3) = 0$ . A factor is an expression, like $x - 3$ , that divides the polynomial evenly. Each zero $c$ gives the factor $x - c$ .

Summary

Remainder theorem: dividing $f(x)$ by $x - c$ leaves remainder $f(c)$ .
Factor theorem: $x - c$ is a factor exactly when $f(c) = 0$ .
For $x + a$ , use $c = -a$ .
Synthetic substitution finds $f(c)$ quickly for big powers.
Each x-intercept $c$ of the graph gives a factor $x - c$ .

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