The Remainder and Factor Theorems | Algebra & Trigonometry

The Remainder Theorem connects division and evaluation: the remainder when a polynomial $f (x)$ is divided by $x - b$ is exactly $f (b)$ . This means synthetic division does double duty as a fast method for evaluating polynomials. Its immediate consequence, the Factor Theorem, characterizes roots in terms of factors and is the foundation for all polynomial factoring.

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Result	Statement
Remainder Theorem	Remainder of $f (x) \div (x - b)$ equals $f (b)$
Factor Theorem	$f (b) = 0 ⟺ (x - b)$ is a factor of $f (x)$
Root test	To check if $b$ is a root of $f$ , divide synthetically and check if $R = 0$
Polynomial evaluation	$f (b)$ can be computed as the remainder of synthetic division by $x - b$
Corollary	If $f (a) = 0$ and $f (b) = 0$ , then $(x - a) (x - b)$ divides $f (x)$
Degree bound	A degree- $n$ polynomial has at most $n$ roots

The Remainder Theorem

Theorem 1

When a polynomial $f (x)$ is divided by $x - b$ , the remainder equals $f (b)$ .

Proof

Let

ϕ (x)

be the quotient and

R

the remainder when

f (x)

is divided by

x - b

. By the division algorithm:

f (x) = ϕ (x) (x - b) + R .

Since

R

has degree less than 1 (the degree of

x - b

R

is a constant. This identity holds for all values of

x

. Setting

x = b

f (b) = ϕ (b) (b - b) + R = ϕ (b) \cdot 0 + R = R .

Therefore

f (b) = R

Evaluating Polynomials via Synthetic Division

The Remainder Theorem makes synthetic division a rapid evaluation tool. To find $f (b)$ , divide $f (x)$ by $x - b$ synthetically; the last entry in the bottom row is $f (b)$ .

Example 1.

Find $f (4)$ where $f (x) = 5 x^{4} - 12 x^{3} - 20 x^{2} - 43 x + 6$ .

Solution

Direct substitution: $f(4) = 5(256) - 12(64) - 20(16) - 43(4) + 6 = 1280 - 768 - 320 - 172 + 6 = 26.$ Synthetic division by $x - 4$ :

\begin{array}{crrrrr} 4 & 5 & - 12 & - 20 & - 43 & 6 \\ + & 20 & 32 & 48 & 20 \\ 5 & 8 & 12 & 5 & 26 \end{array}

The remainder is

26 = f (4)

. Both methods agree, confirming the Remainder Theorem.

The Factor Theorem

Theorem 2

(Factor Theorem) A polynomial $f (x)$ vanishes when $x = b$ (i.e., $f (b) = 0$ ) if and only if $(x - b)$ is a factor of $f (x)$ .

Proof

By the Remainder Theorem, when

f (x)

is divided by

x - b

f (x) = ϕ (x) (x - b) + f (b) .

f (b) = 0

, then

f (x) = ϕ (x) (x - b)

, so

(x - b)

is a factor. Conversely, if

(x - b)

is a factor, the division is exact and the remainder is 0, so

f (b) = 0

Example 2.

If $x^{3} + 3 x^{2} - m$ is exactly divisible by $x - 2$ , find $m$ .

Solution

By the Factor Theorem,

x = 2

must be a root:

f (2) = 8 + 12 - m = 0 ⟹ m = 20.

Exercise 1.

Show that $x^{n} + b^{n}$ is exactly divisible by $x + b$ if $n$ is odd, but not if $n$ is even.

Solution

By the Factor Theorem,

x + b = x - (- b)

is a factor if and only if

f (- b) = 0

where

f (x) = x^{n} + b^{n}

f (- b) = (- b)^{n} + b^{n} .

n

is odd:

(- b)^{n} = - b^{n}

, so

f (- b) = - b^{n} + b^{n} = 0

. Thus

(x + b)

is a factor. If

n

is even:

(- b)^{n} = b^{n}

, so

f (- b) = b^{n} + b^{n} = 2 b^{n}

. This is zero only when

b = 0

. For

b \neq 0

(x + b)

is not a factor.

Consequences of the Factor Theorem

Theorem 3

If $f (x)$ vanishes at $x = a$ and also at $x = b$ (with $a \neq b$ ), then $(x - a) (x - b)$ divides $f (x)$ .

Proof

Since

f (a) = 0

(x - a)

is a factor:

f (x) = (x - a) ϕ_{1} (x)

. Substituting

x = b

0 = f (b) = (b - a) ϕ_{1} (b)

. Since

b \neq a

, we have

b - a \neq 0

, so

ϕ_{1} (b) = 0

. By the Factor Theorem applied to

ϕ_{1}

(x - b)

is a factor of

ϕ_{1} (x)

ϕ_{1} (x) = (x - b) ϕ_{2} (x)

. Therefore

f (x) = (x - a) (x - b) ϕ_{2} (x)

The same reasoning extends to any finite set of distinct roots:

Theorem 4

If $f (x)$ vanishes at $x = a_{1}, a_{2}, \dots, a_{k}$ (all distinct), then $(x - a_{1}) (x - a_{2}) \dots (x - a_{k})$ divides $f (x)$ .

Example 3.

Find a degree-2 polynomial $f (x)$ that equals 0 when $x = 2$ and $x = 3$ , and equals 6 when $x = 4$ .

Solution

Since

f

vanishes at

x = 2

and

x = 3

, the factor theorem gives

f (x) = a_{0} (x - 2) (x - 3)

for some constant

a_{0}

. Using

f (4) = 6

6 = a_{0} (4 - 2) (4 - 3) = 2 a_{0}

, so

a_{0} = 3

. Therefore

f (x) = 3 (x - 2) (x - 3) = 3 x^{2} - 15 x + 18

Example 4.

Find a degree-3 polynomial $f (x)$ that vanishes at $x = 2$ and $x = 3$ , equals 6 when $x = 1$ , and equals 18 when $x = 4$ .

Solution

Since

f

vanishes at

x = 2

and

x = 3

, we have

f (x) = (a_{1} x + a_{0}) (x - 2) (x - 3)

. Using the conditions

f (1) = 6

and

f (4) = 18

6 = (a_{1} + a_{0}) (1 - 2) (1 - 3) = 2 (a_{1} + a_{0}) ⟹ a_{1} + a_{0} = 3.

18 = (4 a_{1} + a_{0}) (4 - 2) (4 - 3) = 2 (4 a_{1} + a_{0}) ⟹ 4 a_{1} + a_{0} = 9.

Subtracting:

3 a_{1} = 6

, so

a_{1} = 2

and

a_{0} = 1

. Therefore

f (x) = (2 x + 1) (x - 2) (x - 3) = 2 x^{3} - 9 x^{2} + 7 x + 6

Degree Bound on the Number of Roots

Theorem 5

A polynomial $f (x)$ of degree $n$ can have at most $n$ distinct roots.

Proof

f

had more than

n

distinct roots, it would be divisible by the product of more than

n

linear factors

(x - a_{i})

. That product has degree greater than

n

, which is impossible for a degree-

n

polynomial.

An important corollary gives a very useful identity principle:

Theorem 6

If two polynomials of degree at most $n$ agree at more than $n$ values of $x$ , then they are identical: all their corresponding coefficients are equal.

This means that if you know a polynomial of degree $\leq n$ by its values at $n + 1$ or more points, you know it completely.

Frequently Asked Questions

What is the quickest way to check if a number is a root of a polynomial?

Use synthetic division. If the remainder (last entry in the bottom row) is 0, the number is a root. This is faster than substituting directly, especially for polynomials of high degree with large coefficients.

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

The Remainder Theorem says the remainder when dividing

f (x)

x - b

f (b)

, regardless of whether

f (b)

is zero or not. The Factor Theorem is the special case where the remainder is zero:

f (b) = 0

if and only if

(x - b)

is a factor of

f (x)

Can a polynomial have more roots than its degree?

No. A nonzero polynomial of degree

n

has at most

n

roots. The only exception is the zero polynomial (all coefficients zero), which satisfies

f (x) = 0

for every

x

and has no well-defined degree.

If I know all n roots of a degree-n polynomial, can I recover it?

Yes, up to a scalar multiple. If the roots are

r_{1}, r_{2}, \dots, r_{n}

and the leading coefficient is

a_{n}

, then:

f (x) = a_{n} (x - r_{1}) (x - r_{2}) \dots (x - r_{n}) .

Knowing the leading coefficient

a_{n}

in addition to all roots completely determines

f (x)

How does the Factor Theorem help with factoring polynomials?

It converts root-finding into a factoring problem. Once you find a root

b

(by testing rational candidates, graphing, or numerical methods), you know

(x - b)

is a factor. Divide

f (x)

(x - b)

to get a lower-degree polynomial, then repeat. This is the basis of the rational root test and complete factorization over the integers.

The Remainder Theorem

Quick Reference

The Remainder Theorem

Evaluating Polynomials via Synthetic Division

The Factor Theorem

Consequences of the Factor Theorem

Degree Bound on the Number of Roots

Frequently Asked Questions

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