Apply the remainder and factor theorems

Ontario Grade 12 Mathematics

Polynomial and Rational Functions

Evaluate a polynomial to find a remainder or test a factor

Polynomial division can take several steps. When the divisor is linear, you can find the remainder by evaluating the polynomial at one value. The factor theorem connects a zero remainder to a factor. This lesson reviews substitution, states both theorems, and applies them to a polynomial. The focus is on divisors such as

x-a

and

x+a

.

What you will learn

Explain how evaluating a polynomial gives the remainder when dividing by a linear expression.
Apply the remainder theorem to find a remainder.
Apply the factor theorem to decide whether a linear expression is a factor of a polynomial.

1. Prerequisite bridge: evaluate a polynomial

A polynomial is an expression made from constants and non-negative whole-number powers of a variable, combined using addition, subtraction, and multiplication. For example,

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

is a polynomial.

To evaluate a polynomial, replace its variable with a number and calculate using the usual order of operations. For example, evaluating

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

at

x=2

gives

P(2)=2^2+3(2)-4=6

. The notation

P(2)

means the polynomial's value when its input is

2

.

Take care with negative inputs. Put parentheses around a substituted negative number. For example,

(-2)^2=4

, while

-2^2=-4

under the usual order of operations. The parentheses show that the square applies to the negative input.

P(a)=the value of P(x) when x=a

Substitution means replacing each occurrence of the variable with the chosen number.
Use parentheses when substituting a negative number.
A polynomial's value at an input is a number.

2. The remainder theorem: evaluation gives the remainder

When a polynomial is divided by a linear expression, the remainder theorem gives the remainder without requiring the full division process. If the divisor is

x-a

, evaluate the polynomial at

a

. That value is the remainder.

The sign in the divisor matters. The divisor

x-4

matches

a=4

, so evaluate at

4

. The divisor

x+4

can be written as

x-(-4)

, so evaluate at

-4

. Matching the divisor to the form

x-a

helps prevent a sign error.

For example, if

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

, division by

x-2

has remainder

P(2)=6

. This is not the quotient. It is the amount left over after dividing.

Remainder when dividing P(x) by x-a=P(a)

For divisor $x-a$ , the remainder is $P(a)$ .
For divisor $x+a$ , use the input $-a$ .
A zero remainder means the divisor divides the polynomial evenly.

3. The factor theorem: a zero remainder means a factor

A factor is an expression that divides another expression evenly. The factor theorem says that

x-a

is a factor of

P(x)

exactly when

P(a)=0

. Use the remainder theorem's evaluation: if it is zero, the divisor is a factor; if it is not zero, the divisor is not a factor.

An input that makes a polynomial's value equal to zero is called a zero of the polynomial. The factor theorem connects the input

a

to the factor

x-a

.

To check a divisor, write it in the form

x-a

, identify

a

, and evaluate the polynomial at that input. For example,

x+2=x-(-2)

, so the relevant input is

-2

.

P(a)=0\iff x-a is a factor of P(x)

If $P(a)=0$ , then $x-a$ is a factor of $P(x)$ .
If $P(a)\ne 0$ , then $x-a$ is not a factor of $P(x)$ .
The input that makes the polynomial zero matches the factor $x-a$ .

4. Apply the theorems carefully

Start by identifying the divisor's form. Rewrite it as

x-a

if needed. Next, evaluate the polynomial at

a

, showing the substitution and arithmetic. Finally, state the result in context: give the remainder, or say whether the divisor is a factor.

The theorems give a quick test, but accurate arithmetic still matters. Substitute into every term. Keep signs attached to their terms, and use parentheses for negative inputs. If the result is zero, connect that result explicitly to the factor theorem.

These theorems answer questions about dividing a polynomial by a linear divisor. Keep the remainder, quotient, and factor conclusion distinct.

Match the divisor to $x-a$ before substituting.
Evaluate every term, then report what the result means.
A zero evaluation establishes a factor; a nonzero evaluation gives a nonzero remainder.

Matching a linear divisor to the evaluation input

Divisor	Write it as $x-a$	Evaluate at
$x-5$	$x-5$	$5$
$x+2$	$x-(-2)$	$-2$
$x-1$	$x-1$	$1$

Worked example

Find a remainder and test for a factor

Let

P(x)=x^3-2x^2-5x+6

. Find the remainder when

P(x)

is divided by

x-3

. Then determine whether

x-3

is a factor.

Match the divisor
The divisor is already written as $x-a$ . Comparing $x-3$ with that form gives $a=3$ , so the remainder theorem tells us to evaluate $P(3)$ .
$a=3$
Substitute the input
Replace every $x$ in the polynomial with $3$ . Evaluate each power and multiplication before combining the terms.
$P(3)=3^3-2(3^2)-5(3)+6$
Calculate the value
The terms evaluate to $27$ , $-18$ , $-15$ , and $6$ . Adding them gives zero. By the remainder theorem, this value is the remainder.
$P(3)=27-18-15+6=0$
State the factor result
Since the remainder is zero, the factor theorem says that $x-3$ is a factor of $P(x)$ . Equivalently, $P(3)=0$ implies that $x-3$ is a factor.

Answer: The remainder is

0

, and

x-3

is a factor of

P(x)

.

Check: Substituting

3

gives

27-18-15+6=0

. The zero remainder agrees with the factor conclusion.

Common mistakes and how to avoid them

For the divisor

x+2

, evaluating at

2

.

Correction: Rewrite

x+2

as

x-(-2)

. Evaluate the polynomial at

-2

.

Calling

P(a)

the quotient.

Correction: By the remainder theorem,

P(a)

is the remainder for division by

x-a

. The quotient is a separate part of polynomial division.

Concluding that

x-a

is a factor when

P(a)

is nonzero.

Correction: The factor theorem requires the value to equal zero. A nonzero value is the remainder and means the divisor is not a factor.

Dropping parentheses when substituting a negative number.

Correction: Write the negative input in parentheses, such as

(-2)^2

, so the square applies to the negative number.

Lesson summary

For a divisor $x-a$ , evaluate the polynomial at $a$ to find the remainder.
For a divisor written with addition, such as $x+a$ , the input is $-a$ .
The factor theorem says $x-a$ is a factor exactly when $P(a)=0$ .
Show the substitution and arithmetic, then state what the result means.

Check your understanding

Question 1

What is the remainder when

Q(x)=x^2-4x+1

is divided by

x-3

?

$-2$
$2$
$-8$
$0$

Show answer and explanation

-2

The divisor is

x-3

, so evaluate at

3

. Then

Q(3)=9-12+1=-2

, which is the remainder.

Question 2

Is

x+1

a factor of

R(x)=x^2+2x+3

?

Yes, because $R(1)=6$ .
Yes, because $R(-1)=2$ .
No, because $R(-1)=2$ .
No, because $R(1)=6$ .

Show answer and explanation

No, because

R(-1)=2

.

Write

x+1

as

x-(-1)

, so test the value at

-1

. Since

R(-1)=1-2+3=2

, the remainder is nonzero and

x+1

is not a factor.

Key terms

Polynomial: An expression formed from constants and non-negative whole-number powers of a variable, combined using addition, subtraction, and multiplication.
Evaluate: Replace a variable with a chosen number and calculate the resulting value.
Remainder: The amount left over after one expression is divided by another.
Factor: An expression that divides another expression evenly.
Zero of a polynomial: An input that makes the polynomial's value equal to zero.

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About this lesson and its review

Published by DoAssignment. This reviewed lesson follows Ontario Grade 12 Mathematics (MHF4U), expectation C3.1. It is a study resource, not an official curriculum publication.

Before publication, content is checked for structure, mathematical or chemical notation, calculations, course boundaries, and readability. Errors can still occur, so corrections are welcomed.

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