# Remainder Theorem

To find the **remainder** of a polynomial divided by some linear factor, we usually use the method of Polynomial Long Division or Synthetic Division. However, the concept of the Remainder Theorem provides us with a straightforward way to calculate the remainder without going into the hassle. Why? Take a look at the summary below.

## Remainder Theorem in a Nutshell

When the polynomial [latex]P\left( x \right)[/latex] is divided by some linear factor in the form of [latex]x – c[/latex], then the remainder is simply the value of [latex]P\left( x \right)[/latex]

In symbol,

To show how this works, let’s go over some examples.

### Examples of Using the Remainder Theorem

**Example 1**: Use Polynomial Long Division to find the remainder of the problem below. Verify using the Remainder Theorem.

Divide the top expression by the bottom expression. If you need a refresher on how to divide polynomials using the Long Method, check out my separate tutorial: Polynomial Long Division. The pattern is rather simple.

**These are the key steps**:

- Divide the first term of the dividend by the first term of the divisor. Place the partial quotient on top.

- Then, go down by multiplying that partial quotient you put up with the terms of the divisor. Place the product below.

- Perform subtraction. Make sure to switch (alternate) the signs of the bottom row. See the change of signs in red.

- Carry down the next “unused” term of the dividend.

- Repeat the process by performing division when you go up, multiplication when you go down, subtraction and carrying down of unused terms until you arrive at the remainder.

**Finding the Remainder using Long Division Method**

The long division of this problem is shown below. The remainder equals [latex]3[/latex].

**Finding the Remainder using the Remainder Theorem**

Now, let’s check if the remainder we found using long division method is indeed correct. To do that, we will need to substitute the value of “[latex]c[/latex]” into the polynomial [latex]P\left( x \right)[/latex], and simplify. The value of “[latex]c[/latex]” is obtained when the linear factor is expressed in the form [latex]x – c[/latex]. Since the divisor is [latex]x + 2[/latex], we have [latex]x – \left( { – 2} \right)[/latex] therefore [latex]c = – \,2[/latex].

Evaluating [latex]c = – \,2[/latex] into the polynomial [latex]P\left( x \right)[/latex]…

It is great to see that the remainder obtained using the long method division, came out to be equal with the remainder found using the Remainder Theorem Method.

**Example 2**: Use Synthetic Division to find the remainder of the problem below. Verify using the Remainder Theorem.

What we want this time is to show that the remainder found using synthetic division matches with the remainder obtained using the theorem.

**These are the key steps involved in synthetic division:**

- Take a look at the polynomial being divided (dividend). Make sure that it is written in standard form which means that the exponents are in decreasing order. Write down only the coefficients and constant of the polynomial in the top row. Don’t forget to include zeroes for the missing terms of the polynomial.

- Next, take a look at the divisor. Switch the sign of the constant and place it in the “box” directly to the left of the list of coefficients found in the previous step. For example, if the divisor is [latex]\left( {x + 2} \right)[/latex] then you write [latex] – \,2[/latex]. If the divisor is [latex]\left( {x – 5} \right)[/latex], then you have [latex]+ 5[/latex].

- Now, we can start with the actual synthetic division process. Bring down the first coefficient below the horizontal line. Multiply that by the number in the “box”. Just above the horizontal line, place the product under the next coefficient in the list.

- Add the column of constants and place the sum below the horizontal line.

- Repeat the process of multiplying the number below the horizontal with the number in the “box” and adding the columns of constants until such time you reach the last column.

- The last number in the bottom row (below the horizontal line) is the remainder!

The steps may sound “confusing” but wait until you see an example. You should agree that it is very simple! It’s time to perform synthetic division to the example above.

The synthetic division for this problem gives us a remainder of [latex] – \,2[/latex].

**Checking our answer using the Remainder Theorem:**

Since our divisor is [latex]\left( {x – 3} \right)[/latex], we have [latex]x – \left( { + \,3} \right)[/latex] and therefore [latex]c = + \,3[/latex].

Now we evaluate [latex]c = + \,3[/latex] into the given polynomial to get the remainder using the theorem.

Again, the remainder values from two different methods are equal!

**Example 3**: Find the remainder of the problem below. Choose the most convenient method.

There are problems where you will be asked to find the remainder without being specifically told what method to use. This is a great opportunity for you to use your previous knowledge and apply them appropriately.

**Option 1: Use Polynomial Long Division**

I hope that you quickly realize how tedious this problem can be using the long division method because the exponent of the leading term is relatively “large”. You may work this out using this method for more practice.

**Option 2: Use Synthetic Division**

I find

In fact, let me show you what I mean.

So the remainder using synthetic division is found to be equal to [latex]10[/latex]. Not bad.

**Option 3: Use Remainder Theorem**

The best method to find the remainder of this problem is the remainder theorem. The number that will be substituted in the polynomial is [latex]{ – 1}[/latex]. The value of [latex]{ – 1}[/latex], when raised to some power, will simply alternate either to positive [latex]1[/latex] or negative [latex]1[/latex].

Notice, that

and

This greatly simplifies the calculation!

Since the divisor is [latex]x + 1[/latex], then we have [latex]x – \left( { – 1} \right)[/latex] which gives us [latex]c = – 1[/latex] to be substituted in the given polynomial.

As you can see, the remainder that came out using remainder theorem is equal to the remainder found by synthetic division. It is nice to see the connection!

**Example 4**: Determine if [latex]x – 1[/latex] is a factor of

To answer this question, remember that when a number is divided by another and the remainder is zero, that means the number dividing the other is a factor of that number. This concept should apply here as well.

In order for [latex]x – 1[/latex] to be a factor implies that the remainder of

**equals zero**.

In other words, applying the remainder theorem we must get [latex]P\left( c \right) = 0[/latex].

Because the divisor is [latex]x – 1[/latex], we have [latex]x – \left( { + 1} \right)[/latex] which gives us the value of “[latex]c[/latex]” to be [latex]c = + 1[/latex].

Evaluating [latex]c = + 1[/latex] into the given polynomial…

By getting an answer of zero, this shows that [latex]x – 1[/latex] is indeed a factor of

.