Algebraic division, or polynomial division, allows you to divide one polynomial expression by another. There are a number of different ways of carrying out algebraic division:

• Algebraic long division
• The grid method
• Equating coefficients

Below are some examples using all three of these methods – chose the method that makes most sense to you, and that you find easiest to remember, and you’ll be comfortable with algebraic division in no time.

To download the content of this blog in PDF or PowerPoint formats, click here. If you’d like to practice some of the prior knowledge required for algebraic division, try this multiple-choice quiz. For more opportunities to practice algebraic division, try this maze activity.

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A rational function is an algebraic fraction where both the numerator and denominator are polynomial expressions. Before carrying out algebraic division on a rational function, its worth checking if it can be simplified – this may make the division easier or, if the numerator is a factor of the denominator, completely unnecessary.

To simplify a rational function, cancel any common factors. This might involve first factorising one or more polynomial.

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Example Question 1

Simplify \(\boldsymbol{\frac{9x^3\,+\,12x^2\,-\,6x}{3x}}\).

Each term in the numerator can be divided by 3𝑥:

\( \begin{aligned}&\frac{9x^3}{3x}=3x^2\\[0.5em]&\frac{12x^2}{3x}=4x\\[0.5em]&\frac{-6x}{3x}=-2\end{aligned}\)

\( \begin{aligned}\frac{9x^2\,+\,12x^2\,-\,6x}{3x}=3x^2+4x-2\end{aligned} \)

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Example Question 2

Simplify \( \boldsymbol{\frac{2x^2\,+\,9x\,-\,18}{x^2\,+\,6x}}\).

There are no obvious common factors, so factorise both the numerator and denominator.

\(\begin{aligned}&2x^2+9x-18=(2x-3)(x+6)\\&x^2+6x=x(x+6)\end{aligned}\)

\( \require{cancel}\begin{aligned}\frac{2x^2+9x-18}{x^2+6x}&=\frac{(2x-3)\cancel{(x+6)}}{x\cancel{(x+6)}}\\&=\frac{(2x-3)}{x} \end{aligned}\)

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Once you’ve simplified your rational function, you may still need to carry out algebraic division. Below are examples of three different methods you can use. Each have their own advantages and disadvantages – use the one that works best for you.

You should be aware of two terms used in the following explanations. The dividend is the expression that will be divided and the divisor is the expression that is being divided by. At this stage, the divisor will always be linear.

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Method 1: Algebraic Long Division

Algebraic long division is similar to dividing integers using the bus stop method. The dividend goes inside the bus stop and the divisor goes on the outside, then you divide the first term in the dividend by the first term in the divisor.

Example Question 3

Find the remainder when the polynomial 2𝑥² – 5𝑥 – 5 is divided by (𝑥 – 3).

1. Divide the first term inside the bus stop by the first term of the divisor:

2𝑥² ÷ 𝑥 = 2𝑥

Put the answer on top of the bus stop.

2. Multiply your answer, 2𝑥, by the divisor. Write this below the first two terms in the dividend.

3. Subtract your multiplied answer (2𝑥² – 6𝑥) from the equivalent terms in the dividend (2𝑥² – 5𝑥):

(2𝑥² – 5𝑥) – (2𝑥² – 6𝑥) = 𝑥

Then, write your answer on the line below:

4. Bring down the next term from the dividend, -5.

5. Divide the first term in ‘𝑥 – 5’ by the first term in your divisor:

𝑥 ÷ 𝑥 = 1

Put your answer on top of your bus stop.

6. Again, multiply your answer, 1, by the divisor. Write this under the bus stop. Make sure the 𝑥 and constant terms line up.

7. Subtract 𝑥 – 3 from 𝑥 – 5:

(𝑥 – 5) – (𝑥 – 3) = -2

Write your answer on the line below.

The remainder in this case is -2. Note, if the remainder is zero then you can say that the divisor is a factor of the dividend.

You can write the solution as a rational expression by using the divisor as a denominator.

\(\begin{aligned}\frac{2x^2-5x-5}{x-3}&=2x+1-\frac{2}{x-3}\end{aligned}\)

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Sometimes, you will need to divide a polynomial which has some missing terms. In this case, write the missing terms with a coefficient of zero and then carry out the usual process.

Example Question 4

Simplify \( \boldsymbol{\frac{4x^3\,+\,6x^2\,-\,5}{2x\,+\,1}}\).

Begin by dividing the dividend by the divisor, including an 𝑥 term with a coefficient of zero.

\(\begin{aligned}\frac{4x^3-6x^2-5}{2x+1}&=2x^2+2x-1-\frac{4}{2x+1}\end{aligned}\)

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Method 2: The Grid Method

The grid method works by reversing the process for multiplying brackets using a grid. It treats algebraic division a bit like a puzzle.

Example Question 5

Work out (2𝑥³ – 3𝑥² – 29𝑥 – 30) ÷ (𝑥 – 5).

Set up the grid with the divisor on the left-hand side.

Your job is to fill in the grid so that the sum of all of the terms inside it form the dividend.

Begin by making sure the dividend is written in order of decreasing powers of 𝑥, then write the first term in the dividend (2𝑥³) in the top-left cell.

To find the term at the top of the first column, divide 2𝑥³ by 𝑥:

2𝑥³ ÷ 𝑥 = 2𝑥²

Next, fill in the gap in the first column by multiplying 2𝑥² by -5: 2𝑥² × -5 = -10𝑥²

The terms inside the grid sum to form the dividend. You know the 𝑥² term in the dividend is -3𝑥², so you can work out what is needed to achieve the next term in the dividend: 7𝑥².

Put this at the top of the next column.

Divide 7𝑥² by 𝑥 to get the term at the top of the second column. Repeat this process until the grid is complete.

Read the result from the top of the table:

(2𝑥³ – 3𝑥² – 29𝑥 – 30) ÷ (𝑥 – 5) = 2𝑥² + 7𝑥 + 6

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Method 3: Equating Coefficients

Equating coefficients uses the relationship between algebraic division and multiplication to form and solve a series of equations. Forming the equations is simpler if you know there is no remainder, for example, if you have used the factor theorem to find your divisor.

Example Question 6

Divide (𝑥³ + 12𝑥² + 47𝑥 + 60) by its factor (𝑥 + 4).

If (𝑥³ + 12𝑥² + 47𝑥 + 60), a cubic expression, is divisible by (𝑥 + 4), a linear expression, then the result must be a quadratic expression. Flipping this around, this means that an unknown quadratic expression, which we can write as (𝑎𝑥² + 𝑏𝑥 + 𝑐), multiplied by (𝑥 + 4) gives (𝑥³ + 12𝑥² + 47𝑥 + 60):

(𝑥 + 4)(𝑎𝑥² + 𝑏𝑥 + 𝑐) = 𝑥³ + 12𝑥² + 47𝑥 + 60

Expanding the brackets on the left-hand side gives:

𝑎𝑥³ + 4𝑎𝑥² + 𝑏𝑥² + 4𝑏𝑥 + 𝑐𝑥 + 4𝑐 = 𝑥³ + 12𝑥² + 47𝑥 + 60

Next, equate the coefficients of each power of 𝑥. You can do this by inspection, but it can help to partially factorise the left hand side:

(𝑎)𝑥³ + (4𝑎 + 𝑏)𝑥² + (4𝑏 + 𝑐)𝑥 + (4𝑐) = 𝑥³ + 12𝑥² + 47𝑥 + 60

Comparing both sides of the equals sign gives:

𝑎 = 1

4𝑎 + 𝑏 = 12

4𝑏 + 𝑐 = 47

4𝑐 = 60

Solve each equation to find the values. Substituting from the top gives:

4(1) + 𝑏 = 12

4 + 𝑏 = 12

𝑏 = 8

Then:

4c = 60

c = 15

You can substitute into the third equation to check your values:

4(8) + 15 = 47

32 + 15 = 47

This is correct, therefore we can say:

𝑥³ + 12𝑥² + 47𝑥 + 60 = (𝑥 + 4)(𝑥² + 8𝑥 + 15)

and so,

(𝑥³ + 12𝑥² + 47𝑥 + 60) ÷ (𝑥 + 4) = 𝑥² + 8𝑥 + 15

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You can also use this method when there may be a remainder by adding an expression to take that into account. If there’s no remainder, that expression will turn out to be 0.

The degree (the greatest power of x) of the remainder will always be less than the divisor. This means if you’re dividing by a quadratic (degree 2) then your remainder will either be linear (degree 1) or a constant (degree 0). It’s sensible to use the largest possible degree for your remainder unless you’ve been specifically told otherwise.

Example Question 7

Find the remainder when 6𝑥⁴ + 2𝑥³ + 4𝑥² – 4𝑥 + 7 is divided by 2𝑥² + 3.

In this example, you are dividing an expression with an x⁴ term (degree 4) by a quadratic (degree 2). This means your answer will also be a quadratic, so write it as:

𝑎𝑥² + 𝑏𝑥 + 𝑐

You also know you have a remainder; this could be a constant or linear expression, as it must have a lower degree than your answer. Write the remainder as:

𝑟𝑥 + 𝑠

This means you can set out your equation as:

(𝑎𝑥² + 𝑏𝑥 + 𝑐)(2𝑥² + 3) + (𝑟𝑥 + 𝑠) = 6𝑥⁴ + 2𝑥³ + 4𝑥² – 4𝑥 + 7

As before, start by expanding brackets on the left:

𝑎𝑥⁴ + 2𝑏𝑥³ + 2𝑐𝑥² + 3𝑎𝑥² + 3𝑏𝑥 + 3𝑐 + 𝑟𝑥 + 𝑠 = 6𝑥⁴ + 2𝑥³ + 4𝑥² – 4𝑥 + 7

Group by coefficients of 𝑥:

(𝑎)𝑥⁴ + (2𝑏)𝑥³ + (2𝑐 + 3𝑎)𝑥² + (3𝑏 + 𝑟)𝑥 + (3𝑐 + 𝑠) = 6𝑥⁴ + 2𝑥³ + 4𝑥² – 4𝑥 + 7

Then compare your coefficients:

𝑎 = 6

2𝑏 = 2

2𝑐 + 3𝑎 = 4

3𝑏 + 𝑟 = -4

3𝑐 + 𝑠 = 7

By inspection, you can already see that 𝑎 = 6 and 𝑏 = 1. Substitute these values into the other equations to find the remaining values:

2𝑐 + 3𝑎 = 4
2𝑐 + 18 = 4
2𝑐 = -14
𝑐 = -7

3𝑏 + 𝑟 = -4
3 + 𝑟 = -4
𝑟 = -7

3𝑐 + 𝑠 = 7
-21 + 𝑠 = 7
𝑠 = 28

This means that your remainder, 𝑟𝑥 + 𝑠, is:

28 – 7𝑥

If needed, you could write the full result as:

\(\begin{aligned}\frac{6x^4+2x^3+4x^2-4x+7}{2x^2+3}&=6x^2+x-7+\frac{28-7x}{2x^2+3}\end{aligned}\)

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Practice Questions

1. Divide (𝑥³ + 8𝑥² + 17𝑥 + 10) by (𝑥 + 5).

2. Divide (𝑥³ + 6𝑥² + 15𝑥 + 14) by (𝑥 + 2).

3. Divide (𝑥³ – 𝑥² – 3𝑥 – 36) by (𝑥 – 4).

4. Divide (3𝑥³ – 4𝑥² + 3𝑥 – 2) by (𝑥 – 1).

5. Divide (4𝑥³ + 16𝑥² + 3𝑥 – 18) by (2𝑥 + 3).

6. Divide (6𝑥⁴ – 14𝑥³ + 13𝑥² + 3𝑥 – 2) by (3𝑥 – 1).

7. Given that 2𝑥 – 1 is a factor of 2𝑥³ + 𝑥² – 25𝑥 + 12, factorise 2𝑥³ + 𝑥² – 25𝑥 + 12 completely.

8. Find the value of the constant which must be added to 𝑥³ + 5𝑥² + 7𝑥 + 10 so that (𝑥 + 4) is a factor.

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