How to Expand a Pair of Brackets - An Algebra Walkthrough

Expanding a pair of algebraic brackets

Expanding pairs of brackets​

In this article, we are going to have a quick look at how to expand pairs of brackets such as (x + 1)(x + 3) or (x − 5)(3x + 2). This is an important skill in high school/GCSE maths and one which often appears in algebraic problem solving, so it's important to have a good understanding of the basics.

What is Pythagoras' theorem?​

One good way to gain an understanding of this topic is to think about what it means when we multiply two brackets. If we look at (x + 1)(x + 3), we have an unknown with one added to it and the same unknown with 3 added to it. These two totals are then being multiplied together. If we compare this to a rectangle with sides of x + 1 and x + 3, the end result would be the area of the rectangle. The picture below shows what this looks like.

(x + 3)(x + 1) as rectangle dimensions

Calculating the area

To find the area, we have split each side of the rectangle into its x component and the integer component. This has created 4 smaller rectangles inside of the original one. The top left rectangle has sides of length x and so an area of x × x = x^2. The top right rectangle has dimensions of 3 and x and so an area of 3x, and so on.
By adding the four separate areas together, we get the area of the whole rectangle = x^2 + 4x + 3.
The area of the rectangle can also be expressed as the product of the two original sides (x + 3)(x + 1). As these expressions both represent the area of the same rectangle, it follows that (x + 3)(x + 1) = x^2 + 4x + 3.

Expanding brackets: The algebraic method

The pictorial method shown above is a great way to picture the problem and gain some understanding of what is happening, but we can be much quicker with an effective algebraic method.
To expand the brackets algebraically, we simply need to make sure that each value in the first bracket is multiplied by each value in the second bracket in the same way as just happened in our pictorial example.
In our example of (x + 3)(x + 1), this means that the x in the first bracket needs to multiply both the x and the 1 in the second bracket, followed by the 3 in the first bracket also multiplying both values in the second bracket. We can add arrows to help us make sure every pair is used.

Multiplying and collecting like terms

We now multiply out each of the pairings as so:
x × x = x^2
x × 1 = x
3 × x = 3x
3 × 1 = 3
Collecting these together gives x^2 + x + 3x + 3.
We now simplify by collecting like terms (any terms with the same power of x) and so add together the x and 3x to give 4x.
We now have a final answer of (x + 3)(x + 1) = x^2 + 4x + 3, just like in our pictorial example.

Using Pythagoras' theorem to find shorter sides

In the example above, we found the length of one of the shorter sides by substituting into the usual equation a^2 + b^2 = c^2 and then rearranging.
It is sometimes preferable to have rearranged the equation before substituting, in which case we get:
a^2 = c^2 - b^2
In short, when finding the hypotenuse we are adding the squares together and when finding a shorter side, we are taking the square of one shorter side away from the square of the hypotenuse.

Using our method with more complicated pairs

The same method works regardless of what terms are in the brackets.
Let's try expanding (2x + 3)(6x − 1). Again, make sure that each term in the first bracket multiplies each term in the second bracket being especially careful with the negative 1.
2x × 6x = 12x^2
2x × –1 = –2x
3 × 6x = 18x
3 × –1 = –3
So (2x + 3)(6x − 1) = 12x^2 + –2x + 18x + –3 = 12x^2 + 16x − 3

What if there are more than two terms in a bracket?

If there are more than two terms in either/both brackets, the same method still applies. Simply make sure that you multiply each item in the first bracket by each item in the second bracket. A good rule of thumb is that the number of separate terms you get after expanding but before collecting like terms, should be equal to the the number of terms in each bracket multiplied together.
For example, if you one bracket had two terms and the other has three terms, you should get 2 × 3 = 6 terms.
Trying this with (3x + 4)(2x + y − 5) we should expect to get six terms.
3x × 2x = 6x^2
3x × y = 3xy
3x × –5 = –15x
4 × 2x = 8x
4 × y = 4y
4 × –5 = –20
Therefore (3x + 4)(2x + y − 5) = 6x^2 + 3xy + –15x + 8x + 4y + –20 = 6x2 + 3xy − 7x + 4y − 20

Expanding more than two brackets

If there are more than two brackets multiplying together, the best way to expand them is to first expand one pair. Once this pair have been expanded, multiply your new answer by the next bracket and so on.
For example, let's expand (x − 3)(2x + 1)(5x + 2)
It doesn't matter which pair we start with; I will start with the first pair.
(x − 3)(2x + 1) = 2x^2 + x − 6x − 3 = 2x^2 − 5x − 3
Now multiply this by the remaining bracket.
(2x2 − 5x − 3)(5x + 2) = 10x^3 + 4x^2 − 25x^2 − 10x − 15x − 6 = 10x^3 − 21x^2 − 25x − 6

Recap

Remember, the golden rule when expanding brackets is make sure that each term in the first bracket is multiplied by each term in the second bracket. Once this is done then collect like terms together and you have your answer.

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