## What you need to know

Multiplying Out Triple Brackets

Previously, we’ve seen how to expand single and double brackets – if you don’t remember, click here (https://mathsmadeeasy.co.uk/gcse-maths-revision/multiplying-single-double-brackets/) – so now we’re going to look at expanding triple brackets. Fortunately, this new kind of expansion isn’t really a new kind at all. The ideas are much the same as when we expanded double brackets, it’s just the messier algebra that makes it tricky. Let’s take a look.

Example: Expand and simplify $(x - 2)(x + 1)(x + 3)$.

To do this, we’re first going to expand the second two brackets, $(x+1)(x+3)$, as this is a familiar process. You can use whichever method you prefer to expand these brackets fully, here we’re going to follow the FOIL method of multiplying the First terms in each bracket, then the Outer terms, then the Inner terms, and finally the Last terms in each bracket. So, we get

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

Now we’ve expanded these two brackets, we can rewrite the original expression, replacing the last two brackets with their expansion. Doing so, we get

$(x-2)(x+1)(x+3) = (x-2)(x^2+4x+3)$

This is now a double bracket expansion, albeit one with a more complicated second bracket than usual. As mentioned, the principle is the same as usual – we must remember to multiply every term in the first bracket by every term in the second. Sadly, there is no handy acronym like FOIL that we can use in this case (why not try to come up with one?), so to make sure that we’ve done all requisite multiplications it helps to draw lines joining each pair of terms you’ve multiplied.

Here we will multiply the $x$ in the left-hand bracket by every term in the right-hand bracket, and then multiply the -2 in the left-hand bracket by every term in the right-hand bracket.

This looks like:

Another way to make sure you’ve done all the requirements is: when you are multiplying a bracket with 2 terms by a bracket with 3, the result should always have 6 terms. Now, we must simplify our answer, so collecting like terms, we get

$x^3+4x^2+3x-2x^2-8x-6=x^3+2x^2-5x-6$.

This is the final result of the expansion.

There are no further methods or tricks you should know for this topic, the process will always be the same as this example. The main rule is to take your time – the algebra is messy and there are plenty of places to mess up, so be careful with it.

Example: Expand and simplify $(a + 5)(2a - 1)(3 + a)$.

Firstly, we expand the second two brackets into a normal quadratic. Using FOIL, we get

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

Then, replacing the second two brackets with their expanded version, we can rewrite the original expression:

$(a+5)(2a-1)(3+a)=(a+5)(2a^2+5a-3)$

We will now expand these two brackets, first multiplying the $a$ in the left-hand bracket by everything in the right-hand bracket, and then multiplying the 5 in the left-hand bracket by everything in the right-hand bracket. Doing so, we get

$(a+5)(2a^2+5a-3)=2a^3+5a^2-3a+10a^2+25a-15$

A quick count shows that this expression has 6 terms, which means we’re probably all good so far. What remains is to collect like terms. Doing so, we get the simplified expansion to be

$2a^3+15a^2+22a-15$

This is the final result of the expansion.

## Example Questions

#### 1) Expand and simplify $(m+8)(m-9)(m+1)$.

Firstly, we expand the second two brackets into a normal quadratic. Using FOIL, we get

\begin{aligned}(m-9)(m+1)&=m^2+m-9m-9 \\ &=m^2-8m-9\end{aligned}

Then, replacing the second two brackets with their expanded version, we can rewrite the original expression:

$(m+8)(m-9)(m+1)=(m+8)(m^2-8m-9)$

We will now expand these two brackets, first multiplying the $m$ in the left-hand bracket by everything in the right-hand bracket, and then multiplying the 8 in the left-hand bracket by everything in the right-hand bracket. Doing so, we get

$(m+8)(m^2-8m-9)=m^3-8m^2-9m+8m^2-64m-72$

A quick count shows that this expression has 6 terms, which is what we want. What remains is to collect like terms. Doing so, we get the simplified expansion to be

$m^3-73m-72$

This is the final result of the expansion.

#### 2) Expand and simplify $(2k-3)(k+4)^2$.

To do this, it helps to recognise that $(k+4)^2$ is the same as $(k+4)(k+4)$. So, the original expression can also be written like

$(2k-3)(k+4)(k+4)$.

This then looks like all the other triple bracket expansions we’ve seen, and we can continue as normal. Firstly, we expand the second two brackets into a normal quadratic. Using FOIL, we get

\begin{aligned}(k+4)(k+4)&=k^2+4k+4k+16 \\ &=k^2+8k+16\end{aligned}

Then, replacing the second two brackets with their expanded version, we can rewrite the original expression:

$(2k-3)(k+4)(k+4)=(2k-3)(k^2+8k+16)$

We will now expand these two brackets, first multiplying the $2k$ in the left-hand bracket by everything in the right-hand bracket, and then multiplying the -3 in the left-hand bracket by everything in the right-hand bracket. Doing so, we get

$(2k-3)(k^2+8k+16)=2k^3+16k^2+32k-3k^2-24k-48$

A quick count shows that this expression has 6 terms, which is what we want. What remains is to collect like terms. Doing so, we get the simplified expansion to be

$2k^3+13k^2+8k-48$

This is the final result of the expansion.

## Multiplying Out Triple Brackets Revision and Worksheets

Multiplying Out Brackets
Level 4-6
Expanding Triple Brackets
Level 6-7

## Multiplying Out Triple Brackets Teaching Resources

If you are searching for multiplying out brackets resources then the worksheets and revision materials above should help! You can also use these as part of a course of tuition or as teaching materials in your Maths class. Whether you are a GCSE Maths tutor in Leeds or an A Level Maths teacher in London, you will find these resources useful.