Algebra

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Expanding Triple Brackets

Expanding triple brackets involves the same process of multiplying each term in each bracket together.

We can multiply two pairs of brackets, but what happens when we have three pairs?

$(x-6)(x+6)(x-5)$

We need to break $(x-6)(x+6)(x-5)$ down.

We multiply two brackets first.

A quick recap! We can use FOIL to multiply two brackets each with two terms.

F - the First term in each bracket O - the Outside term in each bracket I - the Inside term in each bracket L - the Last term in each bracket

Let's take the first two brackets and expand them. Using FOIL, what's the result of multiplying the First term in each bracket? $({\color{orange}{x}}-6)({\color{orange}{x}}+6)$

FOIL tells us the Outside terms are next. $({\color{orange}{x}}-6)(x+{\color{orange}{6}})$

This is $6x$ .

With FOIL we now multiply the Inside terms $(x{\color{orange}{-6}})({\color{orange}{x}}+6)$

The Last terms multiplied together are $(x{\color{orange}{-6}})(x+{\color{orange}{6}})$

We have another negative number so the result is $-36$ .

Bringing this together gives us $x^2+6x-6x-36$ . How can we simplify this?

We now know that $(x-6)(x+6)=x^2-36$

We take this result and multiply by our third bracket, $(x-5)$

We can use FOIL again, what is the first part of our answer? $(x^2-36)(x-5)$

What is the answer to multiplying the Outside terms? $(x^2-36)(x-5)$

Multiplying the Inside terms gives $(x^2-36)(x-5)$

$-36\times x=-36x$

What is the result of multiplying the Last terms? $(x^2-36)(x-5)$

Bringing this all together we have $x^3-5x^2-36x+180$ .

This cannot be simplified further so we are done! 👍

What happens when we expand the brackets here $(a+4)(a-4)(a+12)$

Let's try $(n+2)(n-7)(n+1)$ .

This is slightly trickier.

We start as we did before.

Expand the first two brackets. $(n+2)(n-7)$

Now we need to multiply by our third bracket.

$(n^2-5n-14)(n+1)$

But this is where it gets tricky - our first bracket now has three terms!

$(n^2-5n-14)(n+1)$

Let's work this one through together.

We can use the same principle as FOIL, but we have an extra term.

Let's start with the first term in each bracket. $({\color{orange}{n^2}}-5n-14)({\color{orange}{n}}+1)$

Staying with the first term of the first bracket, now we multiply it by the second term of the second bracket. $({\color{orange}{n^2}}-5n-14)(n+{\color{orange}{1}})$

We now do the same with the second term of the first bracket, multiply by the first term of the second bracket $(n^2{\color{orange}{-5n}}-14)( {\color{orange}{n}}+1)$

This gives us $-5n^2$ .

Staying with the second term in the first bracket, we multiply by the second term in the second bracket. $(n^2{\color{orange}{-5n}}-14)(n+{\color{orange}{1}})$

This is $-5n$ .

We then move onto the third term in the first bracket and multiply with the first term in the second bracket. $(n^2-5n{\color{orange}{-14}})({\color{orange}{n}}+1)$

And finally we multiply the last term in each bracket $(n^2-5n{\color{orange}{-14}})(n+{\color{orange}{1}})$

This gives us $-14$ .

Now we gather all these terms together - there are a lot of them!

$n^3+n^2-5n^2-5n-14n-14$

There is one final step and that is to simplify. $n^3+n^2-5n^2-5n-14n-14$

Great work - that's a lot of steps!🙋🏾‍♂️

$(n+2)(n-7)(n+1)=n^3-4n^2-19n-14$

Have a go at expanding this one $(x+2)(x+3)(x+4)$ , giving your answer in its simplest form.

Summary! We can expand three pairs of brackets.

$(a+1)(a-3)(a+5)$

Break $(a+1)(a-3)(a+5)$ down into steps

First expand two pairs of brackets $(a+1)(a-3)$

We expand $(a+1)(a-3)$ first

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

Take our result $a^2-2a-3$ and multiply by the third bracket $(a+5)$ .

$(a^2-2a-3)(a+5)=a^3+3a^2-13a-15$