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Expanding Triple Brackets

Expanding Triple Brackets

Expanding Triple Brackets

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

1

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

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

2

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

We multiply two brackets first.

3

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

4

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

5

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

This is $6x$.

6

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

7

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

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

8

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

1

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

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

2

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

3

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

4

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

$-36\times x=-36x$

5

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

6

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 $(n+2)(n-7)(n+1)$.

1

This is slightly trickier.

We start as we did before.

2

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

3

Now we need to multiply by our third bracket.

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

4

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

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

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

1

Let's work this one through together.

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

2

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

3

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

4

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 $-5n^2$.

5

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 $-5n$.

6

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

7

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

This gives us $-14$.

8

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

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

9

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

10

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.

1

Summary! We can expand three pairs of brackets.

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

2

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

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

3

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

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

4

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

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