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# Expanding Two Binomial Brackets

### Expanding Two Binomial Brackets

We can find the product of two binomials by multiplying each term in the bracket by each term outside of the bracket.

Which option tells us what binomial brackets are?

Binomial just means two terms.

This is like $x+4$.

We can put them in brackets and multiply by another binomial.

$(x+4)(x+7)$

Let's expand $(x+4)(x+7)$.

Now we know what binomials are

we can multiply them to eliminate the brackets.

We multiply the First term in each bracket together. $({\color{#21affb}{x}}+4)({\color{#21affb}{x}}+7)$

Now we multiply the Outside terms in each bracket. $({\color{#21affb}{x}}+4)(x+{\color{#21affb}{7}})$

Next we multiply the Inside term in each bracket. $(x+{\color{#21affb}{4}})({\color{#21affb}{x}}+7)$

This gives us $4x$.

Finally we multiply the Last term in each bracket. $(x+{\color{#21affb}{4}})(x+{\color{#21affb}{7}})$

We then need to bring this all together:

$(x+4)(x+7)=x^2+7x+4x+28$

What's the last step we need to do with our answer? $x^2+7x+4x+28$

In $x^2+7x+4x+28$ we can combine the terms $7x+4x$, what does this equal?

This gives us our final answer!

$(x+4)(x+7)=x^2+11x+28$

There's a lot to remember here, but there's a trick to help us remember.

**F** - the **F**irst term in each bracket
**O** - the **O**utside term in each bracket
**I** - the **I**nside term in each bracket
**L** - the **L**ast term in each bracket

Let's try it on this one.

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

Using FOIL, we begin with the First term in each bracket. $({\color{#21affb}{a}}+5)({\color{#21affb}{a}}-3)$

We have the first part of our answer! FOIL tells us the Outside terms are next, what's this part of the answer? $({\color{#21affb}{a}}+5)(a{\color{#21affb}{-3}})=a^2...$

Now we have reached I in FOIL, the Inside terms. What is this part of the answer? $(a+{\color{#21affb}{5}})({\color{#21affb}{a}}-3)=a^2-3a...$

We're now at the end of FOIL, the Last terms in brackets. What completes this part? $(a+{\color{#21affb}{5}})(a{\color{#21affb}{-3}})=a^2-3a+5a...$

This gives us $(a+5)(a-3)=a^2-3a+5a-15$.

The final step is to simplify the middle two terms, $-3a+5a=2a$.

Our final answer is

$(a+5)(a-3)=a^2+2a-15$ 😎

Have a go at this one. Expand $(x + 2)(x - 11)$ to its simplest form.
Remember **FOIL**!

Let's try another one together: $(x-3)(x-7)$ Remember **FOIL.**

Multiply the First terms:$({\color{#21affb}{x}} -3)({\color{#21affb}{x}} -7)$

Multiply the Outside terms: $({\color{#21affb}{x}} -3)(x {\color{#21affb}{-7}})$

Multiply the inside terms: $(x {\color{#21affb}{-3}})({\color{#21affb}{x}} - 7)$

Multiply the last terms: $(x {\color{#21affb}{-3}})(x {\color{#21affb}{-7}})$

Simplify by collecting like terms: $x^2-7x-3x+21$

The final answer is $x^2-10x+21$

Good work!

Use FOIL to multiply out $(7-x)(x+4)$.

Practise using FOIL with this one. Make sure you simplify your answer. $(3n+4)(2n-5)$

In summary! A binomial is just two terms.

$x+2$

We can have two binomials in brackets and multiply them.

$(x+2)(x+5)$

We use the acronym FOIL to remind us how to expand the brackets.

**F**irst $({\color{#21affb}{x}}+2)({\color{#21affb}{x}}+5)$

**O**utside $({\color{#21affb}{x}}+2)(x+{\color{#21affb}{5}})$

**I**nside $(x+{\color{#21affb}{2}})({\color{#21affb}{x}}+5)$

**L**ast $(x+{\color{#21affb}{2}})(x+{\color{#21affb}{5}})$

Multiply the First terms:

$({\color{#21affb}{x}}+2)({\color{#21affb}{x}}+5)=x^2...$

Multiply the Outside terms:

$({\color{#21affb}{x}}+2)(x+{\color{#21affb}{5}})=x^2+5x...$

Multiply the Inside terms:

$(x+{\color{#21affb}{2}})(x+{\color{#21affb}{5}})=$ $x^2+5x+2x...$ $(x+{\color{#21affb}{2}})(x+{\color{#21affb}{5}})=$

Multiply the Last terms:

$(x+{\color{#21affb}{2}})(x+{\color{#21affb}{5}})=$ $x^2+5x+2x+10$

Finally gather any like terms together.

$(x+2)(x+5)=$ $x^2+7x+10$

If there are any negatives in the brackets,

make sure to take them into account!