Albert Teen YOU ARE LEARNING:  Expanding Two Binomial Brackets  # 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.

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Which option tells us what binomial brackets are? 2

Binomial just means two terms.

This is like $x+4$.

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

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

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

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Now we know what binomials are

we can multiply them to eliminate the brackets.

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We multiply the First term in each bracket together. $({\color{#21affb}{x}}+4)({\color{#21affb}{x}}+7)$ 3

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

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

This gives us $4x$.

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Finally we multiply the Last term in each bracket. $(x+{\color{#21affb}{4}})(x+{\color{#21affb}{7}})$ 6

We then need to bring this all together:

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

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What's the last step we need to do with our answer? $x^2+7x+4x+28$ 8

In $x^2+7x+4x+28$ we can combine the terms $7x+4x$, what does this equal? 9

This gives us our final answer!

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

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

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

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Let's try it on this one.

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

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Using FOIL, we begin with the First term in each bracket. $({\color{#21affb}{a}}+5)({\color{#21affb}{a}}-3)$ 4

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

Now we have reached I in FOIL, the Inside terms. What is this part of the answer? $(a+{\color{#21affb}{5}})({\color{#21affb}{a}}-3)=a^2-3a...$ 6

We're now at the end of FOIL, the Last terms in brackets. What completes this part? $(a+{\color{#21affb}{5}})(a{\color{#21affb}{-3}})=a^2-3a+5a...$ 7

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

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

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$(a+5)(a-3)=a^2+2a-15$ 😎

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.

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Multiply the First terms:$({\color{#21affb}{x}} -3)({\color{#21affb}{x}} -7)$ 2

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

Multiply the inside terms: $(x {\color{#21affb}{-3}})({\color{#21affb}{x}} - 7)$ 4

Multiply the last terms: $(x {\color{#21affb}{-3}})(x {\color{#21affb}{-7}})$ 5

Simplify by collecting like terms: $x^2-7x-3x+21$ 6

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. $(3n+4)(2n-5)$ 1

In summary! A binomial is just two terms.

$x+2$

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We can have two binomials in brackets and multiply them.

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

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

First $({\color{#21affb}{x}}+2)({\color{#21affb}{x}}+5)$

Outside $({\color{#21affb}{x}}+2)(x+{\color{#21affb}{5}})$

Inside $(x+{\color{#21affb}{2}})({\color{#21affb}{x}}+5)$

Last $(x+{\color{#21affb}{2}})(x+{\color{#21affb}{5}})$

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Multiply the First terms:

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

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Multiply the Outside terms:

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

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Multiply the Inside terms:

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

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Multiply the Last terms:

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

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Finally gather any like terms together.

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

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If there are any negatives in the brackets,

make sure to take them into account!