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Expanding Two Binomial Brackets
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Expanding Two Binomial Brackets

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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.

1

Which option tells us what binomial brackets are?

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2

Binomial just means two terms.

This is like x+4x+4.

3

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

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

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

1

Now we know what binomials are

we can multiply them to eliminate the brackets.

2

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

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3

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

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4

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

This gives us 4x4x.

5

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

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6

We then need to bring this all together:

(x+4)(x+7)=x2+7x+4x+28(x+4)(x+7)=x^2+7x+4x+28

7

What's the last step we need to do with our answer? x2+7x+4x+28x^2+7x+4x+28

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8

In x2+7x+4x+28x^2+7x+4x+28 we can combine the terms 7x+4x7x+4x, what does this equal?

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9

This gives us our final answer!

(x+4)(x+7)=x2+11x+28(x+4)(x+7)=x^2+11x+28

1

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

2

Let's try it on this one.

(a+5)(a3)(a+5)(a-3)

3

Using FOIL, we begin with the First term in each bracket. (a+5)(a3)({\color{#21affb}{a}}+5)({\color{#21affb}{a}}-3)

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4

We have the first part of our answer! FOIL tells us the Outside terms are next, what's this part of the answer? (a+5)(a3)=a2...({\color{#21affb}{a}}+5)(a{\color{#21affb}{-3}})=a^2...

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5

Now we have reached I in FOIL, the Inside terms. What is this part of the answer? (a+5)(a3)=a23a...(a+{\color{#21affb}{5}})({\color{#21affb}{a}}-3)=a^2-3a...

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6

We're now at the end of FOIL, the Last terms in brackets. What completes this part? (a+5)(a3)=a23a+5a...(a+{\color{#21affb}{5}})(a{\color{#21affb}{-3}})=a^2-3a+5a...

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7

This gives us (a+5)(a3)=a23a+5a15(a+5)(a-3)=a^2-3a+5a-15.

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

8

Our final answer is

(a+5)(a3)=a2+2a15(a+5)(a-3)=a^2+2a-15 😎

Have a go at this one. Expand (x+2)(x11)(x + 2)(x - 11) to its simplest form. Remember FOIL!

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Let's try another one together: (x3)(x7)(x-3)(x-7) Remember FOIL.

1

Multiply the First terms:(x3)(x7)({\color{#21affb}{x}} -3)({\color{#21affb}{x}} -7)

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2

Multiply the Outside terms: (x3)(x7)({\color{#21affb}{x}} -3)(x {\color{#21affb}{-7}})

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3

Multiply the inside terms: (x3)(x7)(x {\color{#21affb}{-3}})({\color{#21affb}{x}} - 7)

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4

Multiply the last terms: (x3)(x7)(x {\color{#21affb}{-3}})(x {\color{#21affb}{-7}})

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5

Simplify by collecting like terms: x27x3x+21x^2-7x-3x+21

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6

The final answer is x210x+21x^2-10x+21

Good work!

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

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Practise using FOIL with this one. Make sure you simplify your answer. (3n+4)(2n5)(3n+4)(2n-5)

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1

In summary! A binomial is just two terms.

x+2x+2

2

We can have two binomials in brackets and multiply them.

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

3

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

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

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

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

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

4

Multiply the First terms:

(x+2)(x+5)=x2...({\color{#21affb}{x}}+2)({\color{#21affb}{x}}+5)=x^2...

5

Multiply the Outside terms:

(x+2)(x+5)=x2+5x...({\color{#21affb}{x}}+2)(x+{\color{#21affb}{5}})=x^2+5x...

6

Multiply the Inside terms:

(x+2)(x+5)=(x+{\color{#21affb}{2}})(x+{\color{#21affb}{5}})= x2+5x+2x...x^2+5x+2x... (x+2)(x+5)=(x+{\color{#21affb}{2}})(x+{\color{#21affb}{5}})=

7

Multiply the Last terms:

(x+2)(x+5)=(x+{\color{#21affb}{2}})(x+{\color{#21affb}{5}})= x2+5x+2x+10x^2+5x+2x+10

8

Finally gather any like terms together.

(x+2)(x+5)=(x+2)(x+5)= x2+7x+10x^2+7x+10

9

If there are any negatives in the brackets,

make sure to take them into account!