Lesson video

Hello, Mr. Robson here.

Great choice again to join me for maths.

Our learning outcome is that we'll be able to use distributive law to multiply an expression by a constant.

Keywords that we're going to need today.

A constant is a term that does not change.

It contains no variables.

For example, in the expression x plus five, the x term is a variable, whereas the five is a constant term.

Three parts of today's lesson.

The first part is representations of the distributive law.

First, draw a representation of three multiplied by 5.

Pause this video.

I drew it like this, an array, you may have seen this called in the past, three lots of five.

You can see the 15-ness in the picture I've drawn.

I hope you can see the same in whatever picture you drew.

You might have drawn something that looked like this.

Five, five, five.

Well, that's very three lots of five, isn't it? And we can see again the 15-ness.

You may have drawn a rectangle of dimensions three by five 'cause what is three by five other than three by five? A rectangle, three by five.

In that context, it'll be the area that would be 15.

These two images are called bar models and area models.

They're gonna be really useful for us in today's lesson.

In fact, they might be useful right now because I'm gonna ask you to draw a representation of three lots of 57.

Now I'm not gonna draw an array this time.

It would take me too long, so I'm gonna draw something different and I hope you do too.

Pause this video.

Get drawing.

Did you draw something to represent three lots of 50 plus seven? A bar model perhaps.

50 plus seven, 50 plus seven, 50 plus seven.

Well, very much represents three lots of 57.

In fact, I can see three lots of 50 and three lots of seven, which immediately tells me that sum must be 171.

Well, that's easier than trying to work out my 57 times table or trying to do my three times table 57 times.

No, that visual really helped.

Could we have done it in an area? Three lots of 50 plus seven? Well, this is where distributive law's useful.

I need three lots of 50 and I need three lots of seven.

I need three lots of both things in that bracket.

We need to represent that in our area model.

Three lots of 50, three lots of seven, giving us 150 and 21.

Those area sum to 171, and I think you'd agree that area model helps us again to speed up the multiplication of three lots of 57.

I need to check that you've got that secure, bar models and area models.

So I'd like you to complete these bar and area models for four lots of 39.

Pause this video and get drawing.

Hopefully your bar model looked like 30 and nine four times.

Four lots of 30 plus nine.

That gives you four lots of 30 and it gives you four lots of nine.

That's distributive law.

We've got four lots of everything inside that bracket.

Four lots of 30, four lots of nine.

We can see them when we draw the bar model.

In total, 156, and boy, did that bar model help.

I don't know my 39 times table, if you do, you're sharper than me.

An area model.

I think the area model's even quicker, isn't it? I need four lots of 30 plus nine.

That's gonna mean I need to start with an area of four lots of 30 and add an area of four lots of nine.

Four lots of 30, four lots of nine, and add to get 156 again.

I think that area model helped me do that even quicker, and again, certainly quicker than trying to say my four times table 39 times.

These visual representations are really useful.

They're really useful in number and they're really useful in algebra.

Talking of algebra, here it is.

You noticed, didn't you? Two lots of 40 plus one, two lots of y plus one.

What did you notice? Laura noticed they've got the same form.

It's two lots of something plus one.

That is a really useful observation.

If you can do two lots of 40 plus one, you can expand two lots of y plus one.

Jacob saw it differently.

Jacob said the second shows an algebraic expression multiplied by a constant.

Jacob's right also.

In the second set of brackets or the second expression, we've got an algebraic expression.

Instead of two lots of 40 plus one, two lots of y plus one, a variable.

So what would this look like in a bar model or an area model? Two lots of y plus one in a bar model.

Well, we need y and one two times.

That's two lots of y.

That's two lots of one.

That's just distributive law, isn't it? Two lots of everything in the bracket, two lots of why, two lots of one.

Just a picture of something we already know.

How about an area model? Well, I'm gonna need two lots of y and I'm gonna need two lots of one.

Two lots of y, two lots of one.

This is just distributive law.

We're expanding the brackets two lots of brackets y plus one to give us 2y plus two.

That's what it looks like in our distributive law form.

Two lots of bracket y plus one.

I'd like two lots of everything in that bracket.

I need two lots of y, I need two lots of one.

The bar model and the area model just help us to represent that distributive law.

I'm gonna give a question a go now and then ask you to repeat that skill.

I'm gonna draw a bar model to expand the expression.

To expand the expression.

This bit of algebra you'll see called a few things.

You might be asked to multiply out the bracket, as in to take the expression three lots of bracket x plus five and express an equivalent expression but without the bracket multiply out the bracket.

You might also see it called expanding.

So let me draw the bar model.

I'd like an x and a five, one time, two times, three times, that's three lots of x plus five wouldn't you agree? You would agree.

Three lots of x, three lots of five.

I'll repeat that.

Three lots of x, three lots of five, that's just distributive law, giving us the equivalent expression 3x plus 15.

So we might say we've multiplied out that bracket.

We might say we've expanded the bracket.

We get an equivalent expression 3x plus 15.

Your turn.

Draw a bar model to expand the expression five brackets x plus two.

Pause, draw, go.

Did you draw x and two twice, thrice, four times, five times? Five lots of x plus two.

Your bar model should have looked a little like that.

I can see five lots of x, I can see five lots of two, five x plus 10.

If you've got that, well done.

You've expanded those brackets.

You've multiplied out those brackets.

Another example.

Instead of bar models, we're gonna draw an area model.

Why do you think I'm gonna draw an area model for 12 lots of bracket y plus seven? That's right, I don't want to draw 12 rows in a bar model.

It's quicker and easier to draw this as 12 lots of y, 12 lots of seven.

It says distributive law at the top of the screen.

That's distributive law.

12 lots of y, 12 lots of seven, 12 lots of everything inside that bracket gives us the expression 12y plus 84.

Your turn.

An area model to expand the brackets.

10 lots of bracket y plus six.

Pause, draw.

I hope you drew an area of 10 lots of y and an area of 10 lots of six and filled them in with 10 lots of y, 10 lots of six and gave me an expanded expression 10y plus 60.

So I might say, use distributive law to expand the brackets, and you know that 12 lots of bracket y plus seven means we need 12 lots of everything in the bracket.

You've seen this before in number.

This is the same thing but in algebra.

12 lots of y, 12 lots of seven.

We wouldn't leave it written like that.

We wanna simplify that.

12 lots of y, 12y, 12 lots of seven we can evaluate to 84.

So we've successfully expanded the brackets.

12 lots of bracket y plus seven becomes 12y plus 84.

Your turn.

Use distributive law to expand the brackets eight lots of bracket y plus three.

You needed eight lots of everything in the bracket.

You needed eight lots of y and you needed eight lots of three which you would've written as 8y plus 24.

Well done.

Practise time now.

Question one.

I'd like you to complete the bar model and area model to expand the expression three lots of bracket y plus eight.

Pause this video and get drawing.

Question two.

I'd like a bar model and an area model to represent the expansion of the expression four brackets x plus nine.

Question three.

Sam draws these models to expand the expression three lots of x plus two.

In each case, what has Sam done wrong? Write two sentences to explain Sam's errors.

Question four.

Without the visual representations, can we just write the expansion of these brackets? Pause this video, fill in the blanks.

Feedback now.

For the bar model for three lots of y plus eight, you should have got y plus eight three times, and we can see there our distributed law, three lots of y, three lots of eight.

For the area model, we have one length of three, one length of y plus eight.

That gives us two separate areas.

The first area, three lots of y, three lots of eight.

This gives us the equivalent expression 3y plus 24.

Visual representations for four lots of brackets x plus nine.

Well, we need x plus nine four times.

Four lots of x, four lots of nine, 4x and 36.

Our area model.

Well, I'd like four lots of x and I'd like four lots of nine.

That's 4x and four nines.

Thank you very much.

This means we've expanded the expression four bracket x plus nine into 4x plus 36.

Back to Sam now.

What's wrong with these two visual representations of three lots of bracket x plus two.

In the case of the bar model, I hope you spied that Sam's represented two lots of x plus three rather than three lots of x plus two.

In the case of the area model, Sam shown us three lots of 2x, not three lots of x and three lots of two.

Three lots of x and three lots of two would've correctly applied the distributive law and we would've got 3x and three twos.

We would've got the expression 3x plus six.

Question four, filling in the blanks.

Two lots of bracket x plus eight.

I need two lots of everything in that bracket.

I need two lots of x and I need two lots of eight.

That's why we get the expression 2x plus 16 once we've expanded.

The second row, two lots of bracket x plus 11.

Well, I need two lots of everything in the bracket.

Two lots of x, two lots of 11, giving me 2x plus 22.

On the third row, six lots of x, six lots of 11 gives us 6x and 66.

And the final one, well, we need six lots of everything in that bracket.

Six lots of y, six lots of 11, 6y plus 66.

Part two.

Multiplying an expression by a constant.

Let's go back to number.

I'd like you to quickly work out these two calculations.

Think of some of the efficiencies we saw earlier with our number work.

What efficiencies could you apply here? Pause, give these two questions a go.

Jacob did three lots of 102, and he considered 102 to be 100 add two.

That was smart because then we can use the distributive law.

Three lots of bracket 100 plus two means we've got three lots of 100, and we've got three lots of two.

Three lots of 100 and we add three lots of two.

300 plus six becomes 306.

That was pretty quick.

Well done, Jacob.

Laura tackled three lots of 98, and she spotted that 98 was two away from 100 but in the other direction.

So she's considered 98 to be 100 minus two.

So her bracket looks different.

She's got three lots of 100 minus two.

So when she expands that bracket using distributive law, she has to consider she's got three lots of 100, and she's taking away three lots of two.

So she ends up with 300 minus six giving her 294.

Can you see the similarity and the difference between Jacob's work and Laura's work? That 100 plus two versus the 100 minus two leading us to either 300 plus six or 300 minus six.

We're gonna rewrite these calculations using a being equal to 100.

So Jacob went down the route of 100 plus two.

So in his expression, if we're saying a is equal to 100, Jacob needs to use a plus two.

Three lots of a plus two.

Well, you saw that expansion in the first part of this lesson.

We need three lots of a, three lots of two, it becomes 3a plus six.

If you substituted back in your 100 value for a there, three lots of 100 plus 6, 306.

Same answer, just expressed differently.

Laura's slightly different.

Can you see the difference? Laura's two less than 100.

So her expression is a minus two.

When she expands that bracket, three lots of bracket a and negative two, she ends up with three lots of a, and she's taking away three lots of two.

So her expression is 3a minus six.

Can you see the difference between Jacob's expression, 3a plus six, and Laura's expression, 3a minus six, and why it happened? So let's check you've got that.

Jacob says three lots of a plus two is the same as three lots of a minus two.

What do you think? Do you agree with Jacob? Jacob's wrong because if a equals 100, those two brackets will have different values.

Three lots of a plus two would be three lots of 100 plus two, giving us three lots of 100 plus three lots of 2, 306.

But three lots of a minus two would give us three lots of 100 minus two.

So we'd need three hundreds and we'd need to take away three lots of two, which would give us 294.

They're two different answers.

So three lots of brackets a plus two is different to three lots of brackets a minus two.

Next, I'd like you to have a go at expanding these expressions and then let's see what you notice.

Pause this video.

Give that a go.

We should have this for the first bracket, three lots of a, three lots of two.

For the second one, three lots of a, but we're taking away three lots of two.

For the third, three lots of two, three lots of a.

For the fourth one, three lots of two and we're taking away three lots of a.

For the final one, the constant by which we're multiplying was at the end of the bracket, but it still means we're multiplying by it.

That is brackets a plus two, closed brackets, multiplied by three.

We need three lots of a and we need three lots of two again.

That gives us these expressions.

What have you noticed? Did you notice three of them were the same thing? 3a plus six.

They're the same because addition is commutative.

Similarly, it's important to notice something here.

Three lots of bracket a minus two did not give us the same result as three lots of bracket two minus a.

I hope you appreciate that 3a and negative six are not the same as six and negative 3a.

This is because subtraction is not commutative.

I'm gonna expand the brackets on the left, then I'm gonna give you some problems to have a go at yourself.

When I expand these brackets, six lots of bracket x plus one.

I need six lots of x and six lots of one.

For the second one, six lots of brackets x minus one, I need six lots of x, and I take away six lots of one.

For the third one, six lots of brackets x plus two, I need six lots of x and six lots of two.

The next one's very similar, isn't it? I need six lots of x and six lots of negative two or six lots of x take away two lots of six.

For the final one, six lots of everything in that bracket.

Six lots of two and six lots of negative x, giving me 12 minus 6x.

Your turn.

Pause, have a go at expanding those brackets.

For the top one, you should have got five lots of everything in the bracket, five lots of y, five lots of three, giving you 5y plus 15.

for the next one, five lots of y subtract five lots of three, 5y minus 15.

Two lots of everything in the next bracket.

Two lots of y, two lots of 15, 2y plus 30.

Two lots of everything in the next bracket.

That's two lots of y and two lots of negative 15, 2y minus 30.

For the final one, two lots of 15, two lots of negative y, we should have 30 minus 2y.

Next, three lots of 102, three lots of 98.

Haven't we seen this problem already? Yes we have.

Can we rewrite the calculations three lots of 102, three lots of 98 using algebra, if I say a is 100 and b is two? Give it a go.

Pause this video.

Jacob wrote, well, a equals 100, b equals two.

So 102 can be composed of a plus b.

So I need three lots of brackets a plus b.

I need three lots of a and three lots of b.

My expression is 3a plus 3b.

If we substituted in the value of 100 for a and the value of two for b, we'd get three lots of 100, three lots of two.

That same answer, 306.

Laura tackled three lots of 98.

So her expression needs to be 100 subtract two.

Hence, in her brackets, she's got a minus b.

She needs three lots of that 98, three lots of bracket a minus b.

She needs three lots of a and three lots of negative b or three lots of a subtract three lots of b.

Hence, her expression is 3a minus 3b.

Can you see the difference between Jacob's work and Laura's work? Let's check you've got that.

True or false.

Four lots of bracket x plus y is the same as bracket y plus x close bracket multiplied by four.

Are they the same thing, true or false? Pause this video, tell a person next to you or say it aloud.

It's true.

They're the same thing.

We can help to understand why they're the same thing by justifying that answer.

Is it true because a, order matters when you expand brackets, or b, addition and multiplication are commutative? Which one of those justifications is correct? Pause this video, tell a person next to you or say it aloud.

It was b, addition and multiplication are commutative.

Next, which is a correct expansion of five brackets 3x minus four.

Pause this video, take your pick.

It was a, 15x minus 20.

Five lots of 3x would be 15x, five lots of negative four would be negative 20.

Can we expand this bracket? What's different about the algebraic expression in this bracket? It's got three terms in it, but it's not gonna behave any differently.

We still need four lots of everything in there.

We need four lots of 2a, we need four lots of negative 3b, we need four lots of five.

We can still expand this bracket.

We'll get 8a minus 12b plus 20.

Your turn.

Expand six lots of bracket a plus 3b minus two.

You needed six lots of everything inside that bracket.

You needed six lots of a, six lots of 3b and six lots of negative two.

6a plus 18b minus 12 was your expanded expression.

Let's check you've got that.

Which of these is the correct expansion of four brackets a plus b plus c plus three.

Pause this video, take your pick.

You should have picked c, 4a plus 4b plus 4c plus 12.

It's really important we take four lots of everything inside the bracket.

Four lots of a, four lots of b, four lots of c, four lots of three.

The other three didn't do that.

Practise time now.

I'd like you to match the expressions with their expanded forms. A lot to consider here.

Did you spot this first one? The fact that two lots of y plus six is the same as two lots of six plus y.

We mentioned addition was commutative, so two of our expressions on the left became 2y plus 12.

Two lots of bracket y minus six becomes 2y minus 12, two lots of bracket 5y minus six becomes 10y minus 12, two lots of bracket x minus y, 2x minus 2y.

Eight lots of x minus y becomes eight lots of x minus eight lots of y.

Eight lots of x minus 2y becomes 8x minus 16y, and then that bottom one matches like so.

It was the only one with a z term in it.

Finally, multiplying expressions by a negative.

Spot the difference.

Two lots of x plus five looks like this.

There's x, there's five twice.

We get 2x plus 10.

Negative two lots of x plus five we could draw like this.

Two lots of negative x, negative five twice gives us the expression negative 2x and negative 10.

Notice how the negative two outside the bracket turn that positive x into negative 2x and it turned the positive five into negative 10.

So when we come to expand negative two lots of bracket x plus five, area models are useful.

We can still lay it out like this.

Negative two lots of x, negative 2x.

Negative two lots of five, negative 10.

We can simplify that and write it as negative 2x minus 10.

So I'm gonna do an example and ask you to do an example.

My example is negative five multiplied by bracket x minus seven.

My area model, I need a negative five, I need an x, I need a negative seven, negative five lots of x, that's negative 5x.

Negative five multiplied by negative seven is positive 35.

So my expression becomes negative 5x plus 35.

Yours is negative nine lots of bracket x minus three.

Pause, good luck.

You should have got an area model with negative nine lots of x giving you negative 9x and negative nine lots of negative three giving you positive 27.

Your expression was negative 9x plus 27.

How about these brackets? Well, I need negative five lots of everything in the bracket.

I need negative five lots of 2x and negative five lots of negative b.

That's negative 10x plus 5b.

Let's check you've got that.

Which of these is the correct expansion of negative 13 lots of bracket negative 3a negative 2b.

Pause and take your pick.

c was the correct answer.

You could immediately rule out a and d because negative 13 multiplied by negative 3a gives us a positive.

So it could only have been option b or option c.

It's option c because when we take negative 13 lots of negative 2b, that also becomes a positive, positive 26b.

Practise time now.

Lucas is trying to learn this same skill today.

Write a sentence or two to teach Lucas the difference between these two expansions.

Five lots of bracket x plus six and negative five lots of bracket x plus six.

And for question two, I'd like you to expand those three expressions.

Feedback time.

You could have helped Lucas out by saying the first expression has only positive terms and so expands to have only positive terms and expanded expression 5x and positive 30.

The second expression has positive and negative terms. Multiplying positive and negative terms together produces negative terms, hence, negative 5x and negative 30.

In expanding these expressions, we should have got negative 3x negative 3y negative 24, negative 6x plus 3y minus 24 and 6x plus 9y minus 24.

We're at the end now.

To summarise, we can use our command of the distributive law in number to multiply algebraic expressions by a constant.

This is known as expanding or multiplying out brackets.

Using representations like a bar model or an area model can help us to understand the underlying structure.

Hope you enjoyed learning a bit more algebra today.

I did.

I'll see you again soon.