Lesson video

Hello, my name's Mrs. Hopper and I'm really looking forward to working with you in this lesson on how to apply the distributive law to multiplication.

I wonder if you've come across the distributive law before.

Well, if not, we're gonna learn all about it and how it can really help us to be efficient when we're thinking about multiplications.

So if you're ready, let's make a start.

So in this lesson, we're going to be using knowledge of the distributive law to calculate products beyond our known times tables.

You might have been exploring bits of the distributive law using times tables recently.

We're going to have a look and see how it works when it's beyond our times table knowledge.

We've got three key words or key phrases in our lesson today.

We've got partition, distributive law and partial product.

So I'll take my turn to say them and then it'll be your turn.

Are you ready? My turn.

Partition.

Your turn.

My turn.

Distributive law.

Your turn.

My turn.

Partial product.

Your turn.

Excellent! They may be phrases that you're becoming more familiar with, but let's just double check what they mean 'cause they're gonna be really useful to us in our lesson today.

Partitioning is something I'm sure you've been doing for a long time.

It's the act of splitting an object or value into smaller parts.

The distributive law says that multiplying a number by a group of numbers added together is the same as doing each multiplication separately.

And we've got an example here, four times three.

We can partition our four into two plus two and we can rewrite that multiplication as two lots of three plus two lots of three, and that's the same as four lots of three.

Two times three is six, two times three is six.

Six plus six is equal to 12.

And we know that four times three is also equal to 12 and a partial product, well it's either of those sixes, it's any of the multiplication results we get leading up to an overall result.

So each time we did the two times three in that example above, we got a partial product of six and six plus six is equal to 12.

There are two parts to our lesson today.

In the first part, we're going to be calculating products beyond known times tables, and in part two, we're going to be thinking about efficient methods.

So let's make a start on part one.

And we've got Sam and Jun in our lesson with us today.

They're looking at this multiplication.

What do you think? Sam says, 'Are we supposed to know our 16 times table too? Tricky question I think!' Jun says, 'I think we can use the distributive law to break the question down.

' Good thinking, Jun! 'We could partition 16 and use times tables that we know.

' 'Okay', says Sam, 'Let's start by analysing 16.

' She's gonna look closely at the number 16.

Ah, what she spotted? This shows 16 as 10 and six, so it's been partitioned.

June says we can partition 16 in other ways, but I think this way will be useful.

We know that we can use the distributive law to break down one of our factors, don't we? And breaking it down into 10 and six into two times tables that we do know, it seems to be a good idea.

So I think I agree with Jun.

Let's just check though.

Can you tick all the representations below that show 16.

Pause the video, have a think and when you're ready for some feedback, press play.

Which shows 16 then? Ah, that's right, the 10 frame shows 16 just like we had before and the rekenrek bead bar does.

If we look at all the beads on the left hand side, we've got 10 on the top, six on the bottom.

10 and six is 16.

But our counters, they don't, do they? We've only got a 10 and five ones, so that shows 15 and not 16.

So two of them do and one of them doesn't.

Let's go back to where we were at with Sam and Jun.

So they've partitioned 16 into 10 and six.

Sam says 'Let's just use the counters now.

' So she's got rid of the 10 frames, but we've still got 10 red counters and six blue counters.

And June says, 'We can use this representation to make an array.

' We've broken down our 16, but we still know we've got to multiply all of that by nine.

The expression is nine times 16, so we need nine rows of counters.

Wow, look at all those counters.

So we've got our nine times 10 and our nine times six.

And that's what Sam says, 'This shows nine times 10 and nine times six.

' June says, 'This is an array showing nine times 16.

' What do you think? Who's correct? Sam says, 'I think we're both correct.

' 'Yes', says June, 'but you've used the distributive law.

' It does show nine times 16, but it also does show nine times 10 and nine times six, which is the same overall product but broken down using the distributive law.

Sam says, 'Let's label the array to help us.

' So we've got our nine rows.

We know we've got nine groups of all of that, but we've partitioned our 16 into 10 and six.

So we've got nine groups of 10 and nine groups of six.

Let's check our understanding here.

We've labelled our array.

Can you label this array by filling in the missing numbers? Pause the video, have a go and when you're ready for some feedback, press play.

What did you think? Well this time, we've got seven rows and we've got a group of 10 and a group of five.

So we've got seven times 10 and seven times five.

So our factor of 15 has been partitioned into 10 and five.

This array shows seven times 10 and seven times five, but it also shows altogether seven times 15.

Let's go back to our array where we were looking at 16 times nine.

So Sam says, 'How can we use this to help us to calculate nine times 16 then?' Jun says, 'This is nine times 16 partitioned into two partial products.

' So if we work out how many counters there are in the red area and in the blue area, we'll have two numbers that we can add together.

And those two numbers resulting from our part multiplication are our partial products.

And Jun says, 'Now we only need to know the 10 and the six times tables!' So we don't need to know the 16 times table.

Sam, you're all right.

She says, 'Can we use boxes to make it clearer and create a grid?' So there we go.

We've got boxes around our counters now.

So we can see that the first box represents nine times 10 and the second represents nine times six and that these two arrays are the partial products.

Now we can work out the value of the two arrays using known times table facts and then add them together.

Let's have a look.

And Sam says, 'They're times tables I can recall.

I know the times table nine times 10 is equal to 90.

' So all those red counters represent nine times 10, which is equal to 90.

And she says, 'I know nine times six is equal to 54.

' All those blue counters represent nine times six, which is 54.

So now we've got our partial products, we've done our two multiplications that will help us to calculate the products overall.

Now to add 90 and 54 together, the partial products.

90 plus 54 is equal to 144.

You might have seen that there was a makeup to a hundred there, wasn't there? So we could have maybe partitioned our 54 into 10 and 44 and then we'd have had our 90 plus 10 is equal to a hundred plus another 44 is equal to 144.

So we can say that nine times 16 is equal to 144, but we broke it down using the distributive law so that we could use times table facts that we knew to create partial products that we could then add together.

Ah, we've got another one here.

Sam says, 'Uh oh! Let's start arranging the counters again.

' Jun says, 'We already know this one.

' What's Jun spotted? What were we calculating before? Ah, he says, remember, multiplication is commutative so the factors can swap position and the product will stay the same.

We calculated nine times 16 and that's equal to 16 times nine.

'I see', says Sam, 'so we've just worked out nine times 16 is equal to 144, so it'll have the same product.

' Nine times 16 is equal to 16 times nine and both of them have a product of 144.

Sam's thinking about the stages that they went through to work out nine times 16 and how it helped her to understand how she could use the distributive law.

So first of all, she started thinking about the counters.

She partitioned 16 into 10 and six and created those arrays using the counters.

Then she drew the boxes around the arrays to help her to see the different partial products and then she could remove the counters, work out the partial products, and then add them together to complete the overall multiplication.

June says 'Yes, by using the distributive law, we were able to calculate two partial products from our known times tables and then combine them.

' Sam says, 'Do we need to do an array every time?' What do you think? Well, let's have a look at seven times 17.

Sam says, 'I think the counters will take too long here.

I feel confident enough to use the boxes.

' So here are the two boxes she's drawn.

And you could imagine her putting the arrays and counters in those boxes, couldn't you? So what she's going to do first? She says, 'We'll use the distributive law by partitioning 17 into 10 and seven.

' So there's our 17 partitioned and we know we're multiplying each of those parts of 17 by seven.

10 lots of seven plus seven lots of seven is equal to 17 lots of seven.

June says, 'He knows that seven times 10 is equal to 70 and seven times seven is equal to 49.

' And Sam says, 'I know that 70 plus 49 is equal to 119.

So seven times 17 is equal to 119.

' She says, 'I've seen this strategy before called the grid model or the grid method.

' You might have come across it as well when you've been looking at multiplication and it works because of the distributive law.

She says, 'I think we could have just used jottings and equations to make this even quicker.

' Let's have a look.

She says, 'I'd start with the expression and then write an expression for each partial product added.

' So seven times 17 is what we were trying to work out.

Well we partitioned our 17 into 10 and seven.

So we had seven lots of 10 and seven lots of seven or 10 times seven and seven times seven.

So that gave us partial products of 70 and 49, which she calculated first.

And lastly, she's going to calculate the sum of the partial products, which is equal to the product of seven and 17.

So our overall product is 119.

Time for you to do some practise.

Can you match the following arrays to the expression that represents them and draw a line between to show how they match up? For question two, can you fill in the missing numbers on the arrays and then use them to calculate the product? So you're going to do the partitioning and find those partial products.

For question three, you've just got the boxes.

So we've taken the counters away.

So can you decide how you're going to use the distributive law to create those partial products for A and B? And then for question four, you've just got the equation.

You are going to think about using the distributive law and calculate the partial products.

So you could use equations and jottings.

If you want to draw a grid though, you can, but let's see if we can move our way through from the counters to the grid to just using the equations and jottings.

Pause the video, have a go at those questions and when you're ready for the answers and some feedback, press play.

How did you get on? So in the first one, there are a lot of dots there, aren't there? We've got eight rows and we've got 10 and we've got eight.

So we've got 18 times eight, so we've got 10 times eight and eight times eight.

So that matches with the equation underneath it.

What have we got in the next one? Well, we're multiplying by five.

I can see five rows and I think I've got 10 and two.

So I've got 10 times five plus two times five, which is the same as five times 12 and that matches with that last equation.

So let's just check that this middle one does match with the remaining equation.

Well, we can see that we've got seven rows there, so we're multiplying things by seven here.

So seven times 10 and seven times five, which is equal to seven times 15.

So there we go.

So our middle expression relates to our right hand array and the 18, 12 and 15 were all partitioned using the distributive law.

So in question two you are filling in the missing numbers and then calculating the product.

So we had 15 times five, so we partitioned the 15 into 10 and five and we're multiplying by five.

10 times five is equal to 50.

Five times five is equal to 25.

So 15 times five is equal to 75.

We can combine those partial products.

50 plus 25 is equal to 75.

And in B, we had six times 14.

Well you can see that we've got six rows, so we've partitioned the 14 into 10 and four.

So now we've got six times 10 as one of our partial products, which is 60 and six times four is our other, which is 24.

And then we're going to combine the partial products.

60 plus 24 is equal to 84, so six times 14 is equal to 84.

For question three, we've taken away the counters.

So how can we use the distributive law and find partial products to help us to calculate six times 15? Well, we can partition 15 into 10 and five.

So we have six times 10 which is equal to 60 and six times five, which is equal to 30.

And then we can combine our partial products, 60 plus 30 is equal to 90 because we know that 10 lots of six plus five lots of six is equal to 15 lots of six, which is equal to 90 and 17 times five where we can partition our 17 into 10 and seven.

So we've got 10 lots of five plus seven lots of five, which is 17 lots of five.

We get our partial products 10 times five is 50 and seven times five is 35.

50 plus 35 is equal to 85.

So 17 times five is equal to 85.

And finally you were going to use some jottings this time.

So seven times 18, well we can partition 18 into 10 and eight.

So that gives us seven times 10 and seven times eight.

10 lots of seven plus eight lots of seven is 18 lots of seven.

So seven times 10 is 70.

Seven times eight, my favourite times table fact is 56 and 70 plus 56 is equal to 126.

So seven times 18 is equal to 126.

And then 16 times four or we can partition our 16 into 10 and six.

So we've got 10 lots of four plus six lots of four, which is equal to 16 lots of 4.

40 plus 24, those are our partial products equals 64.

So 16 times four is equal to 64.

Well done if you've got all of those right.

And on into the second part of our lesson, we're going to think about finding efficient methods.

Sam says she's going to solve 18 times four using equations and jottings.

Jun says, 'I'll do the same and then we can compare.

' Let's have a look at what they do.

18 times four.

Sam says, I'm going to partition the 18 into 10 and eight.

So I've got 10 times four plus eight times four, which we know is 18 times four.

Our partial products are 40, 10 times four and 32, 8 times four.

So 18 times four is equal to 72.

Let's have a look at what June did.

June says 18 times four is equal to nine times four plus nine times four.

Well 18 is nine plus nine, so we can partition 18 into nine and nine.

Nine times four is 36, and another nine times four is 36.

36 and 36 is equal to 72.

So they both calculated that 18 times four is equal to 72.

What's the same and what's different? Well, Sam says we both got the same product and we both use the distributive law, but Jun says you partitioned 18 into 10 and eight, whereas I used nine plus nine.

And Jun says that gave us different partial products, but with the same sum.

40 plus 32 is equal to 72 and 36 plus 36 is equal to 72.

Sam says, 'I don't fully understand this.

Let's draw an array to help me.

' She says.

So here are the two arrays.

Both of them show 18 times four, but the first one shows 18 partitioned into 10 and eight, giving our partial products of 40 and 32.

And the second one shows 18 partitioned into nine and nine, giving partial products of 36 and 36.

'Now I see!' says Sam.

'We partitioned differently creating different partial products.

' And the arrays or the grids, the boxes that we drew can really help us to see that different way of partitioning 18.

Time to check your understanding now.

The arrays show that six times 16 is equal to 96, but they show different partial products.

For each array, write the jottings and equations to get the product of 96 using the distributive law.

So pause the video, have a go.

When you're ready for some feedback, press play.

How did you get on? So in the first one, we partitioned our 16 into 10 and six, giving us partial products of 60 and 36, 10 times six and six times six.

So six times 16 is equal to 10 times six plus six times six, and those partial products are 60 and 36 and added together they are equal to 96.

So six times 16 is equal to 96.

In the second array though, we've partitioned our 16 into eight and eight.

So we've got six times eight, which is equal to 48 and six times eight, which is equal to 48.

So six times 16 is equal to six times eight plus six times eight, which is equal to 48 plus 48, which of course is also equal to 96.

So let's go back to looking at 18 times four, because Jun says, 'I thought of another way of doing it.

' He says, 'We could use subtraction instead.

' Oh, I wonder what that's gonna look like.

Sam says, 'Interesting.

Show me what you mean.

' He says, 'Here's how we both use the distributive law to partition the factor 18 last time.

' So they partitioned it into 10 and eight and into nine and nine, and we've got our part-part whole models there to show us, haven't we? So we could see that 10 lots of four plus eight lots of four is equal to 18 lots of four and nine lots of four plus nine lots of four is also equal to 18, lots of four if we think about our whole and if we think about each of those numbers representing four lots of that number.

He says, 'What if we made 18 apart and used 20 as the whole? Then we could use 20 subtract two equals 18 to help.

' Oh, that's interesting! We know that 18 plus two is equal to 20.

So 20 subtract two must be equal to 18.

How could you do this? Well, now we can think of 18 times four as equal to 20 times four, subtract two times four.

We sort of partitioned our 18 into a number bigger and a subtraction.

So 18 is equal to 20 subtract two.

So we've got 20 lots of four, subtract two lots of four, and 20 times four is equal to 80, and two times four is equal to eight and 80 subtract eight is also equal to 72.

That's a really interesting way of doing it, isn't it? 'And there are probably other ways too', says Jun.

'Which was the most efficient do you think?' Well, Sam says, 'I preferred this one because I'm quick with multiples of 10.

' So she knows her 10 times table really well and she knows enough of her other times tables for that to be a good strategy and an efficient strategy for her.

Jun says, 'I preferred this one because I could use doubling to combine the partial products.

' Jun's obviously a good doubler.

So if he's multiplying by an even number, he might well partition it into two equal parts and then he can use doubling.

Neither of them chose the subtraction strategy.

It's a different way of thinking about it, isn't it? But it's sometimes a useful one if you are multiplying and one of your factors is close to a multiple of 10.

Time for you to do some practise now.

Question one, below are three ways of solving seven times 18 with the distributive law.

Which would you prefer to use and why? So look at the three methods that have been used and then evaluate them.

Which do you think would be the most efficient for you? Which would you prefer? And for question two, we've given you the full equation.

15 times nine is equal to 135, and we'd like you to write two different sets of jottings to calculate this equation.

Fill in the part-part whole models to show how the distributive law has been used.

How have you partitioned 15 in different ways to create two different models and then think which was the most efficient and why? And for question three, you're going to use the distributive law to break down these multiplication into partial products of times tables that you know and then solve them.

So pause the video, have a go at the three questions and when you're ready for the answers and some feedback, press play.

How did you get on? So for question one, there were three ways of solving seven times 18.

So in the first one we partitioned the 18 into 10 and eight.

So we had seven lots of 10 and seven lots of eight or 10 times seven and eight times seven, which is 18 times seven.

Our partial products are 70 and 56, and we combine them and we know that our product is 126.

In the second one, Jun liked this one last time, didn't he? It's an even number so we can partition 18 into nine and nine, seven times nine and seven times nine.

Seven times nine and seven times nine or nine times seven and nine times seven is equal to 18 times seven.

Our partial products are 63 this time, both of them, and we can double for 126.

And in the last one we used that subtraction approach, didn't we? We saw 18 times seven as 20 times seven subtract two times seven.

And 20 times seven is 140, two times seven is 14.

140 subtract 14 is equal to 126.

Which one did you prefer and why? I wonder, let's see what Sam and Jun thought.

Sam says 'I preferred the first one because I'm quick with multiples of 10.

' Jun says 'I like the last one because I could use my understanding of place value and subtraction was quick and it was easy to complete.

' But Sam says 'We could have picked any.

' Different people are more efficient with different aspects of their maths, so they will prefer a different strategy.

I wonder which one you picked.

So in question two, we gave you the full equation, 15 times nine is equal to 135.

And you were going to use the distributive law to write two different ways of using jottings and equations to finish this off.

So in the first one, we know we've got a 10 times nine and then a something times nine.

So I think we must have partitioned 15 into 10 and five this time.

So 10 times nine plus five times nine gives us our partial products of 90 and 45 and a sum of 135, which is our product of 15 and nine.

What about the next one? Well, again, we've partitioned the 15, but how might we have partitioned it this time? Lots of different ways, but we could have used a subtraction this time.

So we could have said that we know that if 20 is the whole, 15 is apart and five is apart so 20 subtract five is equal to 15.

So 20 times nine subtract five times nine is equal to 15 times nine.

Our partial products this time are 180 and 45 and 180 minus 45 is of course equal to 135.

Again, which was the most efficient and why? Well, it would absolutely depend on where your strengths lie in your maths.

I think I might have used the addition one this time because it's the 10 and five times tables and I'm good with those.

Ah, and that's what June says.

'I'm good with my fives and tens.

' So that's the one he'd have used.

But you may have partitioned the second one differently.

You may not have used the subtraction.

So again, lots of different ways to do this, but this is how we broke these partial products down and you may have done it differently.

So we partitioned the 14 into 10 and four.

So we've got 10 lots of seven plus four lots of seven.

70 plus 28 is equal to 98.

14 times seven is equal to 98.

Again, we broke 14 down into 10 plus four.

So we've got 10 times eight plus four times eight, 80 plus 32, which is equal to 112.

Can you see what we're doing here? The 14 times table, isn't it? 14 times seven, 14 times eight and 14 times nine.

So again, we've broken our 14 down into 10 and four, so 10 times nine plus four times nine, 90 plus 36 is equal to 126.

The 17 times six, we've partitioned our 17 into 10 and seven.

So our partial products are 10 times six, which is 60 and seven times six, which is 42, which gives us sum of 102.

17 times six is equal to 102.

17 times seven, 10 times seven plus seven times seven so that's 70 plus 49, which is equal to 119.

We're working out the 17 times table this time, aren't we? And 17 times eight is 10 times eight plus seven times eight, which is 80 plus 56, which is equal to 136.

Well done if you've got those right.

And we've come to the end of our lesson.

We've been using knowledge of the distributive law to calculate products beyond our known times tables.

So what have we thought about in this lesson? Well we've revisited again the distributive law that says that multiplying a number by a group of numbers added together is the same as doing each multiplication separately.

And we've used that a lot in this lesson, but we've also looked at the fact that we can use it with subtraction as well.

When faced with a factor outside of times table facts, it can be partitioned to split the multiplication into two partial products, which are known times tables.

We can use an array or a grid model, and that can represent this strategy as well.

And a quicker method once we are confident is to use jottings and equations, partitioning in a way that is most efficient for you.

Thank you for all your hard work and your mathematical thinking in this lesson, and I hope I get to work with you again soon.

Bye-Bye!.