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 thinking about using knowledge of the distributive law to solve two part addition problems. I wonder if you've come across the distributive law.

Well, we're going to be learning about it today together, and we're going to use some very familiar representations to help us to understand what this law is all about, so let's make a start.

So, there are two keywords or phrases in our lesson today.

We've got common factor and distributive law, so I'll take my turn to say them, and then it'll be your turn.

Are you ready? My turn, common factor, your turn.

My turn, distributive law, your turn.

That might be a new one to you, distributive law.

Quite a lot to get our mouths around, isn't it, but I'm sure you're going to get the hang of it during this lesson.

Let's have a look at what they mean 'cause these definitions are going to be really important.

So, you may well have thought about factors before, and when we compare lists of factors of two or more numbers, any factors that are the same are the common factors.

We're also going to think about that in terms of factors in our multiplication equations, and we're going to look for common factors in multiplication equations.

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

So, for example, if we've got four times three, we can say that that is equal to two times three plus another two times three.

And we know that four times three is equal to 12, but we can partition that four into two and two and multiply each of those twos by three, and it will give us the same product as if we just multiplied four by three.

So, we're going to be investigating that in this lesson, and there are two parts to our lesson.

In the first part, we're going to be looking at two part problems with a common factor, and in the second part, two part problems in context.

So, let's make a start on part one, and we've got Aisha and Alex helping us with our learning today.

So, they're looking at an equation, nine times three plus nine times six is equal to something.

What do you spot there? Aisha says, "This is a two part addition problem because there are two multiplication expressions being combined," nine times three and nine times six.

Alex says, "Yes, nine groups of three added to nine groups of six." He says there's a common factor here.

What's the common factor in each expression? So, we've got nine times three and nine times six.

Those are our two multiplication expressions within our equation.

Well, the factor of nine is common to both nine times three and nine times six, so it's a factor in each of our multiplications, and it's a common factor because it's the same.

So, Aisha says, "Six and three are not common factors in the multiplication expressions." Our other factors are different.

We've got nine times three and nine times six, so the nine is common to both, but the three and the six are not common.

So, just let's check that idea of a common factor.

Have a look at the multiplication expressions in the equation below and see if you can work out which is the common factor.

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

How did you get on? So, did you see that our two multiplication expressions were seven times two and seven times eight and the factor that was the same in both of those was seven? That's right, the factor seven is common to both seven times two and seven times eight.

So, we're feeling more confident about what a common factor is.

Let's go back to our equation again, and Alex says, "Let's use an array to represent the equation." So, we've got nine times three and nine times six.

What do you notice? Aisha says, "Both arrays have nine rows because nine is the common factor." So, we've got nine rows of three in the yellow array and nine rows of six in the red array.

And Alex says, "We need to find the sum of the two expressions," so we need to find the sum of nine times three and nine times six, represented by our arrays.

Aisha says, "I will calculate the value of each expression, and then combine the products." Okay, well done, Aisha, off you go.

She says, "Nine times three is equal to 27 and nine times six is equal to 54, and 27, add 54, is equal to 81." Well done, Aisha, I think you've got that right, haven't you? Do you agree with Aisha? Let's see what Alex has to say.

He says, "I would've calculated it differently." Ah, is there another way you can think of to calculate it? Let's see what Alex comes up with.

He says, "I'd start by combining the two arrays together." Ooh, let's have a look.

Ah, so he's put his two arrays together.

What have we got now? He says, "This array can now be calculated with one multiplication." We've put the two arrays together, and because we knew there were nine rows in each one, we can now see that we've got nine lots of nine rows.

We had three lots of nine and six lots of nine, and now we've got nine groups of nine altogether, nine rows with nine counters, and nine times nine is equal to 81.

So, what's the same and what's different in the way they've worked? What can you see? Aisha says, "I worked out both expressions and combined," and you can see Aisha's strategy there on the left of the screen, her two multiplications, and then her addition.

But what did Alex do? Oh, she says, "You combined first, and then calculated one multiplication." So, Alex knew that there were three groups of nine and six groups of nine, and so, altogether, there were nine groups of nine.

He says, "Yours needed more calculations than mine." It did, didn't it? He said, "so my method was more efficient." He says, "I used the distributive law," and that's what the distributive law tells us, that we can add together the number of groups, and then do the multiplication.

So, we knew that we had three groups of nine and six groups of nine, so, altogether, we had nine groups of nine, and it's great if you know that nine times nine is 81, and Alex did, so it did make it a more efficient method for him.

Alex says, "by combining the uncommon factors," that we showed in the boxes, he managed to add first, and then multiply.

Three and six were our uncommon factors.

Time to check your understanding.

Can you spot the mistake in the jottings below? Think about what we just did, think about Alex's method.

What's the mistake? Pause the video, have a go, and when you're ready for the answer and some feedback, press play.

How did you get on? Aisha says, "The factors have been multiplied here together rather than added.

Here's what the jottings should have looked like." It's the uncommon factors that we need to add together.

So, we can see that we've got four seven times and four three times, so, altogether, we've got four 10 times.

And we could read those multiplication expressions the other way around, seven groups of four and three groups of four.

Altogether, we've got 10 groups of four, and four times 10 is equal to 40.

So, we must remember when we're applying the distributive law that we are combining the uncommon factors by adding, and then doing our multiplication.

So, add first, and then multiply.

So, what's the same and what's different here? What can you see? Well, Aisha says, "The factors within the multiplication expressions have swapped position." On Aisha's side, we've got nine times three plus nine times six, and on Alex's side, we've got three times nine plus six times nine.

So, Aisha says, "The common factor of nine has swapped position, but we can still use the distributive law." We've still got those common factors.

Why do we know we can still use the distributive law? Ah, Alex says, "Multiplication is commutative, so the value of each multiplication expression is the same." It doesn't matter which order our factors come in, our value of our expression will be the same, and therefore, we can still find those common factors, and then combine the uncommon factors.

He says, "Multiplication is commutative, so the value of each multiplication expression is the same, whichever factor you start with, the group size or the number of groups," and we can see there that we combined our uncommon factors to create that nine that we're going to multiply our nine by.

In this case, we end up with two factors the same, don't we? What's the same and what's different here? Aisha says, "This time, the common factor of nine appears in different places within the expressions in the different equations." So, we've got nine times three plus nine times six, and then we've got three times nine plus nine times six, but, again, the value of each expression is the same because multiplication is commutative.

So, even though the common factor of nine is positioned differently in each expression, we can still use the distributive law.

Nine is still our common factor, and we can still combine those other factors.

We've still got three groups of nine plus six groups of nine.

We've just swapped the order of the factors around.

Time for you to do some practise.

Can you match the expressions that have the same value from the first column to the second column? There are fewer in the second column, so they'll match to multiple expressions.

So, you'll have more than one of the expressions on the left-hand side matching up with the smaller number on the right-hand side.

Question two, you're going to fill in the missing numbers in the jottings below, and you're going to try and think what's missing? Is it an uncommon factor or a common factor in each side? Think really carefully about the information you've got.

And in part three, you're going to solve some problems using the distributive law, and we've given you some pointers there to begin with in A and B.

Then, in C and D, you're going to find those common factors and those uncommon factors and solve the problems. So, pause the video, have a go at your tasks, and when you're ready for some feedback, press play.

How did you get on? So, we had to do some matching here, didn't we? We had to match those expressions.

So, let's look at the top one, six times two plus six times four.

Well, our common factor is six there, and we've got two lots of six and four lots of six, so we've got six lots of six.

So, it matches with that top expression.

Seven times four plus four times five.

Ooh, we've got our common factors together sort of in the middle there, haven't we? So, the common factor is four, and we've got seven groups of four, and then we've got four five times, or five groups of four.

And seven plus five is equal to 12, so we must have 12 groups of four.

Three times two plus two times eight.

Again, that common factor is in the middle.

We've got three groups of two, and then we've got eight groups of two on the other side, and eight plus three is equal to 11, so we must have 11 groups of two.

Two times six plus six times four.

Again, our common factor there is six, isn't it? So, our expression must be six times six.

Five times four plus seven times four.

Well, our common factor is four.

Five plus seven is equal to 12, so it's 12 times four again.

And two times eight plus two times three.

Well, our common factor is two, and eight plus three is equal to 11, so that matches with our middle expression.

I hope you were successful in finding the common factors and then adding the uncommon factors.

And let's fill in these jottings.

We have to think carefully about this.

So, we've got something times two plus eight times four is equal to eight times something, so we've got eight as a factor there.

Two and four are clearly not our common factors, so that must be a common factor of eight because two and four combine to make six.

So, that must be a missing factor of eight, common to the other eight, and then our simplified equation is eight times six, and eight times six is equal to 48.

So, let's look at B.

We've got something times five plus seven times six.

Again, we've got a seven in our simplified equation, so that has to be our common factor.

Five plus six is equal to 11.

So, our missing factor is seven, the common factor.

Seven times 11 is equal to 77.

Something times nine plus four times nine.

Ah, well, nine is a common factor here, isn't it? So, that must be our missing factor in our shortened equation, and four plus three is equal to seven.

So, it must be three times nine plus four times nine is equal to seven times nine, and that is equal to 63.

What about D, what was the common factor here? We've got something times two plus two times four, so two is our common factor, and four plus two is equal to six.

So, actually, both our missing numbers were twos there.

The uncommon factor was a two, and also, one of the common factors was a two as well.

And in E, something times five plus three times five.

So, five is our common factor there, but three plus five is equal to eight, so our missing factor is also a five.

So, five times five plus three times five is equal to eight times five.

Something times four plus something times three is equal to eight times four.

I think four must be our common factor there, and our other missing factor must be a five 'cause five plus three is equal to eight.

So, five times four plus four times three.

Six times something plus something times four.

Ooh, we've got nine times 10 here, so our missing factor must be the nine that's missing from both of those expressions because six plus four is equal to 10.

So, six times nine plus nine times four is equal to nine times 10, which is equal to 90.

And finally, for H, we've got something times five plus something times two is equal to 12 times two.

So, I think two is the missing common factor, so two times five, and five plus what is equal to 12? Well, that's seven, isn't it? So, two times five plus seven times two is equal to two times 12, which is equal to 24.

Well done for sorting all of those out.

Took a bit of thinking, didn't it, finding out what was going to be the common factor and what was going to be our missing uncommon factor? So, for question three, you were solving the two part addition problems using the distributive law and thinking about what you noticed.

So, in A, six times four and six times five, where we combine the uncommon factors five and four to equal nine, 'cause we've got six as our common factor, and we've got four lots of six and five lots of six, so we've got nine lots of six.

Six times nine, which is 54.

In B, we've got a common factor of seven, and we've got four groups of seven and five groups of seven.

Four plus five is equal to nine, so we've got seven times nine, which is 63.

Are you spotting something here? In the next one, we've got eight times four and five times eight, so we've got a common factor of eight, but, again, our uncommon factors are four and five, which add to equal nine.

Eight times nine is equal to 72.

And we've got four times nine plus nine times five.

We've got a common factor of nine again, and four plus five is equal to nine.

Nine times nine is equal to 81.

What did you notice? Well, Alex says, "This is the nine times table written using a pair of expressions added together." So, we've got four times and five times something each time, which is giving us our nine times table.

So, we've partitioned nine into four and five, and so we've got four lots of six and five lots of six; we've got four lots of seven and five lots of seven; we've got four lots of eight and five lots of eight, and then four lots of nine and five lots of nine.

So, we've partitioned to make the nine times table.

"The uncommon factors are four and five in all the equations, meaning there's always a factor of nine," and, "The common factors increase by one with each question, which produces the nine times table." So, the common factors were six, seven, eight, and nine.

I hope you spotted that, and on into part two of our lesson, looking at some two-part problems in context.

So, "The Oak Academy Owls football team have had five draws and five wins in the first 10 games this season." Well done, Oak Academy Owls.

That's a pretty good record, I think.

How many points have they got so far? Do we know enough? Aisha says, "There's some missing information!" What do we need to know? Alex says, "I agree, we need to know something else." So, what else do they need to know? Ah, that's right, they need to know the points.

They get one point for a draw and three points for a win, so now we can work out their points, can't we? And Aisha says, "We can use the distributive law here." "I agree," says Alex, "let's start by building an equation." So, they've had five draws and five wins.

They get one point for a draw and three points for a win.

So, "What expression would be for the wins," she says.

Can you think? Well, "There were five wins which gained three points each." So, five times three, so there are the wins.

"Now, what about an expression for the draws," says Aisha.

Well, "There were five draws which gained one point each.

So that's five groups of one, which I'd write like this," says Alex.

So, there's our equation now.

Aisha says, "Five is the common factor," and it's the number of groups in each expression.

So, we've got five groups with three points and five groups with one point, and that's our common factor, and it's the number of groups, five groups of three points and five groups of one point.

"Now let's solve it using the distributive law," says Alex.

So, these are our uncommon factors, the three points and the one point, but we know that they scored five of the three points and five times the one point.

So, we've got three points five times and one point five times, so we must have a total of four points five times in all, four times five, and that's equal to 20.

So, "They've got 20 points altogether so far for this season," says Aisha.

Well done, Oak Academy Owls, I hope you go on and have some more wins in the second part of your season.

Another team gets six wins and six draws.

Which of the following equations represents their total number of points? Remember, three points for a win and one per draw.

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

Okay, so this time, it's six wins and six draws.

Let's think.

So, they've got six lots of three points and six lots of one point, so, altogether, they've got six lots of four points.

So, six times four is equal to 24.

Our uncommon factors were three and one.

For the number of points again, we've got three lots of six and one lot of six, so we've got four lots of six altogether, or six times four, which is 24.

Ah, "Alex is collecting battle robot cards.

He buys three packs of cards on Monday, and then two more packs on Tuesday.

How many cards did Alex buy altogether?" Aisha says again, "There's some missing information!" What do we need to know? Have you worked out what else it is they need to know? That's right, we didn't know how many were in each pack, did we? Each packet contains 11 cards.

So, again, we can use the distributive law.

Let's have a go and start with the equation.

"So what's the expression for Monday?" Well, on Monday, he bought three packs of cards, and there were 11, so three groups of 11.

And, "How about Tuesday?" Well, he bought two more packs, didn't he? So, he bought another two groups of 11, which we can write like this.

So, he bought three lots of 11, and then he bought another two lots of 11, so 11 is the common factor this time, the packs of cards, and this time, it represents the number in each group.

He bought three groups of 11 cards, and then two groups of 11 cards.

So, there's our common factor, and this time, the size of the group.

And Alex says, "Yes, that's because there were 11 cards in each packet.

Now let's solve it using the distributive law." So, three and two are our uncommon factors, so we can add those together.

Three groups of 11 plus two groups of 11 gives us five groups of 11, and five times 11 is equal to 55.

So, "Alex, you bought 55 cards altogether." I hope you got the battle robots you wanted in there.

Time to check your understanding.

Aisha buys two packets on Monday and four packets on Tuesday.

Which of these equations represents how many cards she bought altogether? "Remember, 11 cards in each packet," she says.

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

How did you get on? So, this time, there were two groups of 11 plus four groups of 11, which is six groups of 11.

So, six times 11 is equal to 66.

She bought six packets of cards and 11 in each.

So, what did you notice about those problems? So, we have the Oak Academy Owls and their points for their wins and draws and Alex and his battle robot cards in packs of 11.

Well, Aisha says, "The common factor in the first question was the number of groups." We had five groups with three points and five groups with one point.

And Alex says, "The common factor in the second question was the group size." We had three lots of 11 cards, and then two groups of 11 cards.

"So it doesn't matter which factor type is the common factor." The common factor can be the number of groups or it can be the size of each group.

It doesn't matter which way around it is, we can still use the distributive law to solve our problems. So, let's have a look at this one.

There's a word problem below, and it's been solved using the distributive law.

You're going to decide whether the common factor was the group size or the number of groups.

So, "A local hockey team won four games in January, and then another five games in February.

They get three points for a win.

How many points did they get from these games?" So, is the common factor the group size or the number of groups? Pause the video, have a go, and when you're ready for some feedback, press play.

How did you get on? So, the common factor was three points, wasn't it, and that was the group size.

So, they got four lots of three and five lots of three, so it was the group size that was the common factor this time, represented by the number three.

Time for you to do some practise.

So, for each of these contexts, you're going to decide whether the common factor was the group size or the number of groups.

So, in A, Aisha has two sheets of stickers, each with six stickers on, and Alex also has two sheets of stickers, but each sheet only has four stickers.

How many stickers do they have altogether? And in B, Jun has five sheets of stickers with six stickers on, and Izzy has two sheets of stickers, each with six stickers on.

How many stickers do they have combined? So, write out your equations and decide whether the common factor is the group size or the number of groups.

And then, for those same problems, you're going to choose the correct equation and you're going to solve the problem.

For question three, Alex has been playing a game involving throwing a bean bag at a target.

He throws eight times in total.

With half of his throws, he scores five points, and with the other half, he scores two points per throw.

How many points does he score in total, and can you use the distributive law to show your working out? And for question four, some Oak pupils participate in a football penalty competition in pairs.

Each penalty scored gives five points.

The table below shows the pairs and how many penalties they score.

So, for part A, you're going to order the teams into first, second, and third without calculating, ooh, and for B, now calculate their points totals.

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

Okay, let's have a look.

So, for question one, you were deciding on whether the common factor was the group size or the number of groups? So, Aisha had two sheets of stickers with six, and Alex also had two sheets, but he had four on each.

So, what was it this time? We had two times six plus two times four, so the common factor was the two, and it represented the number of groups.

Aisha had two groups of six, and Alex had two groups of four.

In B, though, five sheets of stickers with six and two sheets of stickers, each with six, so we've got five times six and two times six.

This time, the common factor is six, and it represents the group size, how many stickers were on each sheet.

So, our common factor changed its job between A and B, but we can still use the distributive law to solve the problem, which we're going to do next.

So, which equation correctly represented the problem? Well, it was the top one, wasn't it? Two lots of six plus two lots of four, and we can see that our common factor is two.

We can combine our uncommon factors because we've got six lots of two and four lots of two, so we've got 10 lots of two altogether, and that's 20.

And for the second one, yes, our common factor was six this time, so it was the bottom equation.

We had five groups of six plus two groups of six, which is equal to seven groups of six, and seven times six is equal to 42.

So, in part three, Alex had thrown the bean bag eight times.

With half of his throws, he scored five points, and with the other half, he scored two points for each throw, so he scored four lots of five points and four lots of two points.

So, if we think about commutativity, we can say we've got five lots of four and two lots of four.

Our common factor is four, and our uncommon factors are five and two, and we can combine those to equal seven.

So, four times seven, which is equal to 28 points.

Half of the time, he scored five points.

Half of the time, he scored two points.

And finally, the penalty shootout competition.

So, for part A, we had to order the teams into first, second, and third without calculating.

So, each penalty scores five points, which is the common factor, so we needed to find out how many each pair scored.

Alex and Aisha scored four times, Jun and Izzy scored five times.

Lucas and Sam scored six times, so they came first.

Jun and Izzy came second, Alex and Aisha came third, but it was quite close, wasn't it? So, now we've got to calculate their point totals.

So, Alex and Aisha had four scores at five points each, so 20.

Jun and Izzy had five scores at five points each, that's 25, and Lucas and Sam, six scores, five points each, 30 points.

Well done if you got all of that right, and well done for thinking hard about the distributive law.

We've come to the end of our lesson.

We've been using knowledge of the distributive law to solve two part addition problems, so what have we learned about today? We've learned that two part addition problems feature two multiplication expressions that need to be added together.

Identical factors are a common factor, and we can use the distributive law by combining the two not common factors and multiplying that by the common factor.

And both the group size and the number of groups can be the common factors, meaning that this concept can be applied to many contexts.

So, look out for those common factors when you're looking at two part addition problems involving multiplication expressions.

Thank you for all your hard work today.

I hope you've enjoyed the lesson, and I hope I get to work with you again soon.

Bye-bye.