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 multiplication.

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

In this lesson, we're going to be using knowledge of the distributive law to solve problems in different contexts.

I wonder what those contexts are going to be.

Let's make a start and find out.

We've got two key words or a phrase and a word in our lesson today.

So we've got distributive law and polygon.

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

Are you ready? My turn, distributive law, your turn.

My turn, polygon, your turn.

Let's just double check.

They may well be words and phrases that are familiar to you, but let's just look at those definitions 'cause they're going to be really useful to us in our lesson today.

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

So for example, four times three, we can partition the four into two and two and we can do two times three and then add another two times three.

And the product of four and three will be the same as the sum of those partial products of two times three and two times three.

Both of those are worth six and we know that four times three is worth 12, six plus six.

And a polygon is a 2D shape made up of three or more straight lines.

2D means two dimensional, so are completely flat shapes.

They're made up of three or more straight lines that join at the vertices.

So let's have a look at how we're going to think about polygons in a lesson on the distributive law.

There are two parts to our lesson today.

Aha.

The first part is the number of sides in a group of polygons, and the second part is going to be about calculating quantities.

So let's make a start on part one and our polygons.

And we've got Izzy and Andeep helping us in the lesson today.

Andeep and Izzy look at some different polygons.

Wow, look at all of those.

Izzy says, "They are all regular because the sides and are the same length and the angles are equal." Andeep says, "I'm going to label the number of sides." So let's have a look.

4, 5, 6, 7, 8, 9, 10, 11 and 12-sided shapes.

Wonder what they're all called says Izzy.

Do you know what they're all called? Let's have a think.

See if we can name all these shapes.

Well our four-sided shape is a square, isn't it? It's not sitting on one of its sides, it's sitting on its vertex, but it is a square.

All four sides that equal and the angles are all 90 degrees.

The five-sided shape is a Pentagon.

The six-sided shape is a hexagon.

The seven-sided shape is a heptagon, again, or you might call it a septagon.

The eight-sided shape is an octagon.

Nine sides is a nonagon.

Again, 10 sides is a decagon.

11 sides is a hendecagon and 12 sides is a dodecagon.

So we're going to be meeting these shapes as we go through this part of our lesson.

Andeep says, "There's loads here I haven't heard of before." I must admit hen hendecagons are possibly a new one to me.

I'm not sure I've ever named an 11 sided shape before.

And Izzy says, "We don't need to know all of them." Up to 10 sides is pretty good to know, they're useful.

Other than that, they're fun to know, aren't they? They're interesting.

So we've got a whole load of squares here with four sides each.

How many sides would 15 squares have in total? Well we could just count them, but that would take quite a long time and we might lose count.

Izzy says, "I think we can be more efficient using the distributive law." So we're gonna do 15 times four.

We've got 15 groups of four.

Andeep says, "I don't know my 15 times table, so let's partition it." "Okay, how about into seven and eight so we get this equation? Seven lots of four and eight, lots of four.

That will be one complete row and then two extra squares and then the bottom row and the three squares in the middle.

Well it's one way of partitioning, 15.

Is it the most efficient way?" Andeep says, "I don't know my sevens or eights very well either.

I must admit they're not my favourite times tables apart from seven times eight, of course." "Then let's use 10 and five," Izzy says, because you know those times tables really well." So we can partition our 15 into 10 and five.

And if we look at our sort of array of squares, that's the top two rows, 10 times four, and the bottom row is five times four.

"Great," says Andeep.

"Now I can finish this off." 10 times four is equal to 40 and five times four is equal to 20.

And we've got 10 groups of four plus five groups of four.

So we know we've got 15 groups of four altogether.

We've worked out our partial products, 40 and 20 and we can add them together to equal 60.

So 15 squares have a total of 60 sides.

Time to check your understanding.

How many sides are there in 14 hexagons.

Use the distributive law and partition one of the factors to calculate this.

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

How did you get on? So we have 14 groups of six this time.

So we could partition the 14 into 10 and four.

So 10 lots of six plus four, lots of six is equal to 14, lots of six, 10 times six is equal to 64 times six is equal to 24 and 60 plus 24 is equal to 84.

You might have partitioned a different factor.

You might have partitioned the six into five and one, perhaps you knew what five times 14 was.

Or you might have partitioned the 14 in a different way as well, maybe seven and seven.

But whichever way you partitioned it, you should have worked out that 14 hexagons have 84 sides.

Let's have a look at another one.

Do 13 pentagons have more or fewer than 60 sides? Well let's find out.

Here's the equation to begin with.

13 times five.

And Andeep says, "I'm going to partition the 13 into 10 and three." So we've got 10 groups of five plus three groups of five, which gives us 13 groups of five.

"I'll finish this off," then says Izzy." 10 times five is equal to 50 and three times five is equal to 15.

50 plus 15 is equal to 65.

So there are more sides in 13 pentagons than there are in 15 squares.

15 times four is less than 13 times five.

"So let's write an inequality symbol," says Izzy, "to show which is greater." Which symbol should we be writing? Remember, that 15 squares had fewer sides than 13 pentagons.

That's right, so it's the less than symbol that we're going to write between our two equations.

15 times four is less than 13 times five.

Time for you to do some practise.

In question one, you're going to compare the number of sides in these multiples of polygons by placing the correct symbol less than greater than or equals between the equations that you use to work them out.

Use the distributive law to help you.

And then in question two, you've got seven different multiples of polygons.

They can be arranged into pairs with equal numbers of sides except for one.

Can you pair them up and find the odd one out? Use the distributive law again to help you.

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

How did you get on? So we were comparing the number of sides in these multiples of polygons.

So we had 17 squares and 13 hexagons.

So 17 times four, we can use the distributive law to simplify that.

10 lots of four plus seven, lots of four.

40 plus 28 is equal to 68.

So 17 squares have 68 sides.

What about 13 hexagons? 13 times six 'cause there are six sides on a hexagon.

So we can use the distributive law, 10 times six plus three times six is our 13 times six, so that's 60 plus 18 which is equal to 78.

So 13 hexagons have more sides, so that side is greater.

The number of sides on 17 squares is less than the number of sides on 13 hexagons.

B was 13 hendecagons, you remember they had 11 sides and 15 nonagons with nine sides.

So we could have partitioned our 13 into 10 and three, our 15 into 10 and five and multiplied each of those numbers to get a partial product multiplied by 11 for the hendecagons are multiplied by nine for the nonagons.

So you'd have worked out that 13 hendecagons have 143 sides and 15 nonagons have 135 sides.

And 143 is greater than 135.

So our equations would've been worth a greater value.

So for C, 15, heptagons, 15 times seven and 19 times five.

And if you work that out, you'd realise that 15 heptagons have 105 sides and 19 pentagons have 95 sides.

So the 15 heptagons have more sides than the 19 pentagons.

105 is greater than 95.

So in question two, you're trying to find the pairs of multiples of polygons that have the same number of sides using the distributive law to help you.

So you would've worked out that 20 mulligans have 180 sides.

14 heptagons have 98 sides, 20 squares have 80 sides.

I wonder if you needed to use the distributive law, then we had 20 times four.

We know two times four is eight so 20 times four must be 80.

15 hexagons have 90 sides.

16 pentagons have 80 sides, 15 dodecagons, those are the 12-sided shapes have 180 sides and 18 pentagons have 90 sides.

So which was the odd one out? Well there were our pairs.

Our odd one out was the 14 hectagons.

So the 14 seven sided shapes have 98 sides in total.

So that was our odd one out.

And on into the second part of our lesson, we're going to be calculating quantities.

Izzy plays battle robots online against Andeep and she has to buy new robots for her army to try and win in a battle against Andeep's army.

Izzy says, "I have 100 points to spend on new robots." Those are points she's earned in the game I think, aren't they? "I want to buy seven more Ice robots.

Is it possible?" She says.

Well there are the Ice robots in the middle and they cost 15 points.

So 15 times seven.

She says, "I'll partition the 15 into 10 and five using the distributive law." So she's got seven times 15, so she's got seven times 10 and seven times five.

And we can also think of that as 10 groups of seven and five groups of seven, which gives us 15 groups of seven altogether.

So our partial products are 10 times seven, which is 70 and seven times five, which is 35.

And we need to add those together, which gives us a total of 105.

So seven Ice robots cost 105 points.

Oh dear.

She says, "I can't afford it, have to rethink." She says, "I'll try seven doom robots." So they cost 14 points, one point less than the Ice robots.

Ooh.

She says, I'll partition the 14 into 10 and four this time." Andeep says though, "I think there's a more efficient way." What's Andeep thinking of do you think? The Ice robots cost 15 points.

The doom robots cost 14 points.

Andeep says, "Adjacent multiples of a number have a difference of that number." Adjacent means next door.

He says, "Seven times 15 and seven times 14 are adjacent multiples." 15 lots of seven and 14 lots of seven.

So as an equation as he says that looks like this, seven times 14 is equal to seven times 15.

Subtract one group of seven.

We've got 14 lots of seven and we've got 15 lots of seven.

So if we take away one lot of seven, they will be equal.

She says, I already know that seven times 15 is equal to 105.

And 105 subtract seven is equal to 98.

So that must be the value of seven times 14.

"I can afford that," she says.

Only 98 points, she'll have two points spare, won't she? Time to check your understanding.

With 125 points, can Andy afford to buy eight Ice robots? So that's eight times 15.

Pause the video, have a go, use the distributive law or maybe another strategy.

And when you're ready for some answers and feedback, press play.

How did you get on? Well, Andeeps's going to calculate it eight times 15.

So we can partition our 15 into 10 and five.

So we've got 10 groups of eight and five groups of eight, which is 15 groups of eight.

Eight times 10 is 80, eight times five is 40, 80 plus 40 is equal to 120.

He can afford it, he had 125 points.

But you might have partitioned differently or you might have thought about that adjacent multiples rule.

We knew that 15 times seven was equal to 105.

So 15 times eight was one more 15, which would've been 120.

So we might have worked it out using adjacent multiples of 15.

Andeep wants to know how many points his army is worth altogether.

He has a tally chart showing how many of each robot he has.

So he's got five seven Dilla robots.

He's got five Ice robots and he's got six Chuck robots.

So how will he do this? He says, "I need to work out each subtotal and then combine them." So he needs to work out what his Dilla robots are worth, his Ice robots are worth and his Chuck robots are worth and then add them together.

He's going to calculate how many Dilla robots he has.

He says this is seven times 13 because I have $7 worth 13 points each.

Seven times 13.

He says, "Well I can partition the 13 into 10 and three.

So I've got 10 times seven and three times seven.

So I've got 13 times seven altogether." The partial products are 70 and 21 and they add together to equal 91.

So his Dilla robots are worth 91 points.

He's going to calculate how many Ice robots he has this time.

Five times 15 'cause he has five Ice worth 15 points each.

15 times five, well he can use the distributive law to make two multiplication of five times 10 and five times five by partitioning his 15.

And those partial products sent are 50 and 25, which together are equal to 75.

So his Ice robots are worth 75 points in total.

And then he's going to calculate how many Chuck robots he has.

And this is six times 19.

He has six robots and they're worth 19 points each.

So six times 19 he can again partition his 19 into 10 and nine.

So then he's got two partial products of 60 and then six times nine is 54.

He's still got 16 groups of nine altogether he's got 10 groups of six and nine groups of six, which is equal to 19 groups of six.

And when we add those together, they are worth 114 points.

Now what does he need to do? Well, he wants to know how many points his army is worth altogether.

And so he's now gonna have to add these three numbers together.

He says, "I'm going to find the sum using a written method." So he's going to do a column addition.

So 114 plus 91 plus 75.

He's gonna start with the ones four ones plus one ones plus five ones is equal to 10 ones.

So he needs to regroup a 10.

There's this regroup, 10 and a zero to show that we've got no extra ones.

Nine tens, add one ten, that's 10 tens plus another eight.

So he's got 18 tens, so that's 100 and eight tens.

So he is put his eight tens in the tens column and he's got his regroup 100 and then he's got one 100 and another one to equal 200s.

So his army is worth 280 points all together.

Time for you to do some practise.

In question one, you're going to calculate the cost of the quantity of battle robots in each question.

So you're going to calculate all of those different totals and work out which quantity was the most expensive.

In question two, you're going to calculate the total number of points that Izzy's army is worth using the tally chart and the distributive law.

And in question three, Jun and Alex also have armies as shown below, whose army is worth more? Pause the video, have a go at those three questions and when you're ready for the answers and some feedback, press play.

How did you get on? So in question one, you were calculating the different costs of these numbers of robots.

So six method robots, that's 17 points each.

So six times 17.

So we can partition our 17 into 10 and seven.

So we've got 10, lots of six and seven, lots of six.

So that's 60 plus 42, which is 102.

So the method robots are worth 102 points.

For Doom, we're going to do eight times 14.

By using the distributive law, we can work out that they are worth 112 points.

Seven Ice robots, that's seven times 15.

Again using the distributive law.

Seven times 10 and seven times five.

So 70 plus 35 is equal to 105 points.

Four Chuck robots, so that's four times 10, which is 40 and four times nine, which is 36.

So they're worth 76 points.

And the Dilla robots, that's nine times 10, which is 90, nine times three, which is 27.

And if we add those together, we get 117 points.

So by just five points, the nine Dilla robots were the most expensive.

They were worth five points more than each doom.

Robots.

In question two, we're going to calculate the total number of points that Izzy's army is worth.

So we've got her tally of her different robots and we're going to use the distributive law.

So if we do that, we work out that the Dilla robots are worth 78 points.

The Doom robots are worth 70 points.

She doesn't have any Ice robots.

Method robots are worth 136 points.

And the Chuck robots are worth 114 points.

And she's written down how she worked out the chuck robots.

Six times 19.

Ah, do you see what she did this time? She did partition her 19, but she thought of it as 20 groups of six, subtract six.

She knows that six times two is equal to 12.

So six times 20 is 120.

Subtract six is equal to 114.

Now she's got to work out the total number by adding up the values of her robots.

So she's got a column edition with four that ends this time.

And when she adds those up, she finds that her total is 398 points.

Wow, what a lot of points you've got in robots there, Izzy.

Well done.

398 points.

And for question three, we've got to work out whose army is worth more, Alex's or Jun's? So again, when they do their calculating seven times 14 using the distributive law, seven times 10 plus seven times four is equal to 98.

For Ice, we've got six lots of 15, six times 10 plus six times five is equal to 90.

And for Chuck, 19 times five.

So five times 10 plus five times nine is equal to 95.

And when we add those together, we get a total of 283 points.

What about Alex's army? Well, he has nine times 13 for Dilla robots, which is equal to 117.

He's got seven times 17 for his Method robots, which is 119 points and four times 19, which is equal to 76 points.

And when he adds those together, his total is 312 points.

He seemed to have more robots, didn't he? He had nine Dilla robots and seven method robots.

So whose army is worth more? It's Alex's army, isn't it? Well done if you've got all those correct.

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

In this lesson, we've been using our knowledge of the distributive law to solve problems in different contexts.

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

So a factor can be partitioned to split the multiplication into two partial products and then we can add them together.

And this is particularly useful when working out problems involving numbers that are not in our known times tables.

Thank you for all your hard work in this lesson.

I hope you've enjoyed it as much as I have and I look forward to working with you again in another lesson soon.

Bye-Bye.