Lesson video

Hi, and welcome to today's lesson on the Distributive Law With Two or More Binomials.

By the end of today's lesson, you'll be able to use the distributive law when we have two or more binomials.

Our lesson has two parts today, so we're going to start with the first one which is expanding two or more binomial expressions.

How do we expand and simplify when we have two binomial expressions and we want to find the product? Well, we've done this before in our learning.

We did this when we were expanding with algebraic terms. The only difference here is that we now have surd terms, but it works exactly the same way.

Let's use our grid to support this.

Remember, we can write each term, taking up either a space on the row or a space in the columns.

So I've chosen to write five and then root two across the top and the two plus root two down the side to make my rows.

Remember, to fill in the grid, we multiply.

Now, we know this grid method works because we built this up from our area model in our previous learning.

Remember, this came from finding the area of a rectangle that we broke into smaller rectangles.

This grid is just an abstracted version of that area model, so we know why this works.

Let's fill the grid in now.

Two multiplied by five is 10.

Two lots of root two is two root two.

Five lots of root two is five root two and at the end, we have root two multiplied by root two which gives me two.

I can then write out the four terms I've got when I multiplied and here they are.

Now remember, we should always try to simplify, so let's gather our like terms. Two add 10 is obviously 12.

And two root two plus five root two is seven root two, and that's all there is to it.

That's how we expand two binomial expressions.

Let's make sure that we're really confident with this and do some more practise.

I'm going to expand these two expressions and simplify where possible.

So again, I'm going to use the grid to support my understanding here.

I'm going to write root five and the negative one across the top, and two and then two root five down the side.

Let's fill in the grid.

Two multiplied by root five is two root five.

Two multiplied by negative one is negative two.

Now, two root five multiplied by root five will mean two times five, so 10.

And at the end, I'm going to have negative two root five.

I can then write out my terms and look to simplify.

Well, two root five takeaway two root five will give me nothing, so I'm just left with 10 takeaway two which is eight.

It's now your turn.

Expand and simplify where possible, please.

You are very welcome to use the grid to support your multiplication.

Pause the video now and have a go.

Welcome back.

Let's see how you got on.

Did you use a grid? Well, if you did, you can just check your multiplication with mine right now.

If not, check the terms that you got against what I get in the grid.

You can see inside the grid the four terms I got when I multiplied my two binomial expressions.

Whether or not you used the grid, you should still have reached the same four terms. We then need to gather those like terms, so six root seven take away two root seven is four root seven.

And then two takeaway 42 is equal to negative 40.

In this example, we're going to expand and simplify the following expression.

We have three root three subtract four, and we are squaring all of this.

There's nothing different here though to what we did before because we know when we square something, we are multiplying it by itself.

In other words, three root three subtract four, multiplied by three root three subtract four.

Again, you're welcome to use the grid if you'd like to support your multiplication here.

Because this is the same expression, I know that I'm going to have three root three and negative four as both my column headings and my row headings.

Three root three multiplied by three root three is 27.

I've then got negative 12 root three, another negative 12 root three, and then 16.

I then just need to write out my terms and simplify.

Well, I can see from the grid, 27 add 16 is 43, and then I have two lots of negative 12 root three, which is negative 24 root three, giving my final expression of 43 subtract 24 root three.

It's now your turn.

Please expand and simplify two root seven plus three, all squared.

Pause the video and do this now.

Welcome back.

Let's see how you got on.

You should have reached the answer, 37 plus 12 root seven.

You would of course, have written two root seven plus three as both the column headings and the row headings, and then we just needed to multiply.

It's now time for our first task.

For Part A, I'd like you to mark Sara's homework for the answers that are wrong and there will be some.

Please correct it and put in the correct answer.

Pause the video and do this now.

Welcome back.

Let's look at the remaining parts of this task.

For B, C, and D, I'd like you please to find the areas of the shapes.

Now, you are very welcome to draw a diagram to support you in your working if that's what you'd like to do.

I find that a diagram can make things significantly easier as it helps me to visualise what the problem is asking me to do, but it is of course your choice.

You may find you want a diagram for some questions, but not others.

Pause the video and have a go now.

Welcome back.

Let's go through our solutions.

We'll start with Sara's homework.

Well, the middle row was completely correct, so for those, you could have just ticked them.

Unfortunately, she'd got the others wrong.

Let's start with the top row.

We know that when we expand this, in order to get the whole number, I'll have done three multiplied by two, and then negative root two multiplied by positive root two which will leave me the result of negative two.

Six takeaway two is four.

For the root two part, I'll have three root two, subtract two root two, leaving me with just one root two or root two.

Similar working can be used to find the other three results.

Remember, you can always use the grid to support your multiplication.

That can be really handy to make sure that you have multiplied every term by the other respective terms it should be multiplied with.

Let's now look at parts B, C, and D.

You were asked to find the area of a rectangle and the two side lengths were given.

While we know to find the area of a rectangle, we must multiply these two sides together.

We know it's not the same side twice because these two expressions are not equivalent, so let's multiply.

Here, we have our two binomials and we wish to find the product.

Again, you can use the grid to support your multiplication if you wish, or you can multiply and write your terms. Regardless, when we expand these brackets, we achieve the following four terms, eight root two plus six root four, plus 20, plus 15 root two.

Well, we can gather the like terms here.

Let's start with the root twos.

Eight root two plus 15 root two is 23 root two.

Now, you might be wondering where the 32 has come from.

Well, six multiplied by the square root of four is actually just six times two, so 12 and 12 add 20 is 32.

In C, you were told that you had a square with sides four plus three root 20.

Well, if it's a square, we know that all sides are the same length, so I need to square this value in order to get the area of my square, and that's what I've done here.

Expanding those brackets gives me 16 plus 12 root 20, plus 12 root 20, plus 180, or we can gather the like terms. 16 plus 180 is 196 and then 12 root 20 plus 12 root 20 is 24 root 20.

Now root 20 can of course be simplified, so we should do this, meaning we have a final result of 196 plus 48 root five.

It's now time for D.

In this one, we had a rectangle, and one of the side lengths was given and so was the perimeter.

In order to find the area, we need that missing side length.

Luckily, we know that we can use the perimeter to do this.

I'll start by dividing my perimeter by two.

Now remember, the perimeter of a rectangle is two lots of the length plus two lots of the width, or in other words, it's two lots of the length plus the width, so if I divide the perimeter by two, the result I get, in this case, nine plus two root three is equal to the length of my rectangle plus the width of my rectangle.

Well, this means I just need to subtract seven takeaway root 27 in order to get the remaining side length.

Now, I've had to be very careful here because I am subtracting the total which means I am subtracting seven, and I'm also subtracting negative root 27, so I'm being very careful here to make sure that my signs are correct.

Nine takeaway seven is two and two root three subtract negative root 27 means what I'm actually doing is two root three add root 27.

Or root 27 simplifies to three root three, so I end up with two root three at three root three or five root three.

I now have the remaining side length, so I have the two binomial expressions I require to find the area.

Seven takeaway root 27 and this will be multiplied by two plus five root three.

Expanding those brackets gives me a result of 14 add 35 root three minus two root 27, minus five root 81.

I've then gathered the like terms and the square root of 27 can be simplified, remember, into three root three, and the square root of 81 is just nine.

So once I've tidied this all up, I have a result of negative 31 plus 29 root three.

Now, we did talk about how we could expand more than two binomials and we looked at this in our unit on algebraic manipulation.

So where there are three binomial terms, we handle this in the same way as before.

We find the product of two of the binomials.

It's entirely your choice, remember, as to which you multiply together first.

And then we multiply this product by the remaining binomial expression.

Here's an example.

I'm going to start by multiplying the first two expressions together, so that was the one plus root two and the one plus root three.

Multiplying them together led me to the result of one plus root two, plus root three, plus root six.

I must then multiply this by my remaining binomial expression.

Remember, I'm multiplying every term in my second expression by every term in my first expression, so what I did was I went through and multiplied every term by two and then I multiplied every term by negative root five and you can see what it led to.

Goodness me, it's long.

There are no like terms here, and there are no surds that can be simplified, so that is my final expression.

What I'd like you to do is have a go at expanding this.

What do you notice? I recommend expanding the first two brackets and then multiplying by the third bracket.

See what you notice.

Pause the video now and have a go.

Welcome back.

What did you notice? Did you see that if you expand the first two brackets, the result is an integer product? This means that the final product is significantly easier to calculate.

Let's take a look.

One plus root three multiplied by one takeaway root three, led to a result of one plus root three, subtract root three, subtract root nine.

Well, the root threes will cancel leaving me with one takeaway the square root of nine, or one takeaway three which is negative two.

All I need to do now, therefore, is multiply the remaining binomial expression, two subtract root five by my result giving me an answer of negative four plus two root five.

Wow, that was a lot easier.

Glad that I multiplied those first two expressions together, rather than trying to do them in a different order.

In fact, that was particularly helpful that the two surd terms cancelled out.

I wonder if there's something more to that.

Let's move on and look at our second section, and this is on the difference of two squares.

Now, you've heard that before.

It came up in the algebraic manipulation unit, but let's look at it in more detail here.

What do you notice? What about now? Now, does this make a difference? Pause the video and either discuss or write down what do you notice about these expressions and what happens when I expand the brackets? Did you notice that in all four cases, I end up with an integer result? Let's try expanding these brackets.

I'd like you to pause the video and have a go at this now.

Welcome back.

What happened? Did you get an integer result? You should indeed.

You should have reached 22.

Let's delve into this a bit further.

What about this one? Do you think you'll get an integer result again? Pause the video, expand the brackets, and see what you get.

Welcome back.

Did you get an integer result? You should have done.

It should be 19.

Can you see what's happening here? Let's try generalising.

What happens when we expand these brackets? Pause the video and have a go now.

Welcome back.

You should have got to A subtract B squared.

If in fact we'd had A plus B and A subtract B, we would've reached a result of A squared minus B squared which is the result that we saw in our algebraic manipulation unit when we were doing the difference of two squares.

What we're showing here is that that result still works here only if there's a surd term, as long as it's repeated in both brackets, and we have that important change of sign, we will still end up with a difference of two squares formula.

But because we had surd terms rather than seeing the term squared, a surd term squared just leaves us with the radical, and we know that from when we were squaring surds.

What about this particular expression? If I expand these brackets, I'll get an integer answer.

If you think that's true, then tell me what the integer answer will be.

And if you think it's false, tell me what result I'll get instead.

Pause the video and do this now.

Welcome back.

Which one did you go for? It's false.

We can see that what we're actually doing here is squaring the expression, root three plus four.

We don't have a difference of two squares here at all and when we in fact expand these brackets, we'll get 19 plus eight root three.

It's now time for our final task.

I'd like you please to match the binomials, the surd on the left to the difference of two squares on the right.

Now, there will be two blanks.

You'll discover when you try to expand one set of brackets, you won't see the result you need on the right-hand side.

When that happens, you'll need to write down what result you were looking for.

Likewise, you're going to reach the point where you have a result, and you don't have the binomials to go with it, so you'll need to write those yourself.

Pause the video now while you complete this matching task.

Welcome back.

Let's see how you got on.

Starting with the top one.

Well, we know that's going to give us 16 because four times four and then we're gonna be taking away five because root five times root five gives us five as a result.

We have a difference of two squares.

We're gonna see we have a negative and a positive.

This is going to give us a negative product, hence the negative five.

So I'm looking for the expression 16 takeaway five, and there it is.

I'm now gonna skip the second one because at the moment, we wouldn't know what would go there.

And I'm gonna move down and look at the next one.

Two root three plus three and then three subtract two root three, or three times three is nine.

so I'm looking for something that starts with a nine.

Well, we can tell it's gonna be the bottom one, but let's make sure.

Two root three multiplied by two root three is going to give me four times three or 12, and I know it needs to be negative, so I'm looking for nine subtract 12, and that's my bottom one.

Now, either my blank goes to my blank.

Maybe.

Or I'm going to be getting a result here from expanding root six minus root five by root six plus root five, I'll get the 11 minus nine at the top.

Well, let's see.

Root six multiplied by root six is going to be six.

And root five multiplied by root five will be five, and I can tell that term will be negative.

So I need six takeaway five, and that's not what I've got left.

I've got 11 takeaway nine left.

So the six takeaway five must be the missing difference of two squares, so I'm going to fill that into the box on the right.

This means that I need to fill in on the left two binomial expressions that will multiply to give me 11 subtract nine.

Now, nine is a square number, so I know that three must appear in my brackets and because that term is the negative one, I know I have a positive three and a negative three.

Now, what do I multiply by itself to get 11? And that must be root 11, so I know that my expressions will be root 11 plus three and root 11 minus three.

It's now time to summarise what we've done in our lesson today.

Writing the product of two or more binomial expressions has not changed now that surds are involved.

What we mean by this is it works just like it did before.

When multiplying brackets that come from the difference of two squares, the answer will be rational, and that's what we saw in task two.

It is important still to simplify wherever we can.

That hasn't changed.

In fact, today's lesson was really us supplying our knowledge of surds to our knowledge of expanding brackets that we've built up over the years.

Well done.

You've worked really well today, and I know you've put in a lot of effort.

I look forward to seeing you for your next lesson.