Lesson video

Well done for taking the first step and choosing to load this video to learn with today.

We've got lots of exciting algebra to have a look at.

If there's things you've been unsure of in the past, I promise that by the end of this video, you are going to feel more confident with those skills.

My name is Ms. Davies and I'm going to be helping you every step of the way as we work our way through.

Make sure you've got everything that you need to make a good start on this lesson and then join me and we'll get ourselves started.

Welcome to today's lesson on the difference of two squares.

If you haven't already looked at the lesson on the product of two binomials, I suggest you do that before continuing with this lesson.

By the end of this lesson, you'll be able to use this special case where the product of two binomials is the difference of two squares.

If you're not confident already with what a binomial is, please take this time to read the definition and make sure you are happy.

I'm going to talk lots today about a partial product.

A partial product refers to any of the multiplication results that lead up to the overall multiplication.

So we're going to start with a bit of a reminder about multiplying binomials.

We can use an area model to help us find the product of two binomials.

We're going to try it now with x plus four and x plus six.

It does not matter which binomial is written on which length and then we can calculate our partial products.

X multiplied by x is x squared, x multiplied by six is six x four multiplied by x is four x and four multiplied by six is 24.

Our answer then is x squared plus six x plus four x plus 24.

So can we write this product in a simpler way and how do you know? Right, it's that final part then, that's going to lead into what we're looking at today.

So we can write this in a simpler way because we've got like terms, the four x and the six x can be added to give a single term that leaves us with x squared plus 10 x plus 24.

There's no other like terms, so we cannot write that in a simpler way.

Our final expression now has three terms. I'm going to refer to them as the x squared term, the x term, and the constant term.

I'm going to give you a couple more to have a look at.

X plus four x minus, six x minus four x plus six and x minus four x minus six.

How many terms are these going to have when expanded and simplified? Give this a go.

Okay, there's our four partial products.

We've got two like terms which leaves us with x squared minus two x minus 24.

Again, we have four partial products.

When simplified, we've got x squared plus two x minus 24.

And this final one, we have x squared minus 10 x plus 24.

Remember negative four multiplied by negative six is positive 24.

So those are the examples we've just looked at and you can see that all of them had three terms when expanded and simplified.

What I want to know, is do you think this will always be the case? Have a think if you've got someone to discuss it with, discuss it and then share your answers with me.

Okay, so what happened, when multiplying two binomials together? Is you've got four partial products, that will always be the case.

In these examples we ended up with three terms because two of the partial products were like terms and therefore could be simplified to a single term.

We're going to see now whether it's possible to end up with less than three terms to answer the question that we stated is not always going to be the case and we're going to have a look at that now.

Let's try this one.

We've got x plus six and x minus six.

Our four partial products are x squared, negative six x, positive six x and negative 36.

What happens when we collect like terms this time? Can you see it? So what you'll see is the two x terms are additive inverses of each other.

That means they are a zero pair, so when you add them they'll sum to zero.

That means our three terms would be x squared plus zero x minus 36.

Well zero x is nothing, so that can be written as x squared minus 36.

So it's possible for the product of two binomials to have two terms. Aisha thinks she spotted something, in that example there is a six in both binomials.

I think x minus five X minus five will also have two terms when simplified, what do you think? What was really great about what Aisha did is she gave a reason to back up her answer.

What we can do now is actually try it out and that's really important in Math, that we make theories and we try things out and we learn just as much for finding things that are incorrect as we do when we find things that are correct.

So let's have a look.

So we've got x squared, negative five x, negative five x and positive 25.

When we collect like terms, we've got x squared minus five x minus five x plus 25, but they are not a zero pair, so we have x squared minus 10 x plus 25 Aisha was incorrect then, it wasn't just the fact that there was a six in both binomials, there must have been something else going on.

What could we do to change this expression, so that two of the partial products will be zero pairs? Can you see it? That's it, we need to change one of the constant terms to be positive.

So one needs to be positive and one needs to be negative.

Let's try x plus five, x minus five.

That gives us x squared, negative five x, positive five x and negative 25.

This time we do have a zero pair which leaves us with x squared minus 25.

Let me give you four examples now, do you think any of these products can be simplified to an expression with two terms, x plus eight x minus eight, x plus eight x plus eight, x minus eight x plus eight or x minus eight x minus eight, what do you reckon? Let's have a look.

We've got x squared minus 64.

Yes, that one did give us two terms. Two of our partial products were a zero pair, x plus eight x plus eight gave us three terms, x minus eight x plus eight gave us x squared minus 64 again and finally x squared minus 16 x plus 64.

It was those two on the left hand side which gave us an expression with two terms. Actually we know that those two things are identical.

Doesn't matter which way round we multiply our two binomials.

Time for you to have a check.

Which of these products of two binomials do you think will be simplified to an expression with two terms? Make sure you read all four carefully.

They might not be in the same order as the previous example.

Off you go.

So we're looking for two partial products to be a zero pair and then that simplifies to an expression with two terms. So we have x plus three x minus three and x minus three x plus three.

Okay, you've got six examples this time.

Which of these do you think could be simplified to an expression with two terms? Off you go.

Well done, so again, we are looking for a zero pair, not just the fact that one bracket needs to be an addition of two terms and one bracket needs to be a subtraction of two terms. Those terms need to be a zero pair.

So the first one we've got x squared minus 16, second one x squared plus two x minus eight.

Third one x squared minus eight x plus 16.

We've got x squared minus four x plus four, x squared minus six x plus eight and x squared minus four.

It was that first one and that last one that gave us expressions with two terms time for you to have a practise.

So for each one I'd like you to fill in the area models for these products of two binomials.

Once you've done that, I'd like you to write your answer as a simplified expression, thinking all the time about what you are noticing, making sure you're reading the question carefully.

Well done, this time you do not need to calculate the products of two binomials.

Just need to tell me which of them will be able to be simplified to an expression with exactly two terms. Off you go.

And finally, a little bit of a puzzle for you.

I'd like you to match a binomial in the left column with a binomial in the middle column and find its product in the end column.

Notice some of the products have two terms, but some of the products have three terms. Off you go.

Let's have a look.

So the top left, you should have x squared minus one, one multiplied by negative one is negative one for B, x squared minus a hundred, for C, I hope this didn't catch you out.

You should have four partial products and three terms, x squared minus four x minus four.

Notice the terms in the brackets are not a zero pair.

For D, 0.

5 multiplied by negative 0.

5 is negative 0.

25.

If you wanted to write that as a half and negative a half, you might have found the multiplication easier.

A half multiplied by half is a quarter, so a half multiplied by negative a half is negative a quarter.

Your final answer is x squared minus naught 0.

25 or x squared minus quarter.

For this one you are looking at D, x minus 13 x plus 13.

F, x plus 11 x minus 11.

G, x minus 2.

4 x plus 2.

4.

Good spot, if you saw that 11 over two is the same as 5.

5.

So you've got positive 5.

5 and a negative 5.

5.

J, four fifths is eight tenths or naught 0.

8.

So you've got negative naught 0.

8 and positive naught 0.

8.

And then for K and L, if you use your fraction conversion skills, you'll see that those two terms are not zero pairs in K or in L.

And finally let's see if you manage to solve this puzzle.

So A matches with P and the product is X, B matches with K and the product is U, C matched with O and the product was V, D matched with N and the product was T, E matched with M and the product was x squared minus four or S, F matched with I, you've got x squared minus five x minus 36 which was W, G matched with J and you've got a product of Q and H matched with L and you've got a product of R.

Well done if you managed to match all of those up.

So now we're going to look at this term difference of two squares so that we can identify.

So Jun says, could I multiply x plus three y plus three using an area model? Well of course he can.

He just wants to find the product as two by binomials.

So let's have a look at our area model, x multiplied by y is xy, x multiplied by three is three x, three multiplied by y is three y and three multiplied by three is nine.

Just like before we've calculated our four partial products, our final answer is xy plus three x plus three y plus nine, does not matter which order we write those four terms in.

I wonder if you spotted that none of these are like terms, I should have spotted it, this time there are four terms in the simplified expansion.

So, so far we've seen that multiplying two binomials gives us four partial products.

We know examples where they can be simplified into three terms 'cause two of the terms are alike and examples where these are a zero pair, so the final product could be written with two terms. However, if none of the terms are alike then there will be four terms in the expanded form.

Here are some examples.

Think about what we've just said, which of these do you think will have four terms in the expanded form and then we're going to try 'em out.

Let's have a look then, for this first one, you might have noticed that this is going to give us two terms because two of our partial products are going to be a zero pair for this second one we've not got any like terms this time, x squared, negative 10x, xy and negative 10y, there's going to be four terms in our answer.

For the third one, again, we haven't got any like terms, xy, negative 10x, 10y and negative a hundred.

Let's have a look at our final two.

So here we've got x times x is x squared, x times y is xy, y times x is also xy.

We write the variables alphabetically.

It helps us find like terms and y times y is y squared.

We have got two like terms, xy at xy can be written as two xy.

You have x squared plus two xy plus y squared.

And finally, really impressed if you spotted this.

We've got x squared negative xy, positive xy and negative y squared.

So this time we've got a zero pair so we end up with two terms. The number of terms then depends on the nature of the like terms. We are going to focus on the cases which had two terms. So Jun I think squaring a binomial will give a product that can be simplified to two terms. Do you agree? Let's try one, let's try x plus four all squared.

We know we can write that as x plus four multiplied by x plus four.

There are four partial products and no that one does not work.

We still have three terms. Okay, let's try another example.

Let's try a plus b all squared.

Well that's the same as a plus b multiplied by a plus b, a multiplied by a is a squared, a multiplied by b is ab, b multiplied by a is ab and b multiplied by b is b squared, ab add ab is two ab.

So no he was incorrect on both accounts this time, Aisha says I think two of the terms must be additive inverses of each other.

She found this out earlier.

Let's see if she's correct.

Let's try x plus four X minus four, we've got x squared negative four x, positive four x and negative 16.

And of course we've got a zero pair, we have x squared minus 16.

Just pause the video.

Do you agree with Aisha's statement? Have we proved that this is going to work? Okay, I'm going to give you a counter example.

Let's try a plus four b minus four.

Two of the terms are additive inverses of each other.

We've got a four and a negative four.

Let's see what's going to happen there.

a multiplied by b is ab, a multiplied by negative four is negative four a.

Four multiplied by b is four b.

Four multiplied by negative four is negative 16.

Well hang on this time we didn't end up with any like terms. So Aisha is almost correct.

One term must be the same in each bracket and the others must be a zero pair.

Fantastic, let's put this to the test.

So here's some of the examples we've seen so far.

What do you notice about the structure? What has happened? Okay, so some of the things you might have thought about, all these examples have an x squared.

Well that's because each binomial has a term of x.

If I started them all with y and y, they'd all start with y squared.

Okay, you might have noticed the second term in the expanded form is always negative.

We've got minus nine, minus a hundred, minus y squared.

You might have also spotted that nine and hundred are square numbers, so something to do with them being squared.

So let's generalise this and explore the structure a bit more.

So let's just say I had a plus b, a minus b.

a and b can be any terms. If we use our area model, we can see that gives us a squared plus ab minus ab minus b squared and we've got a zero pair.

So that can be written as a squared minus b squared.

So this form of expression can be called the difference of two squares, a squared minus b squared will be the difference of two squares, and that's because it's expressing the difference between a squared and b squared.

Remember in mathematics difference means one value subtracted from another.

So which of these can be written as the difference of two squares? Pause the video, see if you've got this.

Let's have a look.

So we're looking for two terms be the same and the other two to be a zero pair x plus three x minus three can be written as x squared minus three squared or x squared minus nine x plus y, x minus y can be written as x squared minus y squared.

It's the difference between the two squares.

And the last one we've got y squared minus and then two x squared.

Making sure we multiply two x by negative two x that'll give us negative four x squared.

It's still useful to use area models so there's nothing wrong by using an area model to make sure no partial products are missed.

However, it's a useful form to be able to spot at this point because we can use that structure with different mathematical skills later on.

Okay, we're going to have a go then at finding the product when it's the difference of two squares.

I'm going to do this one on the left and then you are going to try one.

So a plus b, a minus b, we'll do a multiplied by a, a multiplied by b, negative b multiplied by a, negative b multiplied by b.

We've got a zero pair.

So our final answer is a squared minus b squared.

Have a go at calculating the product of these two binomials.

Lovely, hopefully you saw this is going to give us the difference of two squares.

We end up with x squared minus y squared.

Okay, I'm going to try a slightly trickier one.

Let's do three a plus b multiplied by three a minus b.

I'm going to use my area model to help me.

I've got nine a squared, three ab, negative three ab, negative b squared.

I can write that final answer as nine a squared minus b squared.

Or notice that that's three a all squared minus b squared.

It's the difference between three a squared and b squared.

I'd like you to do the same for the product of these two binomials.

Off you go.

Well done, so there's our four partial products.

Check that you're happy with those.

And our final answer can be written as 16 x squared minus y squared or four x all squared minus y squared.

It's the difference between four x squared and y squared.

Right, let's have a look at the final example.

I've got a little bit trickier.

Let's see if we can apply what we've learned so far.

So using an area model, six a multiplied by six a, 36 a squared, I've got 30 ab, negative 30 ab and negative 25 b squared.

Our final product could be written as 36 a squared minus 25 b squared or six a all squared minus five b all squared.

Showing you the difference between the squares of six a and five b.

Challenge yourself now, can you have a go at this one on the right hand side, give it a go.

Lovely, check your four partial products.

Then we've got nine x squared minus four y squared or three x all squared minus two y all squared.

Well done, times to check that you've really understood the structure behind the difference of two squares.

So which of these products of two binomials can be written as the difference of two squares? You don't need to calculate it, you just have to identify them.

Off you go.

Well done, now it's time for you to have a go at calculating them.

So I'd like you to fill in the area models and then write your final answer as the difference of two squares.

Don't skip that first step even if you think you know what the final product will be.

'Cause we're really making sure we're confident with this structure.

Off you go.

For this one, you'll see that I filled in some of the partial products for you.

For G and H, I actually haven't told you the original binomials, so you are going to need to use the final answer to fill in those binomial.

Give that a go.

Come back when you're ready for the answer.

Well done, so for question one, you should have C, E, G might have thrown you a little bit.

I've got 10 plus x and 10 minus x.

I've got two terms the same and the other two are zero pair.

That's absolutely fine.

You'd get 10 squared minus x squared.

I, absolutely fine.

You're going to have xy all squared minus three squared.

Now J might have thrown you a little bit, remember what we said two of the terms in the binomial.

So one in each binomial has to be the same and the other one is zero pair.

We haven't got that this time.

Both are zero pairs.

You used an area model, you'll see that we end up with three terms, not two terms. We do not get the difference of two squares and our answers for two A, pause the video and check that you've got all your partial products correct as well as the answers.

Particularly looking at C and D, we've got x squared minus four y squared and a squared minus nine b squared.

For the trickier ones we've got E is nine minus x squared, F is four y squared minus nine x squared.

G, you had to work backwards.

So to get negative 16 must have been negative four multiplied by positive four.

There's other things that multiply to give you negative 16, but not that are a zero pair.

Well done if you did this final one, you are looking for five a plus three b and five a minus three b.

That's what would've given you the 25 a squared and then the negative nine b squared.

Fantastic, we're really confident now with this idea of the difference of two squares.

We've played around with the product of two binomials that produce four terms. We've seen that when they can be simplified to three terms and we've looked at these cases where they can be simplified to two terms because two of the partial products are zero pairs.

We've seen how this happens when one term in each binomial is the same and the other two are zero pairs.

We've looked at why this is called the difference of two squares.

Keep your eye out for the difference of two squares then in the future, I look forward to working with you again.