Lesson video

Well done for making the great decision to learn using this video today.

My name is Ms. Davies and I'm going to be helping you as we work our way through all this exciting algebra.

There's lots to get done.

There's also lots of opportunities to learn new things and to check that we're happy with things that we may have seen before.

Take your time.

It's really, really important with all of our algebra skills that we're writing our working out.

So you'll want to make sure that you have something to write on and you want to make sure you're taking the time to pause the video and check things as you go through to help you spot any mistakes that you might have made.

Thanks again for joining us.

Let's get started.

Welcome.

Today's lesson, we are going to look at more complex binomial products.

We're going to explore all sorts of binomials today and some of the different things that can happen when we find the product of two binomials.

By the end of this lesson, you'll be really proud of yourself 'cause you'll be able to find more complex binomial products.

What I would say with that in mind that you need to be confident with finding the product of two binomials.

So if you haven't already looked at a lesson on that, I recommend you do that first and then come back to this one.

I'm going to talk lots today about binomials, so make sure you're happy with what they are, and I'm going to use this term partial product quite a lot as well.

So we're splitting this into two parts and we're going to start by using equivalent binomials.

Sounds fancy.

Let's see what happens.

So sometimes terms within the binomials are not in the same order.

Often you might have come across things where it's like x plus four and x plus three, but the x is at the beginning and the constant's at the end.

We're going to do something slightly different.

What about if we wanted the product of x plus three, seven plus x? Well, of course, we can rearrange the terms so that the like terms are in the same order.

Well, seven plus x is just the same as x plus seven, so x plus three, seven plus x can be written as x plus three, x plus seven.

Well, that makes this easier.

We've got x squared, 7x, 3x and 21.

Our final product, x squared plus 10x plus 21.

Well, that's all right.

What about if we've got some negative terms? How could we write three minus x, x minus seven so the like terms are in the same order? Well, let's think about three minus x.

That's the same as three add negative x or negative x plus three.

So we've rewritten that so that the xs are at the beginning on both binomials and the constant terms are second in both binomials.

Let's see what happens when we multiply this one? Right, pay attention now 'cause we've got a negative x and a positive x and negative x multiplied by positive x is negative x squared.

Negative x multiplied by negative seven is positive 7x, 3x and negative 21.

Our final expression is negative x squared plus 10x minus 21.

And you might not have come across using a negative x squared before.

That's absolutely fine.

Right, let's see if you can do this first step.

So which of these binomials are equivalent to one minus x? Off you go.

Of course, it is negative x plus one.

One minus x is the same as one add negative x, so you can write those terms the other way around.

Right, some pupils are going to give this a go.

Let's see how well they do.

They want to find the product of x minus five, two minus x.

Lucas' idea is to rewrite this as x minus five, negative x plus two.

Jacob's going to rewrite it as negative five plus x, two minus x.

Jun is doing something different.

Jun says I'm going to find the product without rearranging.

Let's see how that goes.

Do you think all of these methods are going to produce the same answer? Okay, I think they are going to produce the same answer as long as we do it correctly.

Which method do you prefer? Let's have a look then.

So Lucas' method, rewriting so the x terms are first.

Then he's going to do x lots of negative x plus two and add negative five lots of negative x plus two.

Let's try it.

Negative x squared plus 2x plus 5x, negative 10.

Negative x squared plus 7x minus 10.

Some people like having the x terms first 'cause they always like starting with the x squared terms. That's absolutely fine.

Let's try Jacob's method.

So negative five multiplied by two minus x and x multiplied by two minus x, and then we'll add all four partial products together.

Negative 10, positive 5x, 2x and negative x squared.

Exactly the same as what Lucas had, just the terms are in a different order.

If you want to rewrite that, so the x squared terms are first, that's absolutely fine.

Right, we're a bit nervous about Jun doing this without rearranging.

Let's see.

So we've got to do x multiplied by two minus x and add negative five multiplied by two minus x.

So x multiplied by two, 2x, x multiplied by negative x, negative x squared, negative five multiplied by two, negative 10 and negative five multiplied by negative x, 5x.

There's our four partial products and collecting like terms. Fantastic.

Well, Jun is obviously confident enough with his multiplication of terms to be able to do it in that order.

That is absolutely fine.

You've just got to make sure you don't miss any partial products and that you're confident then writing as a simplified expression.

There you go, he's really proud of himself then that he's managed to get the same answer as the other two who did it in a different way.

So just like Jun, you do not have to rearrange before multiplying.

What we did say is it can help you spot patterns and identify mistakes.

The terms in the final answer can be written in any order, but it is common to see the term with the largest exponent first.

So it's often that you'll see x squared terms, then x terms, then constants.

So the product of two binomials in the form a plus b, a minus b can be written as a squared minus b squared.

If you haven't seen this before, we're just going to explore this for a second.

The reason that we end up with two terms in our final answer is because the other two partial products are a zero pair.

We call this form the difference of two squares.

You'll see that we've got a squared and b squared and a subtraction.

So we're finding the difference between the two.

Let's try this with an example.

X plus 7x minus seven would be x squared minus 7x plus 7x minus 49.

And because negative 7x and 7x are a zero pair, we get left with x squared minus 49.

You could also write that as x squared minus seven squared.

What do you think seven plus x, seven minus x will be in expanded form? See if you can see the link between that and our previous question.

Hopefully you didn't fall into the trap of thinking they were the same thing 'cause they are different.

Let's have a look.

We can write that as seven lots of seven minus x plus x lots of seven minus x.

There's our four partial products and we end up with 49 minus x squared.

Notice that we've got negative x squareds this time rather than last time we had positive x squareds.

Lucas says I think x plus seven, seven minus x is the difference of two squares.

Oh, Jacob reckons x minus seven, seven minus x is the difference of two squares.

And Jun reckons that seven plus x, x minus seven is the difference of two squares.

Take time to read those three, see how they're different.

Who do you agree with? Maybe they're all incorrect.

Now what's going to help us is rearranging the terms. So x plus seven, seven minus x, which was Lucas' idea, could be written as x plus seven, negative x plus seven or seven plus x, seven minus x.

I've tried it with the x terms first and then the constants first.

Now, that second form helps us here 'cause we can see that the seven is the same in both brackets and the plus x and the negative x are a zero pair.

This is going to leave us with two partial products that are zero pairs and therefore, seven squared minus x squared or 49 minus x squared will be our final answer.

Lucas was correct, we have got the difference of two squares.

Let's try Jacob's.

If we rearranged to put the x terms first, we've got x minus seven, negative x plus seven.

Putting the constant terms first, we've got negative seven plus x and seven minus x but not the difference of two squares.

Both terms are zero pairs.

Remember, we need one term in each bracket to be identical and the other term to make a zero pair.

If you want to see what that would actually look like, you get negative 49 plus 14x minus x squared.

Let's look at Jun's.

So seven plus x, x minus seven can be written as x plus seven, x minus seven.

Oh, we can see straight away then we've got the difference of two squares, x squared minus seven squared.

One term is the same in both binomials.

The other term makes a zero pair.

So which of these is equivalent to x minus four, four minus x? Off you go.

Well done.

The top one is we've written the constant term first and the bottom one, we've written the x term first.

Right, I'd like you to try and expand this now.

You could use the area model to help if you wish or you can do it without.

However, you need to make sure you don't miss any partial products.

There's nothing wrong with using that area model.

Let's have a look.

I'm going to use the area model 'cause I don't want to make a mistake.

So x multiplied by negative x, negative x squared.

X multiplied by four is 4x.

Negative four multiplied by negative x is 4x.

Negative four multiplied by four is negative 16.

Collecting like terms. Negative x squared plus 8x minus 16.

Time for our practise then.

I've promised you some complex problems. I'm hoping towards the end of this, you'll find some that really make you think.

So I'd like you to expand and simplify these expressions.

It's up to you whether you want to use the area model, but please make sure you're finding all the partial products before trying to simplify.

Off you go.

Fantastic.

This time, I'd like to know which of these expressions can be written as the difference of two squares.

And then for question three, can you match those with the expressions in the box? So start by identifying which can be written as the difference of two squares and then see if you can match any of them with those four expressions in the purple box.

Off you go, come back when you're ready for the answers.

Lovely.

So the first one, negative y squared plus 3y plus 28.

Terms can be in any order.

B, y squared minus 16y plus 64.

C, negative y squared plus 10y plus 75.

Then we've got four terms for D, negative 10x plus 10y, negative xy plus y squared.

So our difference of two squares is A, D, F and H and they're going to match up.

A was 81 minus x squared.

D, x squared minus 81.

F, y squared minus x squared.

And H, x squared minus y squared.

Well done if you got those all correct.

So now we're going to have a look at when we've got coefficients other than one.

So, so far, we've done lots of multiplying x by x.

What about if we have a 2x or a 3x? Let's see what happens.

So Sofia has been finding the product of two binomials with various expressions.

She's been doing lots of practising , she's feeling quite confident.

She reckons she's seen a pattern in the expanded form.

The coefficient of the x term is the sum of the two constants and the constant term is the product of those two constants.

She's going to show you an example to help you see what she means.

So she's saying in x plus five, x plus four, if you add five and four, you get the coefficient of x.

If you multiply five and four, you get the constant term, which is x squared plus 9x plus 20.

Do you agree with Sofia? Take your time to try some examples.

Lovely, I've tried x plus four, x plus two and x plus eight, x minus one, and as long as I'm careful with my positive numbers and my negative numbers, Sofia does seem to be correct.

Multiplying the two constant terms definitely gives us one of the partial products.

And the other two partial products, if they're like terms, you can simplify them by summing the coefficients.

So that does seem to work.

Adding the constants give us the coefficient of x.

Multiplying the constants give us the constant's term.

But did you think of any examples where this doesn't work? Any example whether binomials have different variables that won't work? So if you had x plus four, y plus five, that's not going to work.

Or what about x plus four, x plus y? That's also not going to work.

We've got two constants in the first one.

We haven't got two constants in the second one.

Either way, we're not going to be able to find our three terms like that.

You might have thought of any example where one or more variables is negative, what about x plus four, five minus x or four minus x, five minus x? You might have also thought, a little bit of a hint in the title, that the variables could have coefficients greater than one.

Let's try 2x plus four and 2x plus five.

Using our area model, 2x multiplied by 2x is 4x squared.

Then we've got 10x, 8x and 20.

Our final answer can be written as 4x squared plus 18x plus 20.

So Sofia seems a little bit disappointed.

This doesn't follow her rule.

So there does seem to be some cases where what Sofia has spotted has merit but not all the time.

Now, we've got to give Sofia some credit 'cause spotting patterns is really useful in mathematics.

It allows us to understand the structure of numbers and calculations and Sofia's observation does help us see the structure when we've got binomial products in the form x plus a, x plus b where a and b are constants.

So x plus five, x plus four, x minus six, x plus one but it doesn't work for other binomial products.

What I would recommend is using an area model so that you do not miss any partial products and although it's useful to spot patterns and rules, do not assume that that's going to work for every example.

So let's try 5a plus b, 2a minus 3b.

I promised you some complex binomial products.

I'm going to do it with an area model on the left-hand side and without on the right.

So this is going to be the same as 5a lots of 2a minus 3b, add b lots of 2a minus 3b.

I've chosen to go that way this time.

I could work horizontally as well.

So I've got to do 5a multiplied by 2a, which is 10a squared.

5a multiplied by negative 3b, which is negative 15ab.

So, so far, I've got 10a squared minus 15ab.

Then I need to do b lots of that second bracket.

So 2a multiplied by b is 2ab and b multiplied by negative 3b is negative 3b squared.

So I'm adding 2ab and negative 3b squared.

I do have like terms, I've got 2ab and negative 15ab, so I can write that as 10a squared minus 13ab minus 3b squared.

Fantastic.

So what I'm going to have go at now is I'm going to show you an example on the left-hand side.

I'm going to let you look through it without me narrating.

Then I'd like you to apply the same thing to the right-hand side.

Think carefully as you're watching, what am I doing at every step? Let's go.

Hopefully you agree with my final answer.

I'd like you now to have a go at expanding and simplifying 4x minus eight multiplied by 2x minus three.

Either use an area model or set it out in a similar way to I did.

Off you go.

So we could split that into 4x lots of 2x minus three minus eight lots of 2x minus three or you could say that as add negative eight lots of 2x minus three.

4x multiplied by 2x is 8x squared.

4x multiplied by negative three is negative 12x.

Negative eight multiplied by 2x is negative 16x or we can think about subtracting 16x.

And negative eight multiplied by negative three is positive 24 or we can think of that as subtracting negative 24.

In total, 8x squared minus 28x plus 24.

Well done.

I hope you agree now that we found some complex binomial products and we've shown that the methods that we have used so far can help us with those in exactly the same way.

Let's see if we can do it.

0.

5x minus 4y multiplied by 0.

5x plus 4y.

0.

5 multiplied by 0.

5 is 0.

25.

So I've got 0.

25x squared.

0.

5 multiplied by four is two.

So I've got 2xy.

Negative 4y multiplied by 0.

5x is negative 2xy and negative 4y multiplied by 4y is negative 16y squared.

Collecting like terms then, we have 0.

25x squared minus 16y squared.

What did you notice about our answer? I bet you managed to spot this before we even started multiplying our two binomials.

Of course, it's the difference of two squares.

Both binomials have a term of 0.

5x and the other terms are negative 4y and 4y, which are a zero pair.

So we could have calculated our answer using this observation 'cause we know that would be 0.

5x squared, subtract 4y all squared, which is 0.

25x squared minus 16y squared.

So you can show me now how confident you've become with this.

I'd like you to fill in the area model and the simplified expression for the product of 3x minus 2y, 3x plus 2y.

Off you go.

Lovely.

So we need 3x and negative 2y on our other side of our rectangle and then we can fill in our four partial products.

As expected, we've got a zero pair, so we end up with 9x squared minus 4y squared.

Superb work.

Time to put that into practise.

I would like you to expand and simplify each product of two binomials.

There is nothing wrong with drawing an area model.

When you've done that, I want to draw your attention to a and f.

What do you notice about your answers? Can you explain why this happens? Come back when you're ready to look at the next section.

Lovely.

I've got some more for you to have a go at this time to expand and simplify each product of two binomials.

Use those area models to help you.

Think about rearranging the terms in your brackets if necessary as well to help you.

This time, I want you to think about c and e.

Can you explain what is happening with those two questions? Off you go.

Lovely, so Izzy wants to give this a go.

She wants to find the expanded form of 2x minus 8y all squared.

She's drawn an area model to help her, which is fantastic, but can you spot and correct any mistakes she has made? For question six, can you write two binomials which will have the product that Izzy has written? Special challenge.

Do you think you can find more than one way? Off you go, come back when you're ready for the answers.

Superb.

Pause the video and check that you are happy with all six of those.

And then we're going to have a look at A and F in more detail.

So what you should notice about A and F is that the final answer has two terms. For F, it's quite clear that it's the difference of two squares.

Both binomials have the term of 3x and the other terms are negative six and positive six.

So that's going to be the difference of two squares as we've seen it in lots of examples.

A was a little bit more interesting.

The first binomial, 2x minus six, can actually be written as two lots of x minus three.

You can spot a common factor in 2x and minus six.

So instead, if you write it as two lots of x minus three, what we end up with is two multiplied by x minus three multiplied by x plus three.

Well, the x minus three, x plus three gives the difference of two squares and then you can multiply that all by two.

Your final answer is not the difference of two squares, but it does only have two terms. It's two lots of x squared minus nine or 2x squared minus 18.

Quite subtle that one.

So well done if you figured out where that was coming from.

For three, again, have a look at all of those answers, pause the video to make sure you're happy with the working.

And then well done if you spotted that C and E were the same.

This is because 2x minus 5y, which is the second binomial for C, is the same as negative 5y plus 2x.

So it's actually exactly the same calculation.

Let's see if we can spot Izzy's mistake.

So first of all, she forgot to write the negative 8y.

2x minus 8y all squared could be written as two separate brackets, 2x minus 8y multiplied by 2x minus 8y.

In our area model, she's missed off the negative 8y.

Then we end up with two partial products of negative 8y, no longer a zero pair.

So you should end up with 4x squared minus 16y plus 64y squared, not the difference of two squares.

However, we can find examples where they are the difference of two squares.

So you could have 2x minus 8y, 2x plus 8y and of course, you can move the terms around within the binomials.

So you could have had negative 8y plus 2x and 2x plus 8y.

That's absolutely fine.

Now, the second bit is a little bit interesting and don't worry if you didn't find this example, it's not something we've looked at in detail, but if you want to change yourself, you might want to give this a go.

If I go for negative 2x in both brackets and then negative 8y in one, but positive 8y in the other, this is going to be the difference of two squares again because I've got the same term at the beginning of both brackets and the other terms are a zero pair.

So 2x minus 8y and 2x plus 8y are going to give the same product of negative 2x minus 8y, negative 2x plus 8y, and that's quite a subtle point and quite hard to see why.

You might want to explore that idea if you really want to test your understanding.

Well done today.

Hopefully you agree that I did find some more complex binomial products, but maybe you found that some of them, once you've done a bit of rearranging, weren't actually as complex as they looked.

We have seen how you can rearrange the terms within the binomials so they can have different orders and that sometimes that can be useful, but it's not necessary.

We've also looked at lots of examples now where the coefficients of the variables are not one, and maybe that's a little bit trickier 'cause we can't always see the patterns quite as easy that are going to form.

The important thing is we've got that method of using an area model to support us with our work.

Thank you for joining me with that complex binomial product lesson today and I look forward to seeing you again in the future.