Lesson video

Hi, I'm Mrs. Wheelhouse, and welcome to today's lesson on the product of three binomials.

This lesson is part of our unit on algebraic manipulation.

Now algebra can be incredibly useful, so let's get started and see how we're going to be using it today.

By the end of today's lesson, you'll be able to use the distributive law to find the product of three binomials.

Now, one of the words we're gonna be using today, and we're gonna be using it a lot, is the word binomial.

A binomial is an algebraic expression representing the sum or difference of exactly two unlike terms. Now there's some examples on the screen, so feel free to pause if you want a moment to read through them.

Another key phrase we're gonna be using today is the distributive law.

And the distributive law says that multiplying a sum is the same as multiplying each addend and summing the result.

And again, feel free to pause if you want a moment to read through these.

Our lesson today has two parts to it, and we're gonna begin with part one on the product of three expressions.

We could apply the distributive law to find the product of two binomials.

So for example, multiplying X plus 7 with X subtract 8.

We can now expand and simplify by collecting any like terms. So we end up with X squared subtract 8X plus 7X subtract 56.

And as we know the 2X terms are like terms so we can gather them and write this more simply as X squared subtract X subtract 56.

Because two of the partial products were like terms, our expanded form now has three terms in it, and we call this a trinomial.

How could we write 2X lots of X minus 5 multiplied by X minus 9 in expanded form? Alex suggests that we could multiply the two binomials and then multiply the result by 2X.

And Lucas suggests that we could multiply 2X by X minus 5, and then multiply that result by X minus 9.

What do you think? Well, we read this expression as 2X multiplied by X minus 5 multiplied by X minus 9.

And since multiplication is associative, we can multiply any two of these expressions first, and then multiply the product by the third.

Sometimes multiplying in one order can be easier than another, however the final product will always be the same.

So let's consider Alex's approach.

We'll multiply the binomials.

This leads to the result X squared minus 14X plus 45.

We now need to multiply everything by 2X.

This gives us 2X cubed subtract 28X squared plus 90X.

But what about Lucas's approach? Will this be any easier? We're going to multiply 2X and X minus 5 together first.

This gives us 2X squared subtract 10X.

We now have our two binomial expressions, and we need to expand the brackets here.

So we end up with 2X squared lots of X minus 9, subtract 10X lots of X minus 9, and this simplifies down to 2X cubed minus 28X squared plus 90X.

And we can see that we got the exact same result as Alex did.

Now Lucas points out that he does like Alex's method because the original binomials had a coefficient of one, and they were very easy to multiply together first.

Alex points out that Lucas's method was good because we reduced the product of three expressions, the product of two binomials, in just one step.

So it's very familiar to us.

Now the most efficient method will depend on the nature of the three expressions we are multiplying.

Let's do a quick check.

I'll go first, and then it's your turn.

I'm going to write 5X multiplied by 2X subtract 5 multiplied by 2X plus 5 in expanded form.

I'm gonna choose to expand the brackets first.

Doing that gives me 4X squared minus 25.

Ah, did you spot maybe why I chose to do that first? I then multiply 3 by 5X, and I get to 20X cubed minus 125X.

Did you notice the product of the two binomials was the difference of two squares? In this case it was a lot quicker to expand those first because it left me with only a two-term expression.

It's now your turn.

Please write negative 4X multiplied by 3X subtract 1 multiplied by 3X plus 1 in expanded form.

Pause and do this now.

Welcome back.

Let's see how you got on.

Did you also spot that you had here a difference of two squares? So expanding those brackets first was definitely the quickest thing to do, and you should have reached the point where you have negative 4X, and then inside the brackets 9X squared subtract 1.

Multiplying three by the negative 4X leads to the result negative 36X cubed plus 4X.

Well done if you got this.

What about this situation? We're going to write X squared multiplied by 2X subtract 5 multiplied by 3X subtract 4 in expanded form.

This time round I'm going to opt to multiply the first bracket by the X squared.

This gives me 2X cubed subtract 5X squared, and I'm to now multiply this by 3X subtract 4.

Doing that and then collecting the like terms leads me to the result of 6X to the power of 4, subtract 23X cubed plus 20X squared.

It's now your turn.

Please write X squared multiplied by 3 minus 2X multiplied by 5 minus 2X in expanded form.

Pause and do this now.

Welcome back.

So what did you do first? I opted to multiply the first bracket by X squared, but remember, you could have taken a different approach.

As long as you get to the same answer, it's not a problem.

If you did do it the way I did, then the first bracket becomes 3X squared minus 2X cubed, and we now need to multiply this expression by 5 subtract 2X.

So expanding the brackets and then collecting the like terms leads to the final result of 15X squared minus 16X cubed plus 4X the power of 4.

It's now time for your first task.

For question one, write each of these products in expanded form, and you should simplify where possible.

Now, it is up to you the order you choose to multiply in, so do take time to reflect and look at the question to decide which approach you think will be best for that question.

Pause and do this now.

Welcome back.

Question two.

Sam is expanding negative 3X lots of 5 minus 2X multiplied by 2X plus 5, and Sam thinks that multiplying the binomials first will be the most efficient method.

For part A, please explain why Sam might be correct, and then in B, expand and simplify the expression.

For question three, multiplying the binomials first may not be the most efficient method for the example you see.

Can you please explain why? Please pause and do this now.

Welcome back.

Let's go through some answers.

So question one, you had to write each of these products in expanded form and simplify where possible.

For 1A, you should have reached the final answer of 5X squared minus 5X minus 100.

In B, the final result was 8X cubed minus 20X squared plus 8X.

For 1C, final result was negative 30X cubed minus 62X squared minus 20X.

In 1D, negative 2X cubed minus 24X squared plus 1/3X plus 4.

1E, 16X squared minus 36Y squared.

And then 1F, 27X squared minus 12X the power of 4.

Now if you didn't get any of these right, you're very welcome to pause the video now so you can compare what you wrote down for your working to what I have on the screen.

See if you can spot any errors that might have been made.

Question two, we had to explain why Sam might be correct, where Sam had the idea that multiplying the binomials first might be the most efficient.

Well if we consider the two expressions, we can see that what we have here is the difference of two squares.

And we can make that really clear just by rewriting the second bracket.

This means there'll only be two terms to multiply by negative 3X.

Does seem like it'd be a faster approach.

And then in part B, I asked you to expand and simplify.

Now you didn't have to deal with the binomial first, but we did suggest it might be more efficient.

If you did that, you would've reached the line negative 3X lots of 25 minus 4X squared.

Multiplying 3 by negative 3X gives us the result of negative 75X plus 12X cubed.

Question three, multiplying the binomials first may not be the most efficient method for this example.

I asked you to explain why that might be the case.

Well, if we consider 1/2X minus 5/2, I can see that multiplying through by 2X will leave all terms with integer coefficients.

Multiplying the binomials should then be easier because we don't need to rely on our fraction skills.

Now remember, it wouldn't be a problem if we did, but we do want things to be as easy as possible.

It's now time for the final part of our lesson today, and that's on the product of three binomials.

We can find the product of three binomials using the same method.

For example, what if I wanted to write an expanded form the following: X plus 4 multiplied by X minus 7 multiplied by X plus 2.

Sam says, "Well, can I multiply the X plus 4 and the X plus 2 first?" Izzy he points out, "Of course you can if you think that's easiest.

Then you multiply the product by X minus 7." It does not matter, remember, which two binomials you multiply first.

So sometimes it will be useful to look for features of the binomials, which may make one method much easier than another.

You could of course rearrange the order of binomials if you wish, so that the two that you're multiplying together are next to each other.

I now have a binomial multiplied by a trinomial.

And I can use the distributive law of an area model to see how that works.

So I've got my area model here, and you can see that I've carefully written down the length of X minus 7 and the length of X squared plus 6X plus 8.

I have X lots of X squared plus 6X plus 8, and you can see where that's represented in the area model.

And then I have negative seven lots of X squared plus 6X plus 8, and you can see where that is in the area model.

Now I can use my area model to calculate the partial products, or I can just use distributive law.

Use whichever one you feel is best.

Izzy points out we're going to have six partial products here, so we need to be careful to make sure we get them all correct.

Sam suggesting there'll be like terms though.

Well, he thinks there will be.

And therefore we'll be doing some collecting.

Let's see what we get.

Our X multiplied by X squared is X cubed, 6X multiplied by X is 6X squared, and eight multiplied by X is 8X.

Negative 7 multiplied by X squared is negative 7X squared, negative 7 multiplied by 6X is negative 42X, and negative 7 multiplied by 8 is negative 56.

When we write down all of those partial products, we then look to see if any like terms are there so that they can be collected and simplified.

We can see that we have negative 7X squared and positive 6X squared.

That can be simplified into just negative X squared.

I've got negative 42X and 8X, and that simplifies to negative 34X, and then to the negative 56 on the end.

So we can write this as X cubed minus X squared minus 34X minus 56.

Make sure you leave yourself plenty of space so that you've got clear working here.

Izzy says, "Let's check we get the same thing when we multiply in another order." So now we're gonna start with the X plus 4 and the X minus 7.

Expanding the brackets there leads us to X squared minus 3X minus 28.

And we need to multiply that by X plus 2.

Again, you can use either the area model or the distributive law.

Both will lead you to the same answer.

You can see in the area model, I've written in my partial products.

I now list them.

I look for any like terms. Collecting the like terms will leave the result of X cubed minus X squared minus 34X minus 56, which is the same result we got before.

Let's do some checking.

I'll go first, and then it's your turn.

I'm going to multiply 5X plus 3 by the square of X minus 4.

The first thing I'm going to do is say X minus 4 all squared.

That means X minus 4 times X minus 4.

And now I can see my three binomials.

I'm going to expand the last two binomial terms, and this leads me to X squared minus 8X plus 16, and it's this I'm going to multiply by 5X plus 3.

Doing that and then collecting the like terms leads to the final result of 5X cubed minus 37X squared plus 56X plus 48.

It's now your turn, expand and simplify, X plus 3 all squared by 2X minus 9.

Pause and do this now.

Welcome back.

Let's see how you got on.

Did you start by writing out X plus 3 twice? Well done if you did, because remember, X plus 3 all squared is X plus 3 times X plus 3.

If you expand the first two brackets and simplify, you reach X squared plus 6X plus 9, and you need to multiply this result by 2X minus 9.

Doing that and then collecting the like terms will leave you with the expression 2X cubed plus 3X squared minus 36X minus 81.

Well done if you got that.

Let's try this one.

Now, I'm going to look very carefully at these three binomial expressions.

And you may notice that the first thing I did was write the first expression slightly differently.

I swapped the two terms around.

I'm then going to rearrange so that the third binomial expression and the second binomial expression have changed places.

Why do you think I might have done this? That's right, I can see the difference of two squares here, and I know that expanding those brackets first is going to make things a lot easier.

So that leads me with 1 subtract 4X squared and multiplying all of that by 3X minus 2.

Expanding those final brackets and then collecting the like terms leads to negative 12X cubed plus 8X squared plus 3X minus 2.

So by spotting the difference of two squares, I gave myself a lot less work to do.

It's now your turn.

Please expand and simplify the following expression.

Pause and do this now.

Welcome back.

Did you spot where the difference of two squares was? That's right, it was the first and the last binomial expressions.

So I'm going to choose to expand those two brackets first.

This leads me with a result of 25 minus 9X squared, and it's this I'm going to multiply by 2 plus 3X.

So using my distributive law, and don't forget, you could draw an area model if you prefer, and then collecting the like terms, I reached the result of negative 27X cubed minus 18X squared plus 75X plus 50.

Well done If you got this right.

We can use the fact that the volume of a cuboid is given by the formula V is equal to L times W times H, where we have the length, the width, and the height of our cuboid to calc the expressions for the volume.

Here is a cube with an edge length of 1/2X minus 2.

The volume of the cube can therefore be written as 1/2X minus 2 or cubed.

What would your first line of working be if we wanted to write this in expanded form? Well, the first thing I would do was write out the brackets three times to show that I'm doing 1/2X minus 2 multiplied by 1/2X minus 2 multiplied by 1/2X minus 2.

Keeping an expression means multiplying the expression by itself and then multiplying by itself again.

So once I've done that, I can now expand and simplify as I did before.

As you can see, expanding two of the brackets has led to expression 1/4X squared minus 2X plus 4, but I still need to multiply that by 1/2X minus 2.

Doing that and then collecting the like terms leads to a final result of an 1/8X cubed minus 3/2X squared plus 6X minus 8.

Now writing our working down the page with clear steps helps us find arithmetic mistakes easily and keep track of our method, especially in questions like these where things have got quite long.

Time for a quick check.

Is it true or false that X plus 5 all cubed is equivalent to X cubed plus 5 cubed? Pause the video and select if you think that's true or false.

Welcome back.

Well, you should have said false, but you need to justify your answer now.

So pick either A or B, please, to justify why it's false.

Pause and do this now.

Welcome back.

You should have said it's part A.

Remember, to write X plus 5 all cubed, The first line of working we should do is to remember that cubed means we're multiplying by itself and then by itself again.

So we should have written it out just to be clear.

Let's try this one.

When multiplying X plus 3 by X plus 5 and X minus 3, there is an order of multiplying which will be more efficient than other methods.

Is that true or false? Pause and make your choice now.

You should have said it's true.

Did you spot why though? Well, pick the answer that justifies why you said it was true.

Pause and do this now.

Welcome back.

Should have picked B because of course what we have here is the difference of two squares.

It's now time for your final task.

For question one, please expand and simplify the following.

Pause and do this now.

The second part of question one, again, expanded and simplify, please.

Pause and do this now.

And then finally, parts G and H for question one, please.

Pause and do these now.

Welcome back.

Let's look at question two.

Izzy's looking back through her work and sees this identity.

How can she tell she's made a mistake? And then part B, which would be the most efficient binomials is to multiply first, and then part C, find the correct expansion.

Pause and do this now.

Question three.

Sam knows that the volume of a cuboid can be found by multiplying the length by the width by the height.

In part A, please write an expression for the volume of the cuboid.

In part B, we have a cube with a different edge length.

Please write an expression for the volume of the cube.

And then C, you can see here a solid, which is made from the larger cuboid subtract the cube.

Please show the volume of this solid can be written as follows.

Pause and do this now.

Welcome back.

Let's go through our answers.

For question 1A, you should reach the final form of X cubed plus 6X squared plus 11X plus 6.

In 1B, you should reach X cubed minus 3X squared minus 34X plus 120.

And then 1C should be 6X cubed plus 5X squared minus 13X minus 12.

For 1D, should be 125X cubed minus 150X squared plus 60X minus 8.

1E, 8X cubed plus 36X squared minus 162X minus 729.

1F, 4XY squared minus 16X cubed plus 2Y squared minus 8X cubed.

And then 1G, 8X cubed plus 22X squared plus 7X minus 10, and 1H, 7.

5Y cubed minus 38.

5Y squared plus 40.

2Y plus 13.

6.

Now remember, if you didn't get these answers, do feel free to go back to the video, pause on the screen with the answers so you can check your working against the working you see on the screen.

In question two, how could Izzy tell that her work was not correct? Well, if she considers the constant term, she'll have negative 3 times 3 times 3, which would be negative 27, and she's got negative 18, so she's definitely gone wrong somewhere.

Now, the most efficient binomials to multiply together first would be the 5X minus 3 and the 3 plus 5X.

Now why would it be those? Well, that's because they can be written as the difference of two squares.

Now of course, she could also have done 3 plus 5X and 3 minus 5X because these can also be written as the difference of two squares.

Part C, you had to find the correct expansion, and you should have got to negative 125X cubed plus 75X squared plus 45X minus 27.

Question three, you had to write an expression for the volume of the cuboid.

Now you didn't have to expand the brackets here, so you could have written it as 3X multiplied by X plus 5 multiplied by 5X minus 5.

In expanded form you have 15X cubed plus 60X squared minus 75X.

And in fact, expanded form wouldn't have been a bad idea because actually in part C, you're going to need this.

Part B, you had to do the expression for the volume of the cube.

And again, you didn't have to write in expanded form, although you will find it useful for part C, and you should have reached therefore either 3X minus 4 all cubed, or you could have written out 3X minus 4 multiplied by 3X minus 4 multiplied by 3X minus 4, or you could have expanded and simplified and written 27X cubed minus 108X squared plus 144X minus 64.

You had to show that the volume of the solid could be written as follows.

Well, what I've done here is I've written my two expressions for the volume of the cuboid and the volume of the cube and shown that we're subtracting.

Of course, if I've done my expanded forms, I'm actually already at the penultimate line of working.

So if you look at the line just before the bottom, we have 15X cubed plus 60X squared minus 75X, and then minus, and this is the expression for the volume of my cube, 27X cubed minus 108X squared plus 144X minus 64.

Remember, we're taking away that entire volume for that cube.

So every term has to be subtracted.

This leads us to the result of 168X squared minus 12X cubed minus 219X plus 64, which was what we were asked to show.

It's important with a show question like this that every stage of our working is made very clear.

We can't just go skipping steps here.

It's now time to sum up what we've looked at today.

The distributive law can be used to find the product of three binomials.

This can be done by finding the product of two of the binomials first.

It does not matter which binomials you multiply first.

However, your choice can reduce the complexity.

Being able to spot binal products, which are the difference of two squares, can make your calculations far more efficient.

Well done, you've worked really well today.

I look forward to seeing you for more lessons on algebraic manipulation.