Lesson video

Hi, I'm Mrs. Wheelhouse and welcome today's lesson on checking and securing understanding of the product of two binomials.

This lesson is from our unit on algebraic manipulation.

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

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

Now, some keywords we're going to be using in our lesson today are binomial and partial product.

Now, if these are familiar to you, that's brilliant, but if they're not, you may want to pause the video right now and just have a read through of their definitions.

Now that you're ready, let's get started.

Our lesson today has three parts, and we're going to begin by using algebra tiles.

We can use area models and algebra tiles to find the product of two binomials.

For example, we read the following expression as x + 2 while multiplied by x + 3.

We can use our algebra tiles to help us see what this product will actually be.

So you can see I have here an area model where I've written the two lengths of x + 2 and x + 3 along the respective sides, and then I've multiplied to find all the areas.

I've also placed algebra tiles in there so that you can see what the respective areas will be and what the total area is.

So how are we going to write our final answer? Well, we have x squared, we have 3x, a 2x, and a 6.

So we can write our answer as x squared + 3x + 2x + 6.

But we can see there are like terms there, so we can collect them.

Can you see the like terms in our model? That's right, we have all those x's.

In fact, we have five of them.

So we can in fact write this as x squared + 5x + 6.

By multiplying the binomials, we are writing the expression in a different format.

The new format shows the expression as a sum of terms rather than a product.

We can refer to this as the expanded form.

The process of multiplying the binomials is sometimes referred to as expanding the brackets.

When writing the product of two binomials, it does not matter which way round we write the binomials.

So for example, x + 3 multiplied by x + 1 is exactly the same as x + 1 multiplied by x + 3.

Let's look at expanding these brackets.

So, I have an area model, and I'm going to put x + 3 on one side and x + 1 on the other.

I've also put algebra tiles around the outside just in case I want that visual aid.

So I have x multiplied by x, which is x squared.

I've got 1 multiplied by x, so that's an x.

I have 3 lots of the x tile, and I've got 3 lots of the 1.

So I can see here in total I have x squared + 4 lots of x + 3.

It's now your turn.

Please expand and simplify the following expression.

Pause and do this now.

Welcome back.

So, you might have drawn an area model like I did, and then I've placed the algebra tiles around the outside to help.

I know that here will go the x squared tile 'cause I have x multiplied by x, and then I've got 2 lots of the x tile and then 4 lots of the x tile and then 4 times 2 lots of the ones tile.

So altogether I have x squared + 6x + 8.

Now, there are some key features we can notice when the product of two binomials is in this form, where a and b are constants.

Jun points out that in the example so far where a and b were positive numbers, all the terms in the expanded form were also positive.

Do you think that's always the case? Of course it is, 'cause when we find the product of two binomials, every term must be multiplied by every other term.

This produces four partial products.

And if all terms in the binomials are positive, then all of the partial products will be positive.

So when collecting like terms, we're going to be summing, and the sum of positive values is also positive.

Sophia asks, "So what about when a and b are both negative values?" Well, let's consider this.

I've got a length of x - 2 and a length of x - 3.

Let's do the multiplication, see what happens.

Well, x times x is x squared.

x times -3 is -3 lots of x.

So you can see I've placed 3 of the -x tiles in there.

And then -2 multiplied by x means that's 2 lots of the -x tile, and then -2 times -3 is 6, so I need 6 lots of the ones tile.

So, when I expand this, I get x squared - 5x + 6.

What we saw there was the product of the two negatives was positive, and the two x terms were negative, and the constant and x squared terms were the positive ones.

Aisha says, "I've placed my algebra tiles this way." Hmm, so slightly different to what we saw before but still using the same tiles and still making a rectangle.

Why do you think Aisha might have arranged them this way? So, does this still show the same product? Yes, it does.

I can see my length of x - 2 and I can see my length of x - 3.

But why might Aisha arrange them this way? Well, if we place our algebra tiles on an axis, it will make sense of the negative lengths.

We know that here we have positive y values and positive x values, and here I have negative x values and I have negative y values if I look at the y-axis.

If this was a multiplication grid, where would the products be positive? Well, they'd be positive in these quadrants because a positive value times a positive value gives a positive answer and a negative value times a negative value gives a negative answer.

Where would the products be negative? Exactly, the other two quadrants.

So when Aisha found the product of x - 2 and x - 3, she rearranged the terms in the binomials so that the negative terms were the negative axes.

So you can see where she's placed them.

So we still have a rectangle and it still has the same dimensions.

She's replaced them on the axes to make it a little bit easier to see.

Did you notice that the positive algebra tiles are in the positive quadrants and the negative algebra tiles are in the negative quadrants? Now, it's personal preference whether you want to use Sofia or Aisha's positioning of the algebra tiles, but you may find that considering the positive and negative direction can help you to explore the structure of binomial products even further.

Which of these would be a correct positioning of the binomial product x - 3 and x + 2 on a set of axes? Pause and make your choice now.

Welcome back.

Did you spot it had to be the last one? You need to think about where it's possible for the product to be negative.

In the first one, we've got the -1s, but they're not in the right quadrant, and neither are the x's.

And in the second diagram, our x's are in the wrong quadrant and our -x's are also in the wrong quadrant.

So it's only the final diagram where they're arranged correctly.

So let's use the algebra tiles to write the product of these binomials in expanded form.

I have -3, lots of x but positive 2 lots of x, and I'm taking away 6.

So as simple as possible, that's x squared - x - 6.

We could, of course, draw the diagram this way.

What would x - 2x + 3 look like on the axes? Well, it could look like this.

And we can see that we get the result, x squared + x - 6.

Let's use algebra tiles now to expand and simplify x + 2 multiplied by x - 1.

So I'm going to use my axes here and place the algebra tiles in the correct quadrants.

So there's the x squared, there's the positive 2x, here's my -x, and my -2.

So I can see that when I simplify, I have x squared + x - 2.

It's now your turn.

Please use algebra tiles to expand and simplify x + 1 multiplied by x - 2.

Pause and do this now.

Welcome back.

Did you place your algebra tiles the same way I did? You might have placed them slightly differently but still got to the same answer, which is x squared - x - 2.

It's now time for your first task.

Use algebra tiles to find the products of these binomials, and you can use the axes provided if you wish.

Pause and do this now.

Welcome back.

It's now time for d, e, f.

Same thing as before, pause the video now, please.

Welcome back, it's time for g, h, and i.

Again, same instructions as before.

Pause the video now, please.

So, I've placed the algebra tiles on the axes as follows, being careful to observe where we were positive and where we were negative.

For a, I got to the final answer of x squared + 5x + 4.

In b, I have x squared - 3x - 4.

In c, x squared + 3x - 4.

d, x squared - 2x + 1.

e, x squared - 4x + 4.

And f, x squared - 4.

For g, x squared - 2x - 3.

For h, x squared + 2x - 3.

And then for i, x squared - 4x + 3.

Well done if you got these all right.

It's now time for the second part of the lesson, on using area models.

So it's not always possible to draw area models to scale, but we can still use the model as a representation of what's happening.

For example, x + 7 multiplied by x + 11.

So I've drawn my area model and I've put my lengths on and I've been careful so that each term has its own length.

But the sum of the terms within a bracket goes along one edge of my rectangle.

I can now calculate the four partial products and sum them to get the overall product.

So this is very similar to us doing the algebra tiles, I've just not put those on.

I'm doing it now just from the notation.

So I've got x times x is x squared, 11 times x is 11x, 7 times x is 7x, and 7 times 11 is 77.

Where do you see the like terms in the area model? That's right, on that diagonal.

So I can gather those like terms and write the expanded form as x squared + 18x + 77.

Now, we can use the distributive property to multiply binomials efficiently.

And the area model is there to help us make sure we calculate all four partial products.

So let's consider here.

I'm expanding the brackets for x + 10 and x - 15.

My area model's showing me that I'm calculating x lots of x - 15.

I'm also doing 10 lots of x - 15.

So, I'll have x squared - 15x.

I'll then have 10x - 150.

And I can also see that from my brackets, expanding each of the two brackets will get us to the same four partial products.

I can then collect the like terms. I end up with x squared - 5x - 150.

Let's consider this one.

I'm going to use the area model to help make sure that I calculate all four partial products correctly.

So, I'll start by doing x lots of x - 12, and then I'll move to - 5 lots of x - 12.

x times x is x squared and x times -12 is -12x.

I've then got -5 times x, so -5x, and -5 times -12, which is 60.

I'll then gather the like terms, and I have x squared - 17x + 60.

It's now your turn.

Please expand and simplify x - 4 and x - 11.

Pause and do this now, please.

Welcome back.

Let's see how you got on.

So, when we do x multiplied by x, we get x squared, and x multiplied by -11 is -11x.

We've then got -4 times x is -4x, and -4 times -11 is positive 44.

We then gather the like terms, and we have x squared - 15x + 44.

Well done if you got that right.

Repeated multiplication of the same expression can be represented by an exponent.

So for example, if I'm multiplying x + 21 by itself, I can write that as x + 21 and then square that expression.

When squaring a binomial, it is helpful to write it as two brackets multiplied together, and that's to make sure you correctly identify which values need to be multiplied together so that I can get the four partial products.

So you can see here.

x + 5 while squared is equivalent to x squared + 5x + 5x + 25, meaning that in expanded form, it's x squared + 10x + 25.

Remember, when multiplying two binomials, there are always four partial products.

We've seen examples where these can be simplified to expressions with three terms because two of the terms were alike.

Now, if none of the terms are alike, then there will be four terms in the expanded form.

Let's consider this one.

What happens when we expand and simplify x + 10 and x - 10? Well, I've used my area model to calculate my four partial products, but the result is x squared - 100.

There aren't four terms here and there aren't three.

This time there are only two.

Why might that be the case? And that's because two of the partial products form a zero pair.

Do you think any of these products of two binomials can be simplified to an expression with just two terms? Now, you might have chosen to try and work these all out, but maybe you spotted something that would make this a little bit quicker.

So it's when we had our two x terms make a zero pair that we ended up with an expression with just two terms when expanded.

We can use algebra to generalise this and explore the structure of these special cases.

So if a product of two binomials is in the form a + b multiplied by a - b, where a and b at any terms, we can write it as follows.

And you can see why two of the terms are going to disappear.

It's because they make a zero pair.

This form of expression can be called the difference of two squares.

It expresses the difference between a squared and b squared.

Which of these has a product which can be written as the difference of two squares? So you should have picked these three.

You can see that in them, the first term in each bracket is identical and the second term of each bracket is the same, but the sign of one of them is different.

And you can see what happens when we expand these brackets here.

Now, an area model is, of course, still recommended to make sure that no partial products are missed.

Use area models to expand and simplify each product of two binomials.

Pause and do this now.

Welcome back.

For the first one, we should have reached x squared + 8x + 16.

For the second one, x squared - 16.

The third one is also x squared - 16.

And the fourth one is x squared - 8x + 16.

You may, of course, have spotted that the second and third ones were in fact the same.

All I'd done is just write the brackets around the other way.

I know that doesn't change the product at all, so it's perfectly fine to do this, and of course I'll get the same result.

It's now time for your second task.

For question one, write each of these products of two binomials in expanded form.

And I've given you an area model in case you'd like to use it to check.

Pause and do this now.

Now's the second half of question one.

So again, please write each of these products of two binomials in expanded form, and you are very welcome to use the area model if you wish.

Pause and do this now For question two, which of these products of two binomials can be written as the difference of two squares? For the ones that can be, please write them in the difference of two squares form.

Pause and do this now.

Time to go through the answers.

Question 1a, you should get x squared + 12x + 32.

for b, x squared + 27x + 140.

c, x squared + 6x - 27.

And d, x squared - 13x - 140.

For e, x squared - 6x - 72.

For f, x squared - 18x + 80.

For g, x squared + 16x + 64.

For h, x squared - 16x + 64.

i, x squared + 9x + 14.

And j, x squared + 3x - 40.

Question two, which of these products of two binomials can be written as the difference of two squares? Well that's c, d, and e.

For f, one of the terms in each bracket needs to be the same and the other two should be a zero pair.

And what we've got here are two zero pairs, so this one doesn't work.

When we write out c, we should have x squared - y squared.

d gives us a squared - 0.

5 squared, so that's -0.

25.

And then for e, I have 100 - x squared.

It's now time for the final part of today's lesson, and that's on more complex binomials.

Sometimes terms within the binomials are not in the same order.

So for example, what about finding the product of x + 9 multiplied by 5 - x? Well, you could just set up an area model just like we did before.

And we can calculate the four partial products.

So we'd have 5x, -x squared, 45, and -9x.

We can then gather any like terms. We have -x squared - 4x + 45.

Now, Lucas says, "I would prefer the x terms to be the first terms in the binomial." So how could Lucas rewrite this product so the x term is first in both of the binomials? Well, he could write it like this.

Now let's consider this area model.

After all, having rewritten it, we want to make sure we still get the same expanded form.

Otherwise, this rewrite is not allowed.

So, let's consider that.

Yep, I still get -x squared, 5x, -9x, and 45, so we get the exact same partial products, we just get them calculated in a different order.

I'd like you now to expand x + 8 and x - 8.

And use an area model to help if you wish.

When you've done that, what do you notice about your answer? Pause and do this now.

Welcome back.

So, when we expand this, which you're very welcome to do by making the x term be the first term in the second bracket, or you can leave it as is, it's up to you, you'll still get the same four partial products.

We have -x squared, 8x, <v ->8x, and 64,</v> which leaves us with -x squared + 64 when we've gathered the like terms. Hang on a second, this looks awfully familiar.

It is the difference of two squares, and it might be easier to spot if I write the 64 term first so that we have 64 - x squared.

What we have here is 8 squared - x squared.

Now, the variables in the binomials can have coefficients greater than one.

For example, 2x + 3 multiplied by 3x + 4.

We can still use our area model to find the expanded form.

So, calculating our four partial products gives us 6x squared, 8x, 9x, and 12.

And then we just sum the like terms. So we have 6x squared + 17x + 12.

Now, binomials might also contain multiple variables.

For example, what's the expanded form of 5a + b multiplied by 2a - 3b? Again, our area model can help to make sure we calculate the four partial products correctly.

I'm calculating 5a lots of 2a - 3b and b lots of 2a - 3b.

So I have 10a squared - 15ab.

I also have 2ab - 3b squared.

When I simplify, I get 10a squared - 13 ab - 3b squared.

Let's do a check now.

I'm going to multiply 4x - 6y by 1/2x - 4y.

And I'm going to use an area model to make sure I get the four partial products correct.

Well, 4x multiplied by 1/2x is 2x squared.

4x multiplied by -4y is -16xy.

When I do -6y multiplied by a 1/2x, I get -3xy.

And -4y multiplied by -6y is 24y squared.

Gathering the like terms leads to a result of 2x squared - 19xy + 24y squared.

It's now your turn.

Please expand and simplify the following.

Pause and do this now.

Welcome back.

If you used an area model, you're very welcome to do so, or you may have tried to use the distributive law.

Either approach is fine, but you should reach the four partial products of 2a squared - 32ab - ab + 16b squared.

Then when you gather the like terms, we have 2a squared - 33ab + 16b squared.

It's now time for your final task.

Expand and simplify these expressions, and you're very welcome to use an area model to help if you wish.

Pause and do this now.

For question two, expand and simplify each product of two binomials.

Now, I've not drawn an area model for you here, but if you want to have one, you're very welcome to have it.

So do feel free to draw one if that's what you want to do.

Please pause the video and attempt this question now.

It's time to go through the answers.

So for 1a, you should have -y squared + 4y + 45.

For b, you should have y squared - 12y + 36.

For c, -4y squared + 20y + 75.

And for d, 4x squared - y squared.

Did you spot the difference of two squares for that one? Question 2a, we should have 2x squared + x - 15.

For b, 3x squared + 16x - 35.

For c, 3x squared - 12.

For d, 8x squared - 14x - 15.

For e, 25xy + 10x - 10y - 4.

You'll notice you couldn't combine any terms there because none of them were like.

For f, -4x squared + 9y squared.

Again, did you spot this is a difference of two squares? And then for g, 12xy squared - 6x squared y squared <v ->24y + 12xy.

</v> Well done if you got these all right.

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

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

Algebra tiles and a set of axes can be used to explore the underlying structure.

Area models can be used to find the product of any two binomials, including when the coefficient of the variables is not 1.

If one term in each binomial is the same and the other two terms are zero pairs, then the product is the difference of two squares.

Well done, you've worked really well today, and I'm really looking forward to seeing you for more of our lessons in the algebraic manipulation unit.