Lesson video

Hello, Mr. Robson here.

Well done, maths, you made a good choice, especially seeing as we're doing algebra today.

Let's get started.

Our learning outcome for today's lesson, will be that we'll use the distributive law to multiply an expression by a term.

Keywords I'll be referring to throughout the lesson, a constant, constant.

That's a term that does not change.

It contains no variables.

For example, the five in the expression, 2x+5, is a constant.

A variable term, however, is one that can take multiple varied values.

In that same expression, 2x+5, it's 2x, which is example of the variable term.

Two parts of this lesson, we're gonna start by multiplying an expression by an algebraic term.

You've seen multiplying expressions before, multiplying expressions by constants.

And you've seen many representations that can be used to illustrate the distributive law.

Two lots of (x+1) is equivalent to 2x+2.

We can see that in this area model.

Two on one length, x and one on the other length, giving us two lots of x, two lots of one.

That's the use of the distributive law to expand the brackets.

X+1 multiplied by two, two lots of x, two lots of one.

We could have also used algebra tiles to represent this same problem.

We have two in the height there, x and one in the length across the top.

And we see our two lots of x and our two lots of one.

This alternative area model, using algebra tiles makes this expansion clearer.

This expansion's different.

An algebraic expression, x+1, but not multiplied by a constant this time, multiplied by an algebraic term.

What will that look like? Well, it'll look like this, x multiplied by x+1.

That's x, that's x+1 and the result is we have an area of x-squared and an x.

We've expanded that expression, x lots of (x+1) is equivalent to x-squared plus x.

I'm gonna take my turn to have a go at one of these and then ask you to take your turn.

My expansion is to multiply the expression x+3 by x.

I'm gonna start by drawing a grid.

I'll need an x tile there, I need an x tile and three ones.

Can you see how I'm set up to multiply x by x+3? Complete this, now I'm gonna need an x-squared tile and one, two, three Xs.

So my expansion of x multiplied by (x+3), is x-squared and 3x.

Can you see my x-squared and my 3x and my algebra tiles? Over to you now.

I'd like you to do the same for expanding x by (x+5).

If you've got algebra tiles, by all means, try and lay this out in front of you.

If you haven't, can you draw a diagram that looks the same? Pause this video and give it a go.

So we start with a grid.

We need x, we need x and five on that length, and then we'll populate it with x-squared and one, two, three, four, five Xs.

That completes your expansion.

Your expression, x-squared and 5x.

Well done.

We can use the area model to ensure we multiply all of the terms. We don't always need a visual representation, but they're very useful, especially in this case the area model, it means we won't miss anything.

This is complicated.

Expand 2x by (3-4y+7x).

There's a lot going on there.

Our area model will look like this.

We'll have a height of 2x and a length of three, negative 4y and 7x.

So you can see we've got three multiplications to do here.

2x multiplied by three is 6x.

2x multiplied by negative 4y, negative 8xy.

And 2x multiplied by 7x, that's 14x-squared.

We've expanded those brackets.

We've got the equivalent expression, 6x minus 8xy plus 14x-squared.

I'd like to check you've got that.

I'll ask you to copy this down and then fill in the missing terms in the area model for this expansion, 3y multiplied by (x-7+5y).

Pause this video, give it a go.

Starting with our lengths, we should have populated 3y and negative seven in those positions.

And then populating the areas, 3y multiplied by x.

We call that 3xy.

In the next area, 3y multiplied by negative seven, negative 21y.

Remember, positive multiplied by negative, gives us a negative.

In this final area, 3y multiplied by 5y.

That's three times y times five times y.

I'm gonna (indistinct) that, three times five times y times y.

That's 15y-squared.

Yes! So our expanded expression would've read, 3xy minus 21y plus 15y-squared.

We didn't miss a term, because we accurately used an area model.

This could be useful.

Multiplying an algebraic term is therefore the same method as multiplying by a constant term in terms of the distributive law.

In the top example, two lots of (y+2).

I need two lots of everything inside those brackets.

I need two lots of y, I need two lots of two.

Two lots of y, we don't call it two lots of two.

We like to write things and say things as concisely as possible.

So we call it 2y.

And two lots of two, well, we make that four that's more simple.

So two lots of y+2 expands like so, a lots of (a-3) is going to do the same thing.

I need a lots of everything inside those brackets.

I need a lots of a and I need a lots of three.

But it's a lots of negative three, so I'm gonna subtract that second term.

That'll simplify to a-squared minus 3a.

In the last example, it might look tricky when multiplying algebraic terms by algebraic terms, all the way through here, but it's a lots of everything in the bracket.

It's a lots of a and it's a lots of b.

And I bet you know how that simplifies.

That's right, you are one step ahead of me, a-squared plus ab.

I'd like to check you've got that now.

I'd like you to select the correct multiplication of the expression, e multiplied by (e+3).

Is it a, b or c? Pause this video, take your pick.

I hope you went for a, e multiplied by e and e multiplied by three.

We need e lots of everything inside the bracket.

B didn't do that, b took e lots of e, but it didn't take e lots of three.

And c took three lots of e and three lots of three.

There's no three lots of three.

I need e lots of e.

That was only present in the top expression, e multiplied by e and we need e multiplied by three.

That would simplify to e-squared plus 3e.

We wouldn't leave it in that form if we were expanding this bracket.

It's just useful to have it in that form to understand that this is just distributive law.

You've seen it before.

One more.

Which represents the correct multiplication of the expression, d multiplied by (e+3)? Pause and take your pick.

It was c, d multiplied by e and d multiplied by three.

Of course, we wouldn't leave it in that form.

We'd simplify, d multiplied by e to de and d multiplied by three to 3d.

Task time now.

Question one, I'd like you to fill the algebra tiles to complete the expansion of the expression, x multiplied by x+8.

Write something on each of those tiles and then write your expanded expression.

Pause this video and do that now.

Question two, six expansions there for you.

The first one, an expression multiplied by a constant.

The rest are expressions multiplied by algebraic terms. I'd like you to expand those brackets, giving me equivalent expressions for each.

Question three, let's pretend you are the teacher and you spotted that Andeep has written this in their maths book, x multiplied by (x+5) is equivalent to 2x+5x, which is equivalent to 7x.

What sentence of feedback are you gonna write in Andeep's book to explain and correct their error? Pause this video and give these questions a go! Feedback time now.

The expansion of x multiplied by (x+8) using algebra tiles, we need an x and eight along that length.

X multiplied by x, that's an x-squared tile.

And then one, two, three, four, five, six, seven, eight Xs, giving us x-squared and 8x.

Question two, expansions, three lots of (x+2).

Well, I need three lots of x and three lots of two.

That's 3x+6.

The next one is multiplied by an algebraic term.

I need x lots of everything inside that bracket, x lots of x, x lots of two.

That's x-squared and 2x.

And the next one, x lots of x and x lots of negative two.

That's x-squared minus 2x.

X lots of x, x lots of y.

X lots of x is x-squared again, x lots of y is xy.

That's the expression, x-squared plus xy.

Next one's different again.

It's not just a variable, it's a term with a multiplier, a coefficient of the x, that's 2x multiplied by x, two times x times x, oh, that's 2x-squared.

And two times x times y is 2xy.

And the last one, I've gotta multiply an algebraic term by an expression with three terms in it.

2x lots of x, 2x lots of y, 2x lots of negative 3z, that's gonna give me 2x-squared plus 2xy minus 6xz.

Andeep's book, your explanation to Andeep might have included x lots of x is x-squared, not 2x.

So Andeep should have got x-squared plus 5x and that expression cannot be simplified further.

X-squared terms, x terms, they're not like terms. So x-squared plus 5x was as far as that expansion could go.

Part two, multiplying by an algebraic term with powers.

We can expand terms with exponents in the same way as before.

We can expand terms with exponents.

Spot the exponent.

That's right, it's the little two next to the x.

When I say x-squared, x to the power of two, that two is an exponent.

And this makes the algebra look a whole lot more difficult, but it's not.

It's the same thing as before.

I'm multiplying an expression with two terms by an algebraic expression.

I can use the same area model.

I need x-squared on that length.

I need x and one on that length.

Is this looking familiar? The only thing that's different now, is in that first area, it's x-squared multiplied by x.

We could think of this by unsimplifying x-squared as x multiplied by itself.

So x-squared is x times x.

When we go to times it by x again, we get x-cubed.

So the area of that first space is x-cubed.

The next area is one lot of x-squared, which is one lot of x-squared.

So our expanded expression here is x-cubed plus x-squared.

Can we simplify that further? I hope you are shouting "No," at the screen.

They're not like terms, x-cubed, x-squared.

They're not like terms. So that's simplified as far as it can go.

We can expand the expression using the distributive law.

Now I think I need x-squared lots of everything inside that bracket.

I need x-squared lots of x, x-squared lots of one.

X-squared lots of x, x-squared lots of one, that's just distributive law.

You've seen this a lot before in maths.

X-squared times x is x-cubed.

X-squared times one is x-squared.

We get the same expression, but without the need to draw the area model.

When I was first learning this, I always drew the area models.

I found they helped me stay very accurate.

But we can just consider that this is just distributive law and just take x-squared lots of everything inside the bracket.

Okay, I'd like to check you've got that.

I'd like you to populate this area model to expand the brackets y-squared multiplied by y+4.

Copy that model down and see if you can populate all those gaps.

Pause this video and try that now.

How did we do? Hopefully, we started by putting y-squared there.

And then we put the length of y and four along the top and then we start to populate the areas, y-squared multiplied by y.

That's y times y times y again, y repeatedly multiplied by itself three times, y-cubed.

And then four lots of y-squared is 4y-squared, giving us an expression, y-cubed and 4y-squared.

Did you get that? Lovely.

Next, I'd like you to use distributive law to expand these brackets, a-squared multiplied by (a-5) and a-squared multiplied by (a+b).

So in both cases, you need to take a-squared lots of everything inside those brackets.

Can you copy this down? See if you can populate those blank spaces and expand these brackets.

Pause this video and try that now.

In the first example, we need a-squared lots of a, a-squared lots of negative five.

That'll give us a-cubed and negative 5a-squared.

For the second bracket, a-squared lots of a, a-squared lots of b, which we'd simplify to a-cubed and a-squared-b.

Not like terms, can't be simplified further.

Leave the expression just like that.

Next, this looks tricky.

2a-squared multiplied by (4-3a).

I'm gonna do this one and then I'm gonna ask you to try a similar one.

Okay, start with my distributive law.

2a-squared lots of everything in that bracket, that's 2a-squared lots of four.

It's 2a-squared lots of 3a.

Remembering that the second term's gonna be negative, 2a-squared lots of four is 8a-squared.

2a-squared lots of three, 6a-cubed.

But it's gonna be a negative 6a-cubed.

So I've expanded those brackets to 8a-squared minus 6a-cubed.

Your turn, pay particular attention to when are you multiplying positives by negatives and anything similar to that.

Hmm, I don't wanna say too much, I might give the game away.

'Cause this is your one to try, right? I'd like you to expand, negative 4h-cubed by brackets, seven minus 2h-squared.

Good luck.

Right, I hope you've given it a go.

Now I don't mind if you've made an error here as long as you've given it a go.

We're always gonna make mistakes when we learn mathematics.

And this is a particularly tricky-looking problem.

We need negative 4h-cubed lots of everything in that bracket.

We need negative 4h-cubed lots of seven and we need negative 4h-cubed lots of negative 2h-squared.

Negative 4h-cubed times seven, that's negative 28h-cubed.

Negative 4h-cubed multiplied by negative 2h-squared, when we multiply negative by a negative, we get a positive.

So it's positive 8h-to-the-power-of-five.

H-cubed multiplied by h-squared, that's h times h times h times h times h.

That was h repeatedly multiplied by itself, five times.

Don't panic if you made an error or two there.

Just pause and read through that method and see if you can spot where you went wrong.

Practise time now.

I'd like you to expand these brackets.

Pause this video and give these questions a go.

For question two, Aisha compares these two expressions, x-squared multiplied by x+7, x-squared multiplied by (x+7).

Aisha says, "They are the same thing." In both cases, x-squared is being multiplied by x and seven.

Explain to Aisha why this is wrong.

Pause this video and explain away.

(engine revving) Okay, question one are expansions.

There's your answers.

You'll want to pause now and take a moment to just check that what these answers are, are they same as the ones that you've got, making note where you've made any errors to copy down the correct answer.

So common errors we might see on a, b, c and d, there's a positive term followed by a negative term in all of them.

They should follow that pattern.

In e, 4e-to-the-power-of-three multiplied by e, 4e-to-the-power-of-four.

4e-to-the-power-of-three multiplied by 3e-to-the-power-of-two.

Four times three is 12, e to the power of three, multiplied by e to the power of two, that's e to the power of five.

Some students might get that wrong.

In part f, we're multiplying by a negative algebraic term outside the bracket.

That's a negative multiplied by a positive and a negative multiplied by a positive, that's why both of those terms are negative, negative 8f-cubed, negative 12f-squared.

And then the bottom one is just to show you, we don't always have an integer coefficient outside of the bracket.

On the g-squared there, the coefficient is negative 0.

4, it's non-integer.

That doesn't affect what's going on.

We still need negative 0.

4g-squared lots of everything inside the bracket.

Hence we get negative 0.

8g-cubed and negative, sorry, positive 1.

2g-squared.

I almost got caught out by multiplying a negative by a negative there.

I must remember, multiplying a negative by a negative, gives us a positive.

In question two, I can see why Aisha thinks this.

In both cases, x-squared are being multiplied by x and seven.

They both read x-squared multiplied by x+7, don't they? It's important to note the role of the brackets in the second one, x-squared lots of (x+7).

Everything inside the brackets are being multiplied by x-squared.

Whereas in the first case, it's only the x that's being multiplied by x-squared.

That's where Aisha has gone wrong.

Your explanation might have included, "In the first expression, x-squared is only being multiplied by x.

The seven is not being multiplied by anything.

The presence of brackets in the second expression, makes it different.

Because of the brackets, both the x and the seven are to be multiplied by x-squared." If you understand this example, you've conquered this.

There we are, at the end.

To summarise, we can multiply expressions by variable terms as well as by constant terms and also by variable terms with exponents other than one.

For example, the expression, x+1, we can multiply by the constant two to get 2x+2.

But we can take that same expression, x+1 and multiply it by x to get x-squared plus x.

In the last example there, x+1 is multiplied by x-squared, the exponent other than one and we get x-cubed plus x-squared.

I hope you enjoyed our algebra today.

I hope to see you again soon for more!.