Lesson video

Hi, I'm Mrs. Wheelhouse and welcome to today's lesson on checking and securing understanding of factorising.

This lesson comes from our unit on algebraic manipulation.

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

By the end of today's lesson, you'll be able to use the distributive law to factorise expressions where there is a common factor.

Here are some keywords that we'll be using today in our lesson.

Factor, numerical factor, factorise, and to fully factorise.

Now if you need to, please pause the video at this point so you can read through these.

Now you're ready.

Let's get started.

Our lesson is broken into three parts today and we're going to begin by factorising using algebra tiles.

To factorise an expression is to write it as a product of two or more expressions.

For example, 3 lots of x + 4 is a factorised expression as it is the product of 3 and x + 4.

We can show this product using an area model and algebra tiles.

So here's our area model, and you can see that one length is 3 and the other length is x + 4.

Well, 3 lots of x is 3x, and 3 lots of 4 is 12, and we've shown this with our algebra tiles as we have 3 x tiles and 12 one tiles.

This is how we can show that 3 lots of x + 4 is the factorised form of the expression 3x + 12.

Jacob has made the expression 2x + 6 with algebra tiles.

He could rearrange these into a rectangle to show one way to factorise this expression and we can see that here.

What are the dimensions of his rectangle? That's right, he's got 2 and he's got x + 3.

He could write the expression 2x + 6 as 2 lots of x + 3.

Hmm, now Jacob has added an x tile, so that he's got the expression 3x + 6.

Oh.

He thinks he can write this as 3 lots of x + 3, but can you see a problem here? It's not a rectangle, so the error cannot be written as the product of the two lengths.

Can Jacob rearrange these tiles into a rectangle though? Yes, he can.

As you can see here, we've now made a rectangle using the three x tiles and the six 1 tiles.

What are the dimensions of this rectangle? That's right, we have 3 and we have x + 2.

So 3x + 6 can be written as 3 lots of x + 2.

How could you rearrange these tiles into a rectangle? Well, our three Oak pupils have had a go and they've all done it differently.

Who do you think is correct? Well, Jacob has a rectangle with dimensions 4 and x + 2.

So he thinks the factorised form would be 4 lots of x + 2.

Laura has a rectangle with dimensions 1 and 4x + 8.

So she thinks it factorises into 1 lot of 4x + 8.

And Alex's rectangle has lengths of 2 and 2x + 4.

So he thinks the expression factorises into 2 lots of 2x + 4.

Well, all three are actually correct, although Laura's is probably the least useful.

I mean, all she said is anything is equal to one times itself.

It's true, but it's not exactly useful factorising.

to fully factorise is to factorise to the point that any remaining term or terms cannot be factorised any further.

So let's consider Jacob and Alex.

After all, we knew Laura couldn't have been fully factorised.

She'd only taken out one as a factor, so there was clearly more that could be done.

Which of the two expressions do you think is the fully factorised form of 4x + 8? That's right, it's Jacob's.

Because for Jacob, we can see that inside the bracket we have x + 2, and we cannot factorise those terms any further because other than 1, there is no common factor between x and 2.

Here is the expression 4x + 12.

How can we make sure we have fully factorised this expression? Well, we need to find the highest common factor of all terms. What is the highest common factor of 4x and 12? Well, we can write them as a product of their prime factors.

So for 4, I have 2 X 2, and then I need them to multiply by any variables, so multiplying by x.

And then 12 can be broken down into 2 X 2 X 3.

Do you remember when we dealt with lessons on prime decomposition? So what's the highest common factor of 4x and 12? We can see the highest common factor is four, so one of our dimensions of the rectangle must be four.

Let's rearrange the algebra tiles to find the other.

So we can see here, we've got lengths of 4 and x + 3.

So 4x + 12 fully factorises to 4 lots of x + 3.

We can do the same with negative algebra tiles.

So here's an expression for 6x - 3.

And we'll arrange these into a rectangle.

What are the dimensions? So one of the dimensions is 3 and the other is 2x - 1.

So we can write this as 3 lots of 2x - 1.

3 is the highest common factor of 6x and -3.

So we know this is fully factorised.

Quick check now.

Which of these shows a correct factorisation of 6x - 12? Well done if you chose b, c, and f.

But which represents an expression which is the fully factorised form of 6x - 12? That's right, it's F.

In this one, we have 6 lots of x - 2, and we can see that inside the bracket, there is no common factor between x and -2 other than 1.

Now let's just pause for a moment and look at d.

You might have thought d showed a correct factorisation.

However, we do not know the length of the x tile.

When building rectangles, we should never place the one tile along the edge of an x tile.

So when building rectangles, we should never place ones tiles along the long edge of an x tile.

It's now time for your first task.

Please use algebra tiles to factorise these expressions and write them as the product of two factors.

Pause and do this now.

Question two.

How many different ways can you arrange each set of tiles into a rectangle? Identify which option has been fully factorised.

Pause and do this now.

Let's go through the answers.

So what you can see on the screen is the algebra tiles being rearranged so that they show the factorised form of each expression.

And then underneath, we have what that is written algebraically.

Please pause the video at this point so you can check your answers.

In question two, I asked you how many different ways you could arrange each set of tiles into a rectangle and then to identify which option had been fully factorised.

Again, you can see the arrangements on the screen and I've indicated which of the options is the fully factorised form.

Please pause the video now while you check your work.

It's now time for the second part of our lesson and that's on using the highest common factor.

It is not always efficient to draw area models to scale.

However, the same structure can be used to factorise any expression.

For example, let's factorise the expression 30a + 12.

We need to find a common factor of 30a and 12.

If we want to fully factorise, this needs to be the highest common factor.

So I have broken 30 down into a product of its prime factors.

We have 2 X 3 X 5, and then I consider any variables, so I've also got to multiply by a.

For 12, I've broken this down into 2 X 2 X 3.

And we consider which of these factors are common.

Well, that's the two and the three.

2 X 3 is 6.

So the highest common factor of these two terms is six.

We now know that one dimension of our area will be six.

We can put the two terms inside the area model and then ask ourselves, what do you multiply 6 by to get to 30a and what do you multiply 6 by to reach 12? Well, we multiply 6 X 5a to reach 30a, and multiply 6 X 2 to get to 12.

So therefore our two smaller missing lengths are 5a and 2 and the total length therefore is 5a + 2, so therefore, 30a + 12 when fully factorised is 6 lots of 5a + 2.

And we can factorise expressions with any number of terms. So let's consider this expression 5a + 10b - 20.

What is the highest common factor of all three terms? Well done if you said five.

So five is one length, what about the others? Well, 5 X A gives us 5A, and 5 X 2b gives us 10b, and 5 X -4 causes -20.

So the fully factorised form of 5a + 10b - 20 will be 5 lots of a + 2b - 4.

Alex, Lucas, and Izzy are factorising the expression 9a - 6 + 6b.

Can you explain each of the approaches they have taken? Pause the video and do this now.

Welcome back.

Let's see if you agree.

Well, Alex used algebra tiles and this shows 3 lots of 3a - 2 + 2b.

Lucas used an area model, and this shows the area can be found by multiplying the two lengths, 3 and 3a - 2 + 2b together.

What's Izzy done? Well, she's used the distributive law and this also shows that 9a -6 + 6b can be written as 3 lots of 3a - 2 + 2b.

Now, sometimes the highest common factor can contain a variable.

For example here.

How can we fully factorise 2x squared + 6x + 4xy? Well, let's consider each term.

2x squared can be written as 2 X x X x.

6x can be written as 2 X 3 X x.

And 4xy can be written as 2 X 2 X x X y.

What's the common factors here? That's right.

Each term has a 2 and an x as its factors and that means the highest common factor across all three terms is 2x.

So that means for our area model, one of the lengths must be 2x.

So what factor is needed to complete the area model? Well, 2x X x gives us 2x squared, so that's the first length.

2x X 3 is 6x, that's the second length.

And then 2x X 2y gives us 4xy, and that's our third length.

Now remember, those three lengths combined make the total length of the rectangle.

This means that the fully factorised form of 2x squared + 6x + 4xy is 2x lots of x + 3 + 2y.

Alex has factorised 24b squared + 12ab cubed - 6a squared b squared.

How could we check if Alex is correct? We could write his product in expanded form and see if it's the same as the original expression, and an area model can help with this.

Well, 2b X 12b is 24b squared.

Looking good so far.

2b X 6ab squared gives us 12ab cubed.

Again, still good.

And then 2b X -3a squared b gives us -6a squared b squared.

Would you look at that? Alex is indeed right.

Lucas is pointing out it's not fully factorised.

Now, how do we know that? Well, let's look at the terms in the bracket.

We can see there are still common factors here.

So a further factor of 3b can be taken out and 2b therefore was not the highest common factor of all the original terms. So what we've done is we've looked at the expression inside the brackets and we've factorised that.

So we've taken out a factor of 3b, which is what you can see here.

We've then multiplied what's outside the bracket, so 2b X 3b giving 6b squared.

Since there are no common factors in all three terms of the expression inside the brackets, we know this is fully factorised.

Let's do a quick example.

For the expression 27x squared + 36x, The highest common factor of the two terms is 9x.

Well, 9x X 3x gives us 27x squared, and 9x X 4 gives me 36x.

This means that I can write 27x squared + 36x as 9x lots of 3x + 4.

It's now your turn.

Please fully factorise the expression 16a squared + 24.

Pause and do this now.

Welcome back.

Let's see how you got on.

Well, the highest common factor of the two terms is 8, so that's one of the dimensions of the rectangle.

8 X 2a squared gives us 16a squared and 8 X 3 gives us 24.

And therefore, the fully factorised form should be 8 lots of 2a squared + 3.

Well done if you got that right.

Let's consider this expression where I've got more than two terms. The highest common factor here is 4ab, so that's one dimension of the rectangle.

What will the other one be? Well, 4ab X 4ab is 16a squared b squared, 4ab X b is 4ab squared, and 4ab X -3a squared is -12a cubed b.

This means the fully factorised form is 4ab lots of 4ab + b - 3a squared.

It's your turn now.

Please fully factorise the expression 5xy squared - 15x cubed y squared plus 20 x squared y squared.

Pause and do this now.

Welcome back.

Let's see how you got on.

Well, the highest common factor is 5xy squared.

So that's one dimension, but what about the other? You can see here, you should have reached the fully factorised form, which is 5xy squared lots of 1 - 3x squared + 4x.

Well done if you got this right.

It's now time for our second task.

In question one, Alex is looking through some of his old notes and tasks on factorising.

How does he know that each of these are not fully factorised correctly? Pause and do this now.

Question two, please fully factorise each expression.

Pause and do this now.

Time for answers.

In question one, let's explain how we know these were not fully factorised.

Now, your explanations may be different to mine, but as long as you mean the same thing, that's absolutely fine.

So in the top one, you can see this isn't right because 3 X 7 is not 20, it's 21, so that can't be a correct form.

In b, there's still a common factor of two in the expression, so that's not fully factorised.

For c, there's a negative term in the expanded form, but no negative term in the factorised form, that can't be right.

In 1d, there's a common factor of a left in the expression.

In other words, we should have 4a outside the bracket, not just 4.

For e, the terms in the factorised form are like terms, but the original terms were not.

Hmm, yes, look inside the bracket.

We've got 3a + 2a, that's not going to have been possible.

And then f, we can see there are still common factors left in the expression.

Question two, you had to fully factorise each expression.

Please pause the video while you go through these and check your answers.

Welcome back.

Did you spot that j had a bit of a trick in it? That's right, there were like terms there.

So you could have simplified by collecting the like terms before you factorised.

It's now time for the final part of today's lesson, and that's factorising to suit the situation.

And I wonder what I mean by that.

Let's have a look.

So we can use factorisation to show expressions as multiples of specific values.

So for example, 42x + 63 is a multiple of 7, assuming x is a positive integer.

And we don't need to fully factorise here, we just need to take out a factor of 7 to show this.

So for example, 42x + 63 can be written as 7 lots of 6x + 9.

We've shown that this expression is a multiple of seven and we didn't have to fully factorise to do it.

Let's consider this example.

Show that 8 lots of 4n -3 + 13 lots of 6n + 3 is always a multiple of 5 when n is a positive integer.

Well, let's start with our expression.

The first thing I'm going to do is expand both sets of brackets.

So 8 X 4n is 32n, 8 X -3 is -24, and then we've got 13 X 6n is 78n, and 13 X 3 is 39.

I'll then collect the like terms, so I end up with 110n + 15.

Now I remember I wanted to show that this expression is a multiple of 5.

So what I'm going to do is take out a factor of 5 and write what remains in the brackets.

So in other words, I can write the expression 8 lots of 4n - 3 + 13 lots of 6n + 3 as 5 lots of 22n + 3.

And this shows that that expression is always a multiple of 5 when N is a positive integer.

It's now your turn.

Please show that the expression 3 lots of 2n + 3 plus 5 lots of 4n + 1 is always a multiple of 2 when n is a positive integer.

Pause the video and do this now.

Welcome back.

Let's see how you got started.

So you should have expanded the brackets first and then collected the like terms to reach the expression 26n + 14.

Now remember you need to show this as a multiple of 2, so you should take out a factor of 2, meaning writing the expression as 2 lots of 13n + 7.

Now, it is possible for one of the factors to be negative.

For example, we could write -3x - 18 as an expression multiplied by -3.

So inside the brackets, I know I need to have x and I need to have positive 6.

Sometimes this will allow us to simplify expressions when we do this.

So for example, 5 lots of x + 6 - 3x - 18 could be written like this.

That's 5 lots of x + 6 - 3 lots of x + 6, which is just 2 lots of x + 6.

How else could we write the expression <v ->x squared - 3x + 10?</v> Well, we could factorise with a factor of -1.

So let's think about what would go inside the brackets.

I'd need to have x squared, 3x, and -10.

In other words, we can write the expression <v ->x squared- 3x + 10</v> as -1 lot of x squared + 3x - 10.

Don't forget it is possible for the coefficient of the expression to be a fraction as well.

So for example, let's find an equivalent way of writing 10/3 lots of x + 8/3.

We can write the expression as a multiple of a 1/3.

Doesn't that look a little bit nicer? So we have a 1/3 lots of 10x + 8.

Now the expression in the bracket still has a common factor, so we could write this as 2/3 of 5x + 4.

Let's go through a quick example.

We're going to fill in the missing expression here, so I've taken out a factor of 2/5.

Well, 12x divided by 2/5 is 30x and 5 divided by 2/5 is 25/2 or 12 1/2.

So therefore we could write the expression 12x + 5 as 2/5 of 30x + 12.

5.

Now of course we can't say this is fully factorised as the coefficient of the expression is not an integer.

Let's fill in the missing expression here.

Pause and do this now.

Welcome back.

When we do 10x squared divided by 2/3, we get 15x squared.

And 6y divided by 2/3 is 9y and that means we should reach that 10x squared + 6y is equivalent to 2/3 of 15x squared + 9y.

It's time for our final task now.

Please fill in the gaps in these factorised expressions.

Pause and do this now.

Question two, show that when x is a positive integer, the expression 6 lots of 4x + 5 + 12 lots of x + 2 is always a multiple of 9.

And then in question three, show that the expression can be written as a multiple of 2x - 3.

Pause and do this now.

It's time to go through our answers.

For question one, I asked you to fill in the gaps in the factorised expressions.

I suggest you pause the video so you can check the answer on the screen with what you have.

And here's the second part of question one.

Again, please pause while you check your work.

Question two, you had to show that when x is a positive integer, the expression is always a multiple of nine.

So you should have expanded the brackets, collected the like terms, and then taken out a factor of nine.

Question three, you had to show the expression can be written as a multiple of 2x - 3.

And in order to do this, I could have expanded all the brackets, collected all the like terms, and then factorised, that will work.

But what I've done here is I've shown that each of the terms can be rewritten.

So what I mean by that is the 5 lots of 2x - 3, that was already in the correct form.

It was written as a multiple of 2x - 3, but the next term wasn't.

So what I did was I took out a factor of -1 and this meant that the -3 outside the bracket became a positive 3, and inside the bracket I had 2x - 3.

And then for the final two terms, so the -4 + 6, what I did was I took out a factor of -2.

So I wrote it as 2 lots of 2x - 3, and remember I'm taking this away.

Now I have three terms where 2x - 3 is common to all of them, and therefore I've got 5 lots of 2x - 3 + 3 lots of 2x - 3 subtract 2 lots of 2x - 3.

Which leads us with 6 lots of 2x - 3.

Well done if you got this right.

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

An expression can be factorised by writing it as the product of two or more expressions.

An expression is fully factorised if the highest common factor has been taken out.

The highest common factor may be a variable term.

We can write expressions as multiples of other values which are not the highest common factor to enable us to manipulate expressions to suit our needs.

Well done! Did a great job today.

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