Lesson video

Well done.

Another brilliant decision to join me, Mr. Robson, for more maths.

Common factors in algebra today.

Who's excited? Lots of hands going up.

Let's get going.

Our learning outcome.

We'll be able to use the distributive law to factorise expressions where there is a common factor.

Keyword, keywords.

To factorise to express a term as the product of its factors.

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

You'll see that in context throughout the lesson.

Two parts to the lesson.

We're gonna start by factorising some expressions.

I'd like you to start with a short task, which should look familiar.

I'd like you to find the highest common factor for each pair of these terms, starting with 12 and 20, then 12ab and 20ac.

Just as we can find the highest common factor of numerical terms, we can find the highest common factor of algebraic terms and that's gonna be crucial in enabling us to factorise today.

So pause this video and find the highest common factor for each of those four pairs.

12 and 20.

We can decompose 12 into its prime factors, two times two times three.

We can decompose 20 into its prime factors of two times two times five.

I hope you remember this from all the number work you've done.

What we're looking for is matching common factors.

So I can see a two and another two in both of those lists.

That is why four is the highest common factor of 12 and two.

Common factors, the common prime factors being two and two, the product of two and two is four.

How does it differ with algebraic terms? The difference is when we decompose them, 12ab is made up of two times two times three times a times b, and there on the right-hand side, you can see the decomposition of 20ac.

What we're looking for is the common ones in each of those lists.

I can see a two, a two and an a.

So our highest common factor of 12ab and 20ac is going to be the product of those common factors.

Two times two times a gives us the highest common factor of 4a.

Onto the next one, C, with two terms, 12ab and negative 20ac.

I've put the working out that we did for B at the top of the screen because question C's incredibly similar.

In fact, what's the only difference between the pair in B and the pair in C? Well done, it's a negative 20 coefficient, not positive 20.

So how does that affect the way that we decompose our algebraic term? Well, 12ab will decompose the same way, but negative 20ac, 20's made up of two times two times five.

Instead of saying negative 20's made up of negative two times two times five, we say it's made up of two times two times five multiplied by negative one.

So when we come to find the highest common factor of those, we're again looking for common ones in each list.

There's a two, a two and an a, I think that's it.

It is.

A two, a two and an a, the product of which is 4a.

The highest common factor again is 4a.

For the final one, quite a mouthful, 12a squared b squared, negative 20a cubed c.

What's the highest common factor of those two terms? Again, I've put the working for C at the top of the screen now because it helps us to see the difference between question C and question D.

When we decompose 12a squared b squared, the list gets a little bit longer versus 12ab.

We've got two times two times three times a times a times b times b.

The decomposition of negative 20a cubed c, negative one times two times two times five times a times a times a times c.

And then we're looking for the common ones, two and two, a and a.

The product of those two times two times a times a is 4a squared.

The highest common factor of 12a squared b squared and negative 20a cubed c is 4a squared.

I just wanna check you've got that.

I'd like you to identify the highest common factor of 36 q cubed r squared, and 48qr squared.

I've given you four choices there.

Pause this video, take your pick.

Which one is the highest common factor of those two terms? It was A, 12qr squared.

It couldn't have been B or C because 16 is a factor of 48 but not a factor of 36.

Could it have been C? C's a common factor.

12qr is a common factor of both those terms, however, 12qr squared is a higher common factor, so 12qr could not have been the highest common factor.

Factorising.

Factorise.

Factor.

Factorise 12 plus 20.

If you remember a long time ago when you did expanding brackets, you might have used an area model.

You may have used this in multiplication.

In order to help us factorise, we can undo that area model or do that area model in reverse.

I've put the terms 12 and 20 inside the area and I've put four as the width because four is the highest common factor of those two terms. Remember, the highest common factor of 12 and 20 was four.

Hence I've labelled that as the width.

Now, what I need to do to complete this area model is find four lots of what make 12 or is it better to say 12 divided by four? That would be three.

How do I find the other part of that length? Well, I need what factor of 20 is needed to complete the area model? Four lots of what make 20 or 20 divided by four.

That would be five.

What I've done here is factorised 12 and 20 into four lots of three plus five.

How that would look is so.

Four lots of three plus five in the brackets because we decomposed our 12 into four lots of three and we decomposed our 20 into four lots of five.

This is distributive law.

Understanding that four lots of three and four lots of five is four lots of bracket, three plus five.

You've seen that before in number.

It's gonna be really useful in algebra, algebraic factorising today.

How does it look different for algebraic terms? Does it look different? Not really.

We're going to unpick that area model again.

Do the area model in reverse if you will.

If I'm gonna factorise 12ab and negative 20ac, I need to start with those two terms inside my area model.

What am I putting on the width? The highest common factor of 12ab and negative 20ac.

So my width would be 4a.

The length across the top must be 3b there on account of 4a multiplied by what will make 12ab.

That must be 3b.

And then how many lots of 4a do I need to make negative 20ac? I need negative 5c lots of 4a.

Yes.

4a multiplied by negative 5c will give us that area negative 20ac.

So 12ab minus 20ac became 4a lots of 3b and 4a lots of negative 5c, which we can write as 4a lots of bracket 3b minus 5c.

Your turn.

Draw an area model, find the highest common factor, put it on the width and see if you can find the length across the top of your area model.

Then you would've factorised the expression 12a squared b squared minus 20a cubed c.

Pause this video and give it a go.

So we start with the area model.

The highest common factor, 4a squared, is the width and then how many lots of 4a squared make that first area? 3b squared lots of 4a squared.

How many lots of 4a squared to make that second area? We need negative 5ac.

So 12a squared b squared minus 20a cubed c became 4a squared lots of 3b squared and 4a squared lots of negative 5ac, giving us a factorised expression, 4a squared, brackets, 3b squared minus 5ac.

Okay, here at Oak, some of our students are being really busy.

They're factorising the expression 6a minus 3b plus 12.

Can you identify in each case how this was done? Alex, Alex likes visual representations.

I like visual representations.

They're really, really useful for understanding what's actually happening in mathematics.

So Alex drew this or rather, Alex rearranged his algebra tiles to make that.

Alex says, "I used algebra tiles.

This shows that three lots of bracket 2a minus b plus four is the equivalent expression." Can you see what Alex has done? Three groups of, three rows of 2a minus b and four.

Lucas unpicked the area model.

He put the term 6a, negative 3b and 12 inside the area model and then the highest common factor as the width, three.

He found that the areas were made up of three lots of 2a, three lots of negative b and three lots of four.

Hence he was able to write the expression three lots of bracket, 2a minus b plus four.

Do note Lucas has got the exact same outcome as Alex.

Just a slightly different way.

Lucas's visual representation is the area model.

Izzy, what did you do? Izzy just used distributive law.

She identified for the algebraic expression 6a and negative 3b and 12, the highest common factor is the three.

So she decomposed 6a into three lots of 2a.

She decomposed negative 3b into three lots of negative b and she decomposed 12 into three lots of four.

Therefore, she had three lots of bracket 2a minus b plus four.

She used the distributive law.

You're welcome to use any of those three methods.

They will all work really effectively for you.

We can use algebra tiles to group, we can unpick the area model or we can just use distributive law.

Practise time now.

Question one, fill in the blanks.

Part A is just the expansion of two lots of bracket y plus three.

Part B, part of the expression within the bracket is missing.

One of the terms in the equivalent expression is missing.

I'd like you to work through those four and fill in the blanks.

Pause and give it a go.

Question two, match the correct factorization to the correct expression.

Pause, match them up.

Question three, factorise the expression 12y squared plus 18y minus 15xy.

There's an area model there to help.

Pause this video, give it a go.

Feedback time.

So two lots of y, two lots of three, that's 2y plus six.

Two lots of y would give us 2y in the expanded expression.

Factorising, two lots of what makes 12? That's two lots of six.

For part C, where do we begin? Something lots of y, but the y term's missing in the equivalent expression.

Ah, something lots of two makes 12.

That must be six lots of two.

Therefore, six lots of y give us 6y.

In question D, the expression 30y plus 12.

Well, if I take out a factor of six, I must be left with 5y and two.

Question two, we should have matched 8y plus 12 to four lots of bracket 2y plus three.

12y plus eight matched to four lots of bracket 3y plus two.

Negative 12y plus eight matched to four lots of bracket two minus 3y.

And 8y squared minus 12y matched to 4y lots of bracket 2y minus three.

You can always expand those brackets to check that you're right.

4y lots of 2y's 8y squared.

4y lots of negative three's negative 12y.

I have equivalent expressions.

I know I'm right.

Algebra's lovely for that.

Quite often, our answer can check that we've done it right.

Question three, factorise the expression 12y squared plus 18y minus 15xy, we should have got 3y lots of bracket 4y plus six minus 5x.

How did we get that? 12y squared decomposes like so.

18y decomposers like so.

Negative 15xy decomposes like so.

Common factors in there.

There's three and y in each.

Hence 3y is the highest common factor.

So 3y becomes the width on our area model and then 3y lots of what makes 12y squared? 4y.

3y lots of six make 18y.

3y lots of negative 5x make negative 15xy.

Hence, it factorises to 3y, lots of that width, 4y plus six minus 5x.

Part two now.

Factorising fully.

Aisha, Andeep and Sam are factorising the expression 6y plus 12.

Aisha writes two lots of 3y plus six.

Andeep writes three lots of 2y plus four.

And Sam writes six lots of y plus two.

Who's factorised correctly.

It's worth going back to our keywords now.

To factorised is to express the terms of the product of its factors.

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

A key distinction here between factorising and fully factorising.

So 3a lots of two plus 3b.

That is a fully factorised expression.

There's no common factors greater than one for the terms inside the bracket.

2s and 3b, there's not a common factor there.

That expression cannot be further factorised.

By contrast, a lots of bracket six plus 9b, there's a common factor for six and 9b.

The expression inside the bracket can be further factorised because three is a common factor of both those terms. So Aisha wrote two lots of 3y plus six.

She's factorised the expression and we know she's factorised the expression because when we expand that bracket out, two lots of 3y, two lots of six, we get the equivalent expression 6y plus 12.

So she's factorised, but in 3y plus six, there's a common factor.

So it can be factorised further.

For Andeep, three lots of 2y, three lots of four.

Andeep's also works.

Andeep's factorised the expression.

When we expand three lots of bracket 2y plus four, we get the equivalent expression 6y plus 12.

It's factorised, but in the expression 2y plus four, we have a common factor of two.

So Andeep's can be further factorised.

In Sam's case, however, Sam's fully factorised.

Neither expression can be further factorised.

The expression inside the bracket y plus two, there is no common factor of y and two.

So this is factorised as far as that expression can go.

And we can check, six lots of y, six lots of two.

Yes, it's an equivalent expression.

So Sam has fully factorised.

Let's check you've got that now.

Which of these is fully factorised? Pause this video, take your pick.

It was C.

Six lots of bracket e plus three.

In A, we had a common factor of 2e and six, common factor of two.

We could further factorise the expression within that bracket.

For answer B, 3e plus nine, there's a common factor of three.

We could further factorise that expression.

Another question.

Which of these is fully factorised? Pause and consider each one.

It was A, 3a, brackets 2e minus five.

We know that an expression has been fully factorised if the highest common factor of all terms is one of the factor pairs.

We know that an expression is fully factorised if the highest common factor of all terms is one of the factor pairs.

Hmm, let's have a look at that.

For the expression 6a plus 9ab, 3a is the highest common factor of 6a and 9ab.

So that expression must be fully factorised.

We've broken 6a and 9ab down into two factors.

3a is a factor and two plus 3b is a factor.

3a is the highest common factor of the two terms of which we started.

Hence it's fully factorised.

If we had gone for take out the factor of a and left six plus 9b inside the bracket, it still works.

A lots of six, 6a, a lots of 9b, 9ab.

It's an equivalent expression.

It's factorised.

A is a factor.

Six plus 9b is a factor, but a is not the highest common factor of 6a plus 9ab.

So it's not fully factorised.

Practise time now.

Question one, fill in the blanks and select the expression which is in its fully factorised form.

By the time you finish question one, you will have some factorizations of the expression 10y plus 40.

Only one of them is fully factorised.

For question two, I'd like you to explain why the below is not fully factorised.

12f minus 18 is the equivalent of three lots of bracket 4f minus six.

Not fully factorised, but can you write a sentence to explain why? Pause and give it a go.

Question three, here's a three term expression.

6x cubed plus 12x squared y minus 18xy.

Factorise the expression.

That's part A.

Part B, factorise the expression in a different way.

Part C, fully factorise the expression.

So be careful.

For part A, factorise it, but don't fully factorise it.

For part B, factorise it but don't fully factorise it.

And then for part C, fully factorise it.

Good luck.

Pause and give it a go.

Feedback time.

Fill in the blanks.

Okay, I've taken out the factor of five from the expression 10y plus 40.

Five lots of 2y make 10y.

Five lots of eight make 40.

Five is a factor.

2y plus eight is a factor.

Second one's a bit trickier.

What clues have we got? Aha, how many lots of y make 10y? 10 of them.

10 lots of what make 40? Four.

So we get the expression 10 lots of bracket y plus four.

And the last one, I'm missing the multiplier.

I'm missing the y term.

How can I possibly start this? Aha, how many lots of 20 make 40? Two lots of 20 make 40.

Two lots of 5y make 10.

So all three of those expressions in brackets factorise 10y plus 40.

Which is fully factorised? It's the middle one.

10 lots of bracket y plus four.

How do we know it's fully factorised? Because 10 is the highest common factor of 10y plus 40.

Question two, why is the below not fully factorised? Your explanation might have included because the expression 4f minus six can be factorised further.

We have a common factor of two, which we could take out to factorise further.

What does that look like? So 4f minus six is two lots of bracket 2f minus three.

So we end up with three lots of two lots of 2f minus three.

Three lots of two is six.

Ah.

So by the time we've done all that, we have actually fully factorised it.

Three lots of bracket 4f minus six was not fully factorised but once you take the common factor of 4f and negative six out and multiply it by three, you end up with a fully factorised expression.

So you can always correct your work if you haven't fully factorised in the first place.

Question three, so many ways you could have done this.

Lots of common factors for 6x cubed, 12x squared y and negative 18xy.

So an example could have been take out the common factor of six to leave yourself with inside the brackets, x cubed plus 2x squared y minus 3xy.

Alternatively, we could have taken out a factor of 2x and left inside the bracket 3x squared plus 6xy minus 9y.

I've not fully factorised yet because neither six or 2x is the highest common factor of those three terms in our first expression.

So what would the fully factorised version look like? Aha, 6x is the highest common factor.

When we take out the factor of 6x, we are left with the expression x squared plus 2xy minus 3y.

Here we are, the end of the lesson.

To summarise, algebraic expressions, like 8x plus 40 can be factorised because the terms share the common factor of eight.

8x plus 40, when we take out the factor of eight, we're left with factors of eight and x plus five.

Whilst 8x plus 40 can be factorised in several ways, it's only fully factorised if the remaining expression cannot be further factorised.

So if we factorised 8x plus 40 to two lots of bracket, 4x plus 20, that's not fully factorised because the expression inside the bracket, 4x plus 20, can be further factorised.

So fully factorised, we take out the highest common factor of eight and we're left with the expression eight lots of bracket x plus five.

That is fully factorising.

That is the end of your lesson for today.

I hope you enjoyed it as much as I enjoyed it, and I hope to see you again soon.