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Hello, Mr. Robson here.
You are wise, you've chosen maths again.
Well done, can't wait to get started.
Let's go.
Our learning outcome for today is that we'll be able to identify the highest common factor of two or more algebraic terms. We'll be using a few keywords.
A factor is a term which exactly divides another term.
The highest common factor, you would've heard that before in number.
In algebra, the highest common factor of two or more terms, which can be divided by all other possible common factors.
HCF is an abbreviation of highest common factor, and factorise means to express a term as the product of its factors.
Three parts of today's lesson.
We're gonna begin by listing the factors of algebraic terms. Let's start with something you're very familiar with.
Two times six equals 12.
We look at the language for what we call this.
We've got a factor by a factor making a product.
Factor, factor, product.
Two lots of five, making 10.
Factor, factor, product.
You would've seen this before, so is it any different in algebra? We wouldn't say two multiplied by y.
We'd simplify that to 2y, but it's useful to write this because it's the same as in number.
We've got factor, factor, product.
So we could say that the product of two and y is 2y, or we could say that two is a factor of 2y and y is a factor of 2y.
We can systematically list the factor pairs to ensure we have them all.
So in the case of the number six, rather than just going factors of six, three, two, it's more systematical to go one lot of six, two lots of three.
When we put them into factor pairs, factors that multiply to make six, we don't miss any.
We can do the same thing with that algebraic term, 6a.
Well that would be one lot of 6a, two lots of 3a, three lots of 2a, and six lots of a.
Can you see how listing the factor pairs of six and listing the factor pairs of 6a is actually quite a similar process.
In the case of 6a squared, what do you think's going to change? Will we get more or less factors than 6a, or will it be the same? What do you think? Hypothesise about that.
Make a suggestion.
Say that to the person next to you or say it aloud to yourself.
There's our systematic list for the factor pairs of 6a squared: one times 6a squared, two times 3a squared, three times 2a squared, six lots of a squared.
And then, we need to remember that a lots of 6a will make 6a squared and 2a multiplied by 3a will make 6a squared.
Okay, I'll model one more of those for you and I'll ask you to take your turn.
I'm gonna do the factor pairs of 15x squared.
That's one lot of 15x squared, two is not a numerical factor of 15, three lots of 5x squared.
There we go.
There's all my x squared factor pairs if you will.
I have lots of 3x squared, 15 lots of x squared.
Finished? Well done, we're not.
When I look at 15x squared, x squared is a factor, but x squared is x multiplied by x, so x must also be a factor.
So I need to do the factor pairs for x, that'll be x lots of 15x, and 3x lots of 5x.
That's my full list of factor pairs that make 15x squared.
Your turn, I'd like the factor pairs of 10y squared.
Pause this video and write those down.
Did you get these to start? One lots of 10y squared, two lots of 5y squared, five lots of 2y squared, and 10 lots of y squared.
You know you're not finished.
Well done.
We need our y factors as well.
That'll be y multiplied by 10y.
It would indeed, and 2y multiplied by 5y.
You should have found two, four, six, eight, 10, 12.
There were 12 factors for 10y squared.
Did you get all 12? If not, pause this video and have a look for the ones you missed.
Practise time now, Question one.
Jacob writes down the factors of 8a.
He writes two, four, eight, a, 2a, 4a.
Part A, what factors have they missed? And for part B, what could they have done to make it easier to be certain that they had all the factors? Pause the video.
Try these questions.
Question two, I'd like you to list the factors of each term.
Those terms being 22a, 8a squared, and 8ab.
Okay, feedback time now.
In Jacob's case, writing down the factors of 8a, two, four, eight.
Yeah, they're the factors of eight.
A, 2a, and 4a, has he missed anything? Yes, he's missed one and 8a from his list.
This happens frequently in number.
When we ask students to tell us the factors of 10, two, five.
Don't forget one and 10.
It's the same thing with the term 8a.
Don't forget one and the term itself.
So Jacob didn't log, didn't write in his list one and 8a.
How could he have made sure that he'd listed those factors, all of those factors? Well, he could have listed them systematically in factor pairs.
For question two.
For 22a, you should have found those eight factors.
For 8a squared, you should have found those 12 factors.
And for 8ab, you should have found those 16 factors.
If you didn't find all 16, pause this video and write down the ones you missed.
So we can list factors of algebraic terms now.
Can we find common factors, common factors? June says, I found all the factor pairs of 6c.
Izzy says, I found all the factor pairs of 15c.
Can you think of any factors that both June and Izzy have written down? Pause this video.
Make a suggestion to the person next to you or say out loud to yourself.
There's June's list, the eight factors of 6c.
Here comes Izzy's list.
Hmm, can you see any factors that they both wrote down? There's an obvious one, pardon the pun.
3c was also a common factor that they both wrote down.
Anymore? You're shouting at the screen, aren't you? 3 and c, there are four common factors for 6c and 15c.
We can always find common factors by listing all the factor pairs and then identifying, as we have in this list, all the common ones.
Depending on the term though, there may be a longer list of factor pairs.
For 6ac, how many factor pairs do you reckon there are or how many factors do you think there are? A lot, two, four, six, eight, 10, 12, 14, 16.
Sixteen factors of 6ac.
So we don't really want to have to write down our entire list of factors every single time we're in search of common factors, and this can be a lot of occasions in algebra we're gonna want to spot common factors of two algebraic terms. So is there a quicker way of doing this? If I said find the common factors of 6ac and 15ac, rather than writing down all those factors and there would be a lot, can you think of a quicker way of doing this? Make a suggestion to the person next to you or make a suggestion aloud to yourself.
It doesn't matter if that suggestion is wrong.
Just making suggestions in mathematics is a really powerful thing to do.
Pause this video.
Have a little go.
So there's lots of ways we can do it.
Very logical systematical way is to start by identifying the common numerical factors.
Common numerical factors of six and 15 would be one, two, and three.
One goes into both 6 and 15, two goes into both six and 15, three goes into both 6 and 15.
So once we've got that, next up let's find the common variables.
They're both ac terms. So ac is a common variable, as is a and as is c.
Once we've got these common numerical factors and the common variables, we could systematically list all the common factors now by starting with one, two, and three because there are some common numerical factors.
And then, writing the same for our a term because 3a, 2a and 1a.
Well it is one lot of a, but we don't write one a because it's just a, but we have these common variable terms now: a, 2a, and 3a.
What do you think I'm gonna write next? That's right, c, 2c, and 3c, and to finish ac, 2ac, and 3ac.
That's a really nice logical way to do this.
Find the common numerical factors, the common variables, and then make a systematical list.
I just wanna check that you've got that now.
For these two terms, 8ab and 12bc, could you follow that process? Can you start by finding the common numerical factors of eight and 12, and then find the common variables in these two algebraic terms? And then, I'd like to systematically list the common factors.
Pause this video, give this a go.
Common numerical factors, one, two, and four.
Both factors of eight and 12.
Common variables a times b, b times c, that's just B.
So when we come to systematically list, well we have the numerical factors, one, two, and four.
And then, we'll have the b factors: b, 2b, and 4b.
Practise time now.
Question one, find the common factors of the below expressions: 5x and 20x, 5x and 20 xy, 5xy and 20xy, Have you spotted that line by line? There's only the slightest difference.
If you spot the tiny differences line by line, this might help you to identify the common factors.
Let's go through this.
For 5x and 20x, one and five common numerical factors.
x is a common variable, so we've got x and 5x also.
What's going to change when we change a second term to 20xy? Is there anything more in common? No, same list.
One, five, x and 5x.
And the next line, 5x became 5xy.
We've got xy terms, two of them.
Is that going to change our list of common factors? Yes, it is.
One and five remain the numerical factors.
X, 5x remain the x factors, the y factors are y and 5y and we also have xy factors xy and 5xy.
Next to the list.
I've changed the 5xy to 4xy.
How is that going to affect a list of common factors? There's a change to the numerical factors.
We've now got different numerical factors.
That'll be one, two, and four.
So a list will look like one, two, four, x, 2x, 4x, y, 2y, 4y, xy, 2xy, and 4xy.
They're all the common factors of 4xy and 20xy.
Finally, 5x squared and 20x cubed.
The numerical factors one and five and we'll have variable factors of x and x squared.
So our list will be these six common factors, one and five, x and 5x, x squared and 5x squared.
How did you do? Did you get them all? It's okay if you didn't.
We're not gonna get everything right first time in mathematics.
But it's really important if you didn't get them all, to pause now.
Spot the ones that you missed.
Final part of the lesson now.
Finding the highest common factor, HCF the abbreviation of highest common factor, using Venn diagrams. You would've seen this in number.
If I said find the highest common factor of 60 and 84, you would say, "Mr. Robson, I'm gonna start by finding the prime factor decomposition of 60." 60 is composed of two times two times three times five, and then you find the prime factor decomposition of 84, and then you draw yourself a Venn diagram.
And then you are looking for, do you remember what's next? Well done.
Are there any common prime factors? There are in the lists.
I can see two in both lists or there's another two in both lists.
There's a three.
They go in the intersection of a Venn diagram.
Five is not common with seven.
So that five lives in the area of a Venn diagram, which is 60 and 60 alone, and that's seven will live in the 84 and 84 alone section.
Once we're at this stage, we need to find the highest common factor.
When we do this with numbers, we're multiplying together the numbers in the intersection.
The product of what's in the intersection will give us our highest common factor.
In this case, it's two times two times three, giving us highest common factor of 12.
I like to at this stage just pause and say, "Is that a logical answer? Twelve, is that a factor of 60, a factor of 84?" Yep.
"Does it feel like the highest common factor?" Yeah, pretty logical.
So does it work in algebra? First, we have to decompose 10abc.
We're doing it for 10abc and 15ab by the way.
Decompose 10abc.
So I'll decompose 10 into its prime factors, two lots of five.
I'll then decompose abc into a times b times c.
I'll do the same for 15ab.
Fifteen is three by five and then ab is a times b.
We've decomposed 10abc and we've decomposed 15ab.
We then need a Venn diagram.
Nothing's changed.
We're looking for things in both lists to put in the intersection.
In this case we've got a five, an a, a b, and then we populate the rest of the Venn diagram with the things that are not going to go in the intersection.
And then, the same rule applies.
I need the product of what's in that intersection to gimme the highest common factor of 10abc and 15ab.
In this case it's five times a times b, which we would write more simply as 5ab.
I hope you spotted a really nice simple repeat of what we've done in number but in algebra here.
Let's check that you've got that.
I'd like to fill in this Venn diagram to find the highest common factor 42xy and 140yz.
I will give you the decomposition of 42xy and the decomposition of 140yz.
There's your Venn diagram.
That's all I'm gonna do for you.
Over to you now.
Pause the video, populate the Venn diagram, find the highest common factor.
Populating the Venn diagram then, intersection.
There's a two in both lists, seven in both lists, a y.
And then, three and an x that go in the 42xy only side.
And then, a two, a five, and a z in the 140yz side.
So the highest common factor must be two by seven by y, two by sevens make 14, multiply that by y, and we get 14y.
Highest common factor of 14y.
Alex has been busy.
He's drawn a Venn diagram.
He's trying to find the highest common factor of 60xyz and 140yz.
Looks pretty good.
Ooh, what has Alex done wrong? This is gonna take some searching.
Alex has made a mistake somewhere.
Can you spot it? Pause this video, have a good look.
I think I've got it.
Have you got it? I think I can see more things that Alex should have put in the intersection.
Can you see them? I think we should have had an extra two in the intersection and the z term in the intersection.
That's going to completely change our highest common factor.
Alex has found a common factor.
But without getting that Venn diagram right, it won't be the highest common factor.
So the Venn diagram should have had that two in the intersection, the z in the intersection.
That looks much better now.
I think that's good.
So that'll give us a highest common factor of 20yz.
If you were to compare that answer with the answer Alex had earlier, maybe your mathematical intuition tells you, "This one is the right answer 20yz." It's a far higher factor than the one that Alex had earlier.
Is it possible to find the highest common factor of three algebraic terms? What if I said highest common factor of 2a, 6ab, and 9ab? Well, we need to decompose those terms. 2a is two lots of a.
6ab is two lots of three lots of a lots of b.
There we go.
At this stage, we can put things in the intersection of 2a and 6ab because the two and the a are common factors to both of those terms. 9ab is made up of three lots of three lots of a lots of b.
Yeah, and from there you can spot that there's an a that's both a factor of 2a and 6ab, and also a factor of 9ab, and that goes in the intersection in the centre of our Venn diagram.
We can also put the 3b in the intersection of 9ab and 6ab.
I can't see any other way.
We can simplify that Venn diagram.
So we must be finished and I'm left with just a in the intersection of all three, making a the highest common factor of 2a, 6ab, and 9ab.
Andeep thinks the HCF, the highest common factor, is 18ab.
What mistake has Andeep made? Pause this video.
Say your answer to your partner or say it aloud to yourself.
A very common error.
This one, we see students doing this a lot.
So Andeep you are not alone.
What Andeep has done is he multiplied all the factors together, two times three times three that's 18, times a times b, that's 18ab.
It's not uncommon that we see this multiplying all of those factors together when all we actually want for the highest common factor is the product of what's in the intersection.
Be super careful when you're working with Venn diagrams to get that right.
Practise time now.
Please populate the Venn diagram and then find the highest common factor of 24ac and 30ab.
Once again, I've given you the decomposition of 24ac and the decomposition of 30ab.
Populate them into the Venn diagram, thus finding the highest common factor.
Pause this video, give it a go.
Next, I'd like you to find the highest common factor of the below pairs of expressions, starting with the pair 15x and 10x, and then the pair 15x and 10xy.
You'll notice as we move from question A to B to C to D, moving down the page, there are the slightest changes, the tiniest little changes as we go.
So with that in mind, will that help to speed up the process? We'll see.
Pause this video, Give those questions a go.
Feedback.
Right, populating a Venn diagram for 24ac and 30ab.
I would like all the things that are common factors in the middle.
I can see two in both lists, a three in both lists, and a in both lists.
Done.
Let's populate the rest of 24ac.
Let's populate the rest of 30ab.
We need the product of what's in the intersection, which is two times three times a, thus making the highest common factor of 6a.
Next, feedback on this one.
Highest common factor the below pairs of expressions.
15x and 10x, I could populate the Venn diagram with that.
In the left hand circle, three lots of five lots of x, that's 15x.
In the right hand circle, five lots of two lots of x, that's 10 x.
What's in the intersection? Five lots of x, highest common factor 5x.
The only difference between question A and question B is the term on the right hand side has changed.
10x becomes 10xy.
So I can change my Venn diagram to reflect that by multiplying by y on the right hand side.
That's five times two times x times y.
That's 10xy.
Has the intersection changed? Not at all.
So the highest common factor remains 5x.
What's changing between question B and question C? Well, 15x is becoming x squared and 10xy becoming 20xy.
I can turn 15x into x squared by multiplying by another x.
I can turn 10xy into 20xy by multiplying by another two.
Has anything changed in that intersection? No.
So the highest common factor remains 5x.
To get to the next stage now, the only subtle difference between question C and question D is that 15x squares been multiplied by y.
I can reflect that by multiplying by y.
You spotted it.
Well done.
The y on the left hand side of the Venn diagram, the y on the right hand side of the Venn diagram should now be in the intersection.
So for the first time, we've got a different highest common factor.
We've now got the highest common factor five multiply by x multiply by Y.
That's 5xy.
Question E still 15x squared y, but it's now 20 x squared y the power of five.
So 20xy becoming 20 x squared y power of five, means multiplying it by x, and multiplying it by y four times.
So on the right hand side, multiplying that by x and multiplying by y once, twice, three times, four times.
Yeah, you guessed it.
The x is on both sides of the Venn diagram can now go in the intersection giving us a different highest common factor this time, five times x times x times y, which is 5x squared y.
Last one 15x squared y, 60x squared y to power five.
Well, all I need to do is multiply the term on the right side by three so we can do that.
There we go.
Oh threes, common threes.
There you go in the intersection, giving us the highest common factor of three by five by x, by x, by y, that's 15x squared y.
That's the end of today's lesson.
To summarise, algebraic terms are made up of factors.
For example, the factors of 2ef, or one, two, 2, 2e, f, and 2f.
We can find the highest common factor of two algebraic terms. For example, the highest common factor of two ef and 10fg is 2f.
Thanks for joining me today.
I hope to see you again real soon for some more algebra.