Lesson video

Well done for deciding to learn using this video today.

My name is Ms. Davies and I'm going to be working with you as you work your way through these really exciting algebra topics.

Now that you've got the basics, we are going to start bringing together all the things you know about algebra to tackle some really interesting problems. So hope you're looking forward to it and I hope that you're spending the time to make sure you really understand what is going on.

The more you understand with mathematics, the easier it is to replicate things in the future, even if situations are slightly different to ones you've seen before.

With that in mind, make sure you pause the video when you need to, check your answers, and you can use calculators for bits and pieces, especially to check your answers as well.

Make sure then that you've got everything you need and let's get started.

Welcome to today's lesson on factorising quadratics in the form axe squared + bx + c.

By the end of the lesson you'll be experts on how to factorised quadratics of this form.

With that in mind, you want to make sure that you are confident with factorising quadratics in the form x squared + bx + c.

So can you factorised something like x squared + 5x + six? To factorise is to express a term as the product of its factors.

If you're not sure about that, you might want to go back and have a look at a lesson on that first before having a look at this one.

Today we're going to have a look at using area models and then we're going to have a look at methods for factorising that don't use area models.

We're going to start then by quick reminder on how to factorised quadratics in this form.

If you've done this before, then the skills that you use to factorise quadratics with 1x squared, we can use to factorised quadratics with a coefficient of x squared and that's what we're focusing on today and it's going to be the same methods.

So let's look at factorising x squared + six x - 16.

This could be written as the product of two binomials.

You might want to start then by writing out two brackets.

Now in order to get an x squared, we know that's the product of x and x, so both of our brackets is going to have a term of x.

Then we need to work out the other terms in our binomial.

Now we know from this area model that they're going to have a product of -16.

Well, that gives us a bit of a clue.

We know that one of our values will be positive and one of our values will be negative.

So we can think about things that have a product of -16.

I've put some options up on the screen, so now we need to turn our attention to the other term in our expression, the six x.

So one of our factors of -16 is going to be multiplied by x and the other one's going to be multiplied by x and then they're going to be collected as like terms and that's going to have to give us six x.

So essentially the values that we are looking for multiplied to -16, but are going to add to six.

Well, I dunno if you spotted, that's our option.

If we now fill in the rest of our area model, we can double check that this is going to give us the product we want.

So x + eight x - two.

Always check your answer by expanding and check you get what you want afterwards.

This is even more important now we're looking at these trickier quadratics today.

So quick check and we do end up with what we wanted to.

So now see if we can apply the same thing to quadratics of the form axe squared + bx + c.

So let's look at this example.

So this time we've got 2x squared, so that's going to change a little bit of what's happening in our area model.

To start with, you might want to do this with algebra tiles.

So if you've got some algebra tiles with you or you've got a computer programme where you can use algebra tiles, have a go and see if you can arrange these into a rectangle.

You've got 2x squareds, 11xs, and five.

Pause the video.

Come back when you think you've got a rectangle.

Okay, so to start with, our x squareds could be arranged in a rectangle this way or you could put them horizontally.

Either way, the only way to get a product of 2x squared is x and 2x.

That's if we are using integer coefficients.

So we've got 2x as one dimension and x as the other.

Now our constant is five because that's a prime number that reduces the options we have for how to place our algebra tiles.

In fact, we can either have them one by five or five by one.

Let's try this way first.

We then fill in with our x tiles.

You can see that we've got seven x tiles.

Well that's not going to work because we wanted 11 x tiles.

What you can hopefully see is if we put our five round the other way, let's have a look.

Now, fill in our x values, just check we've got 5, 10, 11, therefore they must go this way around.

Our final product then is x + five to x + one.

It doesn't matter which way round you write the brackets, but it's important that x + five is one binomial and 2x + one is the other binomial.

So we can use an area model and our negative number skills to factorise quadratics with negative terms. So we could factorised 3x squared + 11x - four.

Now you can do this with algebra tiles, but one thing to be aware of is that we may well need zero pairs to fill this in.

What that means is just starting with 11xs might not be enough.

You're probably going to need a few more xs and a few negative xs to get the correct rectangle.

So instead, I'm just going to use an area model.

Now in order to get 3x squared, the only way to do that with in coefficients is x multiplied by 3x.

We've got another prime coefficient of x squared, which makes this a bit easier.

So we've got X and 3x.

Now the other term in the binomials will multiply to negative four.

So we do have options, we've got 1 X negative four, two X negative two and negative one X negative four.

Let's think about how we get that x term then of 11x.

Well, one constant is going to be multiplied by 3x and the other one by 1x before they're added together.

Let's suggest spending some time just looking at the values and seeing if you can spot one that you think is going to give you the right value.

If you think about 11x, that's a reasonably large number.

We know that one of our values is going to be positive, the other is going to be negative, so they're going to be reasonably far apart because they're going to have to have a difference of 11.

So play around with some ideas.

If you think you find one that looks reasonable, then write it on your area model and give it a go.

If it's not right, that's no problem.

You've ruled that one out and yeah, you can try another one.

So let's try negative one and four.

So let's put the four with the 3x, we've got 3x + four and x - one.

And then let's try it out.

We got -3x and 4x.

Oh, hang on.

That didn't give us the x value we wanted.

Well that's no problem.

We'll just try something different.

So Sophia says, try swapping the constants over Laura reckons that's the same thing.

What do you think? Okay, let's try it out then.

So we're going to put the negative one in the binomial with the 3x and then x + four is our other binomial.

Okay, so this time we have got the right product and interestingly it was different to the one before and that's what we said earlier, wasn't it? It does matter which binomial you put the constants in this time.

So our final answer is x + four, 3x - one.

Of course you could write the binomials either way around as long as you've got x + four and one and 3x - one as the other.

So Laura's a bit unsure, but what do I do if I pick the wrong numbers and it doesn't work? Well, the thing is that's absolutely fine when you're using this method to factorised, it's unlikely that you're going to spot the right value straight away every single time.

So you can start using the features of numbers to pick ones that you think are likely to work and then just try 'em and if they're not right, try something else.

And that's the beauty of using an area model is that you can use that to work things out as you go along.

So let's try one together then.

Now five is a prime number.

So my choice is only x and 5x in my binomials.

I'm looking for values that multiply to negative six and one's going to be positive and one's going to be negative.

So those are my options and now I'm thinking that one of them is going to be multiplied by five and the other one's going to be multiplied by one and added together and they're going to have a difference of 13.

So they're going to be reasonably far apart.

So I'm going to do a little bit of an explore.

Once I'm happy that I'm fairly close, I'm going to try it out.

So I'm going to go with x - three and 5x + two.

Fill in the values and I'm happy then that I've got X - three, 5x + two.

When I'm happy I do need to double check.

So I've got x lots of 5x + two + negative three lots of 5x + two.

It's going to be 5x squared + 2x - 15x - six.

And that does give me the trinomial that I wanted.

5x squared - 13x - six.

Right, time for you to have a go.

Can you factorised this quadratic and then check your answer.

Off you go.

Let's have a look.

So I've gone with positive five and negative two and I've written them that way around.

So that gives me -2x and 10x which gives me the 8x that I wanted.

So I think the final answer should be x + five 2x - two.

But let's check it.

So I have x lots of 2x - two + five lots of 2x - two.

2x squared - 2x + 10x - 10.

And that simplifies to what I wanted.

So give yourself a tick if you got that right.

Okay, so Sophia says x - five, 7x + one is the correct factorization of 7x squared - 12x + five.

Without finding the right answer, I want you to think about why Sophia's answer cannot be correct.

Can you use some features of that quadratic? Laura's had a go as well.

I'd like you to use any method you wish to show that that is also incorrect.

And then finally, can you figure out the correct factorization? Use an area model to help you if you wish.

Off you go.

Okay, so I wonder if she said something like this.

The constant needs to be positive when expanded in our quadratic we've got + five on the end.

So the constant terms in both binomials should either be both positive or both negative in order to multiply to a positive number.

In this case because they also have to give us -12X as an x term, they're going to both be negative.

We can use an area model to show that Laura is incorrect.

You can see that she ends up getting -36x.

<v ->36 is a lot further from zero that we want it to be.

</v> So did you find the correct factorization? Actually Laura was close.

She just had a constants in the wrong binomials.

So you should have x - one and 7x - five.

That gives you 7x squared - 5x - 7x + five or 7x squared - 12x + five.

Fantastic time for you to have a practise then.

So you've got three expressions to factorise.

You are welcome to use algebra tiles to help you if you wish.

Make sure you write your final binomials underneath, give that a go, then come back for the next set.

Fantastic, I would like you to have a go now at factorising each of these expressions.

There's quite a few to do, however, you'll notice that there are some similarities between some of them as you go through.

Once you're happy, come back, I want to have a look at the answers.

Let's have a look then if you use algebra tiles, those are the correct rectangles to help you and we end up with 2x + three, x + three.

The first one.

2x + one, x + two.

For the second one.

And 2x + two, x + two for that third one.

Fantastic, I would like you to pause the video and check your answers for question two.

If you didn't get the right answer, I suggest you go back to your working and make sure you check through what you should find with these it's 'cause you are always checking your answers to make sure they're right.

There shouldn't be too many mistakes in your final answers.

Doesn't matter at this stage if there were some mistakes as long as you know how to check them and correct them.

Once you're happy with your answers, come back and we'll move on to the next bit.

So now I'm going to have a look at a similar linked method to factorised and quadratics but no longer with the area models.

When the coefficient of x squared is not prime is not always easy to see the correct binomials.

In all the examples we've looked at so far we've had 2x squared, 3x squared, 5x squared.

So there's not been options as to what our first term in our binomials would be.

Now there's going to be some options.

What that means is there's quite a lot of combinations that we might have to try.

So Alex wants to factorised 4x squared + 2x - six.

He's got his algebra tiles and he's a bit concerned.

He says there's just too many factors of 4x squared and too many factors of six to get the right combination.

You might want to pause the video and have a play around with algebra tiles.

You might find the correct factorization, that's fantastic.

And Aisha's concern as well.

She says we're going to need some zero pairs, we don't know how many we might need to get the correct x term.

Instead of using algebra tiles then we could use the distributive property to help us.

So let's have a look.

What we can do is we can split the x term up into two like terms. If you think about what happens when we use our area model, we end up with two x terms and then we collect like terms. So what we can do instead is work backwards and work out what those x terms were before they were added together.

So let's try this.

What about if we split two x into x and x? So we have 4x squared + x + x - six and that's equivalent to our trinomial.

We're going to factorise the first two terms and then the second two terms. So that gets us x lots of 4x + one and one lot of x - six.

Right, that doesn't work does it? We using distributive property, we want our expressions in the brackets to be the same so we can collect them like light terms but they're not.

So Alex is correct, we need to try something else.

Well for a start that was a bad choice to spit it into x and x because we know that one of our values is going to be positive and one of our values is going to be negative in order to get the product of negative six.

So one and one was not going to be a good decision.

There are still many possibilities, but there's some features that are going to help us out.

For start if we're factor the terms, we want to see if there's a factor.

So it's useful to pick x terms to share a common factor with the constant terms, so in this case six.

So in this case, let's see if we can pick values that are divisible by two and then they'll share a common factor with four and our 4x squared and six in our negative six terms. So I've gone with 4x squared + 4x - 2x - six.

Let's try that on.

We've got 4x lots of x + one - two lots of x + three.

Okay, that's still not quite right.

So if we use some zero pairs, let's go up to 6x and then down to negative 4x.

And let's try that.

We've got 2x lots of 2x + three - two, lots of 2x + three.

Okay, now they have the same expression.

We can add them like like terms. So we've got 2x - two, lots of 2x + three.

Of course we are going to want to check that answer, make sure it's correct.

I suggest you pause the video now have a look at the working and how we got to that answer and double check it.

Make sure you agree that it is right.

So Ally says we still have to try quite a few different decompositions.

He wants to see if there's a pattern.

Now I wonder if you notice this pattern that Aisha is seeing.

So Aisha is spotted that in our decomposition of the x term, so we've ended up with 4x squared + 6x - 4x - six.

That six and four seem to appear quite a lot.

The coefficient of x squared and the constant were four and negative six, whereas the x terms were positive six and negative four.

She's wondering if that is a coincidence.

So let's have a look.

I'm going to show you some examples from earlier on in the lesson.

I'd like you to pause the video and see if you can spot any patterns with how the x term is decomposed.

Then come back and we'll see if we've discovered the same thing.

Okay, so Aisha's observation that the same values are used in the coefficient of x squared and the constant as in the decomposition of the x term does appear quite often.

So if you look at that second one, 3x squared + 4x - seven, you can see we've got three and negative seven and we've got negative three and seven.

So not exactly the same coefficients, but they are the same values just with different signs.

You can see the same thing happens for 2x squared - 13x + 11 and something similar in 2x squared - x - one.

However, if you have a look at some of the others, they're not the same values.

I wonder if you spotted this though.

If you multiply the coefficient of x squared and the constant, so three X negative seven, you get -21.

If you do the same with the decomposition of the x term, you also get -21.

You can check it's the same for the middle one, three X negative seven is -21, negative three X seven is -21.

And let's look at this one, two X 11 and -11 X negative two.

Both of those have a product of positive 22.

So what we've seen then is multiplying the coefficient of x squared and the constant gives us the same values as multiplying the coefficients of the decomposed x terms. Let's see how we can use this now.

So in general, when a quadratic is in the form axe squared + bx + c, the x term can be decomposed into two like terms with a coefficient sum to the value of b but multiply to give the same value as a X c.

Let's put that into practise.

So we've got 12x squared + 11x - 15.

You can use an area model, you might be able to spot the right binomials.

However, there's quite a few options for the values that multiplies give you 12x squared.

So let's try out this method.

So 12 multiplied by -15 is -180.

So we're looking for values that multiply to give negative 180, but some to 11.

Try a few things out.

Hopefully you can see it's going to be negative nine and 20.

So we can now write this out as 12x squared + 20x - 9x - 15.

It does not matter which way round you write those two x terms and then we factorised 'em separately to get 4x lots of 3x + five - three lots of 3x + five.

Finally then we have 4x - three lots of 3x + five.

Of course we're going to check that back and make sure that works.

Pause a video if you'd like to do that now.

Okay, what I'm going to show you now is how to prove this works with algebra.

Now this is really important when you're working in mathematics in general to have an understanding of why things work.

However, if you find it a little bit tricky to follow along with this, that's absolutely fine.

We've already demonstrated that this works with lots of different examples.

We've explored how this is forming.

Okay, so please do look through but do not panic if you're not a hundred percent convinced by this explanation.

So let, so let's just assume that we've got a quadratic axe squared + bx + c, and it's going to be factorizeable into the form px + q X rx + t.

So we've got an x term in both and a constant in both.

So let's expand this and see what we get.

So we get px lots of rx + t + q lots of rx + t, prx squared + ptx + qrx + qt.

We can see clearly if we compare that to what we started with, that pr must be a and that makes sense.

If you look at your binomials, we know that multiplying p and r is going to get you a.

And we can see that the constant is going to be q X t.

And again, that's something we've explored a lot.

So what we're really going to focus on is the x terms, the decomposed x terms. So we can see that the coefficient of the x term is pt + qr.

Now if you look carefully at those, if you'd multiply them together, you'd get pqrt, it doesn't matter which way round we write those.

When we multiply it, we can multiply in any order.

Now if you look, then that is the same as a multiplied by c.

I'm just going to run through the proof again.

So we've expanded our brackets, we've compared them to what we started with in our trinomial and we've seen that the coefficient of the x term is pt + qr.

Noticing that when we multiply those together, we get pqrt, and that's the same as multiplying a and c.

Therefore, when we factorise by decomposing the x term, the coefficient of the two terms will sum to the value of b but multiply to give the same value as a c.

So let's put that all into practise.

I'm going to show you one and then you are going to have a go.

So I'm going to do a multiplied by c, that gives me positive 90.

I'm looking to decompose the x term into two terms where the coefficients add to -91 but multiply to 90.

Hopefully you can see that's going to be negative one and -90.

I'm going to write that with a decomposed x term and factorised separately.

So we've got a common factor of 10 and a common factor of x.

So I've got 10x lots of x - nine and then - one lots of x - nine.

And I know I'm on the right track because my brackets are the same.

That gives me then 10x - one lots of x - nine.

Make sure you check your answer before moving on.

Great to have a go at 4x squared - 11x + six.

When you're happy, come back to see the steps of working Perfect, so four multiplied by six is 24.

We're looking for values which add to -11 but multiplied to 24.

Well done if you spotted that was negative eight and negative three.

If we write that as part of our decomposed expression and factorise separately.

If you chose to write the negative eight and the -3x the other way around, this step might look a bit different, but you will get the same final answer.

We've got 4x - three, x - two.

And make sure that you have checked that as well.

Let's have a look at another example.

This time we've got a negative constant, six multiply by -10 is -60.

So looking for values which add to negative seven but multiply to -60.

We know that one's going to be positive, one's going to be negative, they're going to have a difference of seven.

Well, I dunno if you spotted -12 and five.

Again, it doesn't matter which way round you put the x terms, it will make the next step look slightly different, but you'll get the final answer the same.

Our brackets are the same, so 6x + five, x - two.

And again you want to make sure that you've checked that answer before moving on.

Your turn, have a go at 8x squared + 10x - three.

Off you go.

So, eight multiplied by negative three is -24.

So we want values that add to 10, but multiply to -24.

So we're looking for one positive, one negative.

Well done If you spotted that's 12 and negative two, it needs to be positive 12 and negative two to sum to 10.

Write them out either way around and factorise.

You should have got 4x - one, 2x + three.

Check your answer.

Fantastic.

Right, so we've looked at quite a few different methods there today.

It's entirely up to you which one you want to play around with.

Use these questions to practise.

You'll notice with this first set there's lots of similarities which might help you really explore the structure of these binomials.

Give those a go, come back when you're ready for the next bit.

Well done.

This second set then we've got a few different coefficients of x squared and we've also got a few different variables here.

We've got some Ys and some As and some Bs.

So just pay attention to what variable you are working with.

Give those a go.

Come back when you're ready for the answers.

Well done.

I hope you had some fun exploring some of these structures, playing around with the number system and finding things that work.

It's a really great feeling when you find that one that does work.

Pause the video and check your answers for these because they're all quite similar.

Make sure you pay attention that you've got the right symbols in the right brackets.

Fantastic.

And this second set.

Again, make sure you've got the right variables this time and you've got your terms in your binomials, the correct way round with the correct sign.

Fantastic.

I hope you're feeling really proud of yourself after that lesson.

There was lots of interesting algebra and number skills that you were using to find those factors.

What I love about factorising quadratics is a really clear way to check that you are correct and you get that sense of satisfaction when you did find those values that worked.

So today we've looked at using an area model to find factors and the fact that when our coefficient of x squared is not prime, or if our constant is not prime, these values can be trickier to spot.

So we've looked at how this method of decomposing the x term works and we prove this always going to work.

So we're feeling really confident that we can use this method, knowing where it came from.

We can use key features of our values.

We're looking at that constant and thinking about whether it's going to be two negative values two positive values that are going to have the product of that constant.

And we're looking at values that sum to give us that x term.

We're thinking about how that x term is going to be decomposed.

What you'll find with factorising quadratics is the more you do it, the easier it gets.

So keep practising , keep working hard and I look forward to seeing you again.