Lesson video

Hi, I'm Mrs. Wheelhouse.

And welcome to today's lesson, on factorising using the difference of two squares.

This lesson is part of our unit on algebraic manipulation.

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

By the end of today's lesson, you'll be able to factorised quadratics of the form x squared plus bx plus c, including using the difference of two squares.

In our lesson today, we're going to be using a lot of technical language, and you can see some of the language we'll be using on the screen right now.

Do feel free to pause the video and read through these keywords and their definitions so that you're confident using these today.

Now, you're ready to begin.

Let's start.

We're going to begin by reviewing factorising using distributivity.

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

When we are factorising quadratic expression in the form x squared plus bx plus c, we can use an area model to help us see the factors.

So, for example, if factorising x squared subtract 9x subtract 36, I know I can place the x squared term here and the -36 here.

The other terms in the area model must sum to -9x.

Jacob suggests, "They could be -5x and -4x." Or, "They could be -10x and x." "And there are loads of combinations" says Andeep.

So how can we work out what are correct? Because so far, what Jacob and Lucas have suggested both sum to -9x.

However, we can use the fact that we know the constant term is -36, and this has to come from multiplying one positive and one negative constant.

So, actually, if we consider the factor pairs of 36, bear in mind that one of the two factors must be negative.

I can see all the different possible combinations.

I know that I have to sum to make a -9.

And Lucas points out, "That must mean I am using 3 and -12." Place them in my area model means that I can check the four partial products and make sure that I would, when I simplify, reach the form x squared minus 9x minus 36, because that's what I started with.

I can then see the dimensions of the area model.

So I know the factorised form must be x + 3 multiplied by x - 12.

So what key features of this quadratic will help us to factorise? Well, we know that x squared is the product of x and x.

And 20 is the product of two positive or two negative constants.

Andeep thinks this must be the product of negative constants.

Why do you think he said that? Well done if you spotted that we need the x term to be negative.

So in order for that to happen, at least one of the partial products must be negative.

Well, since either the constants have to both be positive or both be negative, it means they have to both be negative here.

Jacob suggests that we can split the x term into -5x and -4x.

And that's because -5 times -4 is 20, and -5 add -4, is -9.

What we're going to look at here is using the distributive law to factorise.

So what we've done is we've written the -9x, split into -5x and -4x.

In other words, we've decomposed the -9x term.

What we've then done is split the four terms in half.

So we have two terms at either side of the dotted line.

And I'm going to consider these as two separate expressions and factorise each.

So if I consider x squared minus 5x, what's the highest common factor of the two terms? Well, that's x.

So I can write this factorise as x lots of x - 5.

I'll now consider the other two terms. So what's the highest common factor of -4x and +20? Well, that's -4.

So I have -4 lots of x - 5.

What do you notice about the brackets? Is there another way that we can write this whole expression? That's right, the brackets are the same.

So we can treat the expression the bracket as a term and collect like terms. So what we have is x lots of x - 5, take away 4 lots of x - 5.

Or I can say I have a total of x - 4 lots of x - 5.

And that shows how I've used the distributive law to factorise this quadratic.

Jacob says, "Now I'm not sure if this is right.

I think I need to check the answer by calculating the product." So that's what he's going to do.

He's got the factorised form we think is correct, and he's going to use his area model to calculate the four partial products, and he's also going to use the distributive law alongside.

And we can see the four partial products are x squared minus 5x minus 4x plus 20.

When we simplify, we reach x squared minus 9x plus 20.

Looks good to me, and it looks good to Lucas as well.

Let's consider factorising x squared plus 5x minus 14.

Whilst as Lucas 5x could be written as 3x + 2x.

So let's try splitting it up this way.

So again, I'm going to divide the four partial products into two groups.

Let's factorise the first pair of terms. Well, the highest common factor is x, so I have x lots of x + 3.

And for the second terms, highest common factor is 2.

So I have 2 lots of x - 7.

What do you notice? That's right, the brackets are not the same.

We cannot collect like terms in this form.

And that tells us that when we decompose the 5x into 3x and 2x, this was not correct.

As Lucas points out, "Now we know what will happen if we choose the wrong decomposition of the x term." Let's try this.

We're going to go for x squared plus 7x minus 2x minus 14.

Can you see here, I've chosen to decompose into 7x and -2x, and 7 multiplied by -2, gives me -14.

I've got a feeling this is likely to work.

Remember how the area model worked? Same principle here.

So, highest common factor of x squared and 7x is x.

So we have x lots of x + 7, and then for the other two terms we have -2 lots of x + 7.

And now we can see the expressions in the brackets are the same.

So we can collect like terms. So we have x take away 2 lots of x + 7.

How would you factorise x squared plus 6x minus 7? Well, Andeep says, "I like splitting up the x term and using the distributive law, because I get to practise factorising and collecting like terms." And that's what Andeep's doing here.

But Lucas prefers to use an area model, because he finds it easy to spot what the final binomials will be.

And that's Lucas' approach.

Now, Jacob points out that both methods seem the same to him, but he does know that it's important to check our answer by expanding the binomials.

So it doesn't matter which method we pick, but we should absolutely check that we get back to the expanded form.

Andeep is factorising x squared minus x minus 90.

Please check if his answer is correct.

If you think it's not, please, can you correct it? Pause and do this now.

Welcome back.

Did you check if Andeep was correct? Well, remember we can check by expanding the factorised form.

So using our area model or using the distributive law, we can see that we reach the four partial products x squared minus 9x plus 10x minus 90.

Now, when we gather the like terms, we get to x squared plus x minus 90, and that is not what we started with.

So Andeep was not correct.

So, did you spot what he had to change to make it correct? That's right, the +10 and the -9, those weren't the correct factors of -90 to use.

It should, in fact, have been 9 and -10.

Let's see what happens when we calculate the partial products now.

So, we think that factorised form is x + 9 multiplied by x - 10.

And we can see from the partial products, that when we sum the two x terms, we end up with -x, which is what we should get.

It's now time for your first task.

For question 1, fill in the blanks in the methods to factorise x squared plus 18x plus 32.

So you're asked to first use an area model, and then use distributivity.

Pause and do this now.

Welcome back.

Question 2.

Please factorised each quadratic expression.

Pause and do this now.

And now the second part of question 2.

Again, please factorise each quadratic expression.

It's now time to go through the answers.

So, using your area model, you should have shown that the two factors were 2 and 16, and then written that in factorised form as x + 2 multiplied by x + 16.

Or of course, you could have written your brackets the other way round.

And then for part B, you can see how I use the distributive law to complete this factorization.

Feel free to pause if you want to read this through carefully.

Question 2, you had to factorise each quadratic expression.

So the 2A should reach, x + 4 multiplied by x + 1.

2B should be x + 6 multiplied by x + 2.

2C, x + 9 multiplied by x + 3.

2D, x + 6 multiplied by x + 6.

And you may have written this as x + 6 all squared.

2E, x subtract 6 multiplied by x subtract 2.

And 2F, x - 8 multiplied by x - 3.

For the second part of question 2, 2G, we have x + 10 multiplied x - 5.

2H, x + 5 multiplied by x - 10.

2I, x + 7 multiplied by x - 6.

2J, x + 21 multiplied by x - 20.

2K, x + 2 multiplied by x - 13.

And 2L, x - 7 multiplied by x - 4.

Now don't forget for all of these, you could have your brackets written round the other way.

It's now time for the second part of today's lesson, and that's on factorising using the difference of two squares.

This diagram shows the expression x squared minus 16.

And Izzy says, "I cannot write this in factorised form because it has not made a rectangle." And she's right.

But what could we do to make it a rectangle? Is there a way to do that? So in its current form, it's not clear to see what the factorised form would be, because the partial products do not seem to make a rectangle.

So in that sense, Izzy's right.

But is there a way to fix it? She could add zero pairs.

Are we still showing the expression x squared minus 16? We are indeed, because the two x terms are a zero pair, so they sum to 0x.

So what is x squared minus 16 in factorised form? It's x + 4 multiplied by x - 4.

Here, we have the expression x squared minus 9, and Jun thinks we can use zero pairs to form a rectangle again.

and I think he's right.

You can see it here.

Izzy goes, "Well, actually, I think this is a square, and you can write it as x + 3 all squared.

It certainly looks like a square.

Do you agree with Izzy? Well, although it looks like a square, Izzy's incorrect.

Because in order to make the zero pair when multiplied, one term in each binomial needs to make a zero pair.

So the side lengths are x + 3 and x - 3, and the constant needs to be -9, not +9.

Quick check now.

Please complete this model to show the factorization of x squared subtract 4.

Pause and do this now.

Welcome back.

You should have added two zero sum pairs.

So the factorised form is x + 2 multiplied by x - 2.

What features did you notice in those three examples? I'll put them up here to remind you.

So, here with the algebra tiles, along with the expanded form and the factorised form.

So did you spot that the constant term makes a square with the algebra tiles? The absolute value of the constant is a square number.

So 16 is a square number, 9 is a square number, and 4 is a square number.

Did you spot that the constants in the binomials make a zero pair? So 4 and -4, 3 and -3, and 2 and -2.

The factorised form in each case could be written as x + a and x - a, where a is the square root of the absolute value of the constant in the quadratic.

So, for example, if we consider the first one, 16 was our square value, and when I square root that I get +4 and -4.

Izzy says, "I don't think we're going to be able to factorise x squared minus 10 in the same way.

And Jun says, "I also think that x square plus 25 won't work either." Do you agree with them? Explain why.

Well, 10 is not a square number.

So here, Izzy positions the algebra tiles, the x term will never make a zero pair.

And for Jun, 25 is a square number, but here the x term will not cancel because they're not zero pairs.

The constant term would have to be negative in order for this to work.

So we can use algebra to generalise this, and explore the structure of these special cases.

If an expression is in the form a squared minus b squared, where a and b are any terms, then we call this form, the difference of two squares.

And this is because it expresses the difference between a squared and b squared.

This can be written as the product of two binomials.

Quick check.

Which of these area models show a product of two binomials which can be written as the difference of two squares? Pause and make your choice now.

Welcome back.

Did you spot which one it was? That's right, it's only this one.

What expression is represented by this area model? Please write your answer in expanded form and in factorised form.

Pause and do this now.

Welcome back.

In expanded form, we had x squared minus 81, because the 9x and the -9x, there is zero sum pair.

And the factorised form is x + 9 multiplied by x - 9.

Izzy says that "When I see a quadratic, which is the difference of two squares, I can treat it as a normal quadratic with an x term of 0x." So, for example, if we're factorising x squared minus 100, and we don't spot that this is a difference of two squares, then we could write the quadratic like this.

And then we can decompose the 0x into -10x and +10x.

And again, use the distributive law to factorise, and we reached x + 10 multiplied by x - 10.

Now Jun says, "100 is a square number.

And as this is the difference of two squares, we can square root 100 to get the constant terms in the binomials." And this will work.

So, if you can spot that you have a difference of two squares, you can use Jun's far quicker method.

But if you don't spot it, it's not a problem.

You can use Izzy's approach and you will reach the correct answer.

Jun has seen that expressions in the difference of two squares form can be factorised by square rooting each term.

Now, this can help us factorise trickier expressions which are the difference of two squares.

So, for example here, we're going to factorise 9x squared minus a squared b squared.

But first thing we're going to do is to rewrite each term.

9x squared can be written as 3x all squared, and a squared b squared can be written as ab all squared.

This makes it very easy to see what will go in the brackets for our factorised form.

I'll have 3x + ab and 3x - ab, and we can check our answer by expanding, using either distributive law or our area model.

And we can see that we get back to the correct expanded form.

Quick check now.

Please match the expressions in difference of two squares form with their factorised form.

Pause and do this now.

Welcome back.

You should have paired them as follows, x squared minus 4, pairs with x + 2 multiplied by x - 2.

x squared minus 16, pairs to x + 4 multiplied by x - 4.

4 minus x squared, pairs to 4 + x multiplied by 4 - x.

And x squared minus y squared, matches to x + y multiplied by x - y.

Which is the correct factorization of 4x squared minus 64y squared? Pause and make your choice now.

Welcome back.

Which one did you go for? We should have gone for B.

2x multiplied by 2x gives us 4x squared, and 8y multiplied by -8y, gives us -64y squared.

In A, we would need one term in each binomial to be a zero pair.

And what we have are, both of the x terms are positive, and both of the y terms are negative.

So this is no good.

And remember, when we square root 64y squared, we get 8y, not 16y.

It's time now for our final task.

For question 1, spot and explain the mistake made in each of these attempts to factorise.

Pause and do this now.

In question 2, please factorise each of these expressions.

Pause and do this now.

And then the second part of question 2.

Pause and factorise these expressions now.

Question 3.

This shape is made from a square of side length a with a square of side length b taken away.

Part A, please write an expression for the area of this shaded shape.

In part B, we've now split the shape into two rectangles.

Please label the sides of the top rectangle in terms of a and b.

And then we've taken the top rectangle and rotate it, and then moved it to the right of the bottom rectangle.

Please label the sides of this new rectangle and then consider how we can express the area now.

And then for part E, tell us what you've shown.

Pause while you work through question 3 now.

Welcome back.

Question 1, you had to spot and explain the mistake made in each of these attempts.

For A, the constant has been halved instead of square rooted.

In B, the constant terms should be a zero pair, and what we should have is a -4 and a +4.

In C, the constant terms are not a zero pair, so the x terms will not cancel.

In D, it is the x terms which should be the zero pair, not the constant terms. So it should be 3 + x and 3 - x.

For E, this is not the difference of two squares, 10 is not a square number.

And then for F, the constant should have been square rooted.

What we have here, if we expand the bracket, is x squared minus 81.

Question 2, you had to factorise each of these.

So for 2A, we should have x + 6 multiplied by x - 6.

B is x + 12 multiplied by x - 12.

C is x + 7 multiplied by x - 7.

D is x + 1 multiplied by x - 1.

E is 15 + x multiplied by 15 - x.

And F, we have x + y multiplied by x - y.

In G, 2x + 11 multiplied by 2x - 11.

H, 1 + 4y multiplied by 1 - 4y.

In I, 4xy + 14t multiplied by 4xy - 14t.

In J, 2x squared plus 1 multiplied by 2x squared minus 1.

In K, 3abc + 7ab multiplied by 3abc take away 7ab.

And then L, we have 100 + 6x multiplied by 100 - 6x.

Question 3, you had to write an expression for the area of the shaded shape.

Well, if we consider the area of the whole square, that would be a squared, and then we take away the part that is missing, that's a square with length b, so the area is b squared.

So we have a squared take away b squared.

Part B, the shape has been split into two rectangles and you had to label the sides of the top rectangle in terms of a and b.

Well, we know that the height of this rectangle is b, because remember, we took a square of side length b away, and it had the same height as the rectangle we left with, and then the length of the large square was a, and we took away a length of b, so that length of the top rectangle must be a - b.

In part C, we had taken the pink rectangle, rotated it, and placed it to the right of the bottom rectangle.

You had to label the size for the new rectangle.

Well, we know now that the entire length of this shape is a + b, and we knew that the length of the rectangle was a - b.

Well, it's now the height of this new rectangle, but still a - b.

In part D, how do you express the area of this shape? Well, we have a + b multiplied by a - b.

Now in part E, you had to tell me what you noticed.

Well, hang on a second.

What we've shown is that the area of the shape at the top, which was a squared minus b squared, is equivalent to the area of the shape at the bottom, which was a + b multiplied by a - b.

Remember, all I did was cut a bit off and just move it somewhere else.

I didn't add any extra area, and I didn't take any extra area away.

So we have demonstrated the general form for the difference of two squares and its equivalent factorised form.

It's now time to sum what we've learn today.

When the coefficient of the x term in an expression of the form x squared plus bx plus c is 0, it may still be possible to factorise by using zero pairs.

When an expression is in difference of two squares form, it can be written as the product of two binomials with one term the same in each binomial and the other terms making a zero pair.

Being able to identify difference of two squares and using this structure, can help us factorise efficiently.

Well done.

You've worked really well today.

I look forward to seeing you for more lessons in our algebraic manipulation unit.