Expanding expressions in Brackets

The term which is outside the brackets (2 in the first case, and a in the second case) is multiplied by the whole of the contents of the brackets, i.e. every term inside them. This means that we can remove the brackets, providing that we multiply the term before the brackets by each of the terms inside. This is the reason that each term on the right side of the first example starts with a 2, and why each term on the right side of the second example contains an a term.

This is a rather more complicated example. Both the 4 and the m outside the brackets are multiplied by each term inside the brackets. You will notice that the first term inside the brackets already contains an m, so when it is multiplied by the m outside the brackets, it becomes m². Similarly, multiplying the m² in the second term by m turns it into m³. The 4 is also multiplied by each term inside the brackets. This is no problem in the case of the first term - it simply gains a 4 at the start. However, the second term already has a 3 at the start, so the 4 multiplies by this to give 12. Here are a few more examples - work through each one to make sure each is correct:

You will note from that last example that the terms inside the brackets are being multiplied by a negative number (-5 rather than simply 5). This has the effect of altering the signs of all the terms inside the brackets, so 11pq becomes -55p³q⁶, i.e. changes from positive to negative, and -7p³q⁶ becomes + 35p⁵q¹¹, i.e. changes from negative to positive). This is a particularly easy mistake to miss - it has caught me out many times!

Expanding quadratic expressions

Here we have one bracket multiplied by another. How do we deal with this? Well, just for the moment, let's rewrite the first bracket (x + y) as A. This means that the expression becomes:

This we can expand easily. It becomes

Now we replace A by (x + y) again, to give this:

I know this looks rather strange, with the letter outside the brackets being placed afterwards rather than in front, but mathematically it is still the same. This means we can expand it further to give this:

Effectively, we have taken each of the terms in the first bracket and multiplied it by each term in the second bracket:

We multiply the first term in the first bracket by the first term in the second bracket to give xm.
We multiply the first term in the first bracket by the second term in the second bracket (these are called the outer terms) to give xn.
We multiply the second term in the first bracket by the first term in the second bracket (these are called the inner terms) to give ym.
We multiply the second term in the first bracket by the second term in the second bracket (these are called the last terms) to give yn.

To remember each of these four steps, some students think of First - Inner - Last - Outer, or FILO for short. Of course, the order in which you do each of these multiplications makes no difference (providing you make sure that the + or - signs on the result are correct), so some students rearrange this to give FOIL.

You want another example, perhaps something a little more complicated? Try this:

Yuk! Well, the same method applies. Firstly multiply the first terms from each bracket:

In this case, we look at the terms (two of them in this case, although they could be more) and we find something that divides into both of them. This is written outside the brackets, and the rest of each term (with the appropriate + or - sign) is written inside. Here's another example - this time done in two stages:

The first thing that we noticed was that 3 goes into 3 (once) into 6 (twice) and into 12 (four times), so we take the 3 out of all these terms, leaving 1 (which we don't bother to write in front of the first term), 2 and 4.

Then we look at what remains inside the bracket and discover that something else, namely x, can also be taken out of each term. x joins the 3 outside the bracket, and is removed from each term. Now we ask, can anything else be taken out of each and every term inside the brackets - the answer is no! (Although c looks promising - it appears in two of the terms - it doesn't appear in all of them, so no deal!)

If you want to check that this expression has been factorised properly, simply expand your answer:

Which is what we started with! You see from this that factorisation can be done in several small steps instead of one large one. You could, of course, taken 3x out of each term, in which case the factorisation would be done in one large step. However, it isn't necessary, and if you are unsure about factorisation, several small steps would be better.

Here are some more examples. Again, you should check each one to see that it is correct.

4a2bc - 3ab2c + 4abc2 = abc(4a - 3b + 4c) 8dn + 12n2p - 6np = 2n(4d + 6np - 3p)

Quadratic Factorisation

If you expand it, you will find that it comes to this: x² + Mx + Nx + MN

You may get the terms in a different order, of course, but those will be the terms that you should get. The middle two terms (as I have written them), can be factorised into one term, to give this:

Effectively, we have one term which has x² in it, one term with x in it (that complicated term in the middle) and one term at the end without any x in it (we call this the "absolute term"). Let’s try one with numbers:

Of course, we could state this the other way round, and say that x² + 9x + 14 factorises to give (x + 2)(x + 7), which it does. This is called quadratic factorisation, and it is harder than simple factorisation.

In general, whenever you see an expression with an x² term (or any other letter squared, of course - it doesn’t have to be x), and an x term and an absolute term, then you should consider factorising it into the two brackets as you can see above. How do we go about factorising something like this?

Well, if you look at the example with x above, you will notice that the middle term (9x) was formed by adding the two numbers (2 and 7) and the absolute term was formed by multiplying the numbers (2 x 7 = 14). Perhaps we can apply the same logic to the expression above.

Well, 4 + 9 = 13 and 4 x 9 = 36, so perhaps the two numbers are 4 and 9. Let’s try them:

It worked! This is a general rule for this sort of question (it is different if there is a number in front of the squared term) - look for two numbers that add to give the x term, and multiply to give the absolute term. Here’s another example:

This works because 11 + 20 = 31 and 11 x 20 = 220. Of course, you could write the two terms the other way round, as (m + 11)(m + 20), without it making any difference.

There is always a fly in the ointment, and in this case it comes with negative numbers:

In this case, you may think that the rule breaks down – after all, 5 + 7 is not 2! However, remember that you are adding 7 and minus 5, not plus 5, and 7 + (-5) does give 2. Similarly, 7 x (-5) does give –35 (note that minus sign in front of the 35!)

In general, if the number in front of the absolute term is positive (i.e. +) then the two special numbers that you are looking for are both the same sign (i.e. both + or both -). This is because we multiply them to give the absolute term, and you can only get a positive number by multiplying two numbers of the same sign.

Harder Factorisation

The expansion is not affected by the fact that the number in front of the x² is not 1. Of course, it's easy when you have to expand it - factorising it is harder. How would you go about factorising this ...

... if you didn't already know what the answer was? Here is how you would go about it:

1. Multiply the numbers in front of the letter squared (the "x²" term) and the absolute term. In this case, it gives the answer 8 x 3 = 24.

2. Find two numbers which will multiply to give this answer (24) and add to give the number in front of the letter term (here 14, as the x has 14 in front of it). The easiest way is to write the factors of 24 and to spot which pairs of numbers look "likely":

   24	= 1 x 24	Can we get 14 by adding 1 and 24?	No!
   	= 4 x 6	Can we get 14 by adding 4 and 6?	No!
   	= 3 x 8	Can we get 14 by adding 3 and 8?	No!
   	= 2 x 12	Can we get 14 by adding 2 and 12?	Yes! 2 + 12 = 14

   At this point, we have found the two special numbers, 2 and 12, but we may have to list various pairs of factors!

3. Split the middle term (the one with the letter) into two terms using the special numbers that you have discovered. In this example, you get this:
   8x² + 14x + 3
   
   8x² + 12x + 2x + 3

   We know that we can split the middle term up like this because we specially chose the two special numbers so that they would add to give 14. It doesn't matter which way round you write the numbers - you could write them like this:

   8x² + 2x + 12x + 3

   and that would work just as well. Compare the expression just above with the one that we got when we expanded (4x + 1)(2x + 3).

4. Now factorise the first two terms into one set of brackets and the last two terms into a second set of brackets:
   8x² + 2x = 2x(4x + 1)

   12x + 3 = 3(4x + 1)

   so 8x² + 2x + 12x + 3 = 2x(4x + 1) + 3(4x + 1)

   You will notice that we end up with the same expression both brackets - that's a good sign!

5. Finally, collect the terms outside both the brackets to give the contents of the other bracket. Here we have 2x outside one bracket and + 3 outside the other, so the other bracket contains (2x + 3) and the whole expression factorises to give (4x + 1)(2x + 3).

Of course, we could have written the brackets the other way round, to produce (2x + 3)(4x + 1) which would have been just as good!

Suppose we had written the two special numbers the other way round:

If we factorise the terms in pairs, we get

I couldn't take anything out of the last two terms, so I just left them the way they were, and took 1 out as a factor - it seems like cheating, but it is mathematically correct. This time we end up with (2x + 3) as the term that appears twice, and 4x and + 1 as the remaining parts, so the expression still factorises to give

Be careful of minus signs!

Multiplying the first number and the absolute term gives -60. So far, so good! Now we write down the factors of 60, in the hope that one pair of factors will add to give 11:

Adding those factors simply will not give you 11. However, we are forgetting something! We are looking for factors to give -60, i.e. minus 60! One of the factors will have to be negative. Suddenly, we can get 11:

The two numbers are 15 and -4. You should be careful if you spot a minus sign in either of the two terms, and make sure that you don't just carefully drop the minus sign. When we split the middle term, we get:

Factorising the terms in pairs gives:

d(3d - 4) + 5(3d - 4) = (d + 5)(3d - 4) Sorted!

Difference of Two Squares

If you expand this in the same way as for other quadratic expressions, this is what you get

Check it for yourself! But, wait a minute, don't those middle two terms cancel each other? Yes, they do! The + ab - ab terms disappear, as they are the same size but one is positive and the other negative. This means that the final result of the expansion is:

Of course, there is nothing to stop us reversing this logic. If we wanted to factorise

we know that it must be

We don't have to write the terms in that particular order. We could just as easily write it as

or even

There is a general pattern here. What is x² - y² exactly? Well, both these terms are squared terms, and there is a minus sign between them. For this reason, we call an expression like this the difference of two squares. The same thing applies to any two terms which are squared numbers and with a minus sign between them:

Providing that we recognise that 25d⁴ is the square of 5d² and that 49e² is the square of 7e, and that one is subtracted from the other, then we can factorise the expression. Of course, if the expression had been one of these:

then we would not be able to factorise the expressions. None of those expressions is a difference of two squares. In the first case, the first term is not a square (26 is not a square number). In the second case, there is a plus sign between the terms (so it isn't a difference) and in the third case the second term is not a square number (the power is an odd number, not an even number). Here are some more examples of the difference of two squares:

That last one was an interesting example. It turned out that we could apply the difference of two squares rule twice. When we factorised the expression, one of the brackets also contained an expression that was the difference of two squares, so we applied the same logic to that.

Back	Multiple Choice	Random Question