Comprehensive Guide to WJC Level 2 Certificate and Additional Maths Techniques

hey guys in this video the lovely mr b is going to take you through the skills and techniques you need

for your wjc level 2 certificate and additional maths when you're doing this you can use the

free vision guide in combination with this and take off the skills as you are

confident with them if there's any bits that you're not confident in if you identify any gaps in

your knowledge then you can click through and watch the individual video on

that once you're a bit more confident with your skills then open website there is the course

for you where it will track your progress and take you through loads

of questions and in addition there is also a massive workshop [Music]

we have multiple rules to go through so for our first rule let's say we've got 3 to the power of 2 multiplied

by 3 to the power of 4. now the answer to this

is 3 to the power of six so our first rule is firstly we have our base

numbers so in this case our base numbers threes and these have got to be the same so when we generalize this rule let's

assign it a letter so let's call it n so we have n multiplied by another n then we have the

powers now look at what's happening with just the powers we have a 2 a 4 is equal to a 6.

so we've multiplied together the same base but with the powers they're not the same

they're actually adding together two plus four is six so the way we can generalize this

we can assign numbers to these let's call them a and b and that is going to be equal to

those two numbers being added together so our final answer will be n to the power of

a plus b so shortcut if you're multiplying your powers then you add them for our next example we

could have uh 4 to the power of 8 divided by 4 to the power of 5 and that would equal

4 to the power of 3. so what's happening here now again the base numbers are the same we've got 4

and we're dividing it by another four so these numbers can't change you'll notice the different numbers of the

previous examples they can be any numbers you like they could even be letters

so we generalize this through got an n and we're dividing it by another and another number now let's

have a look at what's happening only the powers so we have an eight a five and it equals a three

now eight take away five is three and that's what's happening here so when we generalize this we've got the a

and the b for our powers so in the answer it's gonna be a take away b the shortcut being if

you're dividing with your powers then you just take them away for our next example

let's have uh five to the power of two and then this one's going to be in brackets

and it could be another power on the outside of the brackets and let's have this be a4 so what's this

going to equal well it's going to equal 5 to the power of eight

so again let's have a look at what's happening now in this case we only have one base number

in the brackets so when we generalize we only have one to work with now let's have a look at just the powers

so we have a 2 a four and then an eight so what math would get us there

well two times four is eight so when we generalize it ends in a bracket we'll give our two powers letters let's

call them a and b again one inside the bracket one outside so for our answer it's going to be

a times b again the shock is easy bracket then you're going to be multiplying the

powers now you can see that we've got addition subtraction and multiplication rules so

do we have a division rule where n is going to be a divided by b and we do this does exist so let's have

a look at an example so let's say we've got 7 to the power of 3 and the whole thing

is being square rooted now the answer to that is going to be 7 for the power of three over two

or seven to the power of one and a half and fractions mean the same thing as division so to get the division you

need to have a square root involved we generalize this again we're gonna have our n number

our base number the kind of whole number in this situation they can be decimals as well actually

and again this one's essentially going to be in brackets but rather than being just normal brackets they're going to be

in a square root now let's have a look at the powers now you'll know it is null two you have the

three and the three stayed the same where's the two coming from well the two

comes from the order of which you're rooting this is a square root so it's a two because the

symbol for square is 2. if it will say a cube root you'd have a little 3

next to the root and then it'll be 3 over 3 for this question the denominator will be 3.

so our a is the power of the root and if it has no power it's 2 and a is the power on our base number

now these are the four main rules but there's other rules we can derive from this

let's have a look at a little pattern here to see these rules let's say we've got 2 to the power of 3. that

means 2 times 2 times two it's three twos multiplied together

which is eight then let's look at two to the power of two that's two twos multiplied together

that's four so silver is pretty easy but what about something like two to the power of one or two to the

power of 0. how can we figure these out well we think about 2 to the power of 3 as

3 2s multiplied and 2 to the power of 2 is 2 2s multiplied then 2 to the power of 1 will be 1

2 multiplied you've got one to two you can't multiply of anything you've only got one two so you can write down two

another piece of element for this is looking at the pattern eight divided by two is four

four divided by two is two and then two divided by two is one so you can go through this

pattern dividing by two and if this is not an evidence for you you can think about well

two to the power of four will be sixteen two times two times two times two sixteen divided by two is eight and

eight divided by two is four so this sequence is a divide by two sequence and so we get the odd results

of something to the power of one is itself and something to the power of zero

is one what if we continued what if we had two to the power of negative one and two

to the power of negative two well we'd follow the same sequence we divide by two each time

one divided by two is a half and a half divided by two is a quarter now look at the pattern

with the divisions we've got a one as the numerator

but the denominator we've got two and four now where have we seen two and four again

well two to the power of one is two two to the power of negative one is one over two

same number two to the power of two is four so 2 to the power of negative 2 is 1 over 4. so let's say we wanted 2

to the power of negative 3 then that would be 1 over 8. so we can see

patterns within this now we've looked at negative numbers what about fractions so we had

2 to the power of a half that's kind of working backwards from our fourth floor of indices so

we've got a 1 divided by a2 and the 2 corresponds to it being a square root so 2 to the power of half

is a square root of two then we increase that two to the power of a third would be the

cube root of two so again notice how if it's a square we don't write the power on the uh the

root position there is a two now the last thing to mention is that what if we had

something like n plus an n and we had a power on each an a and a b what would be the rule for

that now with addition and also with truck traction as well because subtraction is

the opposite of division there is going to be actually no rule for that

so if you did get something like that what you'd have to do is find the actual value because all

these things are shortcuts but you know look at our examples you know we've got three to the power six

four to the power three you can actually work out a large number of these so for example

three to the power of six in our first example is 729

so it's a fairly short number but still 3 to the power 6 is a shorter way of writing it but when

you get to addition and subtraction there isn't a way to shorten the way you write it so you just have to work it out

and then maybe be lucky the answer you get as your value maybe that would then be the product

of a different unrelated set of indices and studying to write it twice in my little list of formulas with the no real

one at the top i've written the plus minus symbol now that's not some new kind of maths that

just means that either adding or subtracting is going to be the same there's no real

notes one other thing i want to do i just want to generalize the kind of the other last column

so the key things take away from that is you have n to the power of one then that's going

to give you an answer of one so it's gonna give you an answer

of the number itself if you have n and it's to the power of zero then that's gonna give you an answer

of 1. if you have n and it's the power of a half then that's going to be a square root of

n and if you have an n and it's a negative power so let's say negative a for this then that's going to

give you as your answer one over that ends the a so these are the kind of the key things that you take

away from that list of numbers but as you can see if you forget them particularly the one and the zero and

negative one you can derive those by looking at the pattern and that pattern will work for

any set of numbers you don't need to do it with two like i did if you get the root one

then you can drive that from the fourth laws of indices which was the one with the division

four thirds can have some rules so let's say we have the square root of a and we add

on the square root of b then there is going to be

no rule for that let's change the add symbol to a plus or minus symbol because it doesn't matter if you're adding

there's no real maybe subtracting there's no rule where there is a rule is when you multiply

you've got the square root of a and you multiply it by the square root of b that's going to give

you the square root of a times b with both of them being underneath the cap of the root

alternatively you can write it as square root of a b same thing with division if you have the square root of

a divided by the square root of b then that's going to be the square root of a

divided by b all under the same root symbol which alternatively can be written

as a fraction square root of a over b let's look at some examples so let's say we had the

square root of nine plus the square root of four like i said there's no rule

for this however we can do the square roots square root of nine is going to give us 3 and the square

root of 4 is going to give us 2 and now we can add those together and get

5. so there is a situation where you would be adding these together now let's look at an example with

multiplication so let's say we had the square root of 2 multiplied by the square root

of 18. now we can't do those square roots we can multiply them together to give us

a square root of 36. now we do know the square root of 36 is the square root of 36

is 6. so 6 would be an alternative answer to this see the benefits of using these rules so

you might be able to change some of these roots into nice whole numbers by using this

rules that'd be a similar thing with division as well now there's three main types of

questions i want to look at with serbs that you might need to do the first one is with brackets

you might have something like 2 plus the square root of 3 multiplied by 2 minus

the square root of 3. now this is a quadratic question so you'd use the methods used for

quadratics but you'd have to be careful that you also use the third rules at the same time the 2 times the 2

which would give 4. you would do the 2 times a negative 3 so 2 times a negative root 3

would give you negative 2 lots of root 3. so you can do it the multiplication rule

is only when both of them are roots two isn't a root so you can write two root three then you would do the root

three times the two and that give a positive to root three and then the final one

would be the root three times the root three now this is where we have to use the

rules of absurds here a root three times root three we can multiply those together to give

root nine and since we have a positive root three and a negative log of root 3 then actually

that's going to be a negative root 9. now we can simplify this actually

so we've got the 4 the negative 2 root 3 and the positive 2 root 3 we can add those together and if we add

them together it's going to give us 0 root 3. so that's going to disappear and we know

what root 9 is root 9 3 is a square number actually it gives us

4 take away 3. and we can answer that it's one so double brackets actually cancel out to give us

one let's be useful to go back to our list of servers and add that rule in green so what i'm going to do is let's

go about the one with no real because we don't need to remember this so let's say we've got root a and we've

got another root a and they are being added together the answer

is also going to be a root a what changes is the coefficient the number on the

root a so let's say we had a b and a c on them so we had b root a and c root a

then our answer would be b plus c root a let's put that in brackets actually to make sure we do that

so you can add roots but in a very specific situation and it's not like the multiplication and

the division rules it's quite different so they're going to have the same base whereas when you multiply and divide

they don't need to have the same base under the root now as well as brackets which is kind of a multiplication

there's also kind of a division version of this called rationalizing the denominator so

let's say we've got something like 2 over root 3. now when you rationalize this what we mean is we do not want a

root as a denominator because that throws up probably become more complicated kinds of maths

what we need to do in this situation is we take o2 over root 3 and remember with

fractions you can multiply and divide the top and the bottom by this the same number to kind of simplify or

find the equivalent fractions i'm gonna exploit that fact here so how do we get rid of a root three

well what you're gonna do is you're gonna multiply it by itself it's gonna multiply

the denominator by root three we'll keep it the same fraction we're going to multiply

the numerator by root 3 as well and what that leaves us with is on the top for the numerator 2 times

root 3 is 2 root 3. but for the denominator root 3 times root 3

is root 9 and we can do root 9 root 9 is 3 so we're going to have 2 root 3 all being divided by 3. now you notice i

wrote that in that purple now look at all the purple examples here this is where and

it's only going to be visual with the thirds if you ever see a square number

underneath a root it's usually beneficial to cancel it out into a whole number

because we can do root nine it's three so why not do that and then we've got a rationalized

denominator we have a whole number for the denominator there's one more type of rationalizing

that i need to show you so let's say we've got two over and then we've got two

plus root three now if you tried to multiply all of that by root 3 this would happen you'd have 2 root 3

for the numerator well that's fine you'd be adding on 3 because root 3 times root 3 is group 9 which is 3 but

then you'd also have to multiply the 2 in the denominator by root 3. you have 2 root 3 on your denominator

now we still have a root on our denominator so it's not being properly rationalized

so you can only multiply a root by itself if it's the only thing down there

what if you've got something else statement how do you get rid of the roots

well notice how i've got two root three as my denominator look at the brackets question when two

group three was multiplied by two minus root three we multiplied out the quadratic

all of the roots cancelled out so what you need to do in this situation when you have multiple things for the

denominator if you take the two you take two plus root three tells us putting brackets at

this point i'm gonna multiply the bottom by two minus root three which is the same

numbers but i flipped the sign we'll do the same thing to the numerator and we already know the answer is the

answer to that denominator is the answer to our bracket example it was

over 1. that's what all cancelled down to but you wouldn't necessarily have that in

the exam so you might have to actually multiply the quadratic and get whatever answer it

is but it should be cancelling out to have no roots and then the numerator when we've

simply got two lots of two minus root three so it's kind of okay to write

it like that actually but you might want to multiply out the brackets and multiply both parts by two

so let's do that so 2 times 2 is 4 and 2 times negative root 3 is negative 2 root 3. now sometimes your answer

will look like that in this situation we can actually deal with the denominator the

denominator is a 1 and if you divide by 1 it's going to stay the same so we can

actually get rid of the denominator in this case but you will sometimes have a

denominator another thing you could do if the denominator ended up being a 2 you can still get rid of it because you

can divide the the 4 and the negative 2 by 2 and you can simplify the numerator

so always look for opportunities to cancel down but again you will sometimes have a denominator here

if you can't cancel but in this case we could changing the subject of a formula use

the same principles as solving equations so let's say we have

y is equal to 11x minus four if we were solving the equation what we would do

is try and get x on its own and we would add 4 to both sides and that's what we do

and that gives us y plus 4 is equal to 11x now when you're solving an equation in the position of a y you would

typically have a number that you can add 4 onto now in this situation we've got a y there so we

can't add 4 on to it so just leave it as y plus 4. when we now do the next step

and we divide both sides by 11 to get x on its own on the right hand side again typically the y plus 4 position

would have a number a number that we can divide by 11. in this case we don't have that so how

can we do it well all you do is you write down y plus 4 again

then you make it into a fraction with 11 as the denominator and what this does is it shows that

you're attempting to divide by 11 but at this time you are unable to divide by 11

because you don't have a value for y and so this is the essence for you change the subject because now we have an

x equals and now we have a formula to find x whereas we started off with just a

formula to find y so let's have x take away 9 and all of that is going to be squared

and it's equal to y so again we want to get x on its own so we can kind of unwrap the layers of

mathematics around x to get it of itself out most layer is the squared symbol so we need to root both sides

that will leave us with x minus nine on the left-hand side and then we have to square root the y

once we've done that we can add nine to both sides i get x is equal to the square root of y

plus nine and again typically here y would be a number that you can square root

and then that will give you a number that you can add 9 to and get a value for x

but within the subject our answer is going to be another formula so we started off with the formula to find y

and now we have a formula to find x let's look at something with a few more steps

so we've got 5x plus 2y all divided by 5 is equal to 7. so again when we get x on its own and we

have to kind of unwrap all the layers around it the outermost layer being they divide by five so we've multiplied both

sides by five and that moves the five over to the right hand side seven times five is

thirty-five so in this case there is actually some mass we can do we can do seven times five

so it's not always going to be maths that you can't do next we can take 2y away from both sides and finally we

want to divide through by 5. now x is equal to and then we're going to have 35 minus 2y

and we're dividing all that by five and note that i said we're dividing it by five but i

haven't actually divided it by five because while we could divide the 35 by five

and get seven we can't divide the two by five you could actually do something here so we could do 35 divided by 5 is 7

then we could do 2 divided by 5 is 2 fifths so actually this could actually be a simpler form

so both of these outcomes are fine it's just depending on if you want you know a

large division a large fraction or if you want a smaller fraction involved in this and again

no with the 2 divided by 5 i couldn't do 2 divided by 5 so i just wrote it as a fraction

i'm indicating i want to do 2 divided by 5 but i can't at this time let's look at

one last example so we're going to have 10 plus 14x all over

xy and that's equal to 7. so this question the change here is that sometimes you

are going to get multiple sets of the thing that change subject to unwrap some layers here so

let's multiply both sides by x y and that will give us 10 y plus 14 x is equal to 7 times x y

and our aim here now we've undone the fraction we want to get x on its own we have two sets of x's

we need to get the x's on to the same side so what we could do is we take away the 14x from both sides and

that will give us 10y is equal to 7xy take away 14x and once you've got all

the x's the same side what you want to do now is factorize

both x terms to get a single x so both for the terms and the next thing they both got an x but they've also

both got e7 because we've got seven x and we've got a 14 x so we've got a number factor as well

so we take seven x outside of brackets and we divide three by 7x 7xy divided by 7x will leave us with just y

and 14x divided by 7x 14 divided by 7 is 2 and x divided by x means we're

multiplying by 1. and so there's our next line so by factorizing we now only have one x

what we can do now is we can divide both sides by the bracket and that will mean that the bracket is

going to be dividing the 10 y now and that's going to be equal to 7x now at this point what we also want to

do is get rid of the 7 on the x and we'll be dividing that as well so we've already divided and

we've got another divide to do all we need to do is add the 7 to the denominator as well so

7 is going to join the denominator which could be multiplying the things that are already there and

the their answer last thing mine would do uh you could expand the bracket to be preferred

to that's something you could do so we could have 10y over 7 times y and 7 times negative 2 is equal

to x so there we have three examples between the subject and again it's exactly the same as

solving an equation because sometimes you won't actually be able to simplify it at those points and

so you leave divisions as fractions for example with factorization you should be familiar

with the kind of single bracket factorization we might have something like

4x plus 20 and when you factorize it you think about what the highest common factor of the two numbers are so we've

got 4 and 20. and so the highest common factor of those is going to be the 4

and you remove that outside of a single bracket inside the bracket is going to be the result of dividing

your expression by 4 so 4x divided by 4 is 1x and 20 divided by 4 is 5. this is something that you should

already be confident with what we're going to focus on with this bit of revision

is quadratic factorization so you might have something like x squared plus nine x

plus eighteen i'd be asked to factorize it into a set of brackets so how would you do

that well the key thing is that you want two numbers inside these brackets

that add together to make the nine and they multiply together to make the 18. now you might be able to

figure that out just by looking at it but there's a method if you can't so what you do is you take

the 18 the number you need to multiply for and you list all the possible options of numbers that

multiply together to make 18. so you get 1 times 18 2 times 9 3 times 6 we can't divide it

by 4 or 5 we can divide it by 6 so we finished our list so not how i did that i divided it

by 1 wrote down the answer divided by 2 wrote down the answer

divided by 3 wrote out the answer tried 4 and 5 didn't work and then when i repeated

myself with 6 i knew that the list was complete so what we need to do now is think of which

ones of those numbers are gonna add together to give nine never look through you'll see that three

and six make nine so what do you write in the brackets we need an x in both

because we need to have x times x as x squared then we're simply gonna write in the three and the six in either order it

doesn't matter and they were being added on they're positive numbers we're going to write an

add symbol and that's it we've factorized the quadratic so let's look at some harder ones

next we'll look at x squared plus 4x take 32. so we have some negative numbers here as well

now we still want two numbers which add together to make four and multiply together to make 32 so you

write your list of factors with 32 and divide it by one we could divide it by two we can't divide by three

we can divide by four we can't divide by five or six or seven and we can divide it by

eight we've already got eight written down so we must have finished our list so we

look through the list and think about which of these numbers add together to make four and at first

glance none of them will add together to make four so we know we write the x and the x in the

brackets what else we write in well because the multiplication is a negative it's a

negative 32 we are allowed to use negative numbers so when you look through this

list you think well if i'm allowed to take away how can i get a 4. so 132 you get you

know you get 33 you get 31 you could get you know negative

31 or 33 depending on these negatives or not so that's not going to help us we can eliminate that one

two and 16. we can add them to get 18. we could take them away to get a 14 that's not going to help us and 4 and 8

we can add them for 12. we take them away 4 take away 8 is negative 4.

well this looks promising what about the other way around 8 take away 4 is 4. and so we found our answer

so we write in the four and the eight again in either order it doesn't matter which way around there

what does matter is the where the negative numbers are you think very carefully about how you got a positive

four and it have to be a positive eight and a negative four

eight take away four is four so in the brackets we need a positive eight and a negative four

so always be looking out for negative numbers as well if you can see it in the question i'd be thinking about

takeaways when we talk about you know two numbers that add together to make the middle term

we are allowed to take them away as well or have any kind of combination negative numbers

that you need lastly we'll look at two special cases so what if we had x squared take away nine so we're i need

two numbers you know that multiplied together or two numbers that you know um

multiply when you look at this there's only one number so you know is this going to be an adding or a

multiplication number so if you look back at the previous questions the numbers you have to add

would have an x on and the numbers that you have to multiply are just numbers so it's going to be two

numbers that multiply together to make 9. but what about the addition bit

now it is actually there and it's hidden so x squared minus nine is the same as

saying x squared plus zero x minus nine and so you want two numbers

that multiplied together to make negative nine but they add together to give zero

that's what it means so when you have your double brackets there's only going to be one option for

something like this and it's going to have to be the same number twice so to get 9

what's the same number twice it gets nine that's gonna be three and three so it's a square number

and there's a missing term just gonna square root it and it's gonna have to be a

positive three and a negative three then what's gonna happen is when you add those two numbers together

three and a negative 3 will make the 0 for you one of the special case is rather than having

the addition part missing the x term missing you might have the multiplication term missing

so you might have something like x squared minus x now in that situation this isn't

going to be a quadratic because you want two numbers that multiply together to make

zero so what that means is one of the brackets is gonna be a zero so when you have your

double brackets it's going to act like it's like a linear kind of factorization so you see you've

got an x in both so we take out the x and then we divide both by x so x squared divided by x is x

negative x divided by x is going to be negative one and that's going to be our answer

again coming back to you know what two numbers that multiply together to make zero and add together to make

negative x you know negative one one of the brackets is missing

the x on the outside doesn't have a number with it and that's how you get the zero because

effectively it's x plus zero however rather than overthinking it just treat questions like this with a missing

number on the end as a linear linear factorization now the next stage difficulty after this is a

something like 4x squared minus 18x plus 14. and with this one when you open

up your double brackets before we're writing x and x because x times x gives us our x squared

however with this question it's not an x squared it's a 4x squared so you could have 4x times x

or you could have 2x times 2x so it's going to be one of these two alternatives and you don't

know which one it's going to be so that's the first difficulty the second difficulty

is when we look at factors for two numbers that multiply together to make 14.

so look at the factor list for 14. you can divide it by one and we can divide it by two and we can't

divide it by three four five or six so these are the only possible factors

now we need to add or subtract these to get a negative eighteen and these numbers are all too small you

are not going to be able to get a negative 18 out of these numbers so what on earth do

you do well you are allowed to multiply by the coefficients on the x because when you expand the brackets

these numbers are going to get involved so if it's the first alternative and it's 4x times

x you can multiply one of the numbers by four and you can multiply one of the numbers

by one and this will give us a fresh list of alternatives so we could have four and fourteen you

multiply the first number by four and the second number by one or we could have one

and then fourteen times four which is fifty-six same thing with the two and the seven we can multiply the

2 by 4 to get 8 and 7 by 1 or we can multiply the 2 by 1 and the 7 by 4 to get 28.

so we've got a larger list of possibilities so this is our four in one list let's

just box it off let's make a list for two and two so we multiply the one by two

and the fourteen by two or we can multiply the two by two and these seven by two and since they're

both the same you can't do it the other opposite way around because i'll just give you the same answer

so we've only got uh two alternatives uh for the two and two options so what needs to do now is look through this

list and think well which ones of these could make eighteen so four and fourteen can

eight and seven are gonna make 15 to this that's not gonna work 156 that's way too big

uh 2 and 28 and we could take away the 2's get 26 but that's still too big 2 and 28 we take when we get 26 that's

too big and 4 and 14 again for a different method that could also give us the

18. so we're going to think about which one it's going to be now we want to add them together to get

a negative 18. so we'd have to be a negative 14 and a negative 4

in both cases now we write these into brackets but don't write before and 14 we write the original numbers they came

from so it's going to be the one and the 14. you're going to think really carefully about which way around there

now the 14 stayed as it is so that's going to go in the bracket with the four and the

four the one was multiplied by four to get four so the one is going to be in the

opposite bracket to the four because when you multiply this out to get that one multiplied by

four it can't be in the same bracket as it so we've got that the correct way around

and then of course in the second alternative because they're both being multiplied by two it doesn't matter

which way round you're going to be now the second alternative multiplied from the 2 and the 7 multiplying them

both by 2 so we're going to have the 2 and the 7 in for that alternative

so now we're going to think about how do we get a negative 18 well we've both got to be

negative and so now we've got two possible answers this test is quite unusual to get two possible answers

but there we are we've got two alternatives there's two different ways to factorize it

and whichever one you use depends on what's going to happen next if the question just to factorize either

one will be a correct answer and i'll get four max you need to do something else afterwards

then there'll be some kind of clue as to which of these two is gonna be the most relevant

which brings on to the next thing we'll look at so you could get something like this you have x squared plus x

minus 20 all over x minus four now any question like this what you're going to do

is factorize what you can and you can factorize the quadratic on the top row now i'm not going to go through this in

detail because we've already revised that so the answer to the factorization is

going to be x plus 5 and x minus four and what you'll notice is that one

of the factors is the same as the denominator of the fraction so all you need to do is

divide them by each other if you divide the number by itself you get one if you multiply by one the

answer stays the same so effectively by factorizing we can actually cancel down the fraction and

the answer is x plus 5. now we can also do this with

more complicated ones so let's say we had 10x plus 15 all over 8x squared

plus 4x minus 12. now normally if we were going to do what we did with the 4x squared question earlier

on how many alternatives are there for 8x squared we could have

eight times one you could have uh four times two you know there's a few different options

there and you're gonna get like a long list of possibilities potentially but there's

a clue in the question because it's a fraction you know that it's probably going to end

up being cancelled down now on the numerator we've got 10x plus 15.

now if we have 10x in a bracket that's not going to multiply by anything to get 8x squared

but you can factorize it as a linear factorization and the 10 and the 15 have got a common

factor of 5 so you could write 5 and then 2x plus 3 in the brackets now this is a clue that one of the

factors is going to be 2x plus 3 and this is going to shorten down

the list of possibilities that you have for this question because you know you want to get a 2x plus 3.

so we want to get a negative 12 for example at the end by multiplying together two numbers so

the only way to get a negative 12 is for the other factor to be negative four and then we

x is when we get eight x squared we're gonna multiply two x by something to get eight x it

must be a four x and now what we can do is we can test it out if it works so maybe you can

multiply it out to check four x times two x date x squared negative four

times three is a negative twelve four x times three will give us twelve x and negative four times 2 will be

negative 8 so 12 and negative 8 will give us that positive 4 in the middle looks like it works so now all we need

to do is divide the same factors by each other and cancel them out what we're left by

is 5 over 4x minus 4 and that's your final answer and so you can see by having that clue in the

question it's a fraction we're probably going to be simplifying the fraction

we could guess what one of the factors were going to be and it's short the calculation down first to the point

where we could do it in our head quite easily with linear equations you have to collect all of the x terms on one side

and all the number terms on the other so if for example let's say we have 5x plus 10 is equal to

35. what we want to achieve here is have the x terms on one side

and the number terms on the other side so we have a value like x equals something we're trying to find

the value of the x now the way that you solve these is through kind of

patient systematic step-by-step changes to both sides of the equation so for example

with this one we can take 10 away from both sides and that leaves us with 5x is equal

to 25 and so what we've done is we've made the equation simpler but we haven't changed the

values of anything in the equation because with the same thing to both sides

then our next step we can divide both sides by five and that leaves us with x is equal

to five and now we have the answer that we were looking for sometimes when solving equation you

might have multiple values of x so for example you could have 5x

minus 20 is equal to x minus 12. now in this situation the first step should be to get all of

the x values onto one side so for example here we could take x away from both sides we're getting rid

of the smallest value of x and that will leave us with 5x take away x will give us 4x

so 4x minus 20 is equal to negative 12. now we've done that we can add 20 to both sides

and that will give us 4x is equal to a negative 12 plus 20 will be a positive 8. so now we divide both sides

by 4 and that will give us that x is equal to two

so the important things here are knowing which order to do each inverse operation and knowing what the inverse operations

are generally it's easier to get rid of plus and minus first and then do multiply and divide

however you need to be careful about things like fractions square roots brackets powers which may

alter the order in which you need to do things one way of solving equations which makes

things look very different is when you have a quadratic with an equation so for

example we could have x squared take away 14x plus 49 is equal to 0.

now you should have already covered factorizing quadratics if you happen to need to compromise that

first because i'm not going to go through it here but essentially when you factorize this

you'll get something like this where we have x minus 7 and x minus 7 in brackets

and the difference between factorizing quadratic and solving a quadratic equation is you just

have this equal zero on the end what we don't have here is the kind of

x equals number we're looking for but there is an easy shortcut we can use to do this

and what one look is we want to make one of the brackets be equal to zero because if one of the brackets is equal

to zero then it's multiplying the other bracket by zero and then you get zero so how can

we make the brackets equal to zero well you have to look at the values of x you have to think what could x be to

make each bracket be equal to zero and you might get multiple options so if the first bracket

highlighted in green x would have to be seven to make that bracket equal zero we have seven take

away seven is zero so no matter what the second bracket is you're always multiplying by zero and

you can get zero how are you in pink the second bracket would also need to be seven

for everything to equal zero and so for this one we can see that x is just equal to seven now why have i

bothered to look at each bracket separately when they have the same answer so let's

look at another example let's let's look at x squared minus x minus 56

is equal to zero now again knowledge about factorizing to factorize the quadratic part of this

and if you factorize a quadratic what you should get is x minus eight and x plus seven

so then again we look at each bracket separately we think what could x be to make the bracket be equal to zero

because another one doesn't matter because you multiply by zero you get zero well for the first bracket

x would have to be eight because then you get eight take away is zero but for the second bracket x would have

to be equal to negative seven because negative seven plus seven is zero

so we have two possible answers so which one do we choose well actually all you need to do is that

x is equal to both of them is equal to eight and is equal to negative seven the only

time you have to choose between the two is if this is a practical question

and a negative value will be impossible if the context is some kind of real life problem this method works even

if you have something more complicated like a cubic so let's say we had x cubed let's do three x cubed

plus 27 x squared plus 72 x plus 48 is equal to zero now again i'm not going

to go through how to factorize this but you're probably gonna be using some form of algebraic

long division to do it but you're gonna end up with values in three brackets and it'd be equal to zero

there's more than one option for this one but you could have x plus one x plus four and three x

plus twelve so again what you need to do is look at each value of x and think what could this value be to

make the entire bracket be equal to zero so for the first x it would have to be negative one

because negative one plus one is equal to zero the second one would be equal to negative four

because negative four plus four is equal to zero then the third one's more complicated it's taking the form of an

equation so we have 3x plus 12. so what we want is we want the 3x to be equal to

negative 12 for it to equal zero so sometimes a little more complicated to bigger x's

and if 3x is equal to negative 12 then 1x needs to be equal to negative 4. so now we've got all of our alternatives

we can write down our answer and x we need to be equal to negative 1 or negative 4. and because 2 the

brackets of negative four as a value we only tried it down once just like with our first example like

this both brackets were equal to positive seven so we could just write seven as our

answer so only write down multiple options either do take different values now the

last thing we do here is that we're so much on the page it might not be very clear which ones

are the answers i'm just going to highlight all the answers in purple now first example x was five a second

example x was two in the third example x was seven

in the fourth example x was eight or negative seven and in the final example x was equal to negative one or negative

four algebraic long division is all about factorizing things

are larger than quadratic so for example a cubic so you might be given something like

factorize and be given a cubic so let's say x cubed minus five x squared

plus eight x minus four and you might be given a clue so you might be told that one of

the factors exists so for example that x minus 1 is a factor alternatively random full factorization you might be

asked to show that x minus 1 is a factor but it's going to be the same

math so what we do to solve this is do something called algebraic long division so we set up a division now with this

division we're going to have x minus 1 as a thing we're dividing by

and the thing that we're dividing is going to be the x cubed minus five x squared plus eight x

minus four you might be thinking well this looks horrific you know how on earth are we gonna divide these

and we use a little bit of a trial and error and we go step by step now what we're

gonna do is we're going to focus on dividing by just the x and we'll deal with a minus 1

after we've attempted dividing by x so we're going to go through one to them at a time we're going to start off with the

x cubed then divide x cubed by x which gives us x squared we'll write in x squared we're

not going to write it in the first spot above the cube section we're going to write it in the squared

section so our answer is going to go here now once you have done that what you

need to do is then check that this is actually the right answer because we didn't divide by x we should

have been dividing by x minus one so we can have to work backwards now

and make sure we've done this correct so we take our answer i'm gonna multiply it by the term at the start so we're going

to do x squared times x is x cubed and we're going to do x squared times

negative 1 which will be negative 1 x squared if we've got x squared as our answer the pair of the

calculation we've dealt with is x cubed minus x squared but there's still

a lot of the calculation left to go so what we do now is we work out how much of the calculation is left

and the way we do that is we are going to take the parts away so we're going to do

x cubed take away x cubed zero then negative 5 x squared take away a negative x squared double negative

thirds will be that negative five plus a one will give us negative four

it'll be negative four x squared and then there's nothing to take away for the next few sections

they stay as they are so this is the pair that we've got left and now all we do is we repeat the

process we're going to divide by just the x so our next term is negative 4 x

squared we're going to divide it by x and that gives us negative 4 x write that down in our answer and then

we're going to check that we've done it correctly so we're going to bring that forward i'm going to multiply it by the

thing we're dividing by so we're going to do negative 4x times x which is negative

four x squared and then negative four x times negative one which will be a positive

four x and then we check how much of the equations left over by taking away so negative four x

squared take away negative four x squared will cancel like to give you zero

atex take away 4x is 4x and there's nothing else to take away so the -4 will stay as it is and then

we'll repeat the process again so now we've got a 4x left over at the start we divide that by x

and that gives us four so we write in four we bring that back to the front and four

times negative one is negative four we then check how much we've got left by taking them away

and you can see now that we're going to have zero left over so we have finished and if you have zero

that proves that it's a factor there's no remainder we've completed it if you got an answer

that wasn't zero that would be a remainder and it means that what you have

isn't going to be a factor so now is proven that x minus 1 is the factor how do we

factorize this well we know that x minus 1 is a factor and we know

the answer to our division is a factor as well so that's going to be x squared minus 4x plus 4.

we know that that's also going to be a factor so all we have to do now is take quadratic and factorize

the quadratic as well so you can do that using any method you like but if you factorize the quadratic which

is something to be more comfortable with that should give you should be x minus two in both

and now we've got our full factorization so not only does the algebraic long division

prove that something's a factor by having a remainder of zero but it also will give you for cubic it

will give you a quadratic that you can then factorize for the rest of the factors so now let's have a look

at another example so let's say we need to factorize x

cubed take away five x squared plus three x plus nine and this time we are not going

to be given a clue so how can we factorize it well a good thing to look at first

is to think about the number at the end we've got a nine at the end and it's cubic so we have three brackets one

thing well which three numbers are gonna multiply together to make the nine and you don't have a lot of options

really for some numbers so if they're nine you could have three times three

times one for example now that's not the full story because we know we have a negative number in there so we're

factorizing it we're gonna be getting a negative five x squared so some of these need to be

negative now given that nine is a positive number two of them are going to have to be

negative so then that gives a lot more options we could have negative three times negative three

times one or we could have three times negative three times negative one for example so here

are a few different alternatives and you know no matter which one you choose you could be wrong

so if you can't figure out exactly which one's gonna be correct using logic you're just gonna have to give something

a go so i'm going to have a go and i'm going to use x being equal to 3. so we're going to set up another

algebraic long division we're going to divide through by x plus 3 and we're going to divide

through the full number so like with the previous question we've divided through

by just the number start which is the x and let's free up a little bit of space here so with an

educated guess i'm going to guess that x 3 will be a factor let's check to see if it is so take your first term which is x

cubed and divided by x that gives us x squared so we can write x squared in

as the first part of our answer then we didn't take in total plus three so we're gonna bring that back to the star and

check the plus three so do x squared times x is x cubed and x squared times three which will

give us three x squared so we've not just dealt with the

x cubed we dealt with an x cubed with a bit extra as well we'll need to see what's left over before we carry on

x cubed takeaway x cubed is zero negative five x squared take away three x squared will be a

negative eight x squared and the rest will stay as it is

so now repeat the process now at the front we have negative eight squared we're dividing that by x and

that will give us negative 8x but then of course we haven't dealt with a plus 3 so we need

to multiply that by our original number so negative 8x times

x is negative 8x squared notice how this always gives us the same number at the start

and then negative 8x times 3 will give us negative 24x so we take away to see

what's left over so the first term always going to cancel out and the next term negative 3

take away a negative 24 double negative third that will give us a positive 27. 27x

and we're adding 9 to it that didn't change now we take the 27 we divided by x

and that gives us 27 as our answer so then we again multiply that by the star to see

what we have left over so 27 times x is 27x and 27 times 3 is going to give us a 81

then we take these away from each other what's going to happen is we have a remainder

we're going to have 9 take away 81 which gives us negative 72 as a remainder

since we don't have zero at the end that means that x plus three is not a factor and so we've guessed

wrong here and this might happen in exams so if this happens don't worry about it okay

we've done some work today at least we've proven that we can do algebraic long division