__About me__

I am a chemical engineering student at the University of Bath. Mathematics and the sciences are subjects I have always enjoyed and I especially love applying the knowledge I have learnt in all sorts of areas, such as cooking or music or difficult engineering problems. I teach my younger brother (age 9) mathematics and he is currently working at a year 7 level. Throughout sixth form I also spent time teaching younger children science and so I am fairly experienced at teaching.

I hope to make my tutorials **exciting and interesting for you** and to **show you that there is fun in learning** and that it should not be just for passing an exam, but **I will prioritise helping you get the highest grade you can**. When I started IB HL Maths I was scoring 4s and 5s despite achieving the highest grades in IGCSE Maths and Admaths but I ultimately finished with a 7. Therefore, I know full well how difficult mathematics can be and I am very patient and friendly when it comes to teaching.

__Sessions__

A strong foundation is essential for learning. Mathematics is a cumulative subject and **you** **must** **understand** certain ideas before progressing. This is something I did not appreciate and after I learned of the importance of **understanding** I found maths a lot easier and more enjoyable. I therefore would like to focus on **helping you develop your understanding** of topics. This will allow you to not only complete ‘standard’ exam questions but to tackle any ‘unusual’ questions exams may give.

__Personal Statements__

These can be stressful and painful to write but when it came to writing my own I found it to be the opposite. I’ve done a lot of research into what university admissions are looking for and hope that **you will find my advice helpful**. When I applied, it only took one small redraft before my tutors and I were happy and confident in my personal statement, and I was ultimately invited to an interview at Cambridge.

__Tutoring__

Please let me know before sessions what you would like to cover, and how much of the topic(s) in question you **understand**. I ask this so I can prepare well for sessions to help **you** get the most out of them.

__About me__

I am a chemical engineering student at the University of Bath. Mathematics and the sciences are subjects I have always enjoyed and I especially love applying the knowledge I have learnt in all sorts of areas, such as cooking or music or difficult engineering problems. I teach my younger brother (age 9) mathematics and he is currently working at a year 7 level. Throughout sixth form I also spent time teaching younger children science and so I am fairly experienced at teaching.

I hope to make my tutorials **exciting and interesting for you** and to **show you that there is fun in learning** and that it should not be just for passing an exam, but **I will prioritise helping you get the highest grade you can**. When I started IB HL Maths I was scoring 4s and 5s despite achieving the highest grades in IGCSE Maths and Admaths but I ultimately finished with a 7. Therefore, I know full well how difficult mathematics can be and I am very patient and friendly when it comes to teaching.

__Sessions__

A strong foundation is essential for learning. Mathematics is a cumulative subject and **you** **must** **understand** certain ideas before progressing. This is something I did not appreciate and after I learned of the importance of **understanding** I found maths a lot easier and more enjoyable. I therefore would like to focus on **helping you develop your understanding** of topics. This will allow you to not only complete ‘standard’ exam questions but to tackle any ‘unusual’ questions exams may give.

__Personal Statements__

These can be stressful and painful to write but when it came to writing my own I found it to be the opposite. I’ve done a lot of research into what university admissions are looking for and hope that **you will find my advice helpful**. When I applied, it only took one small redraft before my tutors and I were happy and confident in my personal statement, and I was ultimately invited to an interview at Cambridge.

__Tutoring__

Please let me know before sessions what you would like to cover, and how much of the topic(s) in question you **understand**. I ask this so I can prepare well for sessions to help **you** get the most out of them.

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Elle (Student)

September 18 2016

great help

Elle (Student)

August 28 2016

very helpful

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July 18 2016

So helpful thank you

Elle (Student)

July 11 2016

very helpful

To factorise z^{3}+1 into a linear and quadratic factor, it is best to try the factor theorem first. We set the equation to zero: z^{3}+1=0.

We need to find a value of z such that the above is true. By inspection we can see that if z=-1 then the above is true because (-1)^{3}+1=0 since (-1)^{3}=-1.

Therefore, (z+1) is a factor. The other factor is quadratic. We know, by inspection, that the quadratic factor will look like: (z^{2}+kx+1) where k is a constant. This is because z^{3}+1 can be written as a product of its factors: z^{3}+1=(z+1)(z^{2}+kx+1). The coefficient of the z^{3} term and constant term are 1 so we know the coefficient of the z^{2 }and constant term in the quadratic. By inspection we can also see that k=-1.

The second part of the questions asks us to show that y^{3}=-1. There are several approaches for this. You can square y to obtain y^{2} then multiply it by y again to obtain y^{3} and show that it is equal to -1. You can also write y in polar form, then apply De Moivre’s theorem to obtain y^{3} then convert back into Cartesian form. The final method is as follows:

We want to show that y^{3}=-1. Earlier we wrote: z^{3}=-1 which leads to z^{3}+1=(z+1)(z^{2}-z+1)=0. We know that if y is a cubic root of -1 then it cannot be equal to -1, so for the above to be true then (z^{2}-z+1) must be equal to 0. We simply solve the quadratic (z^{2}-z+1)=0 using the quadratic formula. The formula gives us that z=(1+i√3)/2 after taking the positive result. We note that this is equal to y. Since y=x then y is a cube root of -1.

The third part of the question asks us to show that y^{2}=y-1. Using our working from earlier, we know that y is a root of z^{2}-z+1=0 so y^{2}-y+1=0. Rearranging this we have y^{2}=y-1 as required. Alternatively you can find y^{2} then show that it is equal to y-1.

The final part of the question is to find the value of (1-y)^{6}. Instinctively most people would use the binomial expansion but there is a far more elegant solution. From the previous part we have y^{2}=y-1. If we multiply both sides by -1 then:

(-y^{2} )=1-y

Raising both sides to the power 6 as the question required and using power laws:

(-y^{2})^{6}=(1-y)^{6}

Why would we do this though? Maybe you can see why; there is a piece of information from earlier that we can use, if not, keep following and you will see that this method is very clever.

(-y)^{6*2}=(y)^{12}=(1-y)^{6}

The negative sign disappears as squaring a number makes it positive.

We can manipulate this further:

(y)^{12}=(y)^{3*4}=(y^{3})^{4}

Earlier we had (y^{3})=-1 and so ultimately the value of (1-y)^{6}=(-1)^{4}=1.

To factorise z^{3}+1 into a linear and quadratic factor, it is best to try the factor theorem first. We set the equation to zero: z^{3}+1=0.

We need to find a value of z such that the above is true. By inspection we can see that if z=-1 then the above is true because (-1)^{3}+1=0 since (-1)^{3}=-1.

Therefore, (z+1) is a factor. The other factor is quadratic. We know, by inspection, that the quadratic factor will look like: (z^{2}+kx+1) where k is a constant. This is because z^{3}+1 can be written as a product of its factors: z^{3}+1=(z+1)(z^{2}+kx+1). The coefficient of the z^{3} term and constant term are 1 so we know the coefficient of the z^{2 }and constant term in the quadratic. By inspection we can also see that k=-1.

The second part of the questions asks us to show that y^{3}=-1. There are several approaches for this. You can square y to obtain y^{2} then multiply it by y again to obtain y^{3} and show that it is equal to -1. You can also write y in polar form, then apply De Moivre’s theorem to obtain y^{3} then convert back into Cartesian form. The final method is as follows:

We want to show that y^{3}=-1. Earlier we wrote: z^{3}=-1 which leads to z^{3}+1=(z+1)(z^{2}-z+1)=0. We know that if y is a cubic root of -1 then it cannot be equal to -1, so for the above to be true then (z^{2}-z+1) must be equal to 0. We simply solve the quadratic (z^{2}-z+1)=0 using the quadratic formula. The formula gives us that z=(1+i√3)/2 after taking the positive result. We note that this is equal to y. Since y=x then y is a cube root of -1.

The third part of the question asks us to show that y^{2}=y-1. Using our working from earlier, we know that y is a root of z^{2}-z+1=0 so y^{2}-y+1=0. Rearranging this we have y^{2}=y-1 as required. Alternatively you can find y^{2} then show that it is equal to y-1.

The final part of the question is to find the value of (1-y)^{6}. Instinctively most people would use the binomial expansion but there is a far more elegant solution. From the previous part we have y^{2}=y-1. If we multiply both sides by -1 then:

(-y^{2} )=1-y

Raising both sides to the power 6 as the question required and using power laws:

(-y^{2})^{6}=(1-y)^{6}

Why would we do this though? Maybe you can see why; there is a piece of information from earlier that we can use, if not, keep following and you will see that this method is very clever.

(-y)^{6*2}=(y)^{12}=(1-y)^{6}

The negative sign disappears as squaring a number makes it positive.

We can manipulate this further:

(y)^{12}=(y)^{3*4}=(y^{3})^{4}

Earlier we had (y^{3})=-1 and so ultimately the value of (1-y)^{6}=(-1)^{4}=1.

We want to make y the subject of the equation and so we need to write it in the form y=f(x) where f(x) is a function in terms of x.

When rearranging equations with fractions for a certain subject, factorising will usually be involved. A good start to this would be to put everything over a common denominator (x(x+4)) which is the product of the two different denominators in the original equation. Recall that to maintain **equality** (i.e. the left hand side equals the right hand side), **what you do to one side you must also do to the other side. **This is an incredibly important rule that will help you solve the **majority, if not, all** of maths problems you will come across while avoiding mistakes.

So, to (y/x), we multiply by (x+4)/(x+4). Note that this is equal to 1, as any number divided by itself is 1, and any number multipled by 1 is itself so we aren't actually changing the equation. Simlarly for (2y/(x+4)) we multiply by (x/x). For the right hand side of the equation, we can multiply by (x(x+4))/(x(x+4)).

This gives us:

(y(x+4)+2xy)/(x(x+4))=3(x(x+4))/(x(x+4))

At this point we can multiply everything on both sides of the equation by (x(x+4)) to remove the denominator. We can do this because of the rule mentioned earlier: **what we do to one side of the equation we must also do to the other side of the equation.**

After this step we have:

xy+4y+2xy=3(x(x+4))

We want y as the subject of the equation so on the left hand side we factorise out the y term, as every term on the left hand side has y in it.

y(x+4+2x)=y(4+3x)=3(x(x+4))

Finally, we divide both sides by (4+3x):

y=(x(x+4))/(4+3x)

We want to make y the subject of the equation and so we need to write it in the form y=f(x) where f(x) is a function in terms of x.

When rearranging equations with fractions for a certain subject, factorising will usually be involved. A good start to this would be to put everything over a common denominator (x(x+4)) which is the product of the two different denominators in the original equation. Recall that to maintain **equality** (i.e. the left hand side equals the right hand side), **what you do to one side you must also do to the other side. **This is an incredibly important rule that will help you solve the **majority, if not, all** of maths problems you will come across while avoiding mistakes.

So, to (y/x), we multiply by (x+4)/(x+4). Note that this is equal to 1, as any number divided by itself is 1, and any number multipled by 1 is itself so we aren't actually changing the equation. Simlarly for (2y/(x+4)) we multiply by (x/x). For the right hand side of the equation, we can multiply by (x(x+4))/(x(x+4)).

This gives us:

(y(x+4)+2xy)/(x(x+4))=3(x(x+4))/(x(x+4))

At this point we can multiply everything on both sides of the equation by (x(x+4)) to remove the denominator. We can do this because of the rule mentioned earlier: **what we do to one side of the equation we must also do to the other side of the equation.**

After this step we have:

xy+4y+2xy=3(x(x+4))

We want y as the subject of the equation so on the left hand side we factorise out the y term, as every term on the left hand side has y in it.

y(x+4+2x)=y(4+3x)=3(x(x+4))

Finally, we divide both sides by (4+3x):

y=(x(x+4))/(4+3x)