Currently unavailable: for regular students
Degree: Chemical Engineering (Masters)  Bath University
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.
CRB/DBS Standard 
No 

CRB/DBS Enhanced 
No 
Please get in touch for more detailed availability
Elle (Student) September 18 2016
Elle (Student) August 28 2016
Zoe (Parent) July 18 2016
Elle (Student) July 11 2016
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}=y1. 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}=y1 as required. Alternatively you can find y^{2} then show that it is equal to y1.
The final part of the question is to find the value of (1y)^{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}=y1. If we multiply both sides by 1 then:
(y^{2} )=1y
Raising both sides to the power 6 as the question required and using power laws:
(y^{2})^{6}=(1y)^{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}=(1y)^{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 (1y)^{6}=(1)^{4}=1.
see moreWe 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)
see more