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Degree: Mathematics (Masters) - Warwick University
Hi! I am a second year mathematics student at the University of Warwick, with a love for pure maths - in fact since 14 I've been set on becoming a mathematician. My intense fascination with this field not only motivated me to study new concepts on my own, but also led to my continued experimenting with simple concepts in order to gain a deep and intuitive understanding. I hope to pass this intuition on to all of my tutees, along with useful methods that I have picked up, and most importantly an engagement with the subject, which I feel is the best route to success.
Throughout my time studying for A levels I frequently helped other students to understand and grasp new concepts, often to an extent where I was an impromptu teaching assistant, and in one particular case I taught my A level class about the Taylor expansion. I also spent one term helping to teach a younger class mathematics, during which I learnt how to simplify ideas into understandable chunks. In year 13, I undertook a project in which I wrote a concise guide to a pure branch of mathematics, group theory. I had to write with an intriguing tone, and also make use of pedagogic features like in-depth examples and analogies. All of this experience is still fresh in my memory, and so I am well-placed to tutor GCSE and A Level material at a high standard.
|Further Mathematics||A Level||£20 /hr|
|Maths||A Level||£20 /hr|
|Further Mathematics||GCSE||£18 /hr|
|STEP I||Uni Admissions Test||1|
|Before 12pm||12pm - 5pm||After 5pm|
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Solving cubics is an interesting problem: while there is a formula which can find the roots of every cubic equation, it isn't taught and is not generally worth learning. Instead, exam questions will often give you a root of a cubic, and from that you are expected to fully factorise it, and hence find the roots. Let's look at an example!
Q: Given that -2 is a root of 2x^3 + 9x^2 - 2x - 24, find all roots.
A: Firstly, we know by the factor theorem that if a is a root of a polynomial (a cubic, for instance), then (x - a) will be a factor of that polynomial. Therefore, we know that (x + 2) is a factor of 2x^3 + 9x^2 - 2x - 24. To get the other roots, we could use polynomial division, but there is a way which is quicker and less error-prone. Write this as such:
2x^3 + 9x^2 - 2x - 24 = (x + 2)( )
Now, we know that in the brackets there will be an x^2 term, an x term and a constant. What can the x^2 term be? Well it must be 2x^2, because when we multiply out the brackets, we need to end up with 2x^3, and the only way we get a cubic term here is by multiplying the x by some x^2 term.
2x^3 + 9x^2 - 2x - 24 = (x + 2)(2x^2 )
Similarly, the constant term must be -12, because we need a -24 after multiplying out the brackets, and the only way to get a constant term here is by multiplying the two constant terms.
2x^3 + 9x^2 - 2x - 24 = (x + 2)(2x^2 - 12)
Now the x term. if we start to multiply out, we see that we have 2x^3 + 4x^2 - 12x - 24. We have 4x^2, which we got from multiplying 2 by 2x^2, but we need 9x^2, so we have to add on 5 more. The other way to get an x^2 term is to multiply two x terms. So our x term must be 5x, so that when we multiply it by the x in the (x + 2), we end up with the extra 5x^2.
2x^3 + 9x^2 - 2x - 24 = (x + 2)(2x^2 + 5x - 12)
Finally, we just have to factorise the quadratic in the bracket. Using inspection, or failing that the quadratic formula (though this is more prone to error), we find that:
2x^3 + 9x^2 - 2x - 24 = (x + 2)(2x - 3)(x + 4)
Applying the factor theorem again, we find that the roots are -4, -2 and 3/2.see more