Piotr  W. A Level Maths tutor, IB Maths tutor, GCSE Maths tutor, 13 p...

Piotr W.

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Degree: Mathematics (Masters) - Bath University

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About me

Hi, I’m Piotr and I am currently an undergraduate mathematics student at University of Bath. I have always been enthusiastic about the opportunity to talk about mathematics with others and share my points of view.  In my sessions I put strong emphasis on making the main concepts clear and understandable by providing detailed description, explaining the theory and the key ideas which motivate it. My goal is also to discuss several techniques of solving problems and describe ways in which one can approach them.  I believe that this is the best way for a student to earn as much as possible from a particular subject.  At sixth form I was involved in tutorials held at my school where I taught younger students as well as helped my classmates with the course. During that time I gained vital experience in interacting with people in various age groups developing as a tutor. In addition I can give some advice on the university admission interviews and for those who are interested, I can talk about more advanced areas of mathematics.

Subjects offered

SubjectLevelMy prices
Further Mathematics A Level £20 /hr
Maths A Level £20 /hr
Further Mathematics GCSE £18 /hr
Maths GCSE £18 /hr
Physics GCSE £18 /hr
Further Mathematics IB £20 /hr
Maths IB £20 /hr
Maths 13 Plus £18 /hr
Maths 11 Plus £18 /hr
.STEP. Uni Admissions Test £25 /hr

Qualifications

QualificationLevelGrade
MathematicsA-LevelA*
Further MathematicsA-LevelB
PhysicsA-LevelA
ChemistryA-LevelB
STEPUni Admissions Test2
AEAUni Admissions TestM
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Questions Piotr has answered

How to solve polynomials

First of all let's start with a lttle explanationn of waht a polynomial is. A polynomial is a function of x denoted f(x) which is defined as f(x)=a0+a1x+. . .+anxn which in other words is a sum of x's raised to some positive power multiplied by some number ai  which is called the coefficent ...

First of all let's start with a lttle explanationn of waht a polynomial is.

A polynomial is a function of x denoted f(x) which is defined as

f(x)=a0+a1x+. . .+anxn

which in other words is a sum of x's raised to some positive power multiplied by some number ai  which is called the coefficent

Examples of polynomials:

f(x)=mx+c -> linear function

f(x)=ax2+bx+c->quadratic

f(x)=ax3+bx2+cx+d->cubic

Now, what does it mean to solve a polynomial? It simply means for which values of x we have that f(x)=0, where f is our polynomial. Graphically, in x-y coordniate system if y=f(x) where does the curve cross the x-axis.

How does it work

Case1: Linear function

for f(x)=mx+c we set f(x)=0 in the equation and solve for x.

Case 2:quadratic

f(x)=ax2+bx+c

Before solving the equation we have to check whether there are any.

When we draw f(x) on the x-y plane we we see that ot represents a parabola.

(Normaly here we would draw the picture)

We can also observe that the curve has a lowest point. As we cab see if the lowest point is below the x-axis we have two solutions, has one solution if it touches the axis, or no solutions if it is above the x-axis. Now the question is how to check it without having to draw the graph. What we do, we calculate the discriminant. Notice that the lowest point of the parabola is at x=b/2a (Shown later on by introducing derivatives). When we plug this value into our equation we get

y=a(b/2a)2+b(b/2a)+c

but we know that it can be positive, zero or negative, therefore if y>0 at this point we said earlier that the polynomial has two solutions so

y>0 means 02+b(b/2a)+c

rewritting it slightly we get b2-4ac > 0. Similarly this equals zero or less that zero if it has one or no solutions respectively. Therefore we found the expression for the number of solutions in terms of coefficients. b2+4ac is called the discriminant.

We now want to find the particular solutions of quadratic equation. There are several ways of doing this. The quickest one is to factorise it by using little trick. Observe that we can rewrite quadratic which has solutions p,q in the following form:

y=m(x-p)(x-q) 

Why? Well, if we plug in the values of p and q we get y=0 an remember that polynomials are unique up to value of m. For simplicity let m=1

y=(x-p)(x-q)=x2-(p+q)x+pq

so, all we need to do is to express the coefficent of x in terms of p and q and we are done.

Case 3: cubic

For any cubic polynomial there is always at least one solution, the reason being that x3 is a dominant term and so for large enough positive or negative values the function is above and below x-axis at some point (Picture should be added to visualize the result). There is clear procedure how to solve cubic polynomials however if we can spot one of them we are almost done as we only need to solve the remaining factor which is a quadratic equation and this we know how to solve.

Example:

f(x)=x3-2x2-x+2 can find x=1 as one of the solutions -> f(x)=(x-1)(x2-x-2) -> f(x)=(x-1)(x+1)(x-2)

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7 months ago

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