Ive notice that most of us have problems at our p3 and ill try helping you guys out since p3 is really really close....so...pls enjoy my review !!
P3 Maths (edexcel)
1.1 Partial Fractions
Students should appreciate the usage of partial fractions as these questions will usually be followed by an integration of partial fractions.Thus it is crucial to understand it as it will determine your scores at your integration sections at some of the papers
there are 4 types of partial fractions which can be used
the first would be linear factors as the denominator
= 5+x / (x+2)(x-2)
A/(x+ 2) + B/(x-2)
5 + x = A(X-2) + B(X+2)
let X as 2 to cancel out the A and leaving the B
5 + (2) = B(2+2)
B = 5/2
now you try getting the A
after getting A and B,substitute into the partial fraction to get the final form
The second form would be the
quadratic factor as the
5+x / (x+2)(x²-2)
it would be advised that you should fully notice that the quadratic factor and the linear factors as the denominators are not the same
you should express it as
=A/(X +2) + (Bx + C)/(X²+2)
then use the normal method shown above to solve it
The third form would be the
repeated factor as the denominator
5+x / (x+2)(x-2)²
just as the quadratic factor as the denominator...students should notice that this isnt the same as the linear factor and the quadratic factor
we should express it as
=A(X + 2) + B(x-2) + C(X-2)²
If we were given an equation of
5+x / (x+2)(x-2)³ instead...
we should realise that it would be written in the form of
=A(X + 2) + B(x-2) + C(X-2)² + D (X -2)³
this goes on till the factors reach up to infinity (any number is possible)
The forth and final form would be
Improper algebraic fractions
this would be dividing the numerator using the denominator
and there is however a condition which one must apply...
the denominator must have a varialble degree less than or equals to the numerator
x² + 5 / (x +5) thus it can be divided since the denominator has degrees of 1 and the numerator has the degree of 2...
this method is also similar to the remainder theorem
example of division can be found out from the text book...(cant write it here...not enough tools)
and thats it for partial fractions !! congrats..you`ve completed one part of the syllabus
now for part 2
1.2 Remainder Theorem
The steps of division is similar to those found in the Improper algebraic fractions
However,i am not going to go onto more details on that but i am however going to the more important parts of the exam questions...
given find the remainder when 2x³ +4x² -6x +1 is divided by 2x-1
one might have thought that this would be a very difficult question,since it has to be divided and it was such a hassle for some and it would usually take only 3 points for sucha hideous method..
now...try this on for size..
since the you are using 2x-1 to divide the equation...then just,
2x-1 = 0 and x = 1/2
use x = 1/2 and substitute into the original equation 2x³ +4x² -6x +1
=2(1/2)³ + 4(1/2)² - 6(1/2) + 1
thats the remainder..
but bear in mind that the exam questions will not give straigh forward questions such as above but would rather give the remainder first or maybe give you that (x-1) is a factor and you must prove it to them that there is no remainder...(deciding a factor would give no remainder)
a popular question of the exam schemes are when they ask you to calculate the value of a constant using a denominator and a remainder given by them
now you try getting the value of a
given 2x³ + ax² + x + 1 = f(x)
given that the remainder is -29 when it is divided by x+2
try getting it...the answer is a=-3
there is nothing muich that can be taught here but a few formulas that must be practised very frequently
Binomial expansions are
(1+ax)^n = 1 + nax + n(n-1)/2! (ax²) + n(n-1)(n-2)/3! (ax³) ...............
the exam questions would usually want and expansion up to x³
assuming that you are given the number (2+4x)^-1
you should always remember to put it in the form of (1+ax)^-1
how do you do it?
take out (2^-1) from the equation since you have to convert 2 into 1
and just use the formula given above to deduce the expansion
another popular type of question would be the type where they asked you to add both expansions...
take as an example
=(1) + (-x)(-1) + (-1)(-1-1)/2! (-1x)² + (-1)(-1-1)(-1-2)/3!(-1x)³
= 1 + x + x² +x³
and for the second expansion
=(1) + (-2x)(-1) + (-1)(-1-1)/2! (-2x)² + (-1)(-1-1)(-1-2)/3!(-2x)³
= 1 +2x + 2x² + 2x³
therefore (1-x)^-1 + (1-2x)^-2 = (1) + (2)
=2 + 3x + 3x² + 3x³
there ,simple isnt it???
just expand both equations and add them both..
these can be most frequently asked after the expression of partial fractions...
Sorry if there are mistakes or i might have missed something out up to now..
ill be starting my differentiation topic at about now.
Give me about 1 hour to finish it...
no one wants the p3 review??
i thought you guys wanted it..
awww..anyways...i dont think my revision notes arent gonna anyone anyways..
read if you want to..
but i think its really compact..
and i just printed it for myself for revision right before my exams this wed
oh no worries...
ill be starting an S2 one shorty around next week
just stay tuned then !