This is the 3rd pattern on the topic of partial fraction.
The topic of partial fraction is quite simple but there are still some tricks that the "O" level examiners used in setting questions on partial fractions.
Perhaps you will like to do these questions.
Question 1
Express x^2/(9-x^2) in the form A + B/(3-x) + C/(3+x)
and find the values of A, B and C.
[The trick on this question is on the constant A]
Question 2
Express in partial fractions
(2x^3 - 3x^2 - 7x + 4)/[(x-1)(x+1)]
Thank you for your kind attention.
Regards,
ahm97sic
For the first qn)
A= -1 B=1.5 C=1.5
Hope i am correct XD
2) A= -2 B=-6
Hi Davidche,
There are some mistakes in your answers. It is probably due to careless mistakes. Please re-check your calculations.
Thank you for your kind attention.
Regards,
ahm97sic
XDXD i always make careless mistakes.
There is no trap with question 1 ?
question 2 is -2/(x-1) - 6/(x+1)
Qn 1
-1 + (3/2)/(3-x) + (3/2)/(3+x)
A = -1, B = 3/2, C = 3/2
Qn 2
2x - 3 - 2/(x-1) - 3/(x+1)
Hi Davidche and Secretkiller,
I have used the long-division and cover up method to solve the questions.
However, these two questions can also be solved by using the comparison method or substitution method ie with the use of long division for example 2 too.
Question 1
Express x^2/(9-x^2) in the form A + B/(3-x) + C/(3+x)
and find the values of A, B and C.
[The trick on this question is on the constant A]
Answer
Use Long Division,
x^2/(9-x^2) = -1 + 9/(x^2-1)
= -1 + 9/[(x-3)(x+3)] = A + B/(3-x) + C/(3+x)
Use Cover up method,
= -1 + 1.5/(3-x) + 1.5/(3+x)
Hence, A = -1, B = 1.5 and C = 1.5
Question 2
Express in partial fractions
(2x^3 - 3x^2 - 7x + 4)/[(x-1)(x+1)]
Answer
(2x^3 - 3x^2 - 7x + 4)/[(x-1)(x+1)]
= (2x^3 - 3x^2 - 7x + 4)/(x^2-1)
Use long Division,
= 2x - 3 + (1-5x)/(x^2 -1)
= 2x - 3 + (1-5x)/[(x -1)(x+1)] = 2x-3 + A/(x -1) + B/(x+1)
= 2x -3 -2/(x-1) – 3/(x+1)
Thank you for your kind attention.
Regards
ahm97sic
PS : The objective of Example 1 is to let students to know/realise that it is not correct to
express in partial fractions as
x^2/9-x^2 = x^2/[(3-x)(3+x)] = A/(3-x) + B/(x+3)
it should be
x^2/(9-x^2) = -1 + 9/(x^2-1) = -1 + 9/[(x-3)(x+3)] = A + B/(3-x) and C/(3+x)
The objective of example 2 is let the students to know / realize the trick involved when they use the long division, there is a trick involved. You must do the long division for this question to realise the trick.
Example 2 also requires the concept learnt in example 1 too.
Dear Davidche, you have solved for the values of A, B and C without realizing the trick for example 1 and hence your answer to second example is incorrect as you have not done the long division for example 2 before you proceed to express it in partial fractions.
Dear Secretkiller, well done ! Your answer is perfectly correct.
Originally posted by Ahm97sic:x^2/(9-x^2) = -1 + 9/(x^2- 1 ) = -1 + 9/[(x-3)(x+3)] = A + B/(3-x) and C/(3+x)
shouldnt it be 9?
XD i realise my mistake. I didnt divide it.
Dear Davidche,
It is Ok. Everyone makes mistakes, as long as we learnt from the mistakes.
You should do the long division for question 2 too as there is a trick in it, you will only realize it when you do the long division for this question. Please do it so that you will not be caught unprepared if it comes out in the "O" level exam.
You must be preparing a lot for your coming "O" level exam. Study hard but you must take care of your health too especially make sure you have enough sleep.
Thank you for your kind attention.
Regards,
ahm97sic
By the way, I'm secretliker. Not killer. You've misread my nick.
For both questions, realise that both fractions are improper.
That means, that the polynomial in the numerator is raised to an equal or higher power than of the polynomial in the denominator.
Question 1:
x^2/(9-x^2)
Realise that the term x^2 in numerator is raised to the same power as the term with highest power, i.e x^2, in the denominator.
Therefore long division is required first.
Question 2:
(2x^3 - 3x^2 - 7x + 4)/[(x-1)(x+1)]
Realise that the term 2x^3 in numerator is raised to power 3, but in the denominator, the term with the highest power is only x^2.
Therefore long division is required first.
Hi Secretliker,
Oops, sorry. Please accept my apology.
Regards,
ahm97sic
It's ok.
Originally posted by Ahm97sic:Dear Davidche,
You should do the long division for question 2 too as there is a trick in it, you will only realize it when you do the long division for this question. Please do it so that you will not be caught unprepared if it comes out in the "O" level exam.
Oh yeah i see the trick. While i ask typing in my post that i cant do the long division, a realisation struck my brain. Now i can do. haha XD
Hi Davidche,
Good, now you know the trick so you would not be caught unaware if it comes out in the "O" level exam.
Study you must but relax you must too in your revision for your coming "O" level exam. Go for a healthy and intensive revision.
Regards,
ahm97sic
Originally posted by Ahm97sic:Hi Davidche,
Good, now you know the trick so you would not be caught unaware if it comes out in the "O" level exam.
Study you must but relax you must too in your revision for your coming "O" level exam. Go for a healthy and intensive revision.
Regards,
ahm97sic
I will! XDXD. Im sleeping early now. On monday i slept at 3am which is totally bad.
Dear Davidche,
Yes, try to have enough sleep.
You study best when you have enough sleep.
Take care.
Regards,
ahm97sic