OHMYGOD SIR I FINALLY GOT IT THANK YOU, I have been wstching so many videos and sitting here for hours because i did not know the division trick...u are an angel
Great tutorial thanks. could someone please help me solve this partial fraction to find A and B. Question: 2x^2-2x-3 / (2x-2)(x-1) Where I've got to: 2x^2 - 2x - 3 = (x - 1)(2x - 2) + A(x - 1) + B(2x - 2) As you can see, I need to choose a value of X so that one of the brackets cancels out leaving me with just A or B, however, the only value that can do this is if x=1 and that cancels both A and B out and I can't find either of them.
I cba to do it but sub in two different values of x. You'll then get two simultaneous equations and then voila you should get constants for A and B. Obviously don't sub in one Are you sure the question is actually correct because if x=1 then LHS = -3 HOWEVER RHS=0 so I think the actual question is wrong.
What if the top and bottom have highest power of x^2. How do we know if its improper or not? In my case (x^2 +3x +3)/((x+1)(x+3)) and the solution says its improper. How and why....
can you explain to me why you make it more complicated when doing the partial fraction bit by not simply doing the partial fractions method on 2/(x+1)(x-2)? get the same answer much more easily
[1 + x ] doesn't have any denominators other than 1 which can be disregarded, so when you multiply both sides by (x+1)(x-2) you'll have both these factors as common factors of [1 + x] so when you input x = -1 or x = 2 or whatever value of x that makes the factor 0 you'll notice that since [1 + x] has *both* these factors as common it'll cancel out every time when trying to find constants. Since the quotient will rarely have a fraction that includes the factors of the denominator, I think it's a safe bet that it'll work always if you want to solve for the constants in insolation.
Beautiful. Learnt this in high school about 7 years ago and forgot this.😅😅.Currently doing my masters and this video is helpful
You're welcome! Haha, always good to revisit
OHMYGOD SIR I FINALLY GOT IT THANK YOU, I have been wstching so many videos and sitting here for hours because i did not know the division trick...u are an angel
dude thank you soo much i have an exam tomorrow and it helped so much.
All the best for that
MrFuriousD how was u'r exam?
hello sir, thank you for the video. is there an alternative to long division? something shorter and more intuitive
Excellent.
thank you very much sir 👍
Hi, is this on your website sir?
C4 next month? less gooooo
Can't we take X common from the numerator and take in LHS and after fraction multiplying it back ??
Great tutorial thanks. could someone please help me solve this partial fraction to find A and B.
Question:
2x^2-2x-3 / (2x-2)(x-1)
Where I've got to:
2x^2 - 2x - 3 = (x - 1)(2x - 2) + A(x - 1) + B(2x - 2)
As you can see, I need to choose a value of X so that one of the brackets cancels out leaving me with just A or B, however, the only value that can do this is if x=1 and that cancels both A and B out and I can't find either of them.
I cba to do it but sub in two different values of x. You'll then get two simultaneous equations and then voila you should get constants for A and B. Obviously don't sub in one
Are you sure the question is actually correct because if x=1
then LHS = -3 HOWEVER RHS=0 so I think the actual question is wrong.
thanks a lot sir
You are welcome. Thanks for viewing.
What if the top and bottom have highest power of x^2. How do we know if its improper or not? In my case (x^2 +3x +3)/((x+1)(x+3)) and the solution says its improper. How and why....
It is improper. It is improper if the top is of greater or equal degree to the bottom. In this case they both equal degree 2.
can you explain to me why you make it more complicated when doing the partial fraction bit by not simply doing the partial fractions method on 2/(x+1)(x-2)? get the same answer much more easily
benj osborne It's just a method, it is up to you which way you want to do it. I felt this looked easier and was neater.
ExamSolutions ok thanks, would've failed c4 today if it wasn't for your videos, so thanks v much!
+ExamSolutions will that method work always? if we were to remove the quotient and just take the partial fraction form to solve A and B ?
[1 + x ] doesn't have any denominators other than 1 which can be disregarded, so when you multiply both sides by (x+1)(x-2) you'll have both these factors as common factors of [1 + x] so when you input x = -1 or
x = 2 or whatever value of x that makes the factor 0 you'll notice that since [1 + x] has *both* these factors as common it'll cancel out every time when trying to find constants.
Since the quotient will rarely have a fraction that includes the factors of the denominator, I think it's a safe bet that it'll work always if you want to solve for the constants in insolation.
I have an exam on this @ 9:00, thats in 8 hrs!
Quick question. I want to know your gear for your videos. I sent you an email too