Partial Fractions - Repeated Linear Factors : ExamSolutions Maths Revision
Вставка
- Опубліковано 31 гру 2014
- Revising partial fractions with repeated linear factors.
Go to www.examsolutions.net/ for the index, playlists and more maths videos on partial fractions and other maths topics.
THE BEST THANK YOU: www.examsolutions.net/donation/
cheers. I think i'm going to use the 'any value' method than the 'comparing coefficients' method, just because i feel like there's less chance of me making a silly mistake.
Thank you so much ❤️❤️❤️ it’s so helpful
Thanks a million!
Thank you!
Thanks very much it has help me a lot
Thank you
Very nice
legend
0:52 how does that make sense? I dont get why youd write out the factors out again but 1 with squared and the other without squared
Please please im stuck on that. That the only issue i am having, nobody explains it so please could you help me?
Look at it another way, it may help. Suppose you had to do A/3 + B/9 + C/2 then the lowest common denominator would be 18. Now if I had something over / (3)² (2) then as you can see it came from A/3 + B/3² +C/2 not B/3²+C/2. We must include the factors of 3² in this case 3, in case it was part of the original sum. A may well turn out to be 0 but let the maths show that.
@@ExamSolutions_Maths ohh, thanks for taking your time to explain that, it really helped me
@@misan2002 You're welcome.
@@ExamSolutions_Maths just one more question, why would 3 be one of the factors?
@@ExamSolutions_Maths thank you for the explanation!!(^_^ )
gooddddddddddddddddd
I subscribe
Welcome to the channel Mwila! I'm sure you will find the videos really helpful
Note to self: 6:35
mashallah
why is your B= 2/3 becomes 4/3 in finding the A?
might be very late but we derived 2B from the substitution of x=0 and since we already found that B= 2/3, 2/3 x 2 = 4/3