I also thought this. But from my understanding now, it actually does do what you were suggesting: A (x+2)(x+2)^2 + B (x+1)(x+2)^2 + C (x+1)(x+2) / (x+2)(x+1)(x+2)^2 You are confused about this step. Basically this does not have the desired denominator of (x+1)(x+2)^2 which is from the original equations denominator. We need the original and partial fraction denominators to be the same to cancel out the denominators leaving us with just the numerator. So to make the denominators the same we remove the (x+2) by cancelling it out in the fraction leaving us with the desired denominator: A -(x+2)- (x+2)^2 + B (x+1) -(x+2)^2- ^1 + C (x+1) -(x+2)- / -(x+2)- (x+1)(x+2)^2 This then gives us the videos fraction of A (x+2)^2 + B (x+1)(x+2) + C (x+1) / (x+1)(x+2)^2. There are easier ways of doing this but this is how this method works. I hope this helps.
im linking a video which explains much better than I could: th-cam.com/video/pYUTZD1GVyU/w-d-xo.html , watch it until you get ur head around it - it makes a lot of sense when u understand, if u still dont understand let me know and I'll try and explain the video to you :)
0:45 If we must account for potentially having a "B term" where there is a fraction with the denominator (x+2), then why do we not also account for potentially having a "D term" where there is a fraction with the denominator (x+2)(x+1). Original denominator: (x+1)(x+2)^2 Surely there are 4 factors to this denominator in the original fraction: Factors: (x+1) Which was your "A term" (x+2) Which was your "B term" (x+2)^2 Which was your "C term" (x+1)(x+2) Which I am wondering why wasn't it your "D term"??! Surely (x+1)(x+2) is also a factor of the denominator and could be a denominator of a partial fraction? I am sure I am missing something here :/ plz help.
I've been wondering the same thing ever since I've read this comment... There has to be a 4th possible factor with it's own value, but I'm not sure if it's necessary to put...
so you have to assume there's a B (due to them having a common denominator), and when u do the x=0 part you're basically working out whether or not there is actually a B, so if it equals 0 then there is no B, etc
![ลองสั่งของจากแอพ temu ถูกจริงหรอ?? ( part : 1/2 ) [โกงไหมครับ ep.89 ] | DOM](http://i.ytimg.com/vi/RTcZSQTiDZw/mqdefault.jpg)
I was so confused about partial fractions until I watched this video, thanks so much!!
no problme
at 1:40 why isnt it A (x+2)(x+2)^2 + B (x+1)(x+2)^2 + C (x+1)(x+2)
Ikr
you multiply only by the original denominator roots, so you do not include the denominator of 'B' when multiplying up. hope that helps
I also thought this. But from my understanding now, it actually does do what you were suggesting:
A (x+2)(x+2)^2 + B (x+1)(x+2)^2 + C (x+1)(x+2) / (x+2)(x+1)(x+2)^2
You are confused about this step. Basically this does not have the desired denominator of (x+1)(x+2)^2 which is from the original equations denominator. We need the original and partial fraction denominators to be the same to cancel out the denominators leaving us with just the numerator. So to make the denominators the same we remove the (x+2) by cancelling it out in the fraction leaving us with the desired denominator:
A -(x+2)- (x+2)^2 + B (x+1) -(x+2)^2- ^1 + C (x+1) -(x+2)- / -(x+2)- (x+1)(x+2)^2
This then gives us the videos fraction of A (x+2)^2 + B (x+1)(x+2) + C (x+1) / (x+1)(x+2)^2.
There are easier ways of doing this but this is how this method works. I hope this helps.
Think of it multiplying the (A+B+C) thing by the denominator [(x+1)(x+2)²]
Why do you have to do this for repeated terms?
this made me clear more than he school did . thank you so much sir.
It's the best explanation available.
0:30 why can you have an (x+2) term?
im linking a video which explains much better than I could: th-cam.com/video/pYUTZD1GVyU/w-d-xo.html , watch it until you get ur head around it - it makes a lot of sense when u understand, if u still dont understand let me know and I'll try and explain the video to you :)
WOW.... perfectly taught 😌
How do i figure out that there would be an x-1 term and not just (x-1)²
Thank you that was really helpful-
yes
Thank you so much
God bless you for this.
It was really helpful 👍🥰💞💜❤
Very helpful, thanks!
Thanks❤ for this vdo
Thk yo straight away a plus student 😊😊
This method won't work if there is a x term multiplied in the denominator??
Thank you for thisss🙏
Thank youuuu
This helped alot
0:45
If we must account for potentially having a "B term" where there is a fraction with the denominator (x+2), then why do we not also account for potentially having a "D term" where there is a fraction with the denominator (x+2)(x+1).
Original denominator: (x+1)(x+2)^2
Surely there are 4 factors to this denominator in the original fraction:
Factors:
(x+1) Which was your "A term"
(x+2) Which was your "B term"
(x+2)^2 Which was your "C term"
(x+1)(x+2) Which I am wondering why wasn't it your "D term"??!
Surely (x+1)(x+2) is also a factor of the denominator and could be a denominator of a partial fraction?
I am sure I am missing something here :/ plz help.
I've been wondering the same thing ever since I've read this comment... There has to be a 4th possible factor with it's own value, but I'm not sure if it's necessary to put...
Thanks. This help me a lot
Thanks for sharing
Thank you very much 🙏❤️ that was awesome 💕
How do we know there might be a b term? X+2 is not a factor of x+1
Please respond, please, i iust need this explanation
its cause they have a common denominator with the fraction on the left side
so you have to assume there's a B (due to them having a common denominator), and when u do the x=0 part you're basically working out whether or not there is actually a B, so if it equals 0 then there is no B, etc
thankyou so much sir!
legend
Wow great
I tried this method for my question that is 1/(s-3)^2X(s^2-7s+12) can you help me
Rly helpful thanks a lot
thank you
love from india
Nice
I don't understand why we are cancelling the A and B at 12:25 😭😭😭
±
i still dont get why you have to take away sixteen 2:14
16 x -1 is -16. therefore you takeaway 16