the cover-up method & why it works! (for partial fractions decomposition) - Videos

Hello

Glad to see a new video! Thanks for the free lesson.

Wow 3 videos in a row ! Semi-automatic mode !!

awesome as always ??

title says "why is works" instead of "why it works" 🙂

Great Stamina???!!!Salute Sir??

Okay so something I've never understood about partial fractions is why the numerator always to be one degree lower than the denominator. What's the reason?

I love your videos :')

This would actually be a great shot to learn in Precalc and then carry into calc II

Wait, I've just seen a video exactly like this one. This is just a remake, right? Also, it doesn't answer my question if it works for same factors, but I've figured it out while watching this. If two (or more) denominators were the same, multiplying by it wouldn't leave only one constant, in fact, it would yield sum of two (or more) constants.

You finally back, yey!

I need help. I know the answers to this problem but I have no idea how to solve it:
Sqrt(x^2-5040)+
Sqrt((x-1)^2-5040)+
Sqrt((x-2)^2-5040)=30

How do you manage to spit out so many videos in such a short time? Loving the content!

Gostei muito das tuas aulas e estou recomendando pra todo mundo aqui no Brasil.

Gos ta da metodologia de ensino tambem.

>800 views, >100 likes, 0 dislikes. Good video!

You just made me flashback to, many decades ago, doing contour integrals by merely finding the sum of residues at the enclosed poles.
I suspect that there is similar 'trickery' as to why that 'works'.
Is this something you will ever touch upon in your videos? (hint, hint}

I'm not sure if I would agree on this. A, B and C may be the correct numbers, but the method to determine them lacks something, I think. Of course, x may not be equal to 1, 2 or 3 in the first equation because then you would divide by 0. That means that in every equation after that, the same x still can't be equal to 1, so you can't just let x = 1 and expect everything to work out. On the other hand, if you multiply by x-1 and let x = 1, you essentially multiplied by 0, so you end up with 0 = 0, but that yields no information about A.

Who needs calc 2 with this channel around? Amazing work.
Please consider expanding to calc iii.

love it

But will you get working out marks?

very nice

only works for linear denominators?

I'm almost certain I never learned partial fractions in school – not middle, high school, nor college! It's a hugely useful technique, and I wonder if those of you in the US can confirm or deny you were taught this. Is this a notable hole in our mathematics education?

thank you a lot

Could you do more videos explaining the method for more kinds of factors in the denominators (repeated or second degree), and also explaining why the set-ups look like they do for those more difficult cases? Also, can you do some of these using complex factorizations? (You're building a really useful library of explanations!)

https://en.wikipedia.org/wiki/Heaviside_cover-up_method for more info

Another interesting use of the method can show that (3x+5)/(1-2x)^2 = (13/2)/(1-2x)^2 + (-3/2)/(1-2x)

Am confused. If you multiply out by the denominator of the lhs you get x^2 terms on the rhs. How do these equate with the linear terms on the left?

This irritates me. I HATED the partial fraction decomposition because of those nasty systems we had to solve. Why didn't my profs show me this?