The Cover-up method of Partial fraction decomposition

In this video, I showed how to use the cover-up method of partial fraction decomposition. Best for rational functions with nonrepeated linear factors.

[@kingbeauregard]

I liked the explanation of why it works. I know the technique, but I am always distrustful of it, like it is a trick or I am misremembering it. Hopefully I will be able to trust it more better now.

[@punditgi]

You can't cover up how good Prime Newtons is as a teacher! 🎉😊

[@PrimeNewtons]

This one wins 🏆 🙌 👌 👏 🤣 😂

[@Occ881]

Sure, he's hot too❤

[@RobG1729]

Prime Newton always covers his head so fashionably!

[@ounaogot]

"Let's get into the video" is my new favorite phrase

[@khairyalkhalidy1316]

Man with whole due of respect screw all my teachers they were idiots never did this

[@holyshit922]

In my opinion this cover up has some advantages when roots of the denominator are real and distinct We need to remember which value we have already used If we allow using complex numbers then there is residue method which gives us partial fraction decomposition (Just like for inverse Laplace transform but without this exp(st) factor)

[@holyshit922]

If we use it for integration then we can use polynomial long division and Ostrogradsky method of isolation rational part of integral and then we have integral with integrad which has only distinct roots of denominator

[@elai3147]

10:37 still don't get it, when you make x zero aren't you still dividing by zero even though the x in the denominator on both sides have been cancelled out, why does it work? is it because undefined=undefined works as a valid equation?

[@skyrider53]

Same thing I was wondering

[@kingbeauregard]

To be sure, you couldn't put x=0 into the original equation. When we multiply everything by x, we have made a different equation, the only purpose of which is, it helps us solve the original equation. So while we can't do x=0 in the original equation, we can in the helper equation.

[@joelmacinnes2391]

I guess it's a bit like limits, where you can take a derivative of 0/0 and instead of putting in the numbers, you solve it first and end up with something else

[@NirDagan]

you take the limit where x goes to 0

[@kiro9291]

I'm learning this this semester so yippee

[@Jason-ot6jv]

I guess its the same thing as multiplying the entire equation by the original denominator so that there is no fraction, then just let x equal all the poles until you find values for A,B,C etc

[@ItsJims]

This guy getting me interesting more.. I am 10th grader, I like learning/watching such advanced mathematics and this guy is a good place...

[@nothingbutmathproofs7150]

Nicely done!

[@ismaëlKanouotchiadzo-r2e]

your explanations are so good . thanks sir

[@NirDagan]

Strictly speaking, you can't just plug zero in the last step as it must be that x is different from 0. However you can take the limit of both sides of the equation as x goes to zero which gives you the same result as plugging in zero.

[@ИринаРзаева-ф2с]

Ответ в. Единице.

[@cggf77fffucfucd]

Thank you, Teacher! İ love this video very much..

[@joycetjizu3160]

Oh wow!

[@kragiharp]

Thank you so much! ❤️🙏

[@saiprasadpadhy6832]

Mathematically brilliant teacher.

[@raivogrunbaum4801]

For prime newton this method works only in some cases. Try to use this method for integral (2x^2+3x-1)/(x^2+x+1)^3

[@ounaogot]

He stated that it works for linear ones

[@raivogrunbaum4801]

@@ounaogot linear?? You mean that denominator can express in form (x-a)(x-b)..... and so on?? But (x^2+x+1)=(x+1/2+isqrt3/2)(x+1/2-isqrt3/2). It isnt linear?? Try to use that method and show what you get.

[@vitotozzi1972]

Wonderful method!