Calculus - Partical Fractions
Why is the decomposition of:
10/(5x^2-2x^3) = (A/x)+(B/(x^2))+(C/(5-2x))?
Shouldn't it be --> (A/x^2)+(B/5-2x)?
Where did the (A/x) come from?
February 5, 2013
Calculus - Partical Fractions
Why is the decomposition of:
10/(5x^2-2x^3) = (A/x)+(B/(x^2))+(C/(5-2x))?
Shouldn't it be --> (A/x^2)+(B/5-2x)?
Where did the (A/x) come from?
7 Comments • Newest first
@cchpm @Oogendune
Ooohhh! Okaaayyy! I won't forget thanks!
[quote=GreatRomantic]I'm confused about that part....because the answer doesn't make sense to me..
Look at the denominators again --->
x+(x^2)+(5-2x) = 5x^2-2x^3
^That doesn't seem right to me.....?[/quote]
No, it is not about the denominators. when you factor the equation, it becomes (x^2)+(5-2x), but (x^2) is not linear, it is basically (x*x), so you have to separate them. the rule is making them a/x^2 + b/x
EDIT: it is called "Q(x) is a product of linear factors, some of which are repeated" Try google?
it's because if you have x^2 you have to repeat the denominator. x^2 is a repeated factor of x.
5x^2-2x^3 simplifies to x^2(5-2x)
Partial fraction will be A/x+B/x^2+C/(5-2x)
[quote=cchpm]on x^2, you have to separate it again, that is where the x come from.[/quote]
I'm confused about that part....because the answer doesn't make sense to me..
Look at the denominators again --->
x+(x^2)+(5-2x) = 5x^2-2x^3
^That doesn't seem right to me.....?
@cchpm WHAT? Second rule? I have my textbook in front of me right now, but I dont see a "second rule?
O.O
My textbook is rather difficult to understand....
on x^2, you have to separate it again, that is where the x come from.
EDIT: I think it is the 2nd rule. For example, if you have x^5, then you have to do a/x^5+b/x^4+c/x^3+d/x^2+e/x
Calculate the mass of the Earth.
then who was phone?