Question

Is there any shortcut to find partials if we have a quadratic factor in denominator??/

Answer 1

Example?

Answer 2

Put the denominator in the numerator with a -1 exponent. Then you can use the product and chain rules. That's how I roll.

Answer 3

\[(2x ^{2}+3)/(x ^{2}+4)(x ^{2}+6) \]

Answer 4

Why partials then if you have one variable?

Answer 5

I hope you're talking about the partial with respect to y!

Answer 6

that example should be easy, you just put x^2 = y and solve like simple partial fractions,
i think you are asking for something like
(x^2+2x+4) (x^2+3x+6) in the denominator ?

Answer 7

Normally what we do is equate this expression with \[(Ax+B)/(x ^{2}+4) + (Cx+D)/(x ^{2}+6)\]

Answer 8

Partial fractions?

Answer 9

@hartnn what if we have a linear expression in numerator

Answer 10

ok, like 2x+3 / ()()

Answer 11

yes....linear in numerator and quadratic in denominator

Answer 12

the best method i know of is to take LCM and compare the co-efficients

Answer 13

2x+3 = (Ax+B)(x^2+6) +(Cx+D)(x^2+4)

Answer 14

so, if you compare the co-efficient of x^3, we get A+C = 0

Answer 15

of x^2, >>>> B+D = 0

Answer 16

and so on.... got it ?
if others have an shorter way, they're welcome to share :)

Answer 17

ya..exactly i know this method....but this goes long...

Answer 18

not that i know of any shorter way.....i'll wait for others' responses..

Answer 19

ok