Question

Split (4/(y^3-16y))dy into partial fractions

Answer 1

\[\frac{4}{y(y-4)(y+4)}=\frac{A}{y}+\frac{B}{y-4}+\frac{C}{y+4}\] and you need A,B, C want to do it the snappy way?

Answer 2

to find A, see that the denominator is 0 is y = 0. look at the original expression. cover the "y" with your hand because you cannot divide by 0, then plug in 0 for y
you get
\[\frac{4}{(0-4)(0+4)}=\frac{4}{-16}=-\frac{1}{4}\] so that is A

Answer 3

repeat to find B, this time hold your finger over y - 4 and plug in 4 for y. you get
\[\frac{4}{4(4+4)}=\frac{4}{32}=\frac{1}{8}\]

Answer 4

C is found the same way

Answer 5

\[\frac{4}{y^3-16 y}=-\frac{1}{4 y}+\frac{1}{8 (y+4)}+\frac{1}{8 (y-4)} \]