Question

write the partial fraction decomposition for the rational expression.
(3x^2 - 7x - 2)/ (x^3 - x)

Answer 1

\[\frac{a}{x}+\frac{b}{x+1}+\frac{c}{x-1}\]
\[a(x-1)(x+1)+bx(x-1)+cx(x+1)=3x^2 - 7x - 2\] put \(x=0\) get \(a=-2\) right away

Answer 2

oops i mean put \(x=0\) get \(a=2\) right away

Answer 3

then replace \(x\) by 1 to get \(c\) and \(x\) by -1 to get \(b\)

Answer 4

@satellite73 how do u know which numbers to pick?

Answer 5

i pick the ones that make one of the factors zero, so it drops out

Answer 6

let me write the steps

Answer 7

before i begin, is this like ok
\[a(x-1)(x+1)+bx(x-1)+cx(x+1)=3x^2 - 7x - 2\]?

Answer 8

*line

Answer 9

yeah

Answer 10

ok now if you replace \(x\) by 0 then the \(b\) and \(c\) terms are gone
you get
\[a(0-1)(0+1)+b\times 0(0-1)+c\times 0(0+1)=3\times 0^2 - 7\times - 2\]

Answer 11

and end up with \(-a=-2\) making \(a=2\)
with some practice you can do it with your eyes (although i messed up the first time)

Answer 12

the point is the \(b\) and \(c\) are gone if \(x=0\) so you can really just write
\[a(0+1)(0-1)=-2\]\[-a=-2\]\[a=2\]

Answer 13

then again start with
\[a(x-1)(x+1)+bx(x-1)+cx(x+1)=3x^2 - 7x - 2\] this time make \(x=-1\) to get rid of the \(a\) and \(c\) terms

Answer 14

you get
\[b(-1)(-1-1)=3(-1)^2-7\times -1-2\]
\[2b=8\]
\[b=4\]

Answer 15

and finally replace \(x\) by \(1\) to get rid of the \(a\) and \(b\) terms
although once you know \(a=2\) and \(b=4\) you can find it easily, but this method is still snappy enough
there are other methods as well, but this one is one of the easiest

Answer 16

okay thank you!

Answer 17

yw