Question

Help!!! Decompose into partial fractions. (x^4-3x^3-3x^2+10)/(x+1)^2(x-3)

Answer 1

really? this is a large pain

Answer 2

you need all the work, or just the answer? because it is annoying as hell

Answer 3

I can help. I have it all except for a few parts. let me post it h/o

Answer 4

first you have to divide

Answer 5

because degree of numerator is bigger than degree of denominator, so that is step one

Answer 6

\[(x-2)-(7x+4/(x+1)^2(x-3)\]

Answer 7

Im not getting the next step right.. I get \[Ax^2+2Ax+A+Bx^3-Bx^2-5Bx-3B+Cx-3x=x^3-x^2-5x-3\]


This is from A/(x-3)+B/(x+1)+C(x+1)^2

Answer 8

ok so we have
\[x-2\] we can ignore, and now you have
\[\frac{-7x-4}{(x+1)^2(x-3)}\] i will take your word for it. i know the
\[x-2\] is right
now you need
\[\frac{-7-x}{(x+1)^2(x-3)}=\frac{a}{x-1}+\frac{b}{(x-1)^2}+\frac{c}{x-3}\]

Answer 9

ok i will use your notation
\[\frac{-7-x}{(x+1)^2(x-3)}=\frac{a}{x-3}+\frac{b}{x-1}+\frac{c}{(x-1)^2}\]

Answer 10

\[-7x-4=a(x-1)^2+b(x-3)(x-1)+c(x-3)\]

Answer 11

i would not use the "equate like coefficients" method yet. i would find a right away, but letting x = 3

Answer 12

*by

Answer 13

its "(x+1)" and "(x+1)^2" I believe, not "(x-1)"

Answer 14

so it is...
\[-7x-4=a(x+1)^2+b(x-3)(x+1)+c(x-3)\]

Answer 15

ok i still let x = 3, get
\[-21-4=16a\]
\[a=-\frac{25}{16}\] but that is wrong, so maybe the -7x-4 is wrong

Answer 16

ah it is
\[-7x+4\]!

Answer 17

Im almost positive I got -7x+4 for the remainder when I divided earlier

Answer 18

ok good, so we get
\[-7x+4=a(x+1)^2+b(x-3)(x+1)+c(x-3)\]
\[x=3\]
\[-21+4=16a\]
\[a=-\frac{17}{16}\]

Answer 19

yeah ok, you had a minus sign in front of the whole thing and i didn't check it, but you are right , it is -7x+4

Answer 20

Yup, you did it all right

Answer 21

you had a minus and then parentheses so i thought you meant minus the whole thing

Answer 22

now maybe let
\[x=-1\] and find c

Answer 23

ohhh, my fault. The -17/16 is right for sure

Answer 24

ok thanks.

Answer 25

oh yeah, i have the answer for sure \ http://www.wolframalpha.com/input/?i=partial+fractions+%28x^4-3x^3-3x^2%2B10%29%2F%28%28x%2B1%29^2%28x-3%29%29

Answer 26

i was just trying to get to it

Answer 27

you can also use the "equate like coefficients" method but it requires solving a 3 by 3 system. i think substitution is much easier usually

Answer 28

for example, if i replace x by -1 i get
\[c=-\frac{11}{4}\] pretty much in my head
\[7+4=-4c\]
\[c=-\frac{11}{4}\]

Answer 29

now that you have two out of the three the third one should be easy enough to find. plug in the numbers and see what is missing

Answer 30

This really helped a lot.. I know it involved a lot so thanks a bunch:)

Answer 31

yw