Question

express the following as partial fraction : (6x\[^3\] + 5x\[^2\] + 4x+3 ) / (x\[^2\] + x +1) (x\[^2\] -1)

Answer 1

maybe you could reformat this

Answer 2

\[\frac{6x^3+ 5x^2+ 4x+3 }{ (x^2+ x +1) (x^2-1) }\] is my guess

Answer 3

\[\frac{2 x-1}{x^2+x+1}+\frac{3}{x-1}+\frac{1}{x+1}\] if so

Answer 4

but how you get 2x-1 , 3 and 1 ??

Answer 5

well it is a bit of a pain actually

Answer 6

you have the denominator in factored form, and you want to solve
\[\frac{6x^3+ 5x^2+ 4x+3 }{ (x^2+ x +1) (x^2-1) }=\frac{ax+b}{x^2+x+1}+\frac{c}{x-1}+\frac{d}{x+1}\]

Answer 7

that is solve for a, b, c and d

Answer 8

okay...thanks a lot!!!

Answer 9

so you have lots of work to do. you have to write
\[(ax+b)(x+1)(x-1)+c(x^2+x+1)(x+1)+d(x^2+x+1)(x-1)\]
\[=6x^3+5x^2+4x+3\]

Answer 10

you can replace x by 1 and get c quickly

Answer 11

since the first and third term will be zero. you get
\[c(1+1+1)(1+1)=6+5+4+3\]
\[6c=18\]
\[c=3\]

Answer 12

okay...i get it..thanks!!

Answer 13

yw