Question

partial fraction decomposition

Answer 1

\[\frac{ -9x^2-28x+12 }{ (x+4)(x^2+3x+6) }=\frac{ f(x) }{ x+4 }+\frac{ g(x) }{ x^2+3x+6 }\]

Answer 2

You can use a constant function for \(f\) and a linear one for \(g\).

Answer 3

ya

Answer 4

and that is because something to do with the discriminant correct?

Answer 5

brb 1-2 mins

Answer 6

Well you can use the discriminant of the quadratic factor to see if it's possible to reduce it further, but that's not the case here since the discriminant is negative.

Answer 7

hmm........

Answer 8

ya.......

Answer 9

So do you know how to proceed from here if you let \(f(x)=A\) and \(g(x)=Bx+C\)?

Answer 10

ya thats where i started, i ended up with a bunch of messy algebra

Answer 11

If you have access to the program, Mathematica, the function "Apart" saves a lot of work and time to deal with a problem like this one.\[\text{Apart}\left[\frac{-9 x^2-28 x+12}{(x+4) \left(x^2+3 x+6\right)}\right] \]\[\left\{0.000539,\frac{6-7 x}{x^2+3 x+6}-\frac{2}{x+4}\right\} \]
539 millions of a second on a mid 2010 IMac.

Answer 12

well i suppose having the answer would be nice, but, i got an exam on this tomorrow, and no comps/cellphones allowed...

Answer 13

\[\begin{align*}
\frac{ -9x^2-28x+12 }{ (x+4)(x^2+3x+6) }&=\frac{A}{ x+4 }+\frac{Bx+C}{ x^2+3x+6 }\\\\

-9x^2-28x+12 &=A(x^2+3x+6)+(Bx+C)(x+4)\\\\
&=(A+B)x^2+(3A+4B+C)x+6A+4C
\end{align*}\]
Matching up coefficients gives
\[\begin{cases}
A+B=-9\\
3A+4B+C=-28\\
6A+4C=12
\end{cases}\]

Answer 14

ah i see i factored it differently on the 3rd line there.

Answer 15

and ended up going in circles with the algebra

Answer 16

@SithsAndGiggles how did you get A+B=-9 and 3A+4B+C=-28

Answer 17

think i figured it out nvm..