Question

partial decomp of (-x^2-7x+27)/X(X^2+9)

Answer 1

\[\frac{A}{x}+\frac{Bx+C}{x^2+9}=\frac{-x^2-7+27}{x(x^2+9)}\]

Answer 2

this looks nasty!

Answer 3

could it also be A/x + B/(x+3) + C/(x-3) ?

Answer 4

No. You don't have difference of squares in the denominator.

Answer 5

ahh yes youre right

Answer 6

I got what he first got but now im stuck

Answer 7

multiply left and right sides by x(x^2+9)

Answer 8

A(x\(^2\)+9)+B(x)=-x\(^2\)-7x+27 \(\Rightarrow\)
use the distributive property to distribute and combine like terms

Answer 9

a good start would be to let \(x=0\) and get \(9A=27\) so at least you know \(A\) right away

Answer 10

so x is 3

Answer 11

Yes sir.

Answer 12

how do I get c

Answer 13

no it is not \(x=3\) but rather \(A=3\)

Answer 14

i think there maybe a mistake above

Answer 15

\[\frac{A}{x}+\frac{Bx+C}{x^2+9}=\frac{-x^2-7+27}{x(x^2+9)}\]
\[x(Bx+C)+A(x^2+9)=-x^2-7x+27\]

Answer 16

put \(x=0\) get
\[9A=27\] and so \(A=3\) giving
\[x(Bx+C)+3x^2+27=-x^2-7x+27\]

Answer 17

expand on the left and get
\[Bx^2+Cx+3x^2+27=-x^2-7x+27\] or
\[(B+3)x^2+Cx+27=-x^2-7x+27\] which gives you \(C=-7\) right away

Answer 18

also tells you \((B+3)x^2=-x^2\) and so \(B+3=-1\) making \(B=-4\)
lets check it

Answer 19

yup looks good
http://www.wolframalpha.com/input/?i=partial+fractions++%28-x^2-7x%2B27%29%2F%28x%28x^2%2B9%29%29

Answer 20

\[\frac{A}{x}+\frac{Bx+C}{x^2+9}=\frac{-x^2-7+27}{x(x^2+9)}\]
\[\frac{3}{x}+\frac{-4x-7}{x^2+9}=\frac{-x^2-7+27}{x(x^2+9)}\]