Question

can someone decompose this into partial fractions?
x-7/x(x^2+2)=A/x+Bx+C/x^2+2

Answer 1

that looks right

Answer 2

i need to solve for A, B, and C to get the answer

Answer 3

hmmm. set x equal to zero and solve and see if a variable gives a number.

Answer 4

Does the initial problem look like this?
\[\frac{x-7}{x(x^2+2)}=\frac{A}{x}+Bx+\frac{C}{x^2+2}\]

Answer 5

i want to say that the Bx+C should all be on top of the X^2+2

Answer 6

Oh good, that makes more sense.
\[\frac{x-7}{x(x^2+2)}=\frac{A}{x}+\frac{Bx+C}{x^2+2}\]Find a common denominator, namely x(x^2+2)
\[\frac{x-7}{x(x^2+2)}=\frac{A(x^2+2)}{x(x^2+2)}+\frac{(Bx+C)x}{x(x^2+2)}\]
Clear out the denominator.
\[x-7=Ax^2+2A+Bx^2+Cx\]
So here's what you have. You have a 0x^2 term, a 1x term, and a -7 term.
\[x^2(A+B)=0x^2\]\[Cx=x\]\[2A=-7\]
These equations yield
\[A=-\frac{7}{2}, B=\frac{7}{2}, C=1\]
Double-check my math, but I think I did that right...