Question

Partial fractions:

x^2 + 18
--------
x(x^2+9)

Answer 1

\[\frac{x^2 + 18}{x(x^2 +9)} = \frac{A}{x} + \frac{B}{x^2 + 9} \]

Answer 2

\[\frac{A(x^2+9) + Bx}{(x^2+9)x}\]

Compare the co-efficient! :-D

Answer 3

Oh Wait..

Answer 4

Sorry I was wrong, this is the right method

\[\frac{x^2 + 18}{x(x^2 +9)} = \frac{A}{x} + \frac{Bx+C}{x^2 + 9}
\]

Now multiply(take lcm) and stuff, then compare co-efficient!

Answer 5

I tried that, but it didn't come out right...?

Answer 6

Hmm try this or better use Wolfram for the multiplication, I hate calculation job

\[\frac{x^2 + 18}{x(x^2 +9)} = \frac{A}{x} + \frac{B}{x +3} + \frac{C}{x-3}\]
But I think it's same as the previous one, Let me check this on Wolfram. When I say it's same I mean the basic concept or the final answer is same but the constant values may differ.

http://www.wolframalpha.com/input/?i= \frac{A}{x}+%2B+\frac{B}{x+%2B3}+%2B+\frac{C}{x-3}

See the alternate form

Answer 7

What if you multiplied the right side by the denominator which would give
x^2 +18 = A(x^2+9) + (bx+c)x ?

Answer 8

Using
\[\frac{x^2 + 18}{x(x^2 +9)} = \frac{A}{x} + \frac{Bx+C}{x^2 + 9}\]
I get A=2, B=-1 and C=0
and if you sub these values into the RHS you will get the LHS, so I'm convinced this is right. Are those the values you got order?

Answer 9

not to butt in by
\[x^2+9\] does not factor so you will have to grind it out and solve
\[a(x^2+9)+(bx+c)x=x^2+18\]

Answer 10

you get "a" instantly by letting x = 0 t give
\[9a=18\]
\[x=2\]

Answer 11

a=2*

Answer 12

right and also
\[c=0\] since if a is 2 the left hand side already has 18 as the constant

Answer 13

and since left hand side has
\[2x^2+bx^2\] and right hand side has
\[x^2\] you get b = -1 as callum wrote

Answer 14

you also have the option of multiplying out to get
\[a(x^2+9)+(bx+c)x=x^2+18\]
\[ax^2+9a+bx^2+cx=(a+b)x^2+cx+9a=x^2+18\] and solving
\[a+b=1, c =0, 9a=18\]