## anonymous 4 years ago Partial fractions: x^2 + 18 -------- x(x^2+9)

1. anonymous

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

2. anonymous

$\frac{A(x^2+9) + Bx}{(x^2+9)x}$ Compare the co-efficient! :-D

3. anonymous

Oh Wait..

4. anonymous

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!

5. anonymous

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

6. anonymous

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

7. anonymous

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

8. anonymous

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?

9. anonymous

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

10. anonymous

you get "a" instantly by letting x = 0 t give $9a=18$ $x=2$

11. anonymous

a=2*

12. anonymous

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

13. anonymous

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

14. anonymous

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 $a+b=1, c =0, 9a=18$