# Thread: How do you do Partial Fractions with 2nd degree denomincator?

1. ## How do you do Partial Fractions with 2nd degree denomincator?

How do you do partial fractions if the denominator has 2nd degree terms (I know the basic method)?

My textbook says:
Using partial fractions we get

$\frac{3s}{(s^2+4)(s^2+1)} = \frac{-s}{s^2+4} + \frac{s}{s^2+1}$

Since I have to check every single thing my textbook says (it's in my genes or something):

$\frac{3s}{(s^2+4)(s^2+1)} = \frac{A}{s^2+4} + \frac{B}{s^2+1} = \frac{(A+B)s^2+(A+4B)}{(s^2+4)(s^2+1)}$

Comparing numerator's coefficients implies that:
$A+B=0$
$A+4B=0$

$A=0$
$B=0$

and therefore:
$\frac{3s}{(s^2+4)(s^2+1)} = \frac{0}{s^2+4} - \frac{0}{s^2+1}=0$

which is quite clearly wrong.

What'd I do wrong? How do I get an s term in the numerator?
Thanks.

2. ## Re: How do you do Partial Fractions with 2nd degree denomincator?

Originally Posted by MSUMathStdnt
How do you do partial fractions if the denominator has 2nd degree terms (I know the basic method)?

My textbook says:
Using partial fractions we get

$\frac{3s}{(s^2+4)(s^2+1)} = \frac{-s}{s^2+4} + \frac{s}{s^2+1}$

Since I have to check every single thing my textbook says (it's in my genes or something):

$\frac{3s}{(s^2+4)(s^2+1)} = \frac{A}{s^2+4} + \frac{B}{s^2+1} = \frac{(A+B)s^2+(A+4B)}{(s^2+4)(s^2+1)}$

Comparing numerator's coefficients implies that:
$A+B=0$
$A+4B=0$

$A=0$
$B=0$

and therefore:
$\frac{3s}{(s^2+4)(s^2+1)} = \frac{0}{s^2+4} - \frac{0}{s^2+1}=0$

which is quite clearly wrong.

What'd I do wrong? How do I get an s term in the numerator?
Thanks.

Partial fraction - Wikipedia, the free encyclopedia

$\frac{3s}{(s^2+4)(s^2+1)} = \frac{A \cdot s +B}{s^2+4} + \frac{C \cdot s +D}{s^2+1}$

Find A,B,C and D.

3. ## Re: How do you do Partial Fractions with 2nd degree denomincator?

I'v done it like this:
$\frac{3s}{(s^2+4)\cdot(s^2+1)}=\frac{As+C}{s^2+4}+ \frac{Bs+D}{s^2+1}$
$\Leftrightarrow \frac{(As+C)\cdot (s^2+1)+(Bs+D)\cdot (s^2+4)}{(s^2+4)\cdot(s^2+1)}$
$\Leftrightarrow \frac{As^3+Cs^2+As+C+Bs^3+Ds^2+4Bs+4D}{(s^2+4)(s^2 +1)}$
$\Leftrightarrow \frac{(A+B)s^3+(C+D)s^2+(A+4B)s+(C+D)}{(s^2+4)(s^2 +1)}$

So $A+B=0 \Leftrightarrow A=-B$ and $A+4B=3$ so $A=-1, B=1$ and $C=D=0$

So:
$\frac{3s}{(s^2+4)\cdot(s^2+1}=\frac{-s}{s^2+4}+\frac{s}{s^2+1}$

Check:
$\frac{-s}{s^2+4}+\frac{s}{s^2+1}=\frac{-s(s^2+1)+s(s^2+4)}{(s^2+4)\cdot(s^2+1)}=\frac{-s^3-s+s^3+4s}{(s^2+4)\cdot (s^2+1)}=\frac{3s}{(s^2+4)\cdot(s^2+1)}$

It seems to work.

EDIT:
In my opinion I don't think it's pre-algebra stuff.

4. ## Re: How do you do Partial Fractions with 2nd degree denomincator?

Originally Posted by Siron
I'v done it like this:
$\frac{3s}{(s^2+4)\cdot(s^2+1)}=\frac{As+C}{s^2+4}+ \frac{Bs+D}{s^2+1}$
$\Leftrightarrow \frac{(As+C)\cdot (s^2+1)+(Bs+D)\cdot (s^2+4)}{(s^2+4)\cdot(s^2+1)}$
$\Leftrightarrow \frac{As^3+Cs^2+As+C+Bs^3+Ds^2+4Bs+4D}{(s^2+4)(s^2 +1)}$
$\Leftrightarrow \frac{(A+B)s^3+(C+D)s^2+(A+4B)s+(C+D)}{(s^2+4)(s^2 +1)}$

So $A+B=0 \Leftrightarrow A=-B$ and $A+4B=3$ so $A=-1, B=1$ and $C=D=0$

So:
$\frac{3s}{(s^2+4)\cdot(s^2+1}=\frac{-s}{s^2+4}+\frac{s}{s^2+1}$

Check:
$\frac{-s}{s^2+4}+\frac{s}{s^2+1}=\frac{-s(s^2+1)+s(s^2+4)}{(s^2+4)\cdot(s^2+1)}=\frac{-s^3-s+s^3+4s}{(s^2+4)\cdot (s^2+1)}=\frac{3s}{(s^2+4)\cdot(s^2+1)}$

It seems to work.
Okay, but why?

5. ## Re: How do you do Partial Fractions with 2nd degree denomincator?

Do you ask me why it's working? ...

6. ## Re: How do you do Partial Fractions with 2nd degree denomincator?

Originally Posted by Siron
Do you ask me why it's working? ...
Maybe...

If you had to write $\frac{1}{1+x^3}$ as sum of partial fraction, what would you do? and if it was $\frac{1}{1+x^4}$ ?

7. ## Re: How do you do Partial Fractions with 2nd degree denomincator?

I would rewrite:
$\frac{1}{1+x^3}=\frac{1}{(1+x)\cdot(1-x+x^2)}=\frac{\frac{1}{3}}{1+x}+\frac{\frac{2}{3}-\frac{1}{3}x}{x^2-x+1}$

8. ## Re: How do you do Partial Fractions with 2nd degree denomincator?

@Also sprach zarathrusta:

Or was the question not mentioned to me? ...