How do you do Partial Fractions with 2nd degree denomincator?

**MSUMathStdnt** · August 9th 2011, 12:16 PM

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.

**Also sprach Zarathustra** · August 9th 2011, 12:30 PM

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.

Read the Procedure here:

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.

**Siron** · August 9th 2011, 12:31 PM

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.

**Also sprach Zarathustra** · August 9th 2011, 12:36 PM

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?

**Siron** · August 9th 2011, 12:37 PM

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

**Also sprach Zarathustra** · August 9th 2011, 12:46 PM

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 $\frac{1}{1+x^4}$ ?

**Siron** · August 9th 2011, 01:15 PM

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

**Siron** · August 9th 2011, 01:44 PM

@Also sprach zarathrusta:

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

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