partial fractions for a function

Question

I need help finding the partial fraction decomposition for this function, I am just lost on it, here it is: $(x^2 + x + 1)/(2x^4+3x^2+1)$. the help is appreciated. thank you.

Answer 1

The key to it is how to factor the denominator $2x^4+3x^2+1$. Denote $y=x^2$ and you're left with a quadratic $2y^2+3y+1$ whose roots are $-1$ and $-1/2$. So

$$2x^4+3x^2+1=(y+1)(2y+1)=(x^2+1)(2x^2+1)$$

Now we're looking for

$${x^2+x+1\over 2x^4+3x^2+1}={ax+b\over x^2+1}+{cx+d\over 2x^2+1}$$

Multiply both sides by $x^2+1$ and substitute $x=i$ to get

$${i\over -1}=ai+b$$

and so $a=-1$ and $b=0$

Multiply both sides by $2x^2+1$ and substitute $x=i/\sqrt{2}$ to get

$${-{1\over 2}+{i\over\sqrt{2}}+1\over -{1\over 2}+1}=c{i\over\sqrt{2}}+d\iff {2i\over\sqrt{2}}+1=c{i\over\sqrt{2}}+d$$

and this yields $c=2$ and $d=1$ and so

$${x^2+x+1\over 2x^4+3x^2+1}={2x+1\over 2x^2+1}-{x\over x^2+1}$$

Answer 2

As suggested by one user.

$\frac{(x^2 + x + 1)}{(2x^4+3x^2+1)} = \frac{ax+b}{2x^2+1} + \frac{cx+d}{x^2+1}$

$x^2+x+1 = (ax+b)(x^2+1) + (cx+d)(2x^2+1)$

Comparing terms containing $x^3$,

$0 = a + 2c$

Comparing terms containing $x^2$,

$1 = b + 2d$

Comparing terms containing $x$,

$1 = a + c$

Comparing terms containing constant terms,

$1 = b + d$

Hope you can solve above equations to find a, b, c and d.