# How do you express as a partial fraction  (6x^2 - 3x +1) / ( (4x+1)(x^2 +1) )?

Jul 30, 2015

$\frac{6 {x}^{2} - 3 x + 1}{\left(4 x + 1\right) \left({x}^{2} + 1\right)} = \frac{2}{4 x + 1} + \frac{x - 1}{{x}^{2} + 1}$

#### Explanation:

Start with: $\frac{6 {x}^{2} - 3 x + 1}{\left(4 x + 1\right) \left({x}^{2} + 1\right)}$

The denominator is already factored into irreducible polynomials (over the Reals).

Because ${x}^{2} + 1$ cannnot be factored using Real number coefficients, we need a linear numerator for one of the fractions

We need:

$\frac{A}{4 x + 1} + \frac{B x + C}{{x}^{2} + 1} = \frac{6 {x}^{2} - 3 x + 1}{\left(4 x + 1\right) \left({x}^{2} + 1\right)}$

$\frac{A {x}^{2} + A + 4 B {x}^{2} + 4 C x + B x + C}{\left(4 x + 1\right) \left({x}^{2} + 1\right)} = \frac{6 {x}^{2} - 3 x + 1}{\left(4 x + 1\right) \left({x}^{2} + 1\right)}$

And so:

$\frac{\left(A + 4 B\right) {x}^{2} + \left(B + 4 C\right) x + \left(A + C\right)}{\left(4 x + 1\right) \left({x}^{2} + 1\right)} = \frac{6 {x}^{2} - 3 x + 1}{\left(4 x + 1\right) \left({x}^{2} + 1\right)}$

So we need to solve the system:

$A$ $+ 4 B$ $\text{ " " }$ $=$ $6$

$\text{ " " }$ $B$ $+ 4 C$ $=$ $- 3$

$A$ $\text{ " " " }$ $+ C$ $=$ $1$

Eq3 implies $A = 1 - C$ and substituting in Eq1 and simplifying gets us:

$\text{ " " }$ $4 B$ $- C$ $=$ $5$

Eq2 is
$\text{ " " }$ $B$ $+ 4 C$ $=$ $- 3$

So
$\text{ " " }$ $- 4 B$ $- 16 C$ $=$ $12$

Thus $- 17 C = 17$ and $C = - 1$

Knowing $C$, we can find $A = 1 - \left(- 1\right) = 2$

And $B + 4 \left(- 1\right) = - 3$ gets us $B = 1$

$\frac{A}{4 x + 1} + \frac{B x + C}{{x}^{2} + 1} = \frac{2}{4 x + 1} + \frac{x - 1}{{x}^{2} + 1}$