How do you express as a partial fraction $\frac{6 x^{2} - 3 x + 1}{(4 x + 1) (x^{2} + 1)}$ $(6x^2 - 3x +1) / ( (4x+1)(x^2 +1) )$ ?

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How do you express as a partial fraction $\frac{6 x^{2} - 3 x + 1}{(4 x + 1) (x^{2} + 1)}$ $(6x^2 - 3x +1) / ( (4x+1)(x^2 +1) )$ ?

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Answer 1 · 2015-07-30T23:25:05

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

Explanation:

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

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

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

We need:

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

This lead to:
$\frac{A x^{2} + A + 4 B x^{2} + 4 C x + B x + C}{(4 x + 1) (x^{2} + 1)} = \frac{6 x^{2} - 3 x + 1}{(4 x + 1) (x^{2} + 1)}$ $(Ax^2+A+4Bx^2+4Cx+Bx+C) / ( (4x+1)(x^2 +1) )=(6x^2 - 3x +1) / ( (4x+1)(x^2 +1) )$

And so:

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

So we need to solve the system:

$A$ $A$ $+ 4 B$ $+4B$ $" " " "$ $=$ $=$ $6$ $6$

$" " " "$ $B$ $B$ $+ 4 C$ $+4C$ $=$ $=$ $- 3$ $-3$

$A$ $A$ $" " " " "$ $+ C$ $+C$ $=$ $=$ $1$ $1$

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

$" " " "$ $4 B$ $4B$ $- C$ $-C$ $=$ $=$ $5$ $5$

Eq2 is
$" " " "$ $B$ $B$ $+ 4 C$ $+4C$ $=$ $=$ $- 3$ $-3$

So
$" " " "$ $- 4 B$ $-4B$ $- 16 C$ $-16C$ $=$ $=$ $12$ $12$

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

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

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

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

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How do you express as a partial fraction $\frac{6 x^{2} - 3 x + 1}{(4 x + 1) (x^{2} + 1)}$ $(6x^2 - 3x +1) / ( (4x+1)(x^2 +1) )$ ?

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Explanation:

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