How do you write the partial fraction decomposition of the rational expression $\frac{x - 2}{x^{2} + 4 x + 3}$ $( x-2 ) / (x^2 + 4x + 3)$ ?

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Answer 1 · 2015-12-16T15:58:33

$\frac{5}{2 (x + 3)} - \frac{3}{2 (x + 1)}$ $5/(2(x+3))-3/(2(x+1))$

Factor the denominator.

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

$x - 2 = A (x + 3) + B (x + 1)$ $x-2=A(x+3)+B(x+1)$

If $x = - 3$ $x=-3$ :

$- 3 - 2 = B (- 2)$ $-3-2=B(-2)$
$B = \frac{5}{2}$ $B=5/2$

If $x = - 1$ $x=-1$ :

$- 1 - 2 = A (2)$ $-1-2=A(2)$
$A = - \frac{3}{2}$ $A=-3/2$

Plug the values back in.

$\frac{x - 2}{x^{2} + 4 x + 3} = \frac{5}{2 (x + 3)} - \frac{3}{2 (x + 1)}$ $(x-2)/(x^2+4x+3)=5/(2(x+3))-3/(2(x+1))$

How do you write the partial fraction decomposition of the rational expression ( x-2 ) / (x^2 + 4x + 3)x−2x2+4x+3 ( x-2 ) / (x^2 + 4x + 3)?