How do you write the partial fraction decomposition of the rational expression $\frac{2 x - 2}{(x - 5) (x - 3)}$ $(2x-2)/((x-5)(x-3))$ ?

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Answer 1 · 2015-12-14T21:02:19

$\frac{4}{x - 5} - \frac{2}{x - 3}$ $4/(x-5)-2/(x-3)$

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

$2 x - 2 = A (x - 3) + B (x - 5)$ $2x-2=A(x-3)+B(x-5)$

IF $x = 3$ $x=3$ :

$2 (3) - 2 = B (3 - 5)$ $2(3)-2=B(3-5)$
$B = - 2$ $B=-2$

IF $x = 5$ $x=5$ :

$2 (5) - 2 = A (5 - 3)$ $2(5)-2=A(5-3)$
$A = 4$ $A=4$

Thus,

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

How do you write the partial fraction decomposition of the rational expression 2x−2(x−5)(x−3) (2x-2)/((x-5)(x-3))?