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

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Answer 1 · 2016-01-06T09:15:50

$(x+2)/(x(x-4)) = 3/(2(x-4))-1/(2x)$

$(x+2)/(x(x-4))$

$(x+2)/(x(x-4)) = A/x + B/(x-4)$

$(x+2)/(x(x-4)) =(A(x-4)+B(x))/(x(x-4))$

$(x+2) = A(x-4)+B(x)$

Let $x=0$ This done to remove the $B$
$(0+2) = A(0-4)+B(0)$

$2=-4A$

$2/-4 =(-4A)/-4$
$-1/2 = A$

$A=-1/2$

Now let $x=4$ Which makes $(x-4)$ as zero and thus eliminating $A$ and we can solve for $B$

$(4+2)=A(4-4)+B(4)$
$6=4B$
$6/4 = (4B)/4$
$3/2 = B$

$B=3/2$

Therefore,

$(x+2)/(x(x-4)) = (-1/2)/x + (3/2)/(x-4)$

$(x+2)/(x(x-4)) = -1/(2x) + 3/(2(x-4))$

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