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

Question

1003 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-01-06T09:15:50

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

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

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

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

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

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

$2 = - 4 A$ $2=-4A$

$\frac{2}{- 4} = \frac{- 4 A}{- 4}$ $2/-4 =(-4A)/-4$
$- \frac{1}{2} = A$ $-1/2 = A$

$A = - \frac{1}{2}$ $A=-1/2$

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

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

$B = \frac{3}{2}$ $B=3/2$

Therefore,

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

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

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