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What is the partial-fraction decomposition of $\frac{x + 11}{(x + 3) (x - 5)}$ $(x+11)/((x+3)(x-5))$ ?

1 Answer

Oct 8, 2015

$\frac{x + 11}{(x + 3) (x - 5)} = \frac{2}{x - 5} - \frac{1}{x + 3}$ $(x+11)/((x+3)(x-5))=2/(x-5)-1/(x+3)$

Explanation:

Partial fraction decomposition is the reverse of the process normally used to add fractional expressions with different denominators.

We want to find values $A$ $A$ and $B$ $B$ such that
$XXX \frac{A}{x + 3} + \frac{B}{x - 5} = \frac{x + 11}{(x + 3) (x - 5)}$ $color(white)("XXX")A/(x+3)+B/(x-5) = (x+11)/((x+3)(x-5))$

That is
$XXX \frac{A (x - 5) + B (x + 3)}{(x + 3) (x - 5)} = \frac{x + 11}{(x + 3) (x - 5)}$ $color(white)("XXX")(A(x-5)+B(x+3))/((x+3)(x-5)) = (x+11)/((x+3)(x-5))$

$XXX A (x - 5) + B (x + 3) = x + 11$ $color(white)("XXX")A(x-5)+B(x+3)=x+11$

$XXX A x + B x = x XXX \Rightarrow$ $color(white)("XXX")Ax+Bx=xcolor(white)("XXX")rArr$

[1] $XXX A + B = 1$ $color(white)("XXX")A+B = 1$
and
[2] $XXX - 5 A + 3 B = 11$ $color(white)("XXX")-5A+3B=11$

Rewriting [1] as
[3] $XXX B = 1 - A$ $color(white)("XXX")B=1-A$

Substitute $(1 - A)$ $(1-A)$ for $B$ $B$ in [2]
[4] $XXX - 5 A + 3 (1 - A) = 11$ $color(white)("XXX")-5A+3(1-A)=11$

[5] $XXX - 8 A + 3 = 11$ $color(white)("XXX")-8A+3=11$

[6] $XXX - 8 A = 8$ $color(white)("XXX")-8A=8$

[7] $XXX A = - 1$ $color(white)("XXX")A=-1$

Substituting $(- 1)$ $(-1)$ for $A$ $A$ in [1]
[8] $XXX B = 2$ $color(white)("XXX")B=2$

Referring back to our original equation
$XXX \frac{A}{x + 3} + \frac{B}{x - 5} = \frac{- 1}{x + 3} + \frac{2}{x - 5}$ $color(white)("XXX")A/(x+3)+B/(x-5)= (-1)/(x+3)+(2)/(x-5)$

$XXXXXXXXXXXX = \frac{x + 11}{(x + 3) (x - 5)}$ $color(white)("XXXXXXXXXXXX") = (x+11)/((x+3)(x-5))$

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