What is the partial-fraction decomposition of (x+11)/((x+3)(x-5))?

Oct 8, 2015

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

Explanation:

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

We want to find values $A$ and $B$ such that
$\textcolor{w h i t e}{\text{XXX}} \frac{A}{x + 3} + \frac{B}{x - 5} = \frac{x + 11}{\left(x + 3\right) \left(x - 5\right)}$

That is
$\textcolor{w h i t e}{\text{XXX}} \frac{A \left(x - 5\right) + B \left(x + 3\right)}{\left(x + 3\right) \left(x - 5\right)} = \frac{x + 11}{\left(x + 3\right) \left(x - 5\right)}$

$\textcolor{w h i t e}{\text{XXX}} A \left(x - 5\right) + B \left(x + 3\right) = x + 11$

$\textcolor{w h i t e}{\text{XXX")Ax+Bx=xcolor(white)("XXX}} \Rightarrow$

[1]$\textcolor{w h i t e}{\text{XXX}} A + B = 1$
and
[2]$\textcolor{w h i t e}{\text{XXX}} - 5 A + 3 B = 11$

Rewriting [1] as
[3]$\textcolor{w h i t e}{\text{XXX}} B = 1 - A$

Substitute $\left(1 - A\right)$ for $B$ in [2]
[4]$\textcolor{w h i t e}{\text{XXX}} - 5 A + 3 \left(1 - A\right) = 11$

[5]$\textcolor{w h i t e}{\text{XXX}} - 8 A + 3 = 11$

[6]$\textcolor{w h i t e}{\text{XXX}} - 8 A = 8$

[7]$\textcolor{w h i t e}{\text{XXX}} A = - 1$

Substituting $\left(- 1\right)$ for $A$ in [1]
[8]$\textcolor{w h i t e}{\text{XXX}} B = 2$

Referring back to our original equation
$\textcolor{w h i t e}{\text{XXX}} \frac{A}{x + 3} + \frac{B}{x - 5} = \frac{- 1}{x + 3} + \frac{2}{x - 5}$

$\textcolor{w h i t e}{\text{XXXXXXXXXXXX}} = \frac{x + 11}{\left(x + 3\right) \left(x - 5\right)}$