# How do you decompose (3p-1)/(p^2-1) into partial fractions?

Dec 24, 2016

${p}^{2} - 1$ can be factored as $\left(p + 1\right) \left(p - 1\right)$.

A/(p - 1) + B/(p + 1) = (3p - 1)/((p + 1)(p - 1)

$A \left(p + 1\right) + B \left(p - 1\right) = 3 p - 1$

$A p + A + B p - B = 3 P - 1$

$\left(A + B\right) p + \left(A - B\right) = 3 P - 1$

We now write a system of equations.

$\left\{\begin{matrix}A + B = 3 \\ A - B = - 1\end{matrix}\right.$

$B = 3 - A \to A - \left(3 - A\right) = - 1$
$A + A - 3 = - 1$
$2 A = 2$
$A = 1$

$1 + B = 3$

$B = 2$

Thus, the partial fraction decomposition is $\frac{1}{p - 1} + \frac{2}{p + 1}$.

Hopefully this helps!