# How do I find A and B in the partial fraction decomposition 91x+269=A(x+4)^2+B(x+4)?

## I know how to do PFDs when there are two factors that are different, but how can I do this when there are two factors that are the same? If I set $x = - 4$ then I remove both terms, and then I have nothing left to solve for.

May 19, 2018

$2 x - 16 + \frac{91}{x + 4} - \frac{95}{{\left(x + 4\right)}^{2}}$

#### Explanation:

If you take the original question: \frac{2x^3−5x+13}{(x+4)^2}
and you apply long division to it you do indeed get:

2x-16+\frac{91x+269}{(x+4)^2 and the equation for the partial fraction decomposition is indeed:

$\setminus \frac{91 x + 269}{{\left(x + 4\right)}^{2}} = \setminus \frac{A}{x + 4} + \setminus \frac{B}{{\left(x + 4\right)}^{2}}$

However, once you clear the denominator by multiplying everything by ${\left(x + 4\right)}^{2}$ , you get:

$91 x + 269 = A \left(x + 4\right) + B$, and now subbing in $x = - 4$

you get:

$B = 91 \left(- 4\right) + 269 = - 364 + 269 = - 95$
and to solve for A:

$91 x + 269 = A x + 4 A + B$
$91 x = A x$
$A = 91$

Thus, the partial fraction decomposition of \frac{2x^3−5x+13}{(x+4)^2}

is $2 x - 16 + \frac{91}{x + 4} - \frac{95}{{\left(x + 4\right)}^{2}}$