Question

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.

Answer 1

`2x-16+ frac{91}{x+4} - frac{95}{(x+4)^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:

`\frac{91x+269}{(x+4)^2}= \frac{A}{x+4}+\frac{B}{(x+4)^2}`

However, once you clear the denominator by multiplying everything by `(x+4)^2` , you get:

`91x + 269 = A(x+4)+B`, and now subbing in `x =-4`

you get:

`B = 91(-4) + 269 = -364 + 269 = -95`
and to solve for A:

`91x + 269 = Ax + 4A + B`
`91x = Ax`
`A = 91`

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

is `2x-16+ frac{91}{x+4} - frac{95}{(x+4)^2}`