How do I find AA and BB in the partial fraction decomposition 91x+269=A(x+4)^2+B(x+4)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=-4x=4 then I remove both terms, and then I have nothing left to solve for.

1 Answer
May 19, 2018

2x-16+ frac{91}{x+4} - frac{95}{(x+4)^2}2x16+91x+495(x+4)2

Explanation:

If you take the original question: \frac{2x^3−5x+13}{(x+4)^2}2x35x+13(x+4)2
and you apply long division to it you do indeed get:

2x-16+\frac{91x+269}{(x+4)^22x16+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}91x+269(x+4)2=Ax+4+B(x+4)2

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

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

you get:

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

91x + 269 = Ax + 4A + B91x+269=Ax+4A+B
91x = Ax91x=Ax
A = 91A=91

Thus, the partial fraction decomposition of \frac{2x^3−5x+13}{(x+4)^2}2x35x+13(x+4)2

is 2x-16+ frac{91}{x+4} - frac{95}{(x+4)^2}2x16+91x+495(x+4)2