# Decompose into partial fractions

• Aug 10th 2012, 10:54 AM
Greymalkin
Decompose into partial fractions
${3x^2-11x-26} \over {(x^2-4)(x+1)}$

${-3\over (x-2)}+{2\over (x+2)}+{4\over (x+1)}$

${-1\over (x-2)}+{-10\over (x+2)}+{4\over (x+1)}$

I seem to get the same answer for C, but my system of equations is off:
$(Ax+B)(x+1)+C(x^2-4)$
$Ax^2+Ax+Bx+B+Cx^2-4C$
$(A+C)x^2+(A+B)x+(B-4C)$

3=A+C
-11=A+B
-26=B-4C

Reverse engineering the solutions shows my system to be incorrect, which seems wierd because I followed a text example step by step.
Why am I wrong?
• Aug 10th 2012, 11:33 AM
Plato
Re: Decompose into partial fractions
Quote:

Originally Posted by Greymalkin
${3x^2-11x-26} \over {(x^2-4)(x+1)}$

${-3\over (x-2)}+{2\over (x+2)}+{4\over (x+1)}$
${-1\over (x-2)}+{-10\over (x+2)}+{4\over (x+1)}$
$(Ax+B)(x+1)+C(x^2-4)$
$Ax^2+Ax+Bx+B+Cx^2-4C$
$(A+C)x^2+(A+B)x+(B-4C)$
You work with $A(x+2)(x+1)+B(x-2)(x+1)+C(x^2-4)$.