# Thread: Decompose into partial fractions

1. ## Decompose into partial fractions

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

Book answer is:
${-3\over (x-2)}+{2\over (x+2)}+{4\over (x+1)}$

My answer is:
${-1\over (x-2)}+{-10\over (x+2)}+{4\over (x+1)}$

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?

2. ## Re: Decompose into partial fractions

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

Book answer is:
${-3\over (x-2)}+{2\over (x+2)}+{4\over (x+1)}$
My answer is:
${-1\over (x-2)}+{-10\over (x+2)}+{4\over (x+1)}$
seem to get the same answer for C, but my system of equations is off:
$(Ax+B)(x+1)+C(x^2-4)$
$Ax^2+Ax+Bx+B+Cx^2-4C$
$(A+C)x^2+(A+B)x+(B-4C)$
You work with $A(x+2)(x+1)+B(x-2)(x+1)+C(x^2-4)$.

3. ## Re: Decompose into partial fractions

Wasn't really getting the gist of decomposition until I watched a khan academy video on it, which sort of explained the relatedness of the method. But it was your post that made the lightbulb go on, thank you very much!