# [SOLVED] help partial fractions

• Feb 2nd 2009, 10:02 AM
walk_in_fark
[SOLVED] help partial fractions
hey i need help with this, X^2 + 9x + 8/(X^2 + x - 6)

When I do the polynomial division I get 1 + 8x - 14. But then the answer for the partial fraction turns out all wrong when i check it against the answer giving in the textbook. The answer given in the book is : 1 + 2/(x - 3) + 6/(X + 1). How did they derive this answer? wtf am I doing wrong, its driving me nuts. Thank you.
• Feb 2nd 2009, 10:34 AM
Jester
Quote:

Originally Posted by walk_in_fark
hey i need help with this, X^2 + 9x + 8/(X^2 + x - 6)

When I do the polynomial division I get 1 + 8x - 14. But then the answer for the partial fraction turns out all wrong when i check it against the answer giving in the textbook. The answer given in the book is : 1 + 2/(x - 3)** + 6/(X + 1)**. How did they derive this answer? What am I doing wrong, its driving me nuts. Thank you.

As you said

$1 + \frac{8x - 14}{x^2 + x - 6} = 1 + \frac{8x - 14}{(x+3)(x-2)}$

Now for the last piece look for a decomposition as follows

$\frac{8x - 14}{(x+3)(x-2)} = \frac{A}{x+3} + \frac{B}{x-2}$

multiply by

$(x+3)(x-2)$

giving

$8x - 14 = A(x-2) + B(x+3)$

expanding a group terms depending on whether there's a x or not

$A + B = 8$
$2A - 3B = -14$

Solving for A and B gives A = 2, B = 6 so

$\frac{8x - 14}{(x+3)(x-2)} = \frac{2}{x+3} + \frac{6}{x-2}$

although it doesn't agree with what you have in your denominator (above **).
• Feb 2nd 2009, 10:40 AM
walk_in_fark
Thanks a lot, I now I know that is a type, again thank you.
• Feb 2nd 2009, 10:09 PM
Jhevon
Quote:

Originally Posted by danny arrigo
As you said

$1 + \frac{8x - 14}{x^2 + x - 6} = 1 + \frac{8x - 14}{(x+3)(x-2)}$

Now for the last piece look for a decomposition as follows

$\frac{8x - 14}{(x+3)(x-2)} = \frac{A}{x+3} + \frac{B}{x-2}$

multiply by

$(x+3)(x-2)$

giving

$\color{red} 8x - 14 = A(x+2) + B(x-3)$

expanding a group terms depending on whether there's a x or not

$A + B = 8$
$2A - 3B = -14$

Solving for A and B gives A = 2, B = 6 so

$\frac{8x - 14}{(x+3)(x-2)} = \frac{2}{x+3} + \frac{6}{x-2}$

although it doesn't agree with what you have in your denominator (above **).

you made an error here. in the red line, you should have $8x - 14 = A(x - 2) + B(x + 3)$. you mixed the signs up. as it stands now, your solutions don't work. a minor mistake. the original poster should take note and fix the error him/herself