
PreCalculus PreCalculus Math Forum 
 LinkBack  Thread Tools  Display Modes 
October 29th, 2017, 07:08 PM  #1 
Newbie Joined: Oct 2017 From: cali Posts: 8 Thanks: 0  partial fraction decomposition My issue with this problem is that after factoring out an x, i get x^2x+4 which gives me x(x^2x+4). Normally i would factor it out but i can't do it normally, using quadratic equations which i was told not to do in class. Solving for a is easy enough, as i make x = 0 to cancel out B and C so i can solve for A. But after that i'm pretty lost. Here's a image of the work in class we did above. 
October 29th, 2017, 07:29 PM  #2 
Senior Member Joined: Sep 2015 From: USA Posts: 2,264 Thanks: 1198 
you've got it all solved $A=\dfrac 1 2,~B=\dfrac 1 2,~C=\dfrac 3 2$ $\dfrac{x+2}{x^3x^2+4x} =\dfrac 1 {2x} + \dfrac{x+3}{2(x^2x+4)}$ 
October 29th, 2017, 08:44 PM  #3 
Newbie Joined: Oct 2017 From: cali Posts: 8 Thanks: 0  my apologies, didn't make it clear what i was struggling with. We finished it in class but im' not sure how solving for b and c for example works. I don't know the steps necessary to approach problems like that.

October 30th, 2017, 12:07 AM  #4  
Senior Member Joined: Sep 2015 From: USA Posts: 2,264 Thanks: 1198  Quote:
$f(x) = \dfrac{x+2}{x(x^2x+4)}$ $x^2x+4 \text{ cannot be factored further.}$ we set it up as usual $f(x) = \dfrac{A}{x} + \dfrac{Bx + C}{x^2x+4}$ $\dfrac{x+2}{x(x^2x+4)} = \dfrac{A(x^2x+4) + Bx^2 + Cx}{x(x^2x+4)} = \dfrac{(A+B)x^2+(A+C)x + 4A}{x(x^2x+4)}$ now we just match up the coefficients of the powers of $x$ $x^2:~~0 = A+B$ $x: ~~1 = A + C$ $1:~~2=4A$ solving these $A = \dfrac 1 2$ $C = 1+\dfrac 1 2 = \dfrac 3 2$ $B=A = \dfrac 1 2$ so $\dfrac{x+2}{x(x^2x+4)} = \dfrac{1}{2x} + \dfrac{3x}{2(x^2x+4)}$  
November 5th, 2017, 03:56 AM  #5 
Math Team Joined: Jan 2015 From: Alabama Posts: 3,261 Thanks: 895 
Another way: Once you have $\displaystyle \frac{x+ 2}{x(x^2 x+ 4)}= \frac{A}{x}+ \frac{Bx+ C}{x^2 x+ 4}$, multiply both sides by $\displaystyle x(x^2 x+ 4)$ to clear the fractions and get $\displaystyle x+ 2= A(x^2 x+ 4)+ x(Bx+ C)$ which must be true for all x. In particular, taking x equal to three different values gives three different values gives three different equations to solve for A, B, and C. Letting x= 0, $\displaystyle 2= 4A$ so $\displaystyle A= \frac{1}{2}$. Letting x= 1, $\displaystyle 3= 4A+ B+ C= 2+ B+ C$ so $\displaystyle B+ C= 1$. Letting x= 1, $\displaystyle 1= 6A ( B+ C)= 3+ B C$ so $\displaystyle B C= 2$. Adding those two equations, $\displaystyle 2B= 1$ so $\displaystyle B= \frac{1}{2}$ and then [math]C= \frac{3}{2}. $\displaystyle \frac{x+ 2}{x(x^2 x+ 4)}= \frac{1}{2x}+ \frac{3 x}{2(x^2 x+ 4)}$ as before. Last edited by Country Boy; November 5th, 2017 at 03:59 AM. 

Tags 
decomposition, fraction, partial 
Thread Tools  
Display Modes  

Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
Partial Fraction Decomposition  WWRtelescoping  Complex Analysis  4  April 27th, 2014 05:44 PM 
help with partial fraction decomposition?  strawberryPK  Calculus  1  February 9th, 2011 09:10 PM 
Partial Fraction Decomposition  RMG46  Algebra  16  July 5th, 2010 08:26 PM 
Partial Fraction Decomposition  mathman2  Calculus  2  February 14th, 2010 06:21 AM 
Partial fraction decomposition  mnangagwa  Calculus  3  March 20th, 2009 10:43 PM 