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 October 29th, 2017, 06:08 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^2-x+4 which gives me x(x^2-x+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, 06:29 PM #2 Senior Member     Joined: Sep 2015 From: USA Posts: 2,120 Thanks: 1101 you've got it all solved $A=\dfrac 1 2,~B=-\dfrac 1 2,~C=\dfrac 3 2$ $\dfrac{x+2}{x^3-x^2+4x} =\dfrac 1 {2x} + \dfrac{-x+3}{2(x^2-x+4)}$
October 29th, 2017, 07:44 PM   #3
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 Originally Posted by romsek you've got it all solved $A=\dfrac 1 2,~B=-\dfrac 1 2,~C=\dfrac 3 2$ $\dfrac{x+2}{x^3-x^2+4x} =\dfrac 1 {2x} + \dfrac{-x+3}{2(x^2-x+4)}$
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 29th, 2017, 11:07 PM   #4
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 Originally Posted by materialartist09 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.
ok we'll start it from scratch

$f(x) = \dfrac{x+2}{x(x^2-x+4)}$

$x^2-x+4 \text{ cannot be factored further.}$

we set it up as usual

$f(x) = \dfrac{A}{x} + \dfrac{Bx + C}{x^2-x+4}$

$\dfrac{x+2}{x(x^2-x+4)} = \dfrac{A(x^2-x+4) + Bx^2 + Cx}{x(x^2-x+4)} = \dfrac{(A+B)x^2+(-A+C)x + 4A}{x(x^2-x+4)}$

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

$\dfrac{x+2}{x(x^2-x+4)} = \dfrac{1}{2x} + \dfrac{3-x}{2(x^2-x+4)}$

 November 5th, 2017, 02:56 AM #5 Math Team   Joined: Jan 2015 From: Alabama Posts: 3,261 Thanks: 894 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 02:59 AM.

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