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$\frac{3x^2+17x}{x^3+3x^2+-6x-8}$

I need to find the value of C in the form of $\frac{A}{x+1} + \frac{B}{x-2} + \frac{C}{x+4}$ which is based on the fraction give at the top.

I can get so far to do the following:

$A(x^2+2x-8) + B(x^2+5x+4) + C(x^2-x-2) = 3x^2+17x$

No clue on my next step or even if this is the right step.

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  • $\begingroup$ Divide $x^3+3x^2-6x-8$ by $x+4$ to get the quotient $q(x)$. Then evaluate $\frac{3x^2+17x}{q(x)}$ at $x=-4$ and you get $C$. The reason why this works is that if $\frac{3x^2+17x}{x^3+3x^2-6x-8}=\frac{A}{x+1}+\frac{B}{x-2}+\frac{C}{x+4}$ then $\frac{3x^2+17x}{(x^3+3x^2-6x-8)/(x+4)}=\frac{A(x+4)}{x+1}+\frac{B(x+4)}{x-2}+C$. $\endgroup$ – Tom Feb 19 '15 at 13:55
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    $\begingroup$ The most efficient way to do the division and the evaluation above is to use Ruffini's algorithm. $\endgroup$ – Tom Feb 19 '15 at 14:01
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    $\begingroup$ FYI this is called partial fraction decomposition. $\endgroup$ – Marc van Leeuwen Feb 19 '15 at 14:52
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$$ \begin{align*} \left(A+B+C\right) x^2 &= 3x^2 \\ \left(2A + 5B -C\right)x &= 17x \\ -8A + 4B - 2C &= 0 \end{align*} $$

Dividing the first equation by $x^2$ and the second by $x$ will get you a linear system of equations.

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  • $\begingroup$ There is a simpler way. $\endgroup$ – Yves Daoust Feb 20 '15 at 8:54
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Compare coefficients of both sides polynomials. for example by comparing coefficient next you $x^2$ you will get $A+B+C = 3$. Do the same for $x^1$ and $x^0$ and solve system of 3 linear equations.

The other way is to put arbitrary 3 values of $x$, for example $-1$, $0$ and $1$, and again you will get system of 3 linear aquations with variables $A$, $B$ and $C$.

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$$A(x^2+2x-8) + B(x^2+5x+4) + C(x^2-x-2) = 3x^2+17x.$$

Setting $x=-4$,

$$0A+0B+18C=-20.$$

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Hint

You have $$\frac{3 x^2+17 x}{x^3+3 x^2-6 x-8}=\frac{A}{x+1} + \frac{B}{x-2} + \frac{C}{x+4}$$ So, write $$f(x)=3x^2+17x-\Big(A(x-2)(x+4)+B(x+1)(x+4)+C(x+1)(x-2)\Big)=0$$ Now $$f(2)=3\times 2^2+17 \times 2-B\times(2+1)\times(2+4)=46-18B=0$$ Repeat for $f(-1)$ and $f(-4)$; you just have one linear equation at the time.

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  • $\begingroup$ The question was about $C$ :) $\endgroup$ – Yves Daoust Feb 19 '15 at 14:14
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    $\begingroup$ @YvesDaoust. Yes, I know ! This was on purpose. Cheers :-) $\endgroup$ – Claude Leibovici Feb 19 '15 at 14:37
  • $\begingroup$ Of course, stupid me ! $\endgroup$ – Yves Daoust Feb 19 '15 at 14:51
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if you want to find just $C$ not other two, then look at the behavior of $$ \frac{3x^2 + 17x}{(x+1)(x-2)(x+4)} = \frac{3*(-4)^2+17*(-4)}{(-4+1)(-4-2)(x+4)} +\cdots=-\frac{10/9}{(x-4)}+\cdots$$ therefore $$C = -\frac{10}9. $$

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