How do I decompose the rational expression $\frac{- x^{2} + 9 x + 9}{(x - 5) (x^{2} + 4)}$ $(-x^2+9x+9)/((x-5)(x^2+4))$ into partial fractions?

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Answer 1 · 2014-11-29T00:41:39

$\frac{- x^{2} + 9 x + 9}{(x - 5) (x^{2} + 4)} = \frac{A}{x - 5} + \frac{B x + C}{x^{2} + 4}$ ${-x^2+9x+9}/{(x-5)(x^2+4)}=A/{x-5}+{Bx+C}/{x^2+4}$

by finding the common denominator,

$= \frac{A (x^{2} + 4) + (B x + C) (x - 5)}{(x - 5) (x^{2} + 4)}$ $={A(x^2+4)+(Bx+C)(x-5)}/{(x-5)(x^2+4)}$

by multiplying out the numerator,

$= \frac{A x^{2} + 4 A + B x^{2} - 5 B x + C x - 5 C}{(x - 5) (x^{2} + 4)}$ $={Ax^2+4A+Bx^2-5Bx+Cx-5C}/{(x-5)(x^2+4)}$

by collecting like terms,

$= \frac{(A + B) x^{2} + (C - 5 B) x + (4 A - 5 C)}{(x - 5) (x^{2} + 4)}$ $={(A+B)x^2+(C-5B)x+(4A-5C)}/{(x-5)(x^2+4)}$

by comparing the coefficients of the numerators,

$=>{(A+B=-1 => 5A+5B=-5" ... (Eq. 1)"),(C-5B=9" ... (Eq. 2)"),(4A-5C=9" ... (Eq. 3)"):}$

by adding (Eq. 1) and (Eq. 2), and by multiplying by $5$ ,

$=>5A+C=4 => 25A+5C=20" ... (Eq. 4)"$

by adding (Eq. 3) and (Eq. 4),

$29A=29 => A=1$

by plugging $A=1$ into (Eq. 1),

$=> (1)+B=-1 => B=-2$

by plugging $B=-2$ into (Eq. 2),

$=> C-5(-2)=9 => C=-1$

Hence, we have the partial fractions

${-x^2+9x+9}/{(x-5)(x^2+4)}=1/{x-5}+{-2x-1}/{x^2+4}$

I hope that this was helpful.

How do I decompose the rational expression (-x^2+9x+9)/((x-5)(x^2+4))−x2+9x+9(x−5)(x2+4)(-x^2+9x+9)/((x-5)(x^2+4)) into partial fractions?