Question

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

Answer 1

`{-x^2+9x+9}/{(x-5)(x^2+4)}=A/{x-5}+{Bx+C}/{x^2+4}`

by finding the common denominator,

`={A(x^2+4)+(Bx+C)(x-5)}/{(x-5)(x^2+4)}`

by multiplying out the numerator,

`={Ax^2+4A+Bx^2-5Bx+Cx-5C}/{(x-5)(x^2+4)}`

by collecting like terms,

`={(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}`

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I hope that this was helpful.