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

1 Answer
Nov 29, 2014

#{-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}#


I hope that this was helpful.