How do you express #(x^2+8)/(x^2-5x+6)# in partial fractions?
1 Answer
Apr 21, 2016
Explanation:
#(x^2+8)/(x^2-5x+6)#
#=((x^2-5x+6)+(5x+2))/(x^2-5x+6)#
#=1+(5x+2)/(x^2-5x+6)#
#=1+(5x+2)/((x-2)(x-3))#
#=1+A/(x-2)+B/(x-3)#
#=1+(A(x-3)+B(x-2))/(x^2-5x+6)#
#=1+((A+B)x-(3A+2B))/(x^2-5x+6)#
Equating coefficients we get:
#{ (A+B=5), (3A+2B=-2) :}#
Subtract twice the first equation from the second to find:
#A = -12#
Then substitute this value of
#B = 17#
So:
#(x^2+8)/(x^2-5x+6)=1-12/(x-2)+17/(x-3)#
