How do you express #(2x^4 + 9x^2 + x - 4)/(x^3 + 4x)# in partial fractions?

1 Answer
Jan 27, 2018

#(2x^4+9x^2+x-4)/(x^3+4x)=2x-1/x+(2x+1)/(x^2+4)#

Explanation:

Let us first divide #2x^4+9x^2+x-4# by #x^3+4x# to get the degree of numerator less than that of denominator and we get

#(2x^4+9x^2+x-4)/(x^3+4x)=2x+(x^2+x-4)/(x(x^2+4))#

Let us now get partial fractions of #(x^2+x-4)/(x(x^2+4))#, which will be of the form

#(x^2+x-4)/(x(x^2+4))=A/x+(Bx+C)/(x^2+4)#

or #x^2+x-4=A(x^2+4)+x(Bx+C)=(A+B)x^2+Cx+4A#

andcomparing coefficients of like powers

#4A=-4# or #A=-1#, #C=1# and

#A+B=1# i.e. #B=1-A=2#

Hence #(2x^4+9x^2+x-4)/(x^3+4x)=2x-1/x+(2x+1)/(x^2+4)#