How do you express # (5x^2 - 25x + 28) / (x^2(x-7))# in partial fractions?

Question

1234 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-12-21T08:37:59

#(5x^2-25x+28)/(x^2(x-7)) = 3/x-4/x^2+2/(x-7)#

Given:

#(5x^2-25x+28)/(x^2(x-7))#

#= A/x+B/x^2+C/(x-7)#

#= (Ax(x-7)+B(x-7)+Cx^2)/(x^2(x-7))#

#= ((A+C)x^2+(-7A+B)x+(-7B))/(x^2(x-7))#

So:

#{ (A+C = 5), (-7A+B=-25), (-7B=28) :}#

From the third equation, we find that #B=-4#

Substituting this value of #B# in the second equation, we find #A=3#

Then substituting this value of #A# into the first equation, we find #C=2#

So:

#(5x^2-25x+28)/(x^2(x-7)) = 3/x-4/x^2+2/(x-7)#