How do you express (5x^2 - 25x + 28) / (x^2(x-7))5x2−25x+28x2(x−7) in partial fractions?

1 Answer
George C.
Dec 21, 2017

(5x^2-25x+28)/(x^2(x-7)) = 3/x-4/x^2+2/(x-7)5x2−25x+28x2(x−7)=3x−4x2+2x−7

Explanation:

Given:

(5x^2-25x+28)/(x^2(x-7))5x2−25x+28x2(x−7)

= A/x+B/x^2+C/(x-7)=Ax+Bx2+Cx−7

= (Ax(x-7)+B(x-7)+Cx^2)/(x^2(x-7))=Ax(x−7)+B(x−7)+Cx2x2(x−7)

= ((A+C)x^2+(-7A+B)x+(-7B))/(x^2(x-7))=(A+C)x2+(−7A+B)x+(−7B)x2(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)

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