How do you express $(4x-1)/( x^2(x-4))$ in partial fractions?

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Answer 1 · 2017-02-05T08:14:16

The answer is $=(1/4)/(x^2)+(-15/16)/(x)+(15/16)/(x-4)$

Let's perform the decomposition into partial fractions

$(4x-1)/(x^2(x-4))=A/(x^2)+B/(x)+C/(x-4)$

$=(A(x-4)+B(x(x-4))+C(x^2))/(x^2(x-4))$

As the denominators are the same, we compare the numerators

$4x-1=A(x-4)+B(x(x-4))+C(x^2)$

Let $x=0$ , $=>$ , $-1=-4A$ , $=>$ , $A=1/4$

Let $x=4$ , $=>$ , $15=16C$ , $=>$ , $C=15/16$

Coefficients of $x^2$

$0=B+C$ , $=>$ , $B=-15/16$

Therefore

$(4x-1)/(x^2(x-4))=(1/4)/(x^2)+(-15/16)/(x)+(15/16)/(x-4)$

How do you express (4x-1)/( x^2(x-4)) (4x-1)/( x^2(x-4)) in partial fractions?