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

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Answer 1 · 2018-01-27T16:53:00

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

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

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