How do you write the partial fraction decomposition of the rational expression $(5x-1)/(x^2-x-2)$ ?

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Answer 1 · 2016-01-04T05:48:00

$(5x-1)/(x^2-x-2)=3/(x-2)+2/(x+1)$

General form to break into partial fractions is
$(ax+b)/(cx^2+dx+e)=A/(x+C)+B/(x+D)$

$(5x-1)/(x^2-x-2)=(5x-1)/((x-2)(x+1))$

$(5x-1)/((x-2)(x+1))=A/(x-2)+B/(x+1)=(A(x+1)+B(x-2))/((x-2)(x+1))$

$(5x-1)/((x-2)(x+1))=(Ax+A+Bx-2B)/((x-2)(x+1))$

we can equate the coefficients of numerators of both sides
$A+B=5,A-2B=-1$
Subtracting both gives $3B=6$
or B = 2
And A=3
thus

$(5x-1)/(x^2-x-2)=3/(x-2)+2/(x+1)$

How do you write the partial fraction decomposition of the rational expression (5x-1)/(x^2-x-2)(5x-1)/(x^2-x-2)?