How do you write the partial fraction decomposition of the rational expression $x/ (8x^2−6x+1)$ ?

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Answer 1 · 2016-08-19T13:34:09

$x/(8x^2-6x+1)=(1/2)/(2x-1)-(1/2)/(4x-1)$ .

The $Dr.=8x^2-6x+1=(2x-1)(4x-1)$

Both the factors are linear polys. Hence, the reqd. decomposition

is given by,

$x/(8x^2-6x+1)=A/(2x-1)+B/(4x-1), where, A,B in RR$ .

To determine $A,&,B$ , Heavyside's Cover-up Method is useful :-

$A=[x/(4x-1)]_(x=1/2) =1/2/(4*1/2-1)=1/2$ .

$B=[x/(2x-1)]_(x=1/4)=1/4/(2*1/4-1)=1/4/(-1/2)=-1/2$ .

Therefore,

$x/(8x^2-6x+1)=(1/2)/(2x-1)-(1/2)/(4x-1)$ .

Method II :-

When there are only two linear factors in $Dr.$ , this decomposition can easily be derived as under :

Observe that, $(4x-1)-(2x-1)=2x$

$rArr Nr. =x=1/2{(4x-1)-(2x-1)}$ . Therefore,

$x/(8x^2-6x+1)=1/2[{(4x-1)-(2x-1)}/((4x-1)(2x-1))]$

$=1/2[cancel(4x-1)/(cancel(4x-1)(2x-1))-cancel(2x-1)/((4x-1)cancel(2x-1))]$

$=1/2[1/(2x-1)-1/(4x-1)]$ , as obtained earlier!

Enjoy Maths.!

How do you write the partial fraction decomposition of the rational expression x/ (8x^2−6x+1)x/ (8x^2−6x+1)?