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Resolve `(x^(3)-6x^(2)+10x-2)/(x^(2)-5x+6)` into partial fractions: |
Answer» Here the given function is an improper rational function. On dividing we get `(x^(3)-6x^(2)+10x-2)/(x^(2)-5x+6) = x-1+(-x+4)/(x^(2)-5x+6)`…………..(i) We have, `(-x+4)/(x^(2)-5x+6)= (-x+4)/((x-2)(x-3))` So, let `(-x+4)/((x-2)(x-3)) = A/(x-2)=B/(x-3)`, then `-x+4=A(x-3)+B(x-2)`...........(ii) Putting `x-3=0` or `x=3` in (ii), we get `(-x+4)/((x-2)(x-3))=A/(x-2)+B/(x-3)`, then `-x+4=A(x-3)+B(x-2)`...............(2) puttting x-2=0 or x=2 in eq. (ii), we get `2=A(2-3)rArr A=-2` `therefore (-x+4)/((x-2)(x-3))=-2/(x-2)+1/(x-3)` Hence `(x^(3)-6x^(2)+10x-2)/(x^(2)-5x+6)=x-1-2/(x-2)+1/(x-3)` |
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