How do you split this into partial fractions?

2y/(y^3-y^2+y-1)





i kno the answer but would you do it with work?


thank youHow do you split this into partial fractions?
2y/(y^3-y^2+y-1)


= 2y/[y^2(y-1) + (y-1)]


= 2y/[(y^2+1)(y-1)]


Let 2y/[(y^2+1)(y-1)] = A/(y-1) + (By+C)/(y^2+1)


=%26gt; 2y = A(y^2+1) + (By+C)(y-1)


Compare coefficients


y^2: A+B = 0


y: 2 = C-B


y^0: A-C = 0


Solve the system,


C+C-2 = 0


=%26gt; A = C = 1, B = -1





Answer:


2y/(y^3-y^2+y-1) = 1/(y-1) + (-y+1)/(y^2+1)How do you split this into partial fractions?
Look at the denominator:


y^3-y^2+y-1


This can be factored by grouping:


y^3 - y^2 = y^2(y - 1) then 1(y - 1)


You get:


(y^2 + 1) (y - 1)





Put this back into the denominator:


2y / (y^2 + 1) (y - 1)





The only way you can split it is by doing this:


[ 2y / (y^2 + 1) ] * [ 1 / (y - 1) ]