# Find the partial fraction decomposition for f(x)= (28)/(x^2(x+2)(x^2-x+1))

You need to decompose the given fraction into four irreducible fractions, as many factors as you can identify at denominator, such that:

`28/(x^2(x+2)(x^2-x+1)) = A/x + B/x^2 + C/(x+2) + (Dx + E)/(x^2-x+1)`

You should notice that the denominator x^2-x+1 cannot be decomposed any further, hence, the numerator is a linear indetermined expression.

Bringing the fractions to the right side, to a common denominator, yields:

`28 = Ax(x + 2)(x^2 - x + 1) + B(x + 2)(x^2 - x + 1) + Cx^2(x^2 - x + 1) + (Dx + E)x^2(x + 2)`

`28 = Ax(x^3 - x^2 + x + 2x^2 - 2x + 2) + B(x^3 - x^2 + x + 2x^2 - 2x + 2) + Cx^4 - Cx^3 + Cx^2 + (Dx^3 + Ex^2)(x + 2)`

`28 = Ax^4 + Ax^3 - Ax^2 + 2Ax + Bx^3 + Bx^2 - Bx + 2B + Cx^4 - Cx^3 + Cx^2 + Dx^4 + 2Dx^3 + Ex^3 + 2Ex^2`

`28 = x^4(A + C + D) + x^3(A + B - C + 2D + E) + x^2(-A + B + C + 2E) + x(2A - B) + 2B`

Equating the coefficients of like powers yields:

`{(A + C + D = 0),(A + B - C + 2D + E = 0),(-A + B + C + 2E = 0),(2A - B = 0),(2B = 28):}`

`{(A + C + D = 0),(A - C + 2D + E = -14),(-A + C + 2E = -14),(2A = 14),(B = 14):} => {(C + D = -7),(- C + 2D + E = -21),(C + 2E = -7),(A = 7),(B = 14):} => C + D = C + 2E => D = 2E`

`{(-C + 5E = -21),(C + 2E = -7):} => 7E = -28 => E = -4 => D = -8 => C = 1`

Replacing the values of coefficients `A,B,C,D,E` in equation of function `f(x)` , yields:

`28/(x^2(x+2)(x^2-x+1)) = 7/x + 14/x^2 + 1/(x+2) + (-8x - 4)/(x^2-x+1)`

Hence, decomposing the equation of the function into irreducible fractions, using the partial fraction decomposition, yields `f(x) = 7/x + 14/x^2 + 1/(x+2) + (-8x - 4)/(x^2-x+1).`

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