have The factor x2 − 4x + 8 is irreducible over the reals, as its discriminant is negative. Thus the partial fraction decomposition over the reals has the shape Multiplying through by x3 − 4x2 + 8x, we have the polynomial identity Taking x = 0, we see that 16 = 8A, so A = 2. Comparing the x2 coefficients, we see that 4 = A + B = 2 + B, so B = 2. Comparing linear coefficients, we see that −8 = −4A + C = −8 + C, so C = 0. Altogether, The fraction can be completely decomposed using complex numbers. According to the fundamental theorem of algebra every complex polynomial of degree n has n (complex) roots (some of which can be repeated). The second fraction can be decomposed to: Multiplying through by the denominator gives: Equating the coefficients of and the constant (with respect to ) coefficients of both sides of this equation, one gets a system of two linear equations in and , whose solution is Thus we have a complete decomposition: One may also compute directly and with the residue method (see also example 4 below). Example 3 This example illustrates almost all the "tricks" we might need to use, short of consulting a computer algebra system. After long division and factoring the denominator, we have The partial fraction decomposition takes the form Multiplying through by the denominator on the left-hand side we have the polynomial identity Now we use different values of x to compute the coefficients: Solving this we have: Using these values we can write: We compare the coefficients of x6 and x5 on both side and we have: Therefore: which gives us B = 0. Thus the partial fraction decomposition is given by: Alternatively, instead of expanding, one can obtain other linear dependences on the coefficients computing some derivatives at in the above polynomial identity. (To this end, recall that the derivative at x = a of (x − a)mp(x) vanishes if m > 1 and is just p(a) for m = 1.) For instance the first derivative at x = 1 gives that is 8 = 4B + 8 so B = 0. Example 4 (residue method) Thus, f(z) can be decomposed into rational functions whose denominators are z+1, z−1, z+i, z−i. Since each term is of power one, −1, 1, −i and i are simple poles.