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Partial fraction decomposition
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=== Example 2 === <math display="block">f(x)=\frac{x^3+16}{x^3-4x^2+8x}</math> After [[Polynomial long division|long division]], we have <math display="block">f(x)=1+\frac{4x^2-8x+16}{x^3-4x^2+8x}=1+\frac{4x^2-8x+16}{x(x^2-4x+8)}</math> The factor ''x''<sup>2</sup> β 4''x'' + 8 is irreducible over the reals, as its [[discriminant]] {{math|1=(β4)<sup>2</sup> β 4Γ8 = β16}} is negative. Thus the partial fraction decomposition over the reals has the shape <math display="block">\frac{4x^2-8x+16}{x(x^2-4x+8)}=\frac{A}{x}+\frac{Bx+C}{x^2-4x+8}</math> Multiplying through by ''x''<sup>3</sup> β 4''x''<sup>2</sup> + 8''x'', we have the polynomial identity <math display="block">4x^2-8x+16 = A \left(x^2-4x+8\right) + \left(Bx+C\right)x</math> Taking ''x'' = 0, we see that 16 = 8''A'', so ''A'' = 2. Comparing the ''x''<sup>2</sup> coefficients, we see that 4 = ''A'' + ''B'' = 2 + ''B'', so ''B'' = 2. Comparing linear coefficients, we see that β8 = β4''A'' + ''C'' = β8 + ''C'', so ''C'' = 0. Altogether, <math display="block">f(x)=1+2\left(\frac{1}{x}+\frac{x}{x^2-4x+8}\right)</math> 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: <math display="block">\frac{x}{x^2-4x+8}=\frac{D}{x-(2+2i)}+\frac{E}{x-(2-2i)}</math> Multiplying through by the denominator gives: <math display="block">x=D(x-(2-2i))+E(x-(2+2i)) </math> Equating the coefficients of {{math|''x''}} and the constant (with respect to {{math|''x''}}) coefficients of both sides of this equation, one gets a system of two linear equations in {{math|''D''}} and {{math|''E''}}, whose solution is <math display="block">D=\frac{1+i}{2i}=\frac{1-i}{2}, \qquad E=\frac{1-i}{-2i}=\frac{1+i}{2}.</math> Thus we have a complete decomposition: <math display="block">f(x)=\frac{x^3+16}{x^3-4x^2+8x}=1+\frac{2}{x}+\frac{1-i}{x-(2+2i)}+\frac{1+i}{x-(2-2i)}</math> One may also compute directly {{math|''A'', ''D''}} and {{math|''E''}} with the residue method (see also example 4 below).
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