How do you write the partial fraction decomposition of the rational expression $(x^2+8)/(x^2-5x+6)$ ?

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Answer 1 · 2015-12-15T20:49:24

$(x^2+8)/(x^2-5x+6)=1-(5x+2)/(x^2+8)$

Since the degree of the denominator is not more than the degree of the numerator, we first long divide the expression to obtain that

$(x^2+8)/(x^2-5x+6)=1+(-5x-2)/(x^2+8)$

The denominator $x^2+8$ is an irreducible quadratic factor and hence has a partial fraction form given by

$(-5x-2)/(x^2+8)=(Ax+B)/(x^2+8)$

By comparing terms in the numerator of these equivalent fractions, it is then clear that $A=-5 and B=-2$ .

Hence the partial fraction decomposition for the original expression is

$(x^2+8)/(x^2-5x+6)=1+(-5x-2)/(x^2+8)$

How do you write the partial fraction decomposition of the rational expression (x^2+8)/(x^2-5x+6) (x^2+8)/(x^2-5x+6)?