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

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Answer 1 · 2015-12-15T20:49:24

$\frac{x^{2} + 8}{x^{2} - 5 x + 6} = 1 - \frac{5 x + 2}{x^{2} + 8}$ $(x^2+8)/(x^2-5x+6)=1-(5x+2)/(x^2+8)$

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

$\frac{x^{2} + 8}{x^{2} - 5 x + 6} = 1 + \frac{- 5 x - 2}{x^{2} + 8}$ $(x^2+8)/(x^2-5x+6)=1+(-5x-2)/(x^2+8)$

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

$\frac{- 5 x - 2}{x^{2} + 8} = \frac{A x + B}{x^{2} + 8}$ $(-5x-2)/(x^2+8)=(Ax+B)/(x^2+8)$

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

Hence the partial fraction decomposition for the original expression is

$\frac{x^{2} + 8}{x^{2} - 5 x + 6} = 1 + \frac{- 5 x - 2}{x^{2} + 8}$ $(x^2+8)/(x^2-5x+6)=1+(-5x-2)/(x^2+8)$

How do you write the partial fraction decomposition of the rational expression (x^2+8)/(x^2-5x+6)x2+8x2−5x+6 (x^2+8)/(x^2-5x+6)?