How do you express $\frac{x^{2} + 8}{x^{2} - 5 x + 6}$ $(x^2+8)/(x^2-5x+6)$ in partial fractions?

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How do you express $\frac{x^{2} + 8}{x^{2} - 5 x + 6}$ $(x^2+8)/(x^2-5x+6)$ in partial fractions?

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Answer 1 · 2016-04-21T20:10:19

$\frac{x^{2} + 8}{x^{2} - 5 x + 6} = 1 - \frac{12}{x - 2} + \frac{17}{x - 3}$ $(x^2+8)/(x^2-5x+6)=1-12/(x-2)+17/(x-3)$

Explanation:

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

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

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

$= 1 + \frac{5 x + 2}{(x - 2) (x - 3)}$ $=1+(5x+2)/((x-2)(x-3))$

$= 1 + \frac{A}{x - 2} + \frac{B}{x - 3}$ $=1+A/(x-2)+B/(x-3)$

$= 1 + \frac{A (x - 3) + B (x - 2)}{x^{2} - 5 x + 6}$ $=1+(A(x-3)+B(x-2))/(x^2-5x+6)$

$= 1 + \frac{(A + B) x - (3 A + 2 B)}{x^{2} - 5 x + 6}$ $=1+((A+B)x-(3A+2B))/(x^2-5x+6)$

Equating coefficients we get:

${ (A+B=5), (3A+2B=-2) :}$

Subtract twice the first equation from the second to find:

$A = -12$

Then substitute this value of $A$ into the first equation to find:

$B = 17$

So:

$(x^2+8)/(x^2-5x+6)=1-12/(x-2)+17/(x-3)$

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How do you express $\frac{x^{2} + 8}{x^{2} - 5 x + 6}$ $(x^2+8)/(x^2-5x+6)$ in partial fractions?

1 Answer

Explanation:

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How do you express $\frac{x^{2} + 8}{x^{2} - 5 x + 6}$ $(x^2+8)/(x^2-5x+6)$ in partial fractions?