How do you write the partial fraction decomposition of the rational expression $\frac{X^{2} - 5 x + 6}{X^{3} - X^{2} + 2 X}$ $(X^2-5x+6)/(X^3-X^2+2X)$ ?

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Answer 1 · 2015-12-30T04:28:43

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

Factor out an $x$ $x$ in the denominator

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

Multiply both sides by $x (x^{2} - x + 2)$ $x(x^2-x+2)$

$x^{2} - 5 x + 6 = A (x^{2} - x + 2) + x (B x + C)$ $x^2-5x+6=A(x^2-x+2)+x(Bx+C)$

Distribute on right hand side

$x^{2} - 5 x + 6 = A x^{2} - A x + 2 A + B x^{2} + C x$ $x^2-5x+6=Ax^2-Ax+2A+Bx^2+Cx$

Factor right hand side

$x^{2} - 5 x + 6 = x^{2} (A + B) + x (- A + C) + 2 A$ $x^2-5x+6=x^2(A+B)+x(-A+C)+2A$

Equate coefficients on the left hand side with the right

$2 A = 6$ $2A=6$

$- A + C = - 5$ $-A+C=-5$

$A + B = 1$ $A+B=1$

Solving we get

$A = 3$ $A=3$

$B = - 2$ $B=-2$

$C = - 2$ $C=-2$

Putting it all together

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

Pull out negative in $- 2 x - 2$ $-2x-2$

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

Factor out a $2$ $2$ in $2 x + 2$ $2x+2$

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

How do you write the partial fraction decomposition of the rational expression X2−5x+6X3−X2+2X(X^2-5x+6)/(X^3-X^2+2X)?