Write out the form of the partial fraction decomposition of $\frac { x ^ { 2 } + 5 } { x ^ { 3 } - 2 x ^ { 2 } + 4 x - 8 }$

A) $\frac { A } { x - 2 } + \frac { B x } { ( x - 2 ) ^ { 2 } } + \frac { C } { x + 2 }$

B) $\frac { A x + B } { x - 2 } + \frac { C } { x ^ { 2 } + 4 }$

C) $\frac { A } { x - 2 } + \frac { B x } { ( x - 2 ) ^ { 2 } }$

D) $\frac { A } { x - 2 } + \frac { B x + C } { x ^ { 2 } + 4 }$

E)none

Question

Write out the form of the partial fraction decomposition of $\frac { x ^ { 2 } + 5 } { x ^ { 3 } - 2 x ^ { 2 } + 4 x - 8 }$

A) $\frac { A } { x - 2 } + \frac { B x } { ( x - 2 ) ^ { 2 } } + \frac { C } { x + 2 }$

B) $\frac { A x + B } { x - 2 } + \frac { C } { x ^ { 2 } + 4 }$

C) $\frac { A } { x - 2 } + \frac { B x } { ( x - 2 ) ^ { 2 } }$

D) $\frac { A } { x - 2 } + \frac { B x + C } { x ^ { 2 } + 4 }$

E)none

Quizplus · Accepted Answer

We first factor the denominator using synthetic division or any other method, and obtain $x^3-2x^2+4x-8 = (x-2)(x^2+4).$ Thus we can write $$\frac{x^2+5}{x^3-2x^2+4x-8} = \frac{A}{x-2} + \frac{Bx+C}{x^2+4}.$$ For some constants $A,$ $B,$ $C.$

Clearing the denominators gives $$x^2+5 = A(x^2+4) + (Bx+C)(x-2).$$ Expanding the right side and equating coefficients, we get  \begin{align*}
A+B &= 1, \
-2B+C &= 0, \
4A-2B &= 0, \
4A-2B+C &= 1.
\end{align*} Solving this system, we find $A = \frac18,$ $B = -\frac14,$ and $C = \frac38,$ so $$\frac{x^2+5}{x^3-2x^2+4x-8} = \boxed{\frac{1}{8(x-2)} - \frac{1}{4(x^2+4)} + \frac{3}{8}\cdot \frac{1}{x^2+4}}.$$

Write Out the Form of the Partial Fraction Decomposition Of \frac