# How do you write the partial fraction decomposition of the rational expression (6x+5) / (x+2) ^4?

May 27, 2017

$\frac{6 x + 5}{x + 2} ^ 4 = \frac{A}{x + 2} ^ 4 + \frac{B}{x + 2} ^ 3 + \frac{C}{x + 2} ^ 2 + \frac{D}{x + 2}$

Multiply both sides by ${\left(x + 2\right)}^{4}$:

$6 x + 5 = A + B \left(x + 2\right) + C {\left(x + 2\right)}^{2} + D {\left(x + 2\right)}^{3}$

Let x = -2

$6 \left(- 2\right) + 5 = A$

$A = - 7$

Add 7 to both sides of equation:

$6 x + 12 = B \left(x + 2\right) + C {\left(x + 2\right)}^{2} + D {\left(x + 2\right)}^{3}$

Let x = 0:

[ (2,4,8,|,12), ]

Let x = -1

[ (2,4,8,|,12), (1,1,1,|,6) ]

Let x = 1

[ (2,4,8,|,12), (1,1,1,|,6), (3,9,27,|,18) ]

${R}_{1} \leftrightarrow {R}_{2}$

[ (1,1,1,|,6), (2,4,8,|,12), (3,9,27,|,18) ]

${R}_{2} - 2 {R}_{1} \to {R}_{2}$

[ (1,1,1,|,6), (0,2,6,|,0), (3,9,27,|,18) ]

${R}_{3} - 3 {R}_{1} \to {R}_{3}$

[ (1,1,1,|,6), (0,2,6,|,0), (0,6,24,|,0) ]

${R}_{2} / 2 \to {R}_{2}$

[ (1,1,1,|,6), (0,1,3,|,0), (0,6,24,|,0) ]

${R}_{3} / 6 \to {R}_{3}$

[ (1,1,1,|,6), (0,1,3,|,0), (0,1,4,|,0) ]

${R}_{3} - {R}_{3} \to {R}_{3}$

[ (1,1,1,|,6), (0,1,3,|,0), (0,0,1,|,0) ]

It is obvious that C and D are 0 so we can clear the matrix:

[ (1,0,0,|,6), (0,1,0,|,0), (0,0,1,|,0) ]

$B = 6 \mathmr{and} C = D = 0$

$\frac{6 x + 5}{x + 2} ^ 4 = - \frac{7}{x + 2} ^ 4 + \frac{6}{x + 2} ^ 3$