## Algebra and Trigonometry 10th Edition

$\frac{-3}{x} + \frac{-1}{x+2} + \frac{5}{x-2}$
We must first find the Partial Fraction Decomposition: $\frac{A}{x} + \frac{B}{x+2} + \frac{C}{x-2} = \frac{x^{2}+12x+12}{x^{3}-4x}$ We must then solve for the constants: $A(x+2)(x-2) + B(x)(x-2) + C(x)(x+2) = x^{2} + 12x + 12$ $Ax^{2}-4A +Bx^{2}-2Bx+Cx^{2}+2Cx = x^{2} + 12x + 12$ $A + B + C = 1$ $-2B + 2C = 12$ $-4A = 12$ This can be represented with the following matrix: $\begin{bmatrix} 1 & 1 & 1 & |1\\ 0 & -2 & 2 & |12\\ -4 & 0 & 0 & |12 \end{bmatrix}$ ~ $\begin{bmatrix} 1 & 1 & 1 & |1\\ 0 & -1 & 1 & |6\\ -1 & 0 & 0 & |3 \end{bmatrix}$ ~ $\begin{bmatrix} 1 & 1 & 1 & |1\\ 0 & -1 & 1 & |6\\ 0 & 1 & 1 & |4 \end{bmatrix}$ ~ $\begin{bmatrix} 1 & 1 & 1 & |1\\ 0 & -1 & 1 & |6\\ 0 & 0 & 2 & |10 \end{bmatrix}$ $2C = 10$ $C = 5$ $-B + 5 = 6$ $B = -1$ $A - 1 + 5 = 1$ $A + 4 = 1$ $A = -3$ The Partial Fraction Decomposition is: $\frac{-3}{x} + \frac{-1}{x+2} + \frac{5}{x-2}$ Checking the Partial Fraction Decomposition: $-3(x+2)(x-2) - 1(x)(x-2) + 5(x)(x+2)$ $-3(x^{2} - 4) - 1(x^{2}-2x) + 5(x^{2}+2x)$ $-3x^{2}+12-x^{2}+2x+5x^{2}+10x$ $x^{2}+12x+12$ Therefore, the answer is correct.