Math 20, Fall 2023 1. Find the partial fraction decomposition. (a) $\frac{3 x^{2}-9 x+8}{(x-1)^{2}(x-3)}$ (b) $\frac{2 x}{x^{2}-2}$

Step 1 ：\(\frac{3 x^{2}-9 x+8}{(x-1)^{2}(x-3)} = \frac{A}{x-1} + \frac{B}{(x-1)^2} + \frac{C}{x-3}\)

Step 2 ：Multiply through by the denominator: \(3x^2 - 9x + 8 = A(x-1)(x-3) + B(x-3) + C(x-1)^2\)

Step 3 ：Setting x = 1, we get: \(2 = -2B\)

Step 4 ：So, \(B = -1\)

Step 5 ：Setting x = 3, we get: \(14 = 4C\)

Step 6 ：So, \(C = 3.5\)

Step 7 ：Substitute \(B = -1\) and \(C = 3.5\) into the equation and simplify, we get: \(3x^2 - 9x + 8 = A(x-1)(x-3) - (x-3) + 3.5(x-1)^2\)

Step 8 ：Setting x = 0, we get: \(8 = -3A - 3 + 3.5\)

Step 9 ：So, \(A = -1.5\)

Step 10 ：The partial fraction decomposition of \(\frac{3 x^{2}-9 x+8}{(x-1)^{2}(x-3)}\) is \(\boxed{\frac{-1.5}{x-1} - \frac{1}{(x-1)^2} + \frac{3.5}{x-3}}\)

Step 11 ：\(\frac{2 x}{x^{2}-2} = \frac{A}{x-\sqrt{2}} + \frac{B}{x+\sqrt{2}}\)

Step 12 ：Multiply through by the denominator: \(2x = A(x+\sqrt{2}) + B(x-\sqrt{2})\)

Step 13 ：Setting x = \sqrt{2}, we get: \(2\sqrt{2} = A(2\sqrt{2})\)

Step 14 ：So, \(A = 1\)

Step 15 ：Setting x = -\sqrt{2}, we get: \(-2\sqrt{2} = B(-2\sqrt{2})\)

Step 16 ：So, \(B = 1\)

Step 17 ：The partial fraction decomposition of \(\frac{2 x}{x^{2}-2}\) is \(\boxed{\frac{1}{x-\sqrt{2}} + \frac{1}{x+\sqrt{2}}}\)