1 Express \( \frac{3 x^{3}+6 x^{2}+5}{x^{2}\left(x^{2}+5\right)} \) in partial fractions.

To express \( \frac{3 x^{3}+6 x^{2}+5}{x^{2}\left(x^{2}+5\right)} \) in partial fractions, we start by ensuring that the degree of the numerator is less than that of the denominator. Here, the numerator \(3x^3 + 6x^2 + 5\) has a degree of 3, which is greater than the degree of the denominator \(x^2(x^2 + 5)\) (which is degree 4). Therefore, we can proceed by performing polynomial long division first. 1. Divide \(3x^3 + 6x^2 + 5\) by \(x^2\): - \(3x\) times \(x^2\) gives \(3x^3\), and subtracting this results in \(6x^2 + 5 - 3x^3 = 6x^2 + 5\). - Now, \(3x^3\) is canceled out, and we can focus on \(6x^2 + 5\), leaving the fraction as: \[ 3x + \frac{6x^2 + 5}{x^2(x^2 + 5)} \] Now we can express the remaining fraction in partial fractions: \[ \frac{6x^2 + 5}{x^2(x^2 + 5)} = \frac{A}{x} + \frac{B}{x^2} + \frac{Cx + D}{x^2 + 5} \] where \(A\), \(B\), \(C\), and \(D\) are constants to be determined. 2. Clear the fractions by multiplying through by the denominator \(x^2(x^2 + 5)\): \[ 6x^2 + 5 = A x (x^2 + 5) + B (x^2 + 5) + (Cx + D) x^2 \] Expanding this gives: \[ 6x^2 + 5 = A x^3 + 5A x + B x^2 + 5B + C x^3 + D x^2 \] 3. Combine like terms: \[ = (A + C)x^3 + (B + D)x^2 + 5Ax + 5B \] Setting coefficients equal gives us a system of equations: - Coefficient of \(x^3\): \(A + C = 0\) - Coefficient of \(x^2\): \(B + D = 6\) - Coefficient of \(x^1\): \(5A = 0 \Rightarrow A = 0\) - Constant: \(5B = 5 \Rightarrow B = 1\) Using \(A = 0\) in the first equation: \[ 0 + C = 0 \Rightarrow C = 0 \] And substituting \(B = 1\) in the second equation: \[ 1 + D = 6 \Rightarrow D = 5 \] Now we have the constants: - \(A = 0\) - \(B = 1\) - \(C = 0\) - \(D = 5\) Thus, we can rewrite the original fraction as: \[ \frac{3 x^{3}+6 x^{2}+5}{x^{2}\left(x^{2}+5\right)} = 3x + \frac{1}{x^2} + \frac{5}{x^2 + 5} \] Therefore, the final partial fraction decomposition is: \[ \frac{3 x^{3}+6 x^{2}+5}{x^{2}(x^{2}+5)} = 3x + \frac{1}{x^2} + \frac{5}{x^2 + 5} \]