Decompose as a sum of partial fractions: \[\frac{3x^2 + 2x - 5}{x^4 + 4x^3 + 6x^2 + 5x + 2}\]

To decompose the given fraction into partial fractions, we first need to factor the denominator.

The denominator can be factored as:
\[x^4 + 4x^3 + 6x^2 + 5x + 2 = (x^2 + 2x + 1)(x^2 + 2x + 2)\]

Now, we can rewrite the fraction as:
\[\frac{3x^2 + 2x - 5}{(x^2 + 2x + 1)(x^2 + 2x + 2)}\]

Next, we need to express the fraction as a sum of partial fractions:
\[\frac{3x^2 + 2x - 5}{(x^2 + 2x + 1)(x^2 + 2x + 2)} = \frac{Ax + B}{x^2 + 2x + 1} + \frac{Cx + D}{x^2 + 2x + 2}\]

Multiplying both sides by the denominator, we get:
\[3x^2 + 2x - 5 = (Ax + B)(x^2 + 2x + 2) + (Cx + D)(x^2 + 2x + 1)\]

Expanding and equating coefficients, we get the following system of equations:
\[A + C = 3\]
\[2A + B + 2C + D = 2\]
\[2B + D = -5\]

Solving this system of equations, we find:
\[A = 1, B = -2, C = 2, D = -1\]

Therefore, the original fraction can be decomposed as:
\[\frac{3x^2 + 2x - 5}{x^4 + 4x^3 + 6x^2 + 5x + 2} = \frac{x - 2}{x^2 + 2x + 1} + \frac{2x - 1}{x^2 + 2x + 2}\]

$$\frac{3x^2 2x - 5}{x^4 4x^3 6x^2 5x 2} = \frac{A}{x} + \frac{B}{x^2} + \frac{Cx D}{x^2 1} + \frac{E}{x^3} + \frac{F}{x^4}$$

$$3x^2 2x - 5 = A(x^3 4x^2 6x 5x 2) + B(x^2 4x^3 6x^2 5x 2) + $$

$$(Cx D)(x^4 4x^3 6x^2) + E(x^4 4x^3 6x^2 5x) + F(x^3 4x^2 6x^2)$$

Equating coefficients yields:

$$F = -5$$

$$E = 0$$

$$D = 0$$

$$C = 3$$

$$B = -4$$

$$A = 2$$

$$\therefore \qquad \frac{3x^2 2x - 5}{x^4 4x^3 6x^2 5x 2} = \frac{2}{x} - \frac{4}{x^2} + \frac{3x}{x^2 1} - \frac{5}{x^4}$$