Divide \( \frac{3 x^{3}-2 x}{x+5} \) \( 3 x^{2}+[?] x+\quad+\frac{}{x+5} \)

To divide the polynomial \(3x^3 - 2x\) by \(x + 5\), we can use polynomial long division. First, divide the leading term of the numerator \(3x^3\) by the leading term of the denominator \(x\), which gives us \(3x^2\). Multiply \(x + 5\) by \(3x^2\) to get \(3x^3 + 15x^2\). Subtract this from the original polynomial: \[ (3x^3 - 2x) - (3x^3 + 15x^2) = -15x^2 - 2x \] Next, divide the new leading term \(-15x^2\) by \(x\) to get \(-15x\). Multiply \(x + 5\) by \(-15x\) to get \(-15x^2 - 75x\) and subtract: \[ (-15x^2 - 2x) - (-15x^2 - 75x) = 73x \] Finally, divide \(73x\) by \(x\) to obtain \(73\). Multiply by \(x + 5\) to get \(73x + 365\) and subtract: \[ 73x - (73x + 365) = -365 \] Putting it all together gives us: \[ 3x^2 - 15x + 73 - \frac{365}{x + 5} \] So the missing values are: \( -15 \) for the coefficient of \(x\), \( 73 \) as the constant term, and \( -365 \) in the remainder.