Divide. \[ \left(2 y^{2}-9 y+5\right) \div(2 y+3) \] \( \square \)

To divide \( (2y^2 - 9y + 5) \) by \( (2y + 3) \), you can use polynomial long division. First, divide the leading term of the dividend \( 2y^2 \) by the leading term of the divisor \( 2y \), which gives you \( y \). Multiply the entire divisor \( (2y + 3) \) by \( y \) to get \( 2y^2 + 3y \). Next, subtract \( (2y^2 + 3y) \) from \( (2y^2 - 9y + 5) \): \[ (2y^2 - 9y + 5) - (2y^2 + 3y) = -12y + 5 \] Now, divide the leading term \( -12y \) by the leading term \( 2y \), which gives you \( -6 \). Multiply the divisor \( (2y + 3) \) by \( -6 \) to get \( -12y - 18 \). Now perform the subtraction again: \[ (-12y + 5) - (-12y - 18) = 23 \] Because there are no more terms to bring down, the final result of the division is \( y - 6 \) with a remainder of \( 23 \). Thus, \[ \frac{2y^2 - 9y + 5}{2y + 3} = y - 6 + \frac{23}{2y + 3} \]