Divide. \[ \left(2 y^{2}+4\right) \div(y-3) \]

To divide \( \left(2 y^{2}+4\right) \div(y-3) \), we first need to simplify the numerator. We can factor out the common factor in the numerator: \[ 2y^2 + 4 = 2(y^2 + 2) \] Now we can rewrite the expression as follows: \[ \frac{2(y^2 + 2)}{y - 3} \] Since \(y^2 + 2\) does not factor out any further related to \(y - 3\), we will write the division in a simpler form: \[ 2 \cdot \frac{y^2 + 2}{y - 3} \] Next, for performing polynomial long division on \(y^2 + 2\) divided by \(y - 3\): 1. Divide the leading term: \(y^2 \div y = y\) 2. Multiply \(y\) by \(y - 3\): \(y(y - 3) = y^2 - 3y\) 3. Subtract: \(y^2 + 2 - (y^2 - 3y) = 3y + 2\) 4. Now divide \(3y + 2\) by \(y - 3\): \(3y \div y = 3\) 5. Multiply: \(3(y - 3) = 3y - 9\) 6. Subtract again: \(3y + 2 - (3y - 9) = 11\) Putting it all together, we have: \[ \frac{2(y^2 + 2)}{y - 3} = 2\left(y + 3 + \frac{11}{y - 3}\right) \] Thus, the final result is: \[ 2y + 6 + \frac{22}{y - 3} \] So, the answer to your division is: \[ 2y + 6 + \frac{22}{y - 3} \]