Divide using long division. State the quotient, \( q(x) \), and the remainder, \( r(x) \). \( \left(4 x^{3}+4 x^{2}-23 x-30\right) \div(2 x-5) \) \( \left(4 x^{3}+4 x^{2}-23 x-30\right) \div(2 x-5)=\square+\frac{\square}{2 x-5} \) (Simplify your answers. Do not factor. Use integers or fractions for any numbers in the expressions.)

To perform long division on \( (4x^3 + 4x^2 - 23x - 30) \div (2x - 5) \): 1. Divide the leading term of the dividend \(4x^3\) by the leading term of the divisor \(2x\), which gives \(2x^2\). 2. Multiply \(2x^2\) by the entire divisor \(2x - 5\) resulting in \(4x^3 - 10x^2\). 3. Subtract this from the original polynomial: \[ (4x^3 + 4x^2) - (4x^3 - 10x^2) = (4x^2 + 10x^2) = 14x^2 \] Bringing down the next term gives \(14x^2 - 23x\). 4. Divide \(14x^2\) by \(2x\) to get \(7x\). 5. Multiply \(7x\) by \(2x - 5\) to get \(14x^2 - 35x\). 6. Subtract \(14x^2 - 35x\) from \(14x^2 - 23x\): \[ (-23x + 35x) = 12x \] Bringing down the next term gives \(12x - 30\). 7. Divide \(12x\) by \(2x\) giving \(6\). 8. Multiply \(6\) by \(2x - 5\) resulting in \(12x - 30\). 9. Subtract \(12x - 30\) from \(12x - 30\): \[ 0 \] Thus, we find that the quotient \(q(x)\) is \(2x^2 + 7x + 6\) and the remainder \(r(x)\) is \(0\). Therefore: \[ \left(4 x^{3}+4 x^{2}-23 x-30\right) \div(2 x-5) = 2x^2 + 7x + 6 + \frac{0}{2x-5} \] In summary, Quotient, \( q(x) = 2x^2 + 7x + 6 \) Remainder, \( r(x) = 0 \)