Divide the polynomials 4x^4+4x−10

To divide the polynomial \(4x^4 + 4x - 10\) by \(2x^2 - 3\), we will use polynomial long division.

1. **Setup**: We place \(4x^4 + 4x - 10\) under the long division symbol and \(2x^2 - 3\) outside.

2. **First Division**:
- Divide the leading term \(4x^4\) by \(2x^2\) to get \(2x^2\).
- Multiply \(2x^2\) by \(2x^2 - 3\):
\[
2x^2 \cdot (2x^2 - 3) = 4x^4 - 6x^2.
\]
- Subtract this from \(4x^4 + 4x - 10\):
\[
(4x^4 + 4x - 10) - (4x^4 - 6x^2) = 6x^2 + 4x - 10.
\]

3. **Second Division**:
- Now divide the leading term \(6x^2\) by \(2x^2\) to get \(3\).
- Multiply \(3\) by \(2x^2 - 3\):
\[
3 \cdot (2x^2 - 3) = 6x^2 - 9.
\]
- Subtract this from \(6x^2 + 4x - 10\):
\[
(6x^2 + 4x - 10) - (6x^2 - 9) = 4x - 1.
\]

Now we can't divide \(4x - 1\) by \(2x^2 - 3\) because the degree of \(4x - 1\) is lower than the degree of \(2x^2 - 3\). Thus, we reach the final step.

The quotient from the division is \(2x^2 + 3\), and the remainder is \(4x - 1\).

In summary:

\[
\text{Quotient} = 2x^2 + 3, \quad \text{Remainder} = 4x - 1.
\]