Question

divide the polynomials 4x^4+4x-10by2x^2-4 to determine the quotient and remainder

Answer 1

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

Here’s how the process works step-by-step:

1. **Set up the division**: We are dividing \(4x^4 + 0x^3 + 4x - 10\) (note the \(0x^3\) term to maintain alignment) by \(2x^2 - 4\).

2. **Divide the leading terms**: Divide the leading term of the dividend \(4x^4\) by the leading term of the divisor \(2x^2\):
\[
\frac{4x^4}{2x^2} = 2x^2
\]

3. **Multiply and subtract**: Multiply \(2x^2\) by the entire divisor:
\[
2x^2(2x^2 - 4) = 4x^4 - 8x^2
\]
Now subtract this from the original polynomial:
\[
(4x^4 + 0x^3 + 4x - 10) - (4x^4 - 8x^2) = 0x^4 + 0x^3 + 8x^2 + 4x - 10
\]
Simplifying gives:
\[
8x^2 + 4x - 10
\]

4. **Repeat the process**: Now divide the leading term \(8x^2\) by \(2x^2\):
\[
\frac{8x^2}{2x^2} = 4
\]
Multiply \(4\) by the entire divisor:
\[
4(2x^2 - 4) = 8x^2 - 16
\]
Now subtract this from the current polynomial:
\[
(8x^2 + 4x - 10) - (8x^2 - 16) = 0x^2 + 4x + 6
\]
Simplifying gives:
\[
4x + 6
\]

5. **Determine the remainder**: The degree of the remainder \(4x + 6\) is less than the degree of the divisor \(2x^2 - 4\), so we stop here.

The final result consists of:
- **Quotient**: \(2x^2 + 4\)
- **Remainder**: \(4x + 6\)

Therefore, when dividing \(4x^4 + 4x - 10\) by \(2x^2 - 4\):

\[
\text{Quotient} = 2x^2 + 4
\]
\[
\text{Remainder} = 4x + 6
\]