To divide the polynomial \(4x^4 + 4x - 10\) by \(2x^2 - 3\), we will use polynomial long division.
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Setup: We place \(4x^4 + 4x - 10\) under the long division symbol and \(2x^2 - 3\) outside.
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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 \(4x^4 + 4x - 10\): \[ (4x^4 + 4x - 10) - (4x^4 - 6x^2) = 6x^2 + 4x - 10. \]
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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:
\[ \text{Quotient} = 2x^2 + 3, \quad \text{Remainder} = 4x - 1. \]
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