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:
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Set up the division: We are dividing \(4x^4 + 0x^3 + 4x - 10\) (note the \(0x^3\) term to maintain alignment) by \(2x^2 - 4\).
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Divide the leading terms: Divide the leading term of the dividend \(4x^4\) by the leading term of the divisor \(2x^2\): \[ \frac{4x^4}{2x^2} = 2x^2 \]
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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: \[ 8x^2 + 4x - 10 \]
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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: \[ 4x + 6 \]
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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 \(2x^2 - 4\):
\[ \text{Quotient} = 2x^2 + 4 \] \[ \text{Remainder} = 4x + 6 \]
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