To perform long division on the polynomials \( \frac{x^3 - x^2 + 4}{x + 2} \), we follow these steps:
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Setup the long division: Write \( x^3 - x^2 + 4 \) under the long division symbol and \( x + 2 \) outside.
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Divide the leading terms: Divide the leading term of the dividend \( x^3 \) by the leading term of the divisor \( x \): \[ \frac{x^3}{x} = x^2 \]
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Multiply and subtract: Multiply \( x^2 \) by the entire divisor \( x + 2 \): \[ x^2(x + 2) = x^3 + 2x^2 \] Now subtract this result from the original polynomial: \[ (x^3 - x^2 + 4) - (x^3 + 2x^2) = -x^2 - 2x^2 + 4 = -3x^2 + 4 \]
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Bring down the next term (if any): In this case, we already have \( -3x^2 + 4 \).
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Repeat the process:
- Now divide the leading term \( -3x^2 \) by \( x \): \[ \frac{-3x^2}{x} = -3x \]
- Multiply \( -3x \) by the entire divisor \( x + 2 \): \[ -3x(x + 2) = -3x^2 - 6x \]
- Subtract this from \( -3x^2 + 4 \): \[ (-3x^2 + 4) - (-3x^2 - 6x) = 4 + 6x = 6x + 4 \]
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Repeat again:
- Now divide the leading term \( 6x \) by \( x \): \[ \frac{6x}{x} = 6 \]
- Multiply \( 6 \) by the entire divisor \( x + 2 \): \[ 6(x + 2) = 6x + 12 \]
- Subtract: \[ (6x + 4) - (6x + 12) = 4 - 12 = -8 \]
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Final result: We have completed the division. The result is: \[ x^2 - 3x + 6 \quad \text{with a remainder of } -8 \] Therefore, \[ \frac{x^3 - x^2 + 4}{x + 2} = x^2 - 3x + 6 - \frac{8}{x + 2} \]
So, the answer is: \[ x^2 - 3x + 6 - \frac{8}{x + 2} \]
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