To divide the polynomial \( x^3 - x^2 + 4 \) by \( x + 2 \) using long division, follow these steps:
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Setup the long division: Write \( x^3 - x^2 + 0x + 4 \) (adding \( 0x \) for the missing term) under the long division symbol, and write \( x + 2 \) outside.
____________ x + 2 | x^3 - x^2 + 0x + 4 -
Divide the first 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 \( x + 2 \) and subtract it from the original polynomial: \[ x^2(x + 2) = x^3 + 2x^2 \] Subtract this from the original polynomial: \[ (x^3 - x^2) - (x^3 + 2x^2) = -x^2 - 2x^2 = -3x^2 \] Now, bring down the next term, which is \( 0x \): \[ -3x^2 + 0x = -3x^2 \]
We write:
x^2 ____________ x + 2 | x^3 - x^2 + 0x + 4 - (x^3 + 2x^2) __________ -3x^2 + 0x -
Repeat with -3x^2: Now, divide the leading term \(-3x^2\) by the leading term of the divisor: \[ \frac{-3x^2}{x} = -3x \]
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Multiply and subtract again: \[ -3x(x + 2) = -3x^2 - 6x \] Subtract this from the current remainder: \[ (-3x^2 + 0x) - (-3x^2 - 6x) = 0x + 6x = 6x \]
Now, we have:
x^2 - 3x ____________ x + 2 | x^3 - x^2 + 0x + 4 - (x^3 + 2x^2) __________ -3x^2 + 0x - (-3x^2 - 6x) __________ 6x -
Repeat with 6x: Now, divide \( 6x \) by the leading term of the divisor: \[ \frac{6x}{x} = 6 \]
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Multiply and subtract one last time: \[ 6(x + 2) = 6x + 12 \] Subtract this from the current polynomial: \[ (6x + 4) - (6x + 12) = 4 - 12 = -8 \]
Putting this all together:
x^2 - 3x + 6 ____________ x + 2 | x^3 - x^2 + 0x + 4 - (x^3 + 2x^2) __________ -3x^2 + 0x - (-3x^2 - 6x) __________ 6x + 4 - (6x + 12) __________ -8 -
Final answer: The quotient is \( x^2 - 3x + 6 \) and the remainder is \( -8 \). Thus, we can express the final result of the division as: \[ \frac{x^3 - x^2 + 4}{x + 2} = x^2 - 3x + 6 - \frac{8}{x + 2} \]
So, the answer is: \[ \boxed{x^2 - 3x + 6 - \frac{8}{x + 2}} \]