To perform long division of the polynomial \(3x^3 + x^2 - 10x - 9\) by \(x + 1\):
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Set up the long division: \[ \begin{array}{r|l} & 3x^2 - 2x - 8 \ \hline x+1 & 3x^3 + x^2 - 10x - 9 \ \end{array} \]
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Divide the first term: Divide \(3x^3\) by \(x\) to get \(3x^2\).
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Multiply and subtract: Multiply \(3x^2\) by \(x + 1\): \[ 3x^2(x + 1) = 3x^3 + 3x^2 \] Now subtract this from the original polynomial: \[ (3x^3 + x^2 - 10x - 9) - (3x^3 + 3x^2) = (x^2 - 3x^2) - 10x - 9 = -2x^2 - 10x - 9 \]
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Repeat the process: Now divide \(-2x^2\) by \(x\) to get \(-2x\).
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Multiply and subtract again: Multiply \(-2x\) by \(x + 1\): \[ -2x(x + 1) = -2x^2 - 2x \] Subtract: \[ (-2x^2 - 10x - 9) - (-2x^2 - 2x) = (-10x + 2x) - 9 = -8x - 9 \]
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Repeat again: Divide \(-8x\) by \(x\) to get \(-8\).
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Multiply and subtract: Multiply \(-8\) by \(x + 1\): \[ -8(x + 1) = -8x - 8 \] Subtract: \[ (-8x - 9) - (-8x - 8) = -9 + 8 = -1 \]
Now we have completed the long division and end up with:
- Quotient: \(3x^2 - 2x - 8\)
- Remainder: \(-1\)
Thus, we can express the result as: \[ \frac{3x^3 + x^2 - 10x - 9}{x + 1} = 3x^2 - 2x - 8 + \frac{-1}{x + 1} \]
So, the final result is: \[ q(x) = 3x^2 - 2x - 8,\quad r(x) = -1,\quad b(x) = x + 1 \]
In summary: \[ 3x^3 + x^2 - 10x - 9 = (x + 1)(3x^2 - 2x - 8) - 1 \]
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