To divide \(x^3 + 3x^2 + x + 6\) by \(x - 1\) using long division, follow these steps:
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Setup the Long Division: \[ \begin{array}{r|l} & x^2 + 4x + 5 \ \hline x - 1 & x^3 + 3x^2 + x + 6 \ \end{array} \]
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Divide the leading terms:
- Divide the leading term of the dividend \(x^3\) by the leading term of the divisor \(x\), which gives us \(x^2\).
- Write this \(x^2\) above the long division bar.
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Multiply:
- Multiply \(x^2\) by the divisor \(x - 1\): \[ x^2(x - 1) = x^3 - x^2 \]
- Write this under the dividend and subtract: \[ (x^3 + 3x^2 + x + 6) - (x^3 - x^2) = (3x^2 + x^2) + x + 6 = 4x^2 + x + 6 \]
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Repeat:
- Divide \(4x^2\) by \(x\) to get \(4x\).
- Add \(4x\) above the bar.
- Multiply \(4x\) by \(x - 1\): \[ 4x(x - 1) = 4x^2 - 4x \]
- Subtract: \[ (4x^2 + x + 6) - (4x^2 - 4x) = (x + 4x) + 6 = 5x + 6 \]
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Continue:
- Divide \(5x\) by \(x\) to get \(5\).
- Add \(5\) above the bar.
- Multiply \(5\) by \(x - 1\): \[ 5(x - 1) = 5x - 5 \]
- Subtract: \[ (5x + 6) - (5x - 5) = 6 + 5 = 11 \]
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Final Result:
- We end up with \(11\) as the remainder.
The long division process yields: \[ x^2 + 4x + 5 \quad \text{(quotient)} \] with a remainder of: \[ 11 \]
So, the answer is:
- The quotient is \(x^2 + 4x + 5\) with a remainder of \(11\).
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