To find the additional linear factors of the polynomial \(x^3 - 6x^2 + 11x - 6\) given that \(x - 3\) is a factor, we can perform polynomial long division to divide the polynomial by \(x - 3\).
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Divide \(x^3 - 6x^2 + 11x - 6\) by \(x - 3\):
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Divide \(x^3\) by \(x\) to get \(x^2\).
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Multiply \(x^2\) by \(x - 3\) to get \(x^3 - 3x^2\).
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Subtract: \[ (x^3 - 6x^2 + 11x - 6) - (x^3 - 3x^2) = -3x^2 + 11x - 6 \]
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Now, divide \(-3x^2\) by \(x\) to get \(-3x\).
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Multiply \(-3x\) by \(x - 3\) to get \(-3x^2 + 9x\).
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Subtract: \[ (-3x^2 + 11x - 6) - (-3x^2 + 9x) = 2x - 6 \]
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Now, divide \(2x\) by \(x\) to get \(2\).
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Multiply \(2\) by \(x - 3\) to get \(2x - 6\).
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Subtract: \[ (2x - 6) - (2x - 6) = 0 \]
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The result of the division gives us: \[ x^3 - 6x^2 + 11x - 6 = (x - 3)(x^2 - 3x + 2) \]
Now we factor \(x^2 - 3x + 2\):
- The quadratic \(x^2 - 3x + 2\) can be factored as \((x - 1)(x - 2)\).
Thus, we have: \[ x^3 - 6x^2 + 11x - 6 = (x - 3)(x - 1)(x - 2) \]
The additional linear factors therefore are \((x - 1)(x - 2)\).
So the correct choice is: B. \((x - 2)(x - 1)\).
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