To find the quotient of the polynomial \(3x^4 - 13x^3 + 7x^2 - x - 9\) divided by \(x^2 - 4x + 1\), we can use polynomial long division.
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Setup the long division:
- Divide \(3x^4\) by \(x^2\) to get \(3x^2\).
- Multiply \(3x^2\) by \(x^2 - 4x + 1\): \[ 3x^2(x^2 - 4x + 1) = 3x^4 - 12x^3 + 3x^2 \]
- Subtract this from the original polynomial: \[ (3x^4 - 13x^3 + 7x^2 - x - 9) - (3x^4 - 12x^3 + 3x^2) = -x^3 + 4x^2 - x - 9 \]
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Repeat the process:
- Divide \(-x^3\) by \(x^2\) to get \(-x\).
- Multiply \(-x\) by \(x^2 - 4x + 1\): \[ -x(x^2 - 4x + 1) = -x^3 + 4x^2 - x \]
- Subtract: \[ (-x^3 + 4x^2 - x - 9) - (-x^3 + 4x^2 - x) = -9 \]
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Final step:
- There are no more terms to bring down, and we have reached the end of the division process. The division leaves us with a remainder of \(-9\).
Putting it all together, we have: \[ \text{Quotient} = 3x^2 - x \quad \text{and} \quad \text{Remainder} = -9 \]
Thus, the result of the division is: \[ 3x^2 - x + \frac{-9}{x^2 - 4x + 1} \]
Final Answer: The quotient of \((3x^4 - 13x^3 + 7x^2 - x - 9)÷(x^2 - 4x + 1)\) is \(3x^2 - x\) with a remainder of \(-9\) (or \(3x^2 - x - \frac{9}{x^2 - 4x + 1}\)).