To find the quotient of the polynomial division \((24x^4 + 18x^3 - 9x^2 + 3) ÷ (4x^2 - 3x + 1)\), we can use polynomial long division.
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Set up the division: We want to divide \(24x^4 + 18x^3 - 9x^2 + 0x + 3\) by \(4x^2 - 3x + 1\).
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Divide the leading terms:
- Divide the leading term of the dividend (\(24x^4\)) by the leading term of the divisor (\(4x^2\)): \[ 24x^4 ÷ 4x^2 = 6x^2 \]
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Multiply and subtract:
- Multiply \(6x^2\) by the entire divisor: \[ 6x^2(4x^2 - 3x + 1) = 24x^4 - 18x^3 + 6x^2 \]
- Subtract this from the original polynomial: \[ (24x^4 + 18x^3 - 9x^2 + 0x + 3) - (24x^4 - 18x^3 + 6x^2) = 0 + 36x^3 - 15x^2 + 3 \]
- This simplifies to \(36x^3 - 15x^2 + 3\).
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Repeat the process:
- Divide the leading term of the new polynomial (\(36x^3\)) by the leading term of the divisor (\(4x^2\)): \[ 36x^3 ÷ 4x^2 = 9x \]
- Multiply \(9x\) by the entire divisor: \[ 9x(4x^2 - 3x + 1) = 36x^3 - 27x^2 + 9x \]
- Subtract from the current polynomial: \[ (36x^3 - 15x^2 + 3) - (36x^3 - 27x^2 + 9x) = 0 + 12x^2 - 9x + 3 \]
- This simplifies to \(12x^2 - 9x + 3\).
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Repeat again:
- Divide \(12x^2\) by \(4x^2\): \[ 12x^2 ÷ 4x^2 = 3 \]
- Multiply \(3\) by the divisor: \[ 3(4x^2 - 3x + 1) = 12x^2 - 9x + 3 \]
- Subtract: \[ (12x^2 - 9x + 3) - (12x^2 - 9x + 3) = 0 \]
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Final result: At this point, there is no remainder.
Thus, the quotient of \((24x^4 + 18x^3 - 9x^2 + 3) ÷ (4x^2 - 3x + 1)\) is:
\[ \boxed{6x^2 + 9x + 3} \]