To perform long division on the polynomial \(6x^3 - 11x^2 - 26x + 156\) by \(3x + 5\), we will follow these steps:
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Set up the long division: Write \(6x^3 - 11x^2 - 26x + 156\) under the long division symbol and \(3x + 5\) outside the division symbol.
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Divide the leading term: Divide the leading term of the dividend \(6x^3\) by the leading term of the divisor \(3x\): \[ \frac{6x^3}{3x} = 2x^2 \] This is the first term of the quotient.
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Multiply and subtract: Multiply \(2x^2\) by the entire divisor \(3x + 5\): \[ 2x^2(3x + 5) = 6x^3 + 10x^2 \] Now, subtract this from the original polynomial: \[ (6x^3 - 11x^2 - 26x + 156) - (6x^3 + 10x^2) = -21x^2 - 26x + 156 \]
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Repeat the process: Now divide the leading term of the new polynomial \(-21x^2\) by the leading term of the divisor \(3x\): \[ \frac{-21x^2}{3x} = -7x \] This is the second term of the quotient.
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Multiply and subtract again: Multiply \(-7x\) by the entire divisor: \[ -7x(3x + 5) = -21x^2 - 35x \] Now perform the subtraction: \[ (-21x^2 - 26x + 156) - (-21x^2 - 35x) = 9x + 156 \]
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Repeat the process once more: Divide the leading term \(9x\) by the leading term \(3x\): \[ \frac{9x}{3x} = 3 \] This is the third term of the quotient.
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Multiply and subtract again: Multiply \(3\) by the divisor: \[ 3(3x + 5) = 9x + 15 \] Perform the subtraction: \[ (9x + 156) - (9x + 15) = 141 \]
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Conclusion: Since \(141\) (the remainder) has a lower degree than the divisor \(3x + 5\), we stop here.
The final result of the division is:
\[ 2x^2 - 7x + 3 \quad \text{with a remainder of } 141. \]
Thus, we can express the result as: \[ \frac{6x^3 - 11x^2 - 26x + 156}{3x + 5} = 2x^2 - 7x + 3 + \frac{141}{3x + 5}. \]
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