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Answer:To solve the division \((3x^4 - x^3 - x - 2) \div (3x^2 + x + 1)\) using long division, follow these steps:1. **Divide the first term of the dividend by the first term of the divisor:**   \[   \frac{3x^4}{3x^2} = x^2   \]   Write \(x^2\) as the first term of the quotient.2. **Multiply the entire divisor by \(x^2\):**   \[   (3x^2 + x + 1) \cdot x^2 = 3x^4 + x^3 + x^2   \]3. **Subtract this result from the original polynomial:**   \[   (3x^4 - x^3 - x - 2) - (3x^4 + x^3 + x^2) = -2x^3 - x^2 - x - 2   \]4. **Divide the new first term of the polynomial by the first term of the divisor:**   \[   \frac{-2x^3}{3x^2} = -\frac{2}{3}x   \]   Write \(-\frac{2}{3}x\) as the next term in the quotient.5. **Multiply the entire divisor by \(-\frac{2}{3}x\):**   \[   (3x^2 + x + 1) \cdot \left(-\frac{2}{3}x\right) = -2x^3 - \frac{2}{3}x^2 - \frac{2}{3}x   \]6. **Subtract this result from the polynomial:**   \[   (-2x^3 - x^2 - x - 2) - \left(-2x^3 - \frac{2}{3}x^2 - \frac{2}{3}x\right) = -\frac{1}{3}x^2 - \frac{1}{3}x - 2   \]7. **Divide the new first term of the polynomial by the first term of the divisor:**   \[   \frac{-\frac{1}{3}x^2}{3x^2} = -\frac{1}{9}   \]   Write \(-\frac{1}{9}\) as the next term in the quotient.8. **Multiply the entire divisor by \(-\frac{1}{9}\):**   \[   (3x^2 + x + 1) \cdot \left(-\frac{1}{9}\right) = -\frac{1}{3}x^2 - \frac{1}{9}x - \frac{1}{9}   \]9. **Subtract this result from the polynomial:**   \[   \left(-\frac{1}{3}x^2 - \frac{1}{3}x - 2\right) - \left(-\frac{1}{3}x^2 - \frac{1}{9}x - \frac{1}{9}\right) = -\frac{8}{9}x - \frac{17}{9}   \]So, the quotient of the division \((3x^4 - x^3 - x - 2) \div (3x^2 + x + 1)\) is:\[x^2 - \frac{2}{3}x - \frac{1}{9}\]And the remainder is:\[-\frac{8}{9}x - \frac{17}{9}\]Step-by-step explanation:

Answered by jumongverano | 2024-09-03