1. **Problem statement:** Given that $x=1$ is a root of the cubic equation $$3x^3 - 8x^2 + 3x + 2 = 0,$$ find the remaining roots using polynomial long division. 2. **Formula and rule:** Since $x=1$ is a root, $(x-1)$ is a factor of the polynomial. We can divide the cubic polynomial by $(x-1)$ to get a quadratic polynomial. Then, solve the quadratic to find the other roots. 3. **Perform long division:** Divide $$3x^3 - 8x^2 + 3x + 2$$ by $$x - 1$$. - Divide the leading term: $3x^3 \div x = 3x^2$. - Multiply divisor by $3x^2$: $(x-1)(3x^2) = 3x^3 - 3x^2$. - Subtract: $(3x^3 - 8x^2) - (3x^3 - 3x^2) = -5x^2$. - Bring down $+3x$. - Divide leading term: $-5x^2 \div x = -5x$. - Multiply divisor by $-5x$: $(x-1)(-5x) = -5x^2 + 5x$. - Subtract: $(-5x^2 + 3x) - (-5x^2 + 5x) = -2x$. - Bring down $+2$. - Divide leading term: $-2x \div x = -2$. - Multiply divisor by $-2$: $(x-1)(-2) = -2x + 2$. - Subtract: $(-2x + 2) - (-2x + 2) = 0$ remainder. So, the quotient is $$3x^2 - 5x - 2$$. 4. **Solve the quadratic:** $$3x^2 - 5x - 2 = 0$$. Use the quadratic formula: $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ where $a=3$, $b=-5$, $c=-2$. Calculate discriminant: $$\Delta = (-5)^2 - 4(3)(-2) = 25 + 24 = 49$$. Calculate roots: $$x = \frac{5 \pm \sqrt{49}}{2 \times 3} = \frac{5 \pm 7}{6}$$. - For $+$: $$x = \frac{5 + 7}{6} = \frac{12}{6} = 2$$. - For $-$: $$x = \frac{5 - 7}{6} = \frac{-2}{6} = -\frac{1}{3}$$. 5. **Final answer:** The roots of the equation are $$x=1, x=2, \text{ and } x=-\frac{1}{3}$$.