1. **State the problem:** We need to divide the cubic polynomial $$2x^3 + 2$$ by the binomial $$2x + 2$$ and express the quotient in the form $$Ax^2 + Bx + C$$. 2. **Recall the division formula:** When dividing polynomials, $$\frac{P(x)}{D(x)} = Q(x) + \frac{R(x)}{D(x)}$$ where $$Q(x)$$ is the quotient and $$R(x)$$ is the remainder. Here, since the divisor is degree 1 and dividend degree 3, the quotient will be degree 2. 3. **Set up the division:** $$\frac{2x^3 + 0x^2 + 0x + 2}{2x + 2} = Ax^2 + Bx + C$$ 4. **Multiply both sides by the divisor:** $$2x^3 + 2 = (2x + 2)(Ax^2 + Bx + C)$$ 5. **Expand the right side:** $$= 2x \cdot Ax^2 + 2x \cdot Bx + 2x \cdot C + 2 \cdot Ax^2 + 2 \cdot Bx + 2 \cdot C$$ $$= 2A x^3 + 2B x^2 + 2C x + 2A x^2 + 2B x + 2C$$ 6. **Group like terms:** $$= 2A x^3 + (2B + 2A) x^2 + (2C + 2B) x + 2C$$ 7. **Match coefficients with the left side:** Left side: $$2x^3 + 0x^2 + 0x + 2$$ Right side: $$2A x^3 + (2B + 2A) x^2 + (2C + 2B) x + 2C$$ Equate coefficients: - For $$x^3$$: $$2 = 2A \implies A = 1$$ - For $$x^2$$: $$0 = 2B + 2A = 2B + 2(1) = 2B + 2 \implies 2B = -2 \implies B = -1$$ - For $$x$$: $$0 = 2C + 2B = 2C + 2(-1) = 2C - 2 \implies 2C = 2 \implies C = 1$$ - For constant term: $$2 = 2C = 2(1) = 2$$ (checks out) 8. **Final answer:** $$A = 1, B = -1, C = 1$$ Thus, the quotient is $$x^2 - x + 1$$.