1. **State the problem:** Factor the polynomial $$P(x) = 2x^3 + 13x^2 + 3x - 18$$ given that $$(x + 6)$$ is a factor. 2. **Use the factor theorem:** Since $$(x + 6)$$ is a factor, $$P(-6) = 0$$. 3. **Perform polynomial division:** Divide $$P(x)$$ by $$(x + 6)$$ to find the quotient polynomial. Set up the division: $$\frac{2x^3 + 13x^2 + 3x - 18}{x + 6}$$ 4. **Divide the leading terms:** $$\frac{2x^3}{x} = 2x^2$$. Multiply back: $$2x^2(x + 6) = 2x^3 + 12x^2$$. Subtract: $$\begin{aligned} (2x^3 + 13x^2) - (2x^3 + 12x^2) &= 13x^2 - 12x^2 = x^2 \end{aligned}$$ Bring down the next term: $$+ 3x$$. 5. **Next division:** $$\frac{x^2}{x} = x$$. Multiply back: $$x(x + 6) = x^2 + 6x$$. Subtract: $$\begin{aligned} (x^2 + 3x) - (x^2 + 6x) &= 3x - 6x = -3x \end{aligned}$$ Bring down the next term: $$- 18$$. 6. **Next division:** $$\frac{-3x}{x} = -3$$. Multiply back: $$-3(x + 6) = -3x - 18$$. Subtract: $$\begin{aligned} (-3x - 18) - (-3x - 18) &= 0 \end{aligned}$$ 7. **Quotient polynomial:** $$2x^2 + x - 3$$. 8. **Factor the quadratic:** Find factors of $$2x^2 + x - 3$$. Look for two numbers that multiply to $$2 \times (-3) = -6$$ and add to $$1$$. These numbers are $$3$$ and $$-2$$. Rewrite: $$2x^2 + 3x - 2x - 3$$ Group: $$(2x^2 + 3x) - (2x + 3)$$ Factor each group: $$x(2x + 3) - 1(2x + 3)$$ Factor out common binomial: $$(2x + 3)(x - 1)$$ 9. **Final factorization:** $$P(x) = (x + 6)(2x + 3)(x - 1)$$ This is the fully factored form of the polynomial.