1. **State the problem:** We are given that the polynomial $f[x] = 6 - x - x^2$ is a factor of $g[x] = ax^3 + 5x^2 + bx - 18$. We need to find the constants $a$ and $b$, and then find the remainder when $g[x]$ is divided by $x + 2$. 2. **Rewrite $f[x]$ in standard form:** $$f[x] = -x^2 - x + 6 = 0$$ 3. **Since $f[x]$ is a factor of $g[x]$, the roots of $f[x]$ are also roots of $g[x]$.** 4. **Find roots of $f[x]$:** Solve $$-x^2 - x + 6 = 0$$ or equivalently $$x^2 + x - 6 = 0$$ 5. **Factor or use quadratic formula:** $$x^2 + x - 6 = (x + 3)(x - 2) = 0$$ So roots are $$x = -3$$ and $$x = 2$$. 6. **Since $g[x]$ has these roots, substitute $x = -3$ and $x = 2$ into $g[x]$ and set equal to zero:** For $x = -3$: $$a(-3)^3 + 5(-3)^2 + b(-3) - 18 = 0$$ $$-27a + 45 - 3b - 18 = 0$$ $$-27a - 3b + 27 = 0$$ For $x = 2$: $$a(2)^3 + 5(2)^2 + b(2) - 18 = 0$$ $$8a + 20 + 2b - 18 = 0$$ $$8a + 2b + 2 = 0$$ 7. **Write the system of equations:** $$-27a - 3b + 27 = 0$$ $$8a + 2b + 2 = 0$$ 8. **Solve the system:** Multiply second equation by 3 to align $b$ terms: $$24a + 6b + 6 = 0$$ Add to first equation: $$(-27a - 3b + 27) + (24a + 6b + 6) = 0 + 0$$ $$-3a + 3b + 33 = 0$$ But this seems inconsistent; better to solve by substitution. From second equation: $$2b = -8a - 2 \\ b = \frac{-8a - 2}{2} = -4a - 1$$ Substitute into first equation: $$-27a - 3(-4a - 1) + 27 = 0$$ $$-27a + 12a + 3 + 27 = 0$$ $$-15a + 30 = 0$$ $$-15a = -30$$ $$a = 2$$ Then, $$b = -4(2) - 1 = -8 - 1 = -9$$ 9. **Find the remainder when $g[x]$ is divided by $x + 2$:** By the Remainder Theorem, remainder is $g[-2]$: $$g[-2] = a(-2)^3 + 5(-2)^2 + b(-2) - 18$$ $$= 2(-8) + 5(4) + (-9)(-2) - 18$$ $$= -16 + 20 + 18 - 18$$ $$= 4 + 18 - 18 = 4$$ **Final answers:** $$a = 2, \quad b = -9, \quad \text{remainder} = 4$$