1. Problem: Find the values of $a$ and $b$ if $f(x) = ax^3 + bx^2 + 12$ can be divided exactly by both $(x + 1)$ and $(x - 2)$. 2. Since $(x + 1)$ and $(x - 2)$ are factors, $f(-1) = 0$ and $f(2) = 0$. 3. Use $f(-1) = a(-1)^3 + b(-1)^2 + 12 = -a + b + 12 = 0$. 4. Use $f(2) = a(2)^3 + b(2)^2 + 12 = 8a + 4b + 12 = 0$. 5. Solve the system: $-a + b + 12 = 0$ $8a + 4b + 12 = 0$. 6. From the first, $b = a - 12$. Substitute into second: $8a + 4(a - 12) + 12 = 0$ $8a + 4a - 48 + 12 = 0$ $12a - 36 = 0$ $12a = 36$ $a = 3$. 7. Then $b = 3 - 12 = -9$. 8. Hence, $a=3$, $b=-9$. 9. Now solve $f(x) = 0$: $f(x) = 3x^3 - 9x^2 + 12 = 0$. 10. Factor out 3: $3(x^3 - 3x^2 + 4) = 0$ $x^3 - 3x^2 + 4 = 0$. 11. Try rational roots: test $x=1$: $1 - 3 + 4 = 2 eq 0$, $x=2$: $8 - 12 + 4 = 0$ root found. 12. Divide by $(x - 2)$: Using synthetic division: coefficients 1, -3, 0, 4 Bring down 1, multiply by 2: 2, add to -3: -1, multiply by 2: -2, add to 0: -2, multiply by 2: -4, add to 4: 0 Quotient: $x^2 - x - 2$. 13. Factor quadratic: $(x - 2)(x + 1)$. 14. So $f(x) = 3(x - 2)^2 (x + 1) = 0$. 15. Solutions: $x = 2$ (double root), $x = -1$.