1. **State the problem:** We have a polynomial $$f(x) = 2x^3 + ax^2 + bx - 6$$ and we know the remainders when divided by $$x-1$$ and $$x+4$$ are 12 and -18 respectively. We need to find constants $$a$$ and $$b$$, factorize $$f(x)$$ completely, and find its zeroes. 2. **Use the Remainder Theorem:** The remainder when $$f(x)$$ is divided by $$x-c$$ is $$f(c)$$. 3. **Apply for $$x-1$$:** $$f(1) = 2(1)^3 + a(1)^2 + b(1) - 6 = 2 + a + b - 6 = a + b - 4$$ Given remainder is 12, so: $$a + b - 4 = 12$$ $$a + b = 16$$ 4. **Apply for $$x+4$$:** $$f(-4) = 2(-4)^3 + a(-4)^2 + b(-4) - 6 = 2(-64) + 16a - 4b - 6 = -128 + 16a - 4b - 6 = 16a - 4b - 134$$ Given remainder is -18, so: $$16a - 4b - 134 = -18$$ $$16a - 4b = 116$$ Divide both sides by 4: $$4a - b = 29$$ 5. **Solve the system:** From step 3: $$a + b = 16$$ From step 4: $$4a - b = 29$$ Add the two equations: $$(a + b) + (4a - b) = 16 + 29$$ $$5a = 45$$ $$a = 9$$ Substitute $$a=9$$ into $$a + b = 16$$: $$9 + b = 16$$ $$b = 7$$ 6. **Rewrite $$f(x)$$:** $$f(x) = 2x^3 + 9x^2 + 7x - 6$$ 7. **Factorize $$f(x)$$:** Try rational roots using factors of constant term (-6) over factors of leading coefficient (2): possible roots are $$\pm1, \pm2, \pm3, \pm6, \pm\frac{1}{2}, \pm\frac{3}{2}$$. Test $$x=1$$: $$f(1) = 2 + 9 + 7 - 6 = 12 \neq 0$$ Test $$x=-1$$: $$f(-1) = -2 + 9 - 7 - 6 = -6 \neq 0$$ Test $$x=\frac{1}{2}$$: $$f(\frac{1}{2}) = 2(\frac{1}{2})^3 + 9(\frac{1}{2})^2 + 7(\frac{1}{2}) - 6 = 2(\frac{1}{8}) + 9(\frac{1}{4}) + \frac{7}{2} - 6 = \frac{1}{4} + \frac{9}{4} + \frac{7}{2} - 6 = \frac{10}{4} + \frac{7}{2} - 6 = \frac{5}{2} + \frac{7}{2} - 6 = 6 - 6 = 0$$ So $$x=\frac{1}{2}$$ is a root. 8. **Divide $$f(x)$$ by $$x - \frac{1}{2}$$:** Use synthetic division or polynomial division: $$f(x) = (x - \frac{1}{2})(2x^2 + 10x + 12)$$ 9. **Factor quadratic:** $$2x^2 + 10x + 12 = 2(x^2 + 5x + 6) = 2(x + 2)(x + 3)$$ 10. **Complete factorization:** $$f(x) = 2(x - \frac{1}{2})(x + 2)(x + 3) = (2x - 1)(x + 2)(x + 3)$$ 11. **Zeroes of $$f(x)$$:** Set each factor to zero: $$2x - 1 = 0 \Rightarrow x = \frac{1}{2}$$ $$x + 2 = 0 \Rightarrow x = -2$$ $$x + 3 = 0 \Rightarrow x = -3$$ **Final answers:** $$a = 9, b = 7$$ $$f(x) = (2x - 1)(x + 2)(x + 3)$$ Zeroes: $$x = \frac{1}{2}, -2, -3$$