1. **Problem statement:** Given the polynomial $$f(x) = x^{4} - 2x^{3} - 2x^{2} + a$$ and that it is divisible by $$x^{2} - 4x + 4$$, find the value of $$a$$ and show that $$f(x)$$ is never negative for this value. 2. **Step 1: Understand divisibility condition.** If $$f(x)$$ is divisible by $$x^{2} - 4x + 4$$, then $$x^{2} - 4x + 4$$ is a factor of $$f(x)$$. Note that $$x^{2} - 4x + 4 = (x-2)^2$$. 3. **Step 2: Use factorization condition.** Since $$(x-2)^2$$ divides $$f(x)$$, then $$f(2) = 0$$ and $$f'(2) = 0$$ must hold. 4. **Step 3: Calculate $$f(2)$$.** $$f(2) = 2^{4} - 2 \times 2^{3} - 2 \times 2^{2} + a = 16 - 16 - 8 + a = -8 + a$$ Set $$f(2) = 0$$: $$-8 + a = 0 \implies a = 8$$ 5. **Step 4: Calculate $$f'(x)$$ and evaluate at $$x=2$$.** $$f'(x) = 4x^{3} - 6x^{2} - 4x$$ $$f'(2) = 4 \times 8 - 6 \times 4 - 4 \times 2 = 32 - 24 - 8 = 0$$ This confirms $$(x-2)^2$$ is a factor. 6. **Step 5: Rewrite $$f(x)$$ with $$a=8$$.** $$f(x) = x^{4} - 2x^{3} - 2x^{2} + 8$$ 7. **Step 6: Factor $$f(x)$$ using the known factor $$(x-2)^2$$.** Divide $$f(x)$$ by $$(x-2)^2 = x^{2} - 4x + 4$$: Using polynomial division or synthetic division, quotient is: $$x^{2} + 2x + 2$$ So, $$f(x) = (x-2)^2 (x^{2} + 2x + 2)$$ 8. **Step 7: Show $$f(x)$$ is never negative.** - $$(x-2)^2 \geq 0$$ for all $$x$$ since it is a square. - The quadratic $$x^{2} + 2x + 2$$ has discriminant $$\Delta = 2^{2} - 4 \times 1 \times 2 = 4 - 8 = -4 < 0$$, so it is always positive. Therefore, $$f(x) = (x-2)^2 (x^{2} + 2x + 2) \geq 0$$ for all $$x$$. **Final answers:** (i) $$a = 8$$ (ii) $$f(x)$$ is never negative because it factors into the product of two positive or zero expressions for all real $$x$$.