1. State the problem: Find two possible degree 4 polynomials with negative leading coefficient, exactly 3 real zeros, one zero of order 2, and positive between two zeros. 2. Given zeros: $x=-2$ (order 1), $x=0$ (order 2), $x=2$ (order 1). 3. General form: $$f(x) = a(x+2)^1 (x)^2 (x-2)^1$$ where $a<0$. 4. Set leading coefficient $a = -1$ to satisfy negative leading coefficient. 5. First polynomial: $$f_1(x) = -(x+2)(x)^2(x-2)$$ 6. Expand $f_1(x)$ partially: $$f_1(x) = -(x+2)(x^2)(x-2)$$ 7. Multiply $(x+2)(x-2) = x^2 - 4$: $$f_1(x) = -x^2(x^2 - 4)$$ 8. Distribute: $$f_1(x) = -x^4 + 4x^2$$ 9. Second polynomial: multiply by a positive constant $c$ to get a different polynomial, e.g. $c=2$: $$f_2(x) = -2(x+2)(x)^2(x-2)$$ 10. Expanded form: $$f_2(x) = -2x^2(x^2 - 4) = -2x^4 + 8x^2$$ Final answers: $$f_1(x) = -(x+2)(x)^2(x-2)$$ $$f_2(x) = -2(x+2)(x)^2(x-2)$$