1. The problem is to find the zeros of the function $f(x) = x^3 + 3x^2 - 4$ given that $(x-1)$ is a factor, and then graph the function. 2. Since $(x-1)$ is a factor, we can perform synthetic division to divide $f(x)$ by $(x-1)$: $$\begin{array}{r|rrrr} 1 & 1 & 3 & 0 & -4 \\ & & 1 & 4 & 4 \\ \hline & 1 & 4 & 4 & 0 \\ \end{array}$$ Note: The original polynomial is $x^3 + 3x^2 + 0x - 4$. 3. The quotient is $x^2 + 4x + 4$ and the remainder is 0, confirming $(x-1)$ is a factor. 4. Now, solve $x^2 + 4x + 4 = 0$ to find the other zeros. 5. Factor the quadratic: $$x^2 + 4x + 4 = (x + 2)(x + 2) = (x + 2)^2$$ 6. Set equal to zero: $$0 = (x + 2)^2$$ 7. Solve for $x$: $$x = -2$$ This root has multiplicity 2. 8. Therefore, the zeros of $f(x)$ are: $$x = 1, -2, -2$$ 9. The function can be written as: $$f(x) = (x - 1)(x + 2)^2$$ 10. The graph of $f(x)$ will cross the x-axis at $x=1$ and touch the x-axis at $x=-2$ (because of the repeated root).