1. **State the problem:** Factor and graph the polynomial $f(x) = (x^3 + x^2)(3x^2 - 2x - 1)$.\n\n2. **Understand the expression:** The polynomial is given as a product of two polynomials: $x^3 + x^2$ and $3x^2 - 2x - 1$.\n\n3. **Factor each part:**\n- Factor $x^3 + x^2$ by taking out the common factor $x^2$: $$x^3 + x^2 = x^2(x + 1)$$\n- Factor $3x^2 - 2x - 1$ using the quadratic factoring method. Find two numbers that multiply to $3 \times (-1) = -3$ and add to $-2$. These numbers are $-3$ and $1$. Rewrite the middle term: $$3x^2 - 3x + x - 1$$\nGroup terms: $$(3x^2 - 3x) + (x - 1) = 3x(x - 1) + 1(x - 1)$$\nFactor out $(x - 1)$: $$(3x + 1)(x - 1)$$\n\n4. **Combine all factors:**\n$$f(x) = x^2(x + 1)(3x + 1)(x - 1)$$\n\n5. **Summary of factors:** The fully factored form is $$f(x) = x^2 (x + 1)(3x + 1)(x - 1)$$\n\n6. **Graphing notes:** The roots (zeros) of the polynomial are at $x = 0$ (with multiplicity 2), $x = -1$, $x = -\frac{1}{3}$, and $x = 1$. The multiplicity 2 at $x=0$ means the graph touches the x-axis and turns around there.\n\n7. **Final answer:**\n$$f(x) = x^2 (x + 1)(3x + 1)(x - 1)$$