1. **State the problem:** Find the real zeros of the function $$f(x) = 6(x^2 - 9)(x^2 + 12x + 27)^2$$ and determine their multiplicities. 2. **Recall the formula and rules:** To find real zeros, set $$f(x) = 0$$ and solve for $$x$$. The multiplicity of a zero is the exponent of the factor in the factored form. 3. **Set the function equal to zero:** $$6(x^2 - 9)(x^2 + 12x + 27)^2 = 0$$ Since 6 is a constant and never zero, we focus on: $$ (x^2 - 9)(x^2 + 12x + 27)^2 = 0 $$ 4. **Solve each factor:** - For $$x^2 - 9 = 0$$: $$x^2 = 9$$ $$x = \pm 3$$ - For $$x^2 + 12x + 27 = 0$$, use the quadratic formula: $$x = \frac{-12 \pm \sqrt{12^2 - 4 \cdot 1 \cdot 27}}{2 \cdot 1} = \frac{-12 \pm \sqrt{144 - 108}}{2} = \frac{-12 \pm \sqrt{36}}{2}$$ $$x = \frac{-12 \pm 6}{2}$$ So, $$x = \frac{-12 + 6}{2} = \frac{-6}{2} = -3$$ $$x = \frac{-12 - 6}{2} = \frac{-18}{2} = -9$$ 5. **List all real zeros:** $$x = -9, -3, 3$$ 6. **Determine multiplicities:** - The factor $$x^2 - 9$$ can be factored as $$(x - 3)(x + 3)$$, each with multiplicity 1. - The factor $$(x^2 + 12x + 27)^2$$ has zeros at $$x = -3$$ and $$x = -9$$, each with multiplicity 2 because of the square. 7. **Combine multiplicities for repeated zeros:** - $$x = 3$$ comes from $$(x - 3)$$ with multiplicity 1. - $$x = -3$$ appears in both factors: once in $$(x + 3)$$ with multiplicity 1 and once in $$(x^2 + 12x + 27)^2$$ with multiplicity 2, so total multiplicity is $$1 + 2 = 3$$. - $$x = -9$$ comes from $$(x^2 + 12x + 27)^2$$ with multiplicity 2. **Final answer:** - Real zeros: $$-9, -3, 3$$ - Multiplicities: $$-9$$ has multiplicity 2, $$-3$$ has multiplicity 3, $$3$$ has multiplicity 1.