1. **State the problem:** We are given the polynomial function $$P(x) = -2x^3 + 8x^2 - 8x + 32$$ and asked to find its real root(s), i.e., the value(s) of $$x$$ for which $$P(x) = 0$$. 2. **Set the equation to zero:** $$-2x^3 + 8x^2 - 8x + 32 = 0$$ 3. **Simplify by dividing both sides by -2:** $$\cancel{-2}x^3 + \cancel{-2}(-4)x^2 + \cancel{-2}4x - \cancel{-2}(-16) = \cancel{0}$$ which simplifies to $$x^3 - 4x^2 + 4x - 16 = 0$$ 4. **Try to find rational roots using the Rational Root Theorem:** Possible roots are factors of 16: $$\pm1, \pm2, \pm4, \pm8, \pm16$$. 5. **Test $$x=2$$:** $$2^3 - 4(2)^2 + 4(2) - 16 = 8 - 16 + 8 - 16 = -16 \neq 0$$ 6. **Test $$x=4$$:** $$4^3 - 4(4)^2 + 4(4) - 16 = 64 - 64 + 16 - 16 = 0$$ So, $$x=4$$ is a root. 7. **Divide the cubic by $$x-4$$ to factor:** Using polynomial division or synthetic division: $$x^3 - 4x^2 + 4x - 16 \div (x-4) = x^2 + 0x + 4$$ 8. **Rewrite the factorization:** $$P(x) = (x-4)(x^2 + 4) = 0$$ 9. **Solve the quadratic $$x^2 + 4 = 0$$:** $$x^2 = -4$$ $$x = \pm \sqrt{-4} = \pm 2i$$ (complex roots) 10. **Summary:** - Real root: $$x=4$$ - Complex roots: $$x=2i$$ and $$x=-2i$$ **Final answer:** The function $$P(x)$$ has one real root at $$x=4$$ and two complex roots at $$x=\pm 2i$$.