1. **Problem statement:** Given the polynomial function $f(x) = 2x^3 - 3x^2 - 16x + 24$, we want to find intervals where the function has at least one real zero. 2. **Key idea:** By the Intermediate Value Theorem, if $f(x)$ changes sign between two points, there is at least one real root in that interval. 3. **Step 1: Evaluate $f(x)$ at some points to find sign changes.** - $f(0) = 2(0)^3 - 3(0)^2 - 16(0) + 24 = 24$ (positive) - $f(1) = 2(1)^3 - 3(1)^2 - 16(1) + 24 = 2 - 3 - 16 + 24 = 7$ (positive) - $f(2) = 2(8) - 3(4) - 16(2) + 24 = 16 - 12 - 32 + 24 = -4$ (negative) Since $f(1) > 0$ and $f(2) < 0$, there is at least one root in the interval $(1, 2)$. 4. **Step 2: Check other intervals for sign changes.** - $f(-1) = 2(-1)^3 - 3(-1)^2 - 16(-1) + 24 = -2 - 3 + 16 + 24 = 35$ (positive) - $f(-2) = 2(-8) - 3(4) - 16(-2) + 24 = -16 - 12 + 32 + 24 = 28$ (positive) No sign change between $-2$ and $-1$. - $f(3) = 2(27) - 3(9) - 16(3) + 24 = 54 - 27 - 48 + 24 = 3$ (positive) No sign change between $2$ and $3$. 5. **Step 3: Check $f(-3)$** - $f(-3) = 2(-27) - 3(9) - 16(-3) + 24 = -54 - 27 + 48 + 24 = -9$ (negative) Since $f(-3) < 0$ and $f(-2) > 0$, there is at least one root in $(-3, -2)$. 6. **Summary:** The polynomial has at least one real zero in the intervals $(-3, -2)$ and $(1, 2)$. 7. **Additional note:** Since it is a cubic polynomial, it has exactly 3 roots (real or complex). We found two intervals with real roots; the third root may be real or complex. **Final answer:** The polynomial $f(x)$ has at least one real zero in the intervals $$(-3, -2) \text{ and } (1, 2).$$