To apply the Intermediate Value Theorem (IVT), we first need to verify that the polynomial function has different signs at $x = -4$ and $x = -3$. The IVT states that if a continuous function $f(x)$ changes sign over an interval $[a, b]$, then there exists at least one real zero in that interval.

Given: $f(x) = x^4 - 3x^3 - 15x^2 + 24x - 100$

We need to calculate $f(-4)$ and $f(-3)$ and check their signs.

Step 1: Calculate $f(-4)$

$f(-4) = (-4)^4 - 3(-4)^3 - 15(-4)^2 + 24(-4) - 100$ $= 256 + 192 - 240 - 96 - 100$ $= 256 + 192 - 240 - 96 - 100 = 12$

So, $f(-4) = 12$.

Step 2: Calculate $f(-3)$

$f(-3) = (-3)^4 - 3(-3)^3 - 15(-3)^2 + 24(-3) - 100$ $= 81 + 81 - 135 - 72 - 100$ $= 81 + 81 - 135 - 72 - 100 = -145$

So, $f(-3) = -145$.

Step 3: Apply the Intermediate Value Theorem

We observe that: $f(-4) = 12 > 0 \quad \text{and} \quad f(-3) = -145 < 0$

Since $f(x)$ is continuous (it’s a polynomial), and the function changes sign between $x = -4$ and $x = -3$, the IVT guarantees that there is at least one real zero in the interval $(-4, -3)$.

Conclusion

By the Intermediate Value Theorem, the polynomial function $f(x) = x^4 - 3x^3 - 15x^2 + 24x - 100$ has a real zero in the interval $(-4, -3)$.

Would you like more details on this method or have any questions?

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Relative Questions:

1. How do we know the Intermediate Value Theorem applies to all polynomials?
2. Can there be more than one real zero in the interval $(-4, -3)$?
3. How does the Intermediate Value Theorem work if the function does not change signs?
4. What happens if both $f(-4)$ and $f(-3)$ are positive?
5. How can we approximate the zero between $-4$ and $-3$ more precisely?

Tip: The Intermediate Value Theorem is often most helpful in confirming the existence of roots, not their exact location—other techniques can pinpoint them more precisely.