Factor completely. If a polynomial cannot be​ factored, say it is prime.
xSuperscript 4negative 5xsquarednegative 24

We need to factor the polynomial \( x^4 - 5x^2 - 24 \) completely.

### Step 1: Substitution
Let's first recognize that this polynomial is quadratic in form. We can use a substitution to make it easier to factor. Set \( y = x^2 \). The polynomial then becomes:

\[
y^2 - 5y - 24
\]

### Step 2: Factor the quadratic expression
Next, we factor the quadratic expression \( y^2 - 5y - 24 \). We are looking for two numbers that multiply to \(-24\) and add to \(-5\).

The numbers that work are \(-8\) and \(3\). So, we can write:

\[
y^2 - 5y - 24 = (y - 8)(y + 3)
\]

### Step 3: Substitute back \( y = x^2 \)
Now, substitute back \( y = x^2 \) into the factors:

\[
(x^2 - 8)(x^2 + 3)
\]

### Step 4: Check for further factorization
Neither \( x^2 - 8 \) nor \( x^2 + 3 \) can be factored further using real numbers. Therefore, the completely factored form of the polynomial over the real numbers is:

\[
(x^2 - 8)(x^2 + 3)
\]

### Conclusion
The polynomial \( x^4 - 5x^2 - 24 \) factors completely as \( (x^2 - 8)(x^2 + 3) \) over the real numbers.

Would you like more details on any of the steps, or do you have any other questions?

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**Related Questions:**
1. How would you solve \( x^4 - 5x^2 - 24 = 0 \) for \( x \)?
2. What are the roots of the equation \( x^2 - 8 = 0 \)?
3. What are the roots of the equation \( x^2 + 3 = 0 \)?
4. How do you determine if a polynomial can be factored further?
5. What is the difference between factoring over the real numbers and factoring over the complex numbers?

**Tip:** When factoring polynomials, always look for substitutions that can simplify the expression, especially when dealing with higher-degree polynomials.

Learn how to factor the polynomial x^4 - 5x^2 - 24 completely with step-by-step instructions. This problem involves factoring a quadratic polynomial using substitution and recognizing irreducible factors. Suitable for Grades 10-12.

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