Show that the Fundamental Theorem of Algebra is true for the quadratic polynomial −4x2−24x−36=0

To show that the Fundamental Theorem of Algebra is true for the polynomial \( -4x^2 - 24x - 36 = 0 \), we will use the quadratic formula, which is given by:

\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

For the quadratic polynomial \( -4x^2 - 24x - 36 \), the coefficients are:

• \( a = -4 \)
• \( b = -24 \)
• \( c = -36 \)

First, we need to calculate \( b^2 - 4ac \):

\[ b^2 = (-24)^2 = 576 \]

\[ 4ac = 4 \cdot (-4) \cdot (-36) = 576 \]

Now we find the discriminant:

\[ b^2 - 4ac = 576 - 576 = 0 \]

Since the discriminant is zero, we can conclude that there is one repeated (identical) solution.

Now, we will compute the solution using the quadratic formula:

\[ x = \frac{-(-24) \pm \sqrt{0}}{2 \cdot -4} = \frac{24 \pm 0}{-8} = \frac{24}{-8} = -3 \]

Thus, the solution to the equation \( -4x^2 - 24x - 36 = 0 \) is \( x = -3 \), and this is a repeated solution.

Conclusion

The original polynomial has one repeated solution:

• There are two identical solutions of \( x = -3 \).

Therefore, the correct statement describing the solution set is:

There are two identical solutions.