To determine the real zeros of the polynomial [tex]\(x^3 + 4x^2 - 9x - 36\)[/tex], we need to find the values of [tex]\(x\)[/tex] that satisfy the equation [tex]\(x^3 + 4x^2 - 9x - 36 = 0\)[/tex].
Let's denote the polynomial by [tex]\(P(x) = x^3 + 4x^2 - 9x - 36\)[/tex]. We need to find the values of [tex]\(x\)[/tex] where [tex]\(P(x) = 0\)[/tex].
The real zeros of the polynomial are:
1. [tex]\(x = -4\)[/tex]
2. [tex]\(x = -3\)[/tex]
3. [tex]\(x = 3\)[/tex]
Thus, we have found the roots of the polynomial to be [tex]\(-4, -3,\)[/tex] and [tex]\(3\)[/tex].
Therefore, the correct answer is:
[tex]\[ \boxed{-4, -3, 3} \][/tex]
And looking at the options provided:
A. [tex]\(1, 2, 3\)[/tex] – This is incorrect.
B. [tex]\(-6, 2, 3\)[/tex] – This is incorrect.
C. [tex]\(-4, -3, 3\)[/tex] – This is the correct answer.
D. [tex]\(-1, 2, 18\)[/tex] – This is incorrect.
The correct choice is:
[tex]\[ \boxed{C} \][/tex]
