select all of the zeros of f(x)=x^3+4x^2-9x-36.
A x=3
B x=4
C x=2
D x=-4
E x=-3
F x=-2
To find the zeros of the function, we need to find the values of x that make the function equal to zero.
Setting f(x) to zero, we have:
0 = x^3 + 4x^2 - 9x - 36
Now we can factor the equation or use synthetic division to find the zeros.
Using synthetic division, testing x = 3:
3 | 1 4 -9 -36
| 3 21 36
|____________________
1 7 12 0
Since the remainder is zero, x - 3 is a factor of f(x), and we have reduced the equation to:
0 = (x - 3)(x^2 + 7x + 12)
Now we can solve for x by setting each factor to zero:
x - 3 = 0 -> x = 3
x^2 + 7x + 12 = 0
Factoring the quadratic equation:
(x + 3)(x + 4) = 0
Setting each factor to zero gives us:
x + 3 = 0 -> x = -3
x + 4 = 0 -> x = -4
Therefore, the zeros of f(x) = x^3 + 4x^2 - 9x - 36 are:
A) x = 3
B) x = -3
C) x = -4