Algebra-1

To find a function that has roots at -2, 3, and 6, we need to check each of the given options by substituting these values into the function and verifying if they result in f(x) equal to zero.

Option A:
f(x)= (x + 2) (x - 3) (x - 6)
When we substitute x = -2, we get: f(-2) = (-2 + 2)(-2 - 3)(-2 - 6) = 0 * -5 * -8 = 0.
When we substitute x = 3, we get: f(3) = (3 + 2)(3 - 3)(3 - 6) = 5 * 0 * -3 = 0.
When we substitute x = 6, we get: f(6) = (6 + 2)(6 - 3)(6 - 6) = 8 * 3 * 0= 0.
Option A satisfies the condition.

Option B:
f(x)= (x - 2) (x + 3) (x + 6)
When we substitute x = -2, we get: f(-2) = (-2 - 2)(-2 + 3)(-2 + 6) = -4 * 1 * 4 = -16.
When we substitute x = 3, we get: f(3) = (3 - 2)(3 + 3)(3 + 6) = 1 * 6 * 9 = 54.
When we substitute x = 6, we get: f(6) = (6 - 2)(6 + 3)(6 + 6) = 4 * 9 * 12= 432.
Option B does not satisfy the condition.

Option C:

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