Algebra-1

To find a function that has roots at 2, 3, and -1, 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 + 1)
When we substitute x = 2, we get: f(2) = (2 - 2)(2 - 3)(2 + 1) = 0 * -1 * 3 = 0.
When we substitute x = 3, we get: f(3) = (3 - 2)(3 - 3)(3 + 1) = 1 * 0 * 4 = 0.
When we substitute x = -1, we get: f(-1) = (-1 - 2)(-1 - 3)(-1 + 1) = -3 * -4 * 0 = 0.
Option A satisfies the condition and has roots at 2, 3, and -1.

Option B:
f(x) = (x - 1)(x - 2)(x - 3)
When we substitute x = 2, we get: f(2) = (2 - 1)(2 - 2)(2 - 3) = 1 * 0 * -1 = 0.
When we substitute x = 3, we get: f(3) = (3 - 1)(3 - 2)(3 - 3) = 2 * 1 * 0 = 0.
When we substitute x = -1, we get: f(-1) = (-1 - 1)(-1 - 2)(-1 - 3) = -2 * -3 * -4 = 24.
Option B does not satisfy the condition, as f(-1) ≠ 0. Therefore, it does not have the roots at 2, 3, and -...