State all the x-intercepts for the function [math: y=x3−2x2−3x] y=x^3-2x^2-3x .

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To find the x-intercepts of y = x^3 − 2x^2 − 3x, we set the equation equal to zero and solve for x.
Option A: 'a. 0, -1, and 3' — This matches the roots found by factoring the polynomial (x)(x^2 − 2x − 3) and further breaking down the quadratic into (x − 3)(x + 1). The set {0, −1, 3} correctly lists all x-values where the function crosses the x-axis.
Option B: 'b. 0, -1, and -3' — The value x = -3 is not a root here; substituting -3 gives (-27) − 18 + 9 ≠ 0, so this option is incorrect.
Option C: 'c. −1 and −3' — This omits the root x = 0, which is clearly a factor (x) of the polynomial, making this choice incomplete and false.
Option D: 'd. 1 and 3' — The numbers 1 and 3 are not the full set of intercepts: while x = 3 is a root, x = 1 does not satisfy x^3 − 2x^2 − 3x = 0 (it yields 1 − 2 − 3 = −4). Therefore this option is incorrect.
Overall, factoring the polynomial into x(x − 3)(x + 1) reveals the intercepts at x = 0, x = 3, and x = −1, which aligns with the values listed in option A.

State all the x-intercepts for the function [math: y=x3−2x2−3x] y=x^3-2x^2-3x .

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