Final answer:
To verify the zeros of the cubic polynomial p(x) = x^3 - 2x^2 - 5x + 6, we substitute the values 3, -2, and 1 into this polynomial. The relation between the zeros and coefficients of a cubic polynomial can be found using Vieta's formulas.
Step-by-step explanation:
To verify that 3, -2, and 1 are the zeros of the cubic polynomial p(x) = x^3 - 2x^2 - 5x + 6, we can substitute these values into the polynomial and check if the result is equal to zero.
For x = 3:
p(3) = (3)^3 - 2(3)^2 - 5(3) + 6 = 27 - 18 - 15 + 6 = 0
For x = -2:
p(-2) = (-2)^3 - 2(-2)^2 - 5(-2) + 6 = -8 - 8 + 10 + 6 = 0
For x = 1:
p(1) = (1)^3 - 2(1)^2 - 5(1) + 6 = 1 - 2 - 5 + 6 = 0
Therefore, 3, -2, and 1 are indeed the zeros of the cubic polynomial p(x).
The relation between the zeros and coefficients of a cubic polynomial can be found using Vieta's formulas. For a cubic polynomial in the form ax^3 + bx^2 + cx + d = 0 with zeros p, q, and r, the following relations hold:
p + q + r = -b/a
pqr = -d/a
In this case, the zeros are 3, -2, and 1, so the relation between the zeros and coefficients is:
3 + (-2) + 1 = -(-2)/1
3(-2)(1) = -6/1