Final answer:
The zeros of the cubic polynomial x³ + 3x² - 11x - 21 are found to be -3, (1 + √29)/2, and (1 - √29)/2 by using synthetic division and the quadratic formula.
Step-by-step explanation:
To find all zeros of the cubic polynomial x³ + 3x² - 11x - 21, one can use various methods such as synthetic division, factoring, or graphing. However, for polynomials of higher degrees without an obvious factor, the Rational Root Theorem can be helpful in identifying potential rational zeros.
Firstly, we can try to identify any rational zeros by checking factors of the constant term (-21) divided by factors of the leading coefficient (1). The possible rational zeros are ±{1, 3, 7, 21}.
Using synthetic division or polynomial division, we can test these potential zeros and find that x = -3 is a zero since the remainder when dividing the polynomial by x + 3 is 0.
After factoring out x + 3, we are left with a quadratic equation (x + 3)(x² - x - 7) = 0. The quadratic factor can then be solved for x using the quadratic formula, x = (-b ± √(b² - 4ac))/(2a), giving us two additional zeros for the polynomial.
By solving the quadratic, we find that the zeros of the cubic polynomial are x = -3, x = (1 + √29)/2, and x = (1 - √29)/2.