Question

What are the rational zeros of the polynomial p(x) = x³ - 6x² - 32?

1) -4, -2, and 4
2) -4, -2, and 8
3) -4, -2, and 16
4) -4, -2, and 32

Answer 1

The rational zeros of the polynomial p(x) = x³ - 6x² - 32 are -4, -2, and 4.

Step-by-step explanation:

To find the rational zeros of the polynomial p(x) = x³ - 6x² - 32, we can use the Rational Zeros Theorem. According to the theorem, any rational zero of a polynomial equation is a factor of the constant term divided by a factor of the leading coefficient. In this case, the constant term is -32 (the coefficient of the highest degree term) and the leading coefficient is 1. The factors of 32 are 1, 2, 4, 8, 16, and 32, and the factors of 1 are 1 and -1.

Therefore, the possible rational zeros of the polynomial are:

1. x = 1
2. x = -1
3. x = 2
4. x = -2
5. x = 4
6. x = -4

After testing these possible zeros using synthetic division or another method, we can determine that the rational zeros of the polynomial p(x) = x³ - 6x² - 32 are x = -4, -2, and 4.