Answer:
{-3, √3, -√3, 13/3}
Explanation:
Since the polynomial has rational coefficients, any irrational zeros must come in conjugate pairs. So, if √3 is a zero, then so is its conjugate, -√3.
We can write the polynomial with these zeros as:
p(x) = a(x + 3)(x - √3)(x + √3)(x - 13/3)
where a is some constant coefficient. Multiplying out the factors, we get:
p(x) = a(x + 3)(x^2 - 3)(x - 13/3)
To find the remaining zeros, we need to solve for x in the expression p(x) = 0. So we set up the equation:
a(x + 3)(x^2 - 3)(x - 13/3) = 0
This equation is true when any of the factors is equal to zero. We already know three of the zeros, so we need to solve for the fourth:
(x + 3)(x^2 - 3)(x - 13/3) = 0
Expanding the quadratic factor, we get:
(x + 3)(x - √3)(x + √3)(x - 13/3) = 0
Canceling out the (x - √3) and (x + √3) factors, we get:
(x + 3)(x - 13/3) = 0
Solving for x, we get:
x = -3 or x = 13/3
Therefore, the other zeros are -3 and 13/3.
The complete set of zeros is {-3, √3, -√3, 13/3}.
Hope it helps^^