find the real zeros of the polynomial:

Question

P(x) = 24x^3 - 76x^2 + 2

Answer 1

so we are solving
24x^3 - 76x^2 + 2 = 0
dividing by 2 helps a bit

12x^3 - 38x^2 + 1 = 0

after a few trials with ±1, ±1/2, ±1/3 I found that 1/6 works, so (6x-1) is a factor.
Using long division gave me
12x^3 - 38x^2 + 1 = (6x-1)(2x^2 - 6x - 1)

I will leave it with you to find the 2 zeros you get from solving the quadratic
2x^2 - 6x - 1 = 0

(hint, one of them is appr. 3.158)

Answer 2

Thank you I just think I got overwhelmed!