For the polynomial [tex] f(x) = 6x^3 - 29x^2 - 6x + 5 [/tex] you can determine the possible rational zeroes as fraction c/d, where c and d are integer numbers that are respectively, divisors of 5 (the last coefficient) and 6 (the first coefficient).
Divisors of 5: [tex] \pm 1, \pm 5; [/tex]
Divisors of 6: [tex] \pm 1, \pm 2, \pm 3, \pm 6. [/tex]
Then zeros can be among fractions:
[tex] \pm\dfrac{1}{1}=\pm 1; \pm\dfrac{1}{2}; \pm\dfrac{1}{3}; \pm\dfrac{1}{6}; \pm\dfrac{5}{1}=\pm 5; \pm\dfrac{5}{2}; \pm\dfrac{5}{3}; \pm\dfrac{5}{6}. [/tex]
You can use syntetic division to determine which fractions fit, or you can simply count:
[tex] f(5)=6\cdot 5^3 - 29\cdot 5^2 - 6\cdot 5 + 5=750-725-30+5=0 [/tex] this means that x=5 is a polynomial zero.
Then
[tex] f(x) = 6x^3 - 29x^2 - 6x + 5=(x-5)(6x^2+x-1). [/tex]
Trinomial [tex] 6x^2+x-1 [/tex] is quadratic, then [tex] 6x^2+x-1=6(x-x_1)(x-x_2)=6(x+\dfrac{1}{2})(x-\dfrac{1}{3}) [/tex]
and
[tex] f(x) = 6x^3 - 29x^2 - 6x + 5=6(x-5)(x+\dfrac{1}{2})(x-\dfrac{1}{3}) . [/tex]
Zeros are [tex] 5, -\dfrac{1}{2}, \dfrac{1}{3} . [/tex]
Answer: correct choice is A.
