Given z(x)=6x^3+bx^2-52x+15, z(2)=35, and z(-5)=0, algebracically determine all the zeros of z(x)

Z(x) = 6x^3 + bx^2 - 52x + 15

z(2) = 6(2)^3 + b(2)^2 - 52(2) + 15 = 6(8) + 4b - 104 + 15 = 4b - 41 = 35
4b = 35 + 41 = 76
b = 76/4 = 19

Thus, z(x) = 6x^3 + 19x^2 - 52x + 15
Since z(-5) = 0, x + 5 is a factor of z(x)

Dividin z(x) by x + 5 gives 6x^2 - 11x + 3 = 6x^2 - 2x - 9x + 3 = 2x(3x - 1) - 3(3x - 1) = (2x - 3)(3x - 1) = 0
x = 3/2 and x = 1/3

Therefore, the zeros of z(x) are -5. 1/3 and 3/2

The answer 
z(x)=6x^3+bx^2-52x+15 can be written 
z(x)=6x^3+bx^2-52x+15=(x+5)(ax²+bx+c)
because z(-5)=0, 
and since z(2)=35, we have 6.2^3+4b-104+15=35, we can find the value of b
4b=76 so b=19
we can write z(x)=6x^3+19x^2-52x+15=(x+5)(ax²+bx+c)=az^3+(b+5a)x²+5bx+10c
by identification, a=6, 15=10c, c =3/2

finally z(x)=(x+5)(6x²+19x+3/2)
let 's solve 6x²+19x+3/2=0
this equation has x= -19/12-5/12√13 and -19/12+5/12√13

the zero are x=-5, x= -19/12-5/12√13 and -19/12+5/12√13

It should be a or d i’m not sure