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